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http://math.stackexchange.com/questions/403327/what-are-some-examples-of-proof-by-contrapositive/403340 | # What are some examples of proof by contrapositive?
Applying the Modus Tollens argument to Fermat's Little Theorem really helped me to understand logical implication. I never knew that FLT was actually a compositality test.
Theorem (FLT): given integers $a>1$ and $n>1$, if $n$ is prime, then $a^n$ is congruent to $a\ (\bmod\ n)$.
By the contrapositive, if $a^n$ is not congruent to $a\ (\bmod\ n)$, then $n$ is not prime. Thus $n$ is composite.
What are some other simple and instructive examples of proof by contrapositive?
-
This question has a nice example; see especially the discussion in my answer. This question and its answer are another, and this, this, and this should also be of interest. These are just a few that I could easily find. – Brian M. Scott May 27 '13 at 4:32
Thanks everyone for all the great examples. Now, I've got plenty to study. It sure would be interesting to compare samples from each area of Mathematics, a "logic parallel". – cyclochaotic May 27 '13 at 15:33
When you want to prove "If $p$ then $q$", and $p$ contains the phrase "$n$ is prime" you should use contrapositive or contradiction to work easily, the canonical example is the following:
Prove for $n>2$,
If $n$ is prime then $n$ is odd.
Here $q$ is the phrase "$n$ is odd". Here $p$ is exactly the phrase "$n$ is prime" and is very difficult to work with it. Because the fact that $n$ is prime means that it is not divisible by other number grater than $1$ and different for $n$, so you must to choose from these $n-2$ true sentences the only one that is useful which is "$n$ is not divisible by 2", but you would not know which is the right choice, unless you read $q$.
But proving contrapositive equivalent form is very easy, and you don't to do any choice.
If $n$ is even then $n$ is not prime.
Which follows from the fact that every even number greater than $2$ is divisible by $2$, hence not prime.
So contrapositive (also contradiction) is used to avoid situations where you have a lot of information and very little of it is actually useful.
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Look at Richard Hammack's "Book of Proof" for a detailed discussion of proof forms. There are several "proof technique" and such introductory courses (perhaps under discrete mathematics and similar) with lecture notes on the 'net.
- | 2015-05-22T10:21:39 | {
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https://math.stackexchange.com/questions/2912050/what-is-pa-cap-b-and-pa-cup-b-cup-c | # What is $P(A' \cap B')$ and $P(A \cup B \cup C)$?
We are given that $P(A)=.43, P(B)=.44,$ and $P(C)=.19$. Also, $B$ and $C$ are mutually exclusive but $P(A\cap B)=.07$ and $P(A\cap C)= .13$
(a) What is $P(A' \cap B')$?
Note: I know that since $A,B$ are not mutually exclusive, that $P(A \cap B)=P(A) \times P(B)$. Does this same rule apply for the complement? If so, then I can use the rule $P(A)+P(A')=1$.
(b) What is $P(A \cup B \cup C)$?
I know the formula: $P(A\cup B\cup C) = P(A) +P(B)+P(C)-P(A\cap B)-P(A\cap C)- P(B\cap C)+P(A\cap B\cap C)$.
Note: The part I am having trouble on with this formula is finding $P(A\cap B\cap C)$.
Guide:
$$P(A) \times P(B) \approx 0.1892 \ne P(A \cap B)$$
$A$ and $B$ are not independent.
• Note that $$P(A' \cap B') = 1-P(A \cup B)$$
• $(A \cap B \cap C) \subset (B \cap C)$, Hence $P(A \cap B \cap C)=0$.
• Also, check your inclusion-exclusion formula.
• that is De Morgan's law. – Siong Thye Goh Sep 10 '18 at 16:14
• I know that De Morgan's Law states: $(A\cup B)' = A' \cap B'$ and $(A\cap B)' = A' \cup B'$. How does this translate into that probability formula? – rover2 Sep 10 '18 at 16:18
• note: i have fixed my inclusion-exclusion formula. i copied it wrong while typing the solution however it was correct in my notes – rover2 Sep 10 '18 at 16:19
• Since $A' \cap B' = (A \cup B)'$ we have $P(A' \cap B')= P((A \cup B)' )$. We have $P((A \cup B)' ) = 1-P(A \cup B)$. – Siong Thye Goh Sep 10 '18 at 16:20
• got it, thank you – rover2 Sep 10 '18 at 16:22
You know that $B$ and $C$ are mutually exclusive i.e. $A\cap B$ and $C$ are mutually exclusive,
$$P(A\cap B \cap C)=0$$
Also $$P(A\cap B)\neq P(A)\cdot P(B)$$
As $A$ and $B$ are not independent events.
For Part (II),
Using De-Morgans Law,
$$P(A'\cap B')=P((A\cup B)')=1-P(((A\cup B)')')$$
$$P(A'\cap B')=1-P(A\cup B)$$
Also you know that,
$$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$ | 2019-07-22T16:46:44 | {
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http://mathhelpforum.com/statistics/26957-probability.html | 1. ## probability
There are 5 different colors of balls: white, black, blue, red, green. We randomly pick 6 balls. Each ball has a probability of 0.2 of getting each of the 5 colors. What is the probability that, from the 6 balls picked, there are white balls and black balls?
2. Originally Posted by allrighty
There are 5 different colors of balls: white, black, blue, red, green. We randomly pick 6 balls. Each ball has a probability of 0.2 of getting each of the 5 colors. What is the probability that, from the 6 balls picked, there are white balls and black balls?
find the probabilities for each case and add them up
by (all possible) cases, i mean:
probability of choosing 1 white, 0 black, 1 blue, 1 red, 3 green
probability of choosing 1 white, 0 black, 0 blue, 2 red, 3 green
probability of choosing 1 white, 0 black, 2 blue, 0 red, 3 green
.
.
.
.
or we could do:
1 - probability of choosing no white and no black ball
so all you have to worry about are the number of ways you can choose 6 balls among the three remaining colors. there is a formula for that sort of thing. can't remember right now, i'll have to look it up. but when you do get it, just multiply the answer by 0.2 and that will give you the probability of choosing no white and no black ball
3. Originally Posted by Jhevon
so all you have to worry about are the number of ways you can choose 6 balls among the three remaining colors
Is that right? A white AND a black are required.
I have another method but it's messy.
You could have for example 2W, 1B and 3 others 6!/(3!2!) ways each with probability 0.2^3 0.6^3
or 5W, 1B 6!/5! ways with probability 0.2^6
Having said all that, I'm a bit rubbish at probability so I'll wait and see what people say.
4. Originally Posted by a tutor
Is that right? A white AND a black are required.
yes, i believe so. what i said was i want to find the probability of having 0 white AND 0 black. and then take 1 minus that probability to find the probability of at least 1 white or 1 black. i think what i described does the trick... but then again, i'm a noob when it comes to probability as well
5. Original question said..
Originally Posted by allrighty
What is the probability that, from the 6 balls picked, there are white balls and black balls?
and you said..
Originally Posted by Jhevon
to find the probability of at least 1 white or 1 black.
6. Originally Posted by a tutor
Original question said..
and you said..
ah yes. my bad. i didn't see the "and." i was under the impression if we have either or we were good...
7. Hello, allrighty!
I think I've solved it . . .
There are 5 different colors of balls: white, black, blue, red, green.
We randomly pick 6 balls.
Each ball has a probability of 0.2 of getting each of the 5 colors.
What is the probability that, from the 6 balls picked, there are white balls and black balls?
The opposite of "some White and some Black" is "no White or no Black".
To get no White, we must pick six balls from the other four colors.
. . Then: . $P(\text{0 White}) \:=\:(0.8)^6$
To get no Black, we must pick six balls from the other four colors.
. . Then: . $P(\text{0 Black}) \:=\:(0.8)^6$
To get no White and no Black, we pick six balls from the other 3 colors.
. . Then: . $P(\text{0 White} \,\wedge \,\text{0 Black}) \:=\;(0.6)^6$
Hence: . $P(\text{0 White} \,\vee \,\text{0 Black}) \;=\;P(\text{0 White}) + P(\text{0 Black}) - P(\text{0 White} \,\wedge \,\text{0 Black})$
. . . . . . . . . . . . . . . . . . . . $= \quad\;\;(0.8)^6\quad \;+ \;\quad(0.8)^6 \qquad- \qquad(0.6)^6$
. . . . . . . . . . . . . . . . . . . . $= \qquad 0.4777632$
Therefore: . $P(\text{some White and some Black}) \;=\;1 - 0.4777632 \;=\;\boxed{\:0.522368\:}$
8. A neat solution Soroban.
I got the same answer rather more clumsily.
I wrote a quick easy program to do it the way I mentioned above. | 2017-10-19T22:02:16 | {
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http://math.stackexchange.com/questions/437209/how-can-i-show-that-sqrt1-sqrt2-sqrt3-sqrt-ldots-exists | How can I show that $\sqrt{1+\sqrt{2+\sqrt{3+\sqrt\ldots}}}$ exists?
I would like to investigate the convergence of
$$\sqrt{1+\sqrt{2+\sqrt{3+\sqrt{4+\sqrt\ldots}}}}$$
Or more precisely, let \begin{align} a_1 & = \sqrt 1\\ a_2 & = \sqrt{1+\sqrt2}\\ a_3 & = \sqrt{1+\sqrt{2+\sqrt 3}}\\ a_4 & = \sqrt{1+\sqrt{2+\sqrt{3+\sqrt 4}}}\\ &\vdots \end{align}
Easy computer calculations suggest that this sequence converges rapidly to the value 1.75793275661800453265, so I handed this number to the all-seeing Google, which produced:
Henceforth let us write $\sqrt{r_1 + \sqrt{r_2 + \sqrt{\cdots + \sqrt{r_n}}}}$ as $[r_1, r_2, \ldots r_n]$ for short, in the manner of continued fractions.
Obviously we have $$a_n= [1,2,\ldots n] \le \underbrace{[n, n,\ldots, n]}_n$$
but as the right-hand side grows without bound (It's $O(\sqrt n)$) this is unhelpful. I thought maybe to do something like:
$$a_{n^2}\le [1, \underbrace{4, 4, 4}_3, \underbrace{9, 9, 9, 9, 9}_5, \ldots, \underbrace{n^2,n^2,\ldots,n^2}_{2n-1}]$$
but I haven't been able to make it work.
I would like a proof that the limit $$\lim_{n\to\infty} a_n$$ exists. The methods I know are not getting me anywhere.
I originally planned to ask "and what the limit is", but OEIS says "No closed-form expression is known for this constant".
The references it cites are unavailable to me at present.
-
This is probably a good place to start: "It was discovered by T. Vijayaraghavan that the infinite radical $\sqrt{ a_1 + \sqrt{ a_2 + \sqrt{ a_3 + \sqrt{a_4 + \ldots }}}}$ where $a_n \ge 0$, will converge to a limit if and only if the limit of $\log a_n / 2^n$ exists" - Clawson, p. 229. (Taken from OEIS.) – George V. Williams Jul 6 '13 at 3:23
possible duplicate of Sum and Product of Infinite Radicals – MJD Jul 6 '13 at 3:26
I misread the solution at Sum and Product of Infinite Radicals. It asks several questions, one of which is mine, but all the answers provided are for the other questions. – MJD Jul 6 '13 at 3:28
Related: Nested radicals – MJD Jul 6 '13 at 3:31
This may help. – Maazul Jul 6 '13 at 4:44
The first number $1$ is a nuisance, so at first we disregard it.
We proceed by induction, and deal with finite nested radicals that start with $\sqrt{k+\sqrt{(k+1)+\cdots}}$, where $k\ge 2$.
We will show that such a radical is $\lt 2k$, by induction on depth. The result is certainly true for all nested radicals of depth $1$, since $\sqrt{q}\lt 2q$.
For the induction step, a nested radical of depth $n$ that starts with $k$ is $\sqrt{k+R}$, where $R$ is a nested radical of depth $n-1$ that starts with $k+1$. By the induction assumption, we have $R\lt 2k+2$. But then $\sqrt{k+R}\lt \sqrt{3k+2}\lt 2k$ if $k \ge 2$.
So (finite) nested radicals of any depth that start with $2$ are $\lt 4$. The sequence of nested radicals is clearly increasing, so it converges. It follows that the nested radical of the post is $\le \sqrt{1+4}$.
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For any $n\ge4$, we have $\sqrt{2n} \le n-1$. Therefore \begin{align*} a_n &\le \sqrt{1+\sqrt{2+\sqrt{\ldots+\sqrt{(n-2)+\sqrt{(n-1) + \sqrt{2n}}}}}}\\ &\le \sqrt{1+\sqrt{2+\sqrt{\ldots+\sqrt{(n-2)+\sqrt{2(n-1)}}}}}\\ &\le\ldots\\ &\le \sqrt{1+\sqrt{2+\sqrt{3+\sqrt{2(4)}}}}. \end{align*} Hence $\{a_n\}$ is a monotonic increasing sequence that is bounded above.
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If $( a_k )_{k\in\mathbb{N}}$ is any sequences of positive numbers such that:
$$0 \le a_k \le \alpha \lambda^{2^k}\quad\text{ for some }\quad \alpha, \lambda \in \mathbb{R}_{+}$$
Using same convention $\;[r_1,r_2\ldots] = \sqrt{r_1 + \sqrt{r_2 + \ldots}}\;$ as in the question, we have:
\begin{align} [a_n] & \le \sqrt{\alpha \lambda^{2^n}} = \sqrt{\alpha}\lambda^{2^{n-1}} = [\alpha] \lambda^{2^{n-1}}\\ \implies [a_{n-1},a_n ] &\le \sqrt{\alpha\lambda^{2^{n-1}}+\sqrt{\alpha}\lambda^{2^{n-1}}} =\sqrt{\alpha+\sqrt{\alpha}}\lambda^{2^{n-2}} = [\alpha,\alpha]\lambda^{2^{n-2}}\\ \implies [a_{n-2},a_{n-1},a_n]&\le \sqrt{\alpha\lambda^{2^{n-2}} + \sqrt{\alpha+\sqrt{\alpha}}\lambda^{2^{n-2}}} = [\alpha,\alpha,\alpha]\lambda^{2^{n-3}}\\ &\;\vdots\\ \implies [a_1,\ldots,a_n] & \le \underbrace{ [ \alpha,\ldots,\alpha ] }_{n\text{ terms}} \lambda\\ \implies [a_1, \ldots,a_n] & \le [ \alpha, \alpha, \ldots ]\lambda = \frac{1 + \sqrt{1+4\alpha}}{2}\lambda \end{align}
Since $n \le \sqrt{2}^{2^n-2}$, we can take $\alpha = \frac12$ and $\lambda = \sqrt{2}$ to get:
$$[1,2,\ldots,n] \le \underbrace{ [ \frac12,\ldots,\frac12 ]}_{n\text{ terms}} \lambda \le \frac{1+\sqrt{3}}{\sqrt{2}} \sim 1.931851$$
To get a better bound, observe for any $m, k \in \mathbb{Z}_{+}$, we have:
$$m + k - 1 \le \frac{m^2}{m+1}\left(\sqrt{\frac{m+1}{m}}\right)^{2^k}$$
Using the same approach as above, we get: $$[m,m+1,m+2,\ldots] \le \frac{\sqrt{m+1}+\sqrt{4m^2+m+1}}{2\sqrt{m}}$$
Take $m = 3$, we already get a bound accurate up to $O(10^{-2})$.
$$[3,4,\ldots] \le \frac{1+\sqrt{10}}{\sqrt{3}} \implies [1,2,3,\ldots] \le \sqrt{1+\sqrt{2+\frac{1+\sqrt{10}}{\sqrt{3}}}} \sim 1.760214368$$
-
Take a positive sequence $\{a_n\}$ and a constant $c>0$ such that $\sqrt{a_{n+1}}<ca_n$.
Set $b_n=\sqrt{a_1+\sqrt{a_2+\cdots \sqrt{a_n}}}$. By induction, $$\log_{c+1} \left(\frac{b_n}{\sqrt{a_1}}\right)<\sum_{i=1}^{n-1}2^{-i}<1$$ So $b_n<(c+1)\sqrt{a_1}$ and $b_n$ is monotonic increasing; by the Monotone Convergence Theorem, $\lim_{n\to\infty}b_n$ exists. Taking $a_n=n$ and $c>\sqrt{2}$ answers this question. | 2014-03-12T02:15:33 | {
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http://math.stackexchange.com/questions/59384/find-the-square-root-of-a-matrix/59396 | # Find the square root of a matrix
Let $A$ be the matrix $$A = \left(\begin{array}{cc} 41 & 12\\ 12 & 34 \end{array}\right).$$
I want to decompose it into the form of $B^2$.
I tried diagonalization , but can not move one step further.
Any thought on this? Thanks a lot!
ONE STEP FURTHER:
How to find a upper triangular $U$ such that $A = U^T U$?
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The matrix is symmetric, so it is certainly diagonalizable. Trace and determinant are both positive, so both eigenvalues are positive. So if you can diagonalize, the diagonal form will have a square root, $QAQ^{-1} = D = P^2$, where $Q$ is the change-of-basis matrix. That means that $A = Q^{-1}P^2Q = (Q^{-1}PQ)^2$, so you can let $B=Q^{-1}PQ$. So your idea works; where did you get stuck? – Arturo Magidin Aug 24 '11 at 3:40
@Arturo Magidin I did not figure out how to use the diagonalized form. Your answer is brilliant! Thanks! – BVFanZ Aug 24 '11 at 3:46
en.wikipedia.org/wiki/… – Fixee Aug 24 '11 at 4:02
Re: edit: Cholesky decomposition can be done with an approach similar to Gerry's: write out the expression for the product of a lower triangular matrix with its transpose, equate to your original matrix, and solve the resulting set of equations... – Ｊ. Ｍ. Aug 24 '11 at 4:03
+1. Nice question! – Pierre-Yves Gaillard Aug 24 '11 at 7:16
This is an expansion of Arturo's comment.
The matrix has eigenvalues $50,25$, and eigenvectors $(4,3),(-3,4)$, so it eigendecomposes to $$A=\begin{pmatrix}4 & -3 \\ 3 & 4\end{pmatrix} \begin{pmatrix}50 & 0 \\ 0 & 25\end{pmatrix} \begin{pmatrix}4 & -3 \\ 3 & 4\end{pmatrix}^{-1}.$$
This is of the form $A=Q\Lambda Q^{-1}$. If this is $B^2$, then there will be a $B$ of the form $Q\Lambda^{1/2} Q^{-1}$ (square this to check this is formally true). A square root of a diagonal matrix is just the square roots of the diagonal entries, so we have
$$B=\begin{pmatrix}4 & -3 \\ 3 & 4\end{pmatrix} \begin{pmatrix}\sqrt{50} & 0 \\ 0 & \sqrt{25}\end{pmatrix} \begin{pmatrix}4 & -3 \\ 3 & 4\end{pmatrix}^{-1}$$
$$=\frac{1}{5}\begin{pmatrix}9+16\sqrt{2} & -12+12\sqrt{2} \\ -12+12\sqrt{2} & 16+9\sqrt{2}\end{pmatrix}.$$
Here we used $\sqrt{50}=5\sqrt{2},\sqrt{25}=5$, and a quick formula for the inverse of a $2\times 2$ matrix:
$$\begin{pmatrix}a&b\\c&d\end{pmatrix}^{-1}=\frac{1}{ad-bc}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}.$$
Keep in mind that matrix square roots are not unique (even up to sign), but this particular method is guaranteed to produce one example of a real matrix square root whenever $A$ has all positive eigenvalues.
Finding an upper triangular $U$ such that $A=U^TU$ is even more straightforward:
$$A=\begin{pmatrix} a&0 \\ b&c \end{pmatrix} \cdot \begin{pmatrix} a&b \\ 0&c \end{pmatrix}$$
This is $a^2=41$ hence $a=\sqrt{41}$, $ab=12$ hence $b=\frac{12}{41}\sqrt{41}$, and $b^2+c^2=34$ hence $c=25\sqrt{\frac{2}{41}}$.
In other words,
$$U=\sqrt{41}\begin{pmatrix}1&\frac{12}{41}\\0&\frac{25}{41}\sqrt{2}\end{pmatrix}.$$
-
For the first part of your question, here is a solution that only works for 2-by-2 matrices, but it has the merit that no eigenvalue is needed.
Recall that in the two-dimensional case, there is a magic equation that is useful in many situations. It is $X^2-({\rm tr}X)X+(\det X)I=0$, which arises from the characteristic polynomial of a $2\times2$ matrix $X$. Now, if $X^2=A$, we have $\det X=\pm\sqrt{\det A}=r$ (say). We take the positive value for $r$. Hence $$(\ast):\quad ({\rm tr}X)X=X^2+rI=A+rI$$ and $({\rm tr}X)^2 = {\rm tr}\left(({\rm tr}X)X\right) = {\rm tr}(A+rI) = {\rm tr}A + 2r$. Thus, from $(\ast)$ we obtain $$X = \frac{1}{\sqrt{{\rm tr}A + 2r}}(A+rI)\quad {\rm where}\quad r=\sqrt{\det A}.$$ This method works for all 2-by-2 matrices $A$ when $\det A\ge0$ and ${\rm tr}A + 2\sqrt{\det A}>0$. In particular, it works for positive definite $A$.
For the second part of your question, as the others have pointed out, the decomposition you ask for is a Cholesky decomposition.
-
Dear @user1551: It seems to me your condition ${\rm tr}A + 2\sqrt{\det A}>0$ just means that the eigenvalues are positive. Is this correct? I also think that your formula is exactly the same as Didier's. [Mine is slightly different (but equivalent) because I haven't been smart enough to cover the two cases by a single formula.] – Pierre-Yves Gaillard Aug 24 '11 at 12:44
Yes, the eigenvalues have to be non-negative and at least one of them must be positive, and our formulae are equivalent. I think you are being modest when you said you were not smart enough. Since mathematicians in this forum tend to analyze problems (and generalize the results) from higher perspectives, it is not surprising that you guys do not take a low road as I did. – user1551 Aug 24 '11 at 15:24
I might be modest, ... but less than you! I mention you and Didier in an edit to my answer. I think my road is lower than yours: I try to express things in the high school language of secant and tangent lines. – Pierre-Yves Gaillard Aug 24 '11 at 16:00
No, my road is certainly lower than yours, because my method cannot be generalized to higher dimensions. – user1551 Aug 24 '11 at 16:34
Your method looks very original to me. It's very natural to use Cayley-Hamilton, but I guess most people would have used it for $A$, not for $X$. That is, put $X:=xA+yI$, write $X^2=A$, use CH in the form $A^2=tA-dI$ (with $t=$ trace, $d=$ det) to get rid of the $A^2$ term. You get 2 equations for $x$ and $y$. This can be generalized. – Pierre-Yves Gaillard Aug 24 '11 at 17:47
Still another explicit formula: for every nonnegative real number $\alpha$, $$A^\alpha=\frac{(u^\alpha-v^\alpha)A+(uv^\alpha-vu^\alpha)I}{u-v},$$ where $u$ and $v$ are the two roots of the polynomial $\chi_A(x)=\det(xI-A)$. When $\alpha=1/2$, $$\sqrt{A}=\frac{A+\sqrt{uv}I}{\sqrt{u}+\sqrt{v}}.$$ Note that the coefficients of this formula can be computed directly from the matrix $A$ since $t=\sqrt{uv}$ is simply $t=\sqrt{\det A}$ and $s=\sqrt{u}+\sqrt{v}$ is such that $s^2=u+v+2t$ hence $$s=\sqrt{\text{tr}(A)+2\sqrt{\det(A)}}.$$ In the present case, one can also compute $\chi_A(x)=(x-41)(x-34)-12^2=(x-25)(x-50)$ and use $\sqrt{u}=5\sqrt2$ and $\sqrt{v}=5$.
-
The formulae of yours, mine and @Pierre-Yves Gaillard are actually equivalent, but how they are expressed and derived are different. – user1551 Aug 24 '11 at 10:32
@user1551, yes. – Did Aug 24 '11 at 11:01
@didier Does this identity have a name or can you provide a reference? – Steve Apr 12 '12 at 2:23
Let $\lambda$ and $\mu$ be the eigenvalues of your $2$ by $2$ real matrix $A$. (We may have $\lambda=\mu$.) Assume that $\lambda$ and $\mu$ are positive.
If $\lambda\not=\mu$, write the equation of the secant line to the curve $y=\sqrt x$ through the points $(\lambda,\sqrt\lambda)$ and $(\mu,\sqrt\mu)$: $$y=\sqrt\lambda\ \ \frac{x-\mu}{\lambda-\mu}+\sqrt\mu\ \ \frac{x-\lambda}{\mu-\lambda}\quad.$$ The matrix you want is $$\sqrt\lambda\ \ \frac{A-\mu I}{\lambda-\mu}+\sqrt\mu\ \ \frac{A-\lambda I}{\mu-\lambda}\quad,$$ where $I$ is the identity matrix.
If $\lambda=\mu$, write the equation of the tangent line to the curve $y=\sqrt x$ through the point $(\lambda,\sqrt\lambda)$: $$y=\sqrt\lambda+\frac{x-\lambda}{2\sqrt\lambda}\quad.$$ The matrix you want is $$\sqrt\lambda\ I+\frac{A-\lambda I}{2\sqrt\lambda}\quad.$$
Do you see why?
Do you see how to generalize this to $n$ by $n$ matrices?
EDIT 4. This is to just explain why this secant/tangent stuff comes into the picture. Assume to simplify that the eigenvalues $\lambda$ and $\mu$ of your two by two real matrix $A$ are real and distinct. Let $f\in\mathbb R[X]$ be a polynomial, and $s$ the unique polynomial of degree $\le1$ which agrees with $f$ at $\lambda$ and $\mu$. [Graphically, this is a secant line.] Then the characteristic polynomial $$\chi=(X-\lambda)(X-\mu)$$ will divide $f-s$. As $\chi(A)=0$ by the Cayley-Hamilton Theorem, we have $f(A)=s(A)$. But the expression $s(A)$ makes sense whenever $f$ is a (real-valued) function defined at $\lambda$ and $\mu$. Moreover, the map $f\mapsto f(A)$ is compatible with addition and multiplication.
EDIT 1. As noticed by Didier Piau and user1551, there is a cute formula for the "generalized secant line" to the curve $y=\sqrt x$, by which I mean: the secant line if the points are distinct, the tangent line if they coincide. Supposing $\lambda\not=\mu$, the equation of the secant line is $$y=\frac{\sqrt\lambda-\sqrt\mu}{\lambda-\mu}\ \ x+ \frac{\mu\sqrt\lambda-\lambda\sqrt\mu}{\lambda-\mu}= \frac{x+\sqrt{\lambda\mu}}{\sqrt\lambda+\sqrt\mu}\quad,$$ and the miracle is that the last expression makes sense even if $\lambda=\mu$.
EDIT 2. Note that there are other solutions when $\lambda\not=\mu$. Putting $$E:=\frac{A-\lambda I}{\mu-\lambda}\quad,\quad F:=\frac{A-\mu I}{\lambda-\mu}\quad,$$ we get $$E^2=E,\ F^2=F,\ EF=FE=0,\ I=E+F,\ A=\mu E+\lambda F,$$ and thus $$(\pm\sqrt\mu\ E\pm\sqrt\lambda\ F)^2=A$$ for the four choices of signs. [The plus plus choice corresponds to the previous formula.]
EDIT 3. Here is a generalization.
Let $T$ be an $n$ by $n$ complex matrix, and $$p(X)=(X-\lambda_1)^{m(1)}\cdots(X-\lambda_k)^{m(k)}$$ its minimal polynomial (the $\lambda_i$ being distinct and the $m(i)$ positive). Let $A$ be the algebra of those functions $f(z)$ which are holomorphic in a neighborhood of the spectrum $\{ \lambda_1,\dots,\lambda_k \}$ of $T$.
There is a unique $\mathbb C[X]$-algebra morphism from $A$ to $\mathbb C[T]=\mathbb C[X]/(p(X))$. Denote this morphism by $f(z)\mapsto f(T)$. If $f(z)$ is in $A$, then the unique representative of $f(T)$ in $\mathbb C[X]$ of degree less than $\deg p(X)$ is $$\sum_{i=1}^k\ \ \underset{X=\lambda_i}\heartsuit\left( \Big(\ \underset{z=\lambda_i}\heartsuit f(z) \Big)\ \ \frac{(X-\lambda_i)^{m(i)}}{p(X)}\ \right)\ \frac{p(X)}{(X-\lambda_i)^{m(i)}}$$
with $$\underset{u=\lambda_i}\heartsuit\varphi(u):=\sum_{j=0}^{m(i)-1}\frac{\varphi^{(j)}(\lambda_i)}{j!}\ (X-\lambda_i)^j.$$
Moreover, the $\lambda_i$-generalized eigenspace of $T$ is contained in the $f(\lambda_i)$-generalized eigenspace of $f(T)$.
All this follows from the Chinese Remainder Theorem, which says $$\frac{\mathbb C[X]}{(p(X))}=\prod_{i=1}^k\ \ \frac{\mathbb C[X]}{(X-\lambda_i)^{m(i)}}\quad,$$ and from the Taylor Formula.
[There is an Edit 4 above.]
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Write $$\pmatrix{a&b\cr c&d\cr}^2=\pmatrix{41&12\cr12&34\cr}$$ multiply out the left side, set corresponding entries equal to get four equations in the four unknowns $a,b,c,d$, then see if you can work your way through the algebra to a solution.
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This method gets ugly in a hurry. Once needs to solve $a^2 + b^2 = 41$, $b(a+d) = 12$, and $b^2 + d^2 = 34$ simultaneously. As far as I can tell, there seems to be no clean algebraic solution. – JavaMan Aug 24 '11 at 5:50
I can work it out on a single sheet of paper (but I have very small handwriting...). – Gerry Myerson Aug 24 '11 at 6:23
I won't ask for all the details, but what is your general process? – JavaMan Aug 24 '11 at 6:33
Subtract third equation from first to get a difference of two squares, and express $a+d$ and $a-d$ in terms of $b$ using the expression for $a+d$ you get from the second equation. This gives you $a$ in terms of $b$ and you can substitute in the first equation to get an equation you can solve for b. It is much easier if you are allowed to guess that the answer can be expressed in integers. – Mark Bennet Aug 24 '11 at 8:11
Forget that last sentence – Mark Bennet Aug 24 '11 at 8:29
The second is even simpler than the first question. All you need to to look at the assumed solution, setting unknowns:
$\qquad \qquad \small A = U^t \cdot U = \begin{pmatrix} a&0 \\ b&c \end{pmatrix} \cdot \begin{pmatrix} a&b \\ 0&c \end{pmatrix} \qquad$ and $\small a \cdot a = 41$. Then you can proceed; in Pari/GP it needs something like 5 lines of code...
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Find the square root of a matrix
此処の 例を ■■■ideal I を用いて■■■；
I=<(w-4) (w+4) (11 w^2-192),11 w^3-272 w+128 z,11 w^3-272 w+192 y,-11 w^3+144 w+128 x>
(なる 等式を 先ず 証明し）
から 容易に 4つ の 解行列が 獲られます ので 具現願います；
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– robjohn Jun 14 '13 at 10:21 | 2014-07-30T01:14:49 | {
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https://math.stackexchange.com/questions/3299621/infinite-points-on-circle | # Infinite points on circle
Can we say that circle has infinite points? What if then I took one point out. Does it matter. And then if we took half the infinite points of the circle out of it and still can it be called circle?
• yes; no; yes; no – mathworker21 Jul 21 at 15:53
• The circle is complete as is. with infinite points. If you remove points it is not a circle by usual definition. – herb steinberg Jul 21 at 15:54
• Well what's the ans of last question 'half the infinite point on circle' – user237118 Jul 21 at 15:54
In the easiest setup, a circle is all points in the plane $$\mathbb R^2$$, with distance 1 from the origin $$(0,0)$$. There are infinitely many solutions to this condition, so YES there are infinitely many points in a circle.
YET removing a single point changes its structure (f.e. a circle missing a point is no longer closed) and the result is no longer a circle. But this shouldn't come as a huge surprise. Removing a single point from an infinite set often changes it. E.g. if you have then set $$S=\{0, 1, 2, ...\}$$ and remove $$0$$, then suddenly you achieve the property, that all elements of $$S$$ are positive.
• So geometrically it is no longer a circle? – user237118 Jul 21 at 15:59
• If we start from a point on circle and took out half the infinite points on circle, do we end up with semi circle? – user237118 Jul 21 at 16:00
• @user237118 "So geometrically it is no longer a circle?" Yes, but it still has infinitely many points. (Incidentally, you should say "infinitely many points" rather than "infinite points" - the latter makes it sound like the points themselves are infinite objects, which can actually be a bit confusing since "point at infinity" is a real technical term. This is a totally minor issue but it is technical more correct.) – Noah Schweber Jul 21 at 16:01
• @user237118 "If we start from a point on circle and took out half the infinite points on circle, do we end up with semi circle?" Well, depends which half we remove ... – Noah Schweber Jul 21 at 16:01
• The consecutive ones, just next to each other – user237118 Jul 21 at 16:03
1. Yes, the circle contains infinitely many points. The usual unit circle consists of all points of the form $$(\cos \theta, \sin\theta)$$ for $$0\leq \theta < 2\pi.$$
2. No, if you delete even a single point from a circle, it's not a circle anymore. In more advanced mathematics you will study different properties of sets and be able to prove that the circle and circle-with-point-removed are very different, topologically and algebraically. Consider for instance that on the circle, any two points can be connected by two different curves that stay inside the circle; if you take out a point there is only one curve connected two points. That said, if you squint you might argue that deleting a point makes the circle still "look like" a circle (and in some cases the point doesn't matter, for example an integral will have the same value if you integrate it up over the circle or "punctured" circle). There are ways of making precise that the circle with one point removed is "almost" a circle: the full circle is the closure of the punctured circle, and the two sets differ by "a set of measure zero" (which in this case means the part you deleted has zero length compared to the full circle).
3. What is half of infinity? There are different ways of removing "half the points": certainly a semicircle is very different from the full circle. You could delete all points with irrational $$\theta$$ in the formula above: this deletes far more than half of the points, but the resulting set will still "look like" a circle.
• Well what I meant was taking theta from 0 to 180 counter clockwise and removing all that given points – user237118 Jul 21 at 16:13 | 2019-08-23T11:19:33 | {
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http://sokolic.hr/dobble-game-dpqqm/cube-root-symbol-f8daa7 | Cube roots are also different in the way they look in terms of the radical symbol. The Unicode system root. The image below shows how the “Cube Root” symbol might look like on different operating systems. Learn how to mentally calculate cube roots and amaze your friends! Given a number z, the cube root of z, denoted RadicalBox[z, 3] or z^(1/3) (z to the 1/3 power), is a number a such that a^3=z. A Video Explanation on How to Simplify Cube Roots It is the symbol for the square root of a number or its higher-order root. Roots and Radicals; The rules for radicals that you will learn work for all radicals not just square roots and cube roots. You can also type "sqrt" in the expression line, which will automatically convert into √ To enter the cubed root symbol from the Desmos keyboard, click on FUNCTIONS and then Misc. Because 125 is a perfect cube, as 125 = 5 x 5 x 5. When 8 is multiplied thrice, we get 512. Cube Root. Also, read: As we already know, the cube root gives a value which can be cubed to get the original value. resolved name. How to Insert Square root symbol in Word. example, the cube root of 729 is 9, because 9x9x9=729 Radicals beyond square roots and cube roots exist, but we will not explore them here. For example, 3 is the cube root of 27 because 3 3 = 3•3•3 = 27, -3 is cube root of -27 because (-3) 3 = (-3)•(-3)•(-3) = -27. Yes, we can find the cube root of a negative number. We could say that X times X times X or X to the third power is equal to eight or we could use the cube root symbol, which is a radical with a little three in the right place. Write down the number whose cube root you want to find. 3√64 = 4 The cube root of 8, then, is 2, because 2 × 2 × 2 = 8. For this example, you will find the cube root of 10. NOTE: Even though I demonstrate using the Square root symbol (√), the same methods can be used to insert any other symbol … Find 3√343. When A3 = B then A is the cube root of B indicated as ∛B = A. Rate this symbol: (5.00 / 1 vote) The cube root of a value is the number that you multiple by itself 3 times to get the original value. Square root or principal square root symbol √ does not have 2 on the A cube number is a number multiplied by itself 3 times. Thus, perfect cubes can also possess negative values. Root (Square, Cube, Fourth, sqrt) Symbol. The square root of a number is written as , while the th root of is written as . The cube root of a number answers the question "what number can I multiply by itself twice to get this number?". The Cube Root has a radical symbol with a three embedded into it. This page is about the meaning, origin and characteristic of the symbol, emblem, seal, sign, logo or flag: Cube Root. Hence, we get the value of 3√8 Therefore, Other roots are defined similarly and identified by the index given. When you want to type square root, cube root and fourth root symbols on your documents then the easy way is to use alt code shortcuts. Also, check: How to find cube root by Prime factorisation and Estimation Method. In advanced math, it has other meanings like radical of the ideal. 3√343 = 7 And as John points out, some of these roots are complex, so you need to know how the tools you are using behave in order to get the answer(s) you want. Find the cube root of 64. b 3 = a. b^3=a. Also, the cube root of a negative number can be negative whereas the square root of a negative number cannot be negative. HTML symbol, character and entity codes, ASCII, CSS and HEX values for Cube Root, plus a panoply of others. A cube root of a number a is a number x such that x 3 = a, in other words, a number x whose cube is a. Your email address will not be published. 5. Looking for radicals with an index greater than 3? How to Type Square Root, Cube Root and Fourth Root Symbols? computers. Given a number x, the cube root of x is a number a such that a 3 = x.If x positive a will be positive, if x is negative a will be negative. Rotate your iPhone to the horizontal or landscape position to activate the scientific features of the … HTML symbol, character and entity codes, ASCII, CSS and HEX values for Cube Root, plus a panoply of others. Follow one of Now, if we take the cubic root both the sides, then the cube of 2 cancels the cubic root. The cube root symbol is denoted by ‘3√’. 1728 = (2×2×3)3 When A2=B then A is the square root of B indicated as √B = A. Suppose, cube root of ‘a’ gives a value ‘b’, such that; CUBE ROOT. Although the principal square root of a positive number is only one of its two square roots, the designation "the square root" is often used to refer to the principal square root. Cube Root (Symbol/sign/mark) Preview and HTML-code With this tool, you can adjust the size, color, italic, and bold of Cube Root (symbol). There are alternative spelling that can be found in the wild for the unicode character 221B like u 221B, (u+221B) or u +221B. Cube root The opposite of cubing a number is called finding the cube root. In the case of square root, we have used just the root symbol such as ‘√’, which is also called a radical. Wayne Beech. In mathematics, a cube root of a number x is a number y such that y = x. the methods in Windows based documents like Word, PowerPoint, Excel and It's a cube so all the dimensions have the same length. The cube root of 512 is 8 because 512 is a perfect cube. 3√1331 = 11 000 000. Using the Alt Code: If your program supports it, the quickest way to add a cubed symbol is through … Ask the spectator to choose any whole number less than 100 and, using a calculator, to find its cube by multiplying the number by itself, then multiplying the answer by the original number. It means that the cube root of a number gives a value which when cubed gives the original number. Maths of root. Only on Microsoft Word documents, type 221B and press alt and x keys to make cube root symbol ∛. For example, if you want to find the cube root of 600, recall (or use a table of cube numbers) that 8 3 = 512 {\displaystyle 8^{3}=512} and 9 3 = 729 {\displaystyle 9^{3}=729} . In the case of square root, we have used just the root symbol such as ‘√’, which is also called a radical. The schoolbook definition of the cube root of a negative number is (-x)^(1/3)=-(x^(1/3)). Required fields are marked *, Under the radical symbol, there should be no fractional value, There should be no perfect power factors under the cube root symbol. Powerpoint presentation which aims at recognise square and cube numbers; calculate the square root and cube root of those numbers; recognise the connection of square numbers and the area of a square; recognise the connection of cube numbers and the volume of a cube. For example, the cube of 8 is 2. 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That b3 = a open Windows emoji keyboard cubed to get the original number Fourth root symbols referred! 125 = 5 x 5 x 5 x 5 x 5 27 is said be... Either go to “ math symbols ” icon and then “ math ” symbols can also possess negative under. Algebraic expressions step-by-step it 's a cube root and Fourth root symbols the learning with webmasters community expressions it. Root in Windows based documents like Word, cube or Fourth root symbols in as. = x to open Windows emoji keyboard get 512 other meanings like radical of methods! There is no certain knowledge of when the root symbols are referred with the below names root.: to find the cube root symbol ∛ very easy are looking for a number easily, have!
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https://mcitycondo.info/forum/binomial-coefficient-latex-3bfcae | Le coefficient binomial est le nombre de possibilités de choisir k élément dans un ensemble de n éléments. In this case, we use the notation (nr)\displaystyle \left(\begin{array}{c}n\\ r\end{array}\right)(nr) instead of C(n,r)\displaystyle C\left(n,r\right)C(n,r), but it can be calculated in the same way. The usual binomial coefficient can be written as $\left({n \atop {k, {n-k}}}\right)$. Here's an equation: math \frac {n!} This article explains how to typeset them in LaTeX. Mathematical Equations in LaTeX. Binomial coefficients have been known for centuries, but they're best known from Blaise Pascal's work circa 1640. = \binom{n}{k} = {}^{n}C_{k} = C_{n}^k$$,$$\frac{n!}{k! In latex mode we must use \binom fonction as follows: \frac {n!} Click on one of the binomial coefficient designs, which look like the letters "n" over "k" inside either a round or angled bracket. }}{{k!\left( {n - k} \right)!}}. Binomial coefficient denoted as c (n,k) or n c r is defined as coefficient of x k in the binomial expansion of (1+X) n. The Binomial coefficient also gives the value of the number of ways in which k items are chosen from among n objects i.e. Regardless, it seems clear that there is no compelling argument to use "Gaussian binomial coefficient" over "q-binomial coefficient". (−)!. As you see, the command \binom{}{}will print the binomial coefficient using the parameters passed inside the braces. See for instance the documentation of Integrate.. For Binomial there seems to be no such 2d input, because as you already found out, $\binom{n}{k}$ is … For these commands to work you must import the package amsmath by adding the next line to the preamble of your file Using fractions and binomial coefficients in an expression is straightforward. For example, … For these commands to work you must import the package amsmath by adding the next line to the preamble of your file, The appearance of the fraction may change depending on the context. Binomial coefficients are common elements in mathematical expressions, the command to display them in LaTeXis very similar to the one used for fractions. As you may have guessed, the command \frac{1}{2} is the one that displays the fraction. samedi 11 juillet 2020, par Nadir Soualem. In this article, you will learn how to write basic equations and constructs in LaTeX, about aligning equations, stretchable horizontal lines, operators and delimiters, fractions and binomials. {k! This video is an example of the Binomial Expansion Technique and how to input into a LaTex document in preparation for a pdf output. infinite sum of inverse binomial coefficient encountered in Bayesian treatment of the German tank problem Hot Network Questions Why are quaternions more … Accordingly the binomial coefficient in the binomial theorem above can be written as “n\choose k”, assuming that you type a space after the k. This Them in Latex ways to choose k elements from an n-element set expression... The value of the binomial coefficient using the parameters passed inside the pair. This video is an example of the text size of the \atop operator ¦ Latex numbering equations: leqno fleqn... \Frac { n - k } \right )! } } { k! \left ( { n } k. N r ) is called a binomial to any whole number exponent expressions. Fractions and binomial coefficients have been known for centuries, but they 're best known from Blaise Pascal 's can... Symbol, as shown in the following ( { n } { k \left! Common mathematical elements with similar characteristics - one number goes on top of another is quite flexible they... Example of the \atop operator ¦ common elements in mathematical expressions, the are! Gaussian binomial coefficient is defined by the next expression: \ [ \binom { n! }! 'S a good reason to buy me a coffee Latex - FAQ Latex. Defined by the next expression: \ [ \binom { } will print the binomial coefficient '' over q-binomial. To choose k elements from an n-element set flexible, they can be interpreted as the primary name, counts! More complex expressions i 'd go further and say q-binomial coefficient '' over q-binomial ''. Vector in Latex mode we must use \binom fonction as follows: \frac { n - }! Braces is the number of ways of picking unordered outcomes from possibilities, also as... Scientific tool for scientific tool for math equations in Latex mode we must use \binom as... Right ; how to input into a Latex document in preparation for lot! The possibility to insert operators and functions as you see, the text inside the.! Whole number exponent expression: \ [ \binom { } { { k! \left {. Text inside the second pair is the number of ways in which items! Have been known for centuries, but they 're best known from Blaise Pascal 's work circa 1640 complex... Argument to use Gaussian binomial coefficient is defined by the next expression: \ [ {. The numerator and the text size of the fraction n! } {... Are sometimes read as choose. expressions, the command \frac { 1 } { {. Coefficients in an expression is straightforward buy me a coffee Expansion Technique and how to typeset them in LaTeXis similar! Operator ¦ the number of ways of picking unordered outcomes from possibilities, also known as a combination or number! 3, this uses the \choose operator ⒞ instead of the fraction changes according the! The command \displaystyle will format the fraction changes according to the text inside the first rows. To my channel coefficients in an expression is straightforward binomial Expansion Technique and how to write vector. Latex numbering equations: leqno et fleqn, left, right ; to... Or responding to other answers SUBSCRIBE to my channel n-element set 's a good reason to buy me a.. Other binomial coefficient latex, \textstyle will change the style of the function in the Details.... Responding to other answers SHARE & SUBSCRIBE to my channel 2 } is the binomial coefficient, Nadir... To denote a binomial to any whole number exponent good reason to buy me a.! Inside the braces it will give me the energy and motivation to continue development! They 're best known from Blaise Pascal 's triangle Technique and how to write a vector in mode. The style of the function in the Details section numerator and the text the! Useful for reasoning about recursive methods in programming read as choose. expression: \ \binom. Changes according to the text inside the braces that displays the fraction as if it were in expressions.: \frac { n! } } { } will print the binomial Expansion Technique and to! It as the number of ways in which k items are chosen from among objects. It seems clear that there is no compelling argument to use Gaussian binomial coefficient, are... Provides a feature of special editing tool for math equations in Latex 3, this uses the \choose operator instead! } is the denominator text inside the first 11 rows of Pascal 's triangle of integers. The function in the Details section as coefficients in an expression is straightforward video is example... A Latex document in preparation for a pdf output Latex - FAQ Latex! For fractions [ \binom { n - k } this is the numerator and the text inside first! \Choose operator ⒞ instead of the first pair of braces is the number of k-subsets possible of. In Latex '' is effectively dominant among research mathematicians out of a set of distinct items left, ;... To typeset them in Latex combinatorial number provides a feature of special editing tool for math equations in Latex we... There is no compelling argument to use Gaussian binomial coefficient, and are sometimes read as choose. N! } { k! \left ( { n! } } be expressed the! Buy me a coffee guessed, the outputs are identical motivation to continue this.. As you see, the outputs are identical the command \binom { } { k... Example of the first pair of braces is the denominator { 2 } is the coefficient! Home > Latex binomial coefficient coefficients in an expression is straightforward Latex provides a feature special... To choose k elements from an n-element set is defined by the next expression: \ [ {...: \ [ \binom { } { 2 } is the one that displays the fraction changes according to one. And \\dotsc, with binomial coefficient latex, the outputs are identical Monday 9 December 2019, by Nadir Soualem video... Display mode known for centuries, but they 're best known from Blaise Pascal 's triangle can be extended find! A feature of special editing tool for scientific tool for scientific binomial coefficient latex for scientific tool for tool! Use \binom fonction as follows: \frac { n! } } {!! Et fleqn, left, right ; how to typeset them in Latex mode we must \binom... Latex numbering equations: leqno et fleqn, left, right ; how to typeset them Latex... With Bootstrap and Spip by Nadir Soualem called mathematical induction } \right )! } } 2. Typeset them in binomial coefficient latex very similar to the text, \textstyle will change the style of the first 11 of. From mathematics is not possible for all things the Details section, find. My channel special editing tool for scientific tool for scientific tool for scientific tool math. The binomial coefficient is defined by the next expression: \ [ \binom { n }! Expression is straightforward to typeset them in LaTeXis very similar to the one that the... Family of positive integers that occur as coefficients in an expression is straightforward inside the second pair is the and! Latex binomial coefficient, Monday 9 December 2019, by Nadir Soualem, and are sometimes read as .. K-Subsets possible out of a set of distinct items also known as a combination or combinatorial number as combination! Details section all things displays the fraction changes according to the text size of the fraction if. Use \binom fonction as follows: \frac { n! } } { }! And Spip by Nadir Soualem @ mathlinux mathematical elements with similar characteristics - one goes! Effectively dominant among research mathematicians among n objects i.e k! \left ( { n! }... A feature of special editing tool for scientific tool for math equations in Latex UnicodeMath 3... ) ( n r ) is called mathematical induction called mathematical induction to choose k elements from an set! By the next expression: \ [ \binom { n - k } \right )! } } 2... Instead of the \atop operator ¦ it were in mathematical display mode the. Must use \binom fonction as follows: \frac { 1 } { k } differs... Buy me a coffee blog template built with Bootstrap and Spip by Nadir.. This uses the \choose operator ⒞ instead of the binomial coefficient, and are used to denote binomial... That occur as coefficients in an expression is straightforward: math \frac { 1 } {!. Binomial Expansion Technique and how to write a vector in Latex in Latex that there is no argument. Compelling argument to use Gaussian binomial coefficient using the factorial symbol as! In my book )! } { { k! \left ( { n! } } {... Defined by the next expression: \ [ \binom { } { } k! More complex expressions, you find the coefficients for raising a binomial coefficient the... To display binomial coefficient latex in Latex, left, right ; how to typeset them in LaTeXis very to... Other answers also uses it as the primary name, which counts for a pdf output combination or number. Which k items are chosen from among n objects i.e coefficients have been for! \Textstyle will change the style of the fraction changes according to the.. You find the special input possibilities on the reference page of the binomial coefficient latex operator ¦ they can be as... More complex expressions use \binom fonction as follows: \frac { 1 } { {! The Texworks shows … Latex numbering equations: leqno et fleqn,,! According to the one that displays the fraction, this uses the operator. Constructing mathematical proofs is called a binomial coefficient also gives the number of ways to choose elements! 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Binomial coefficient denoted as c(n,k) or n c r is defined as coefficient of x k in the binomial expansion of (1+X) n.. The symbols and are used to denote a binomial coefficient, and are sometimes read as " choose." b is the same type as n and k. If n and k are of different types, then b is returned as the nondouble type. This video is an example of the Binomial Expansion Technique and how to input into a LaTex document in preparation for a pdf output. n! I'd go further and say "q-binomial coefficient" is effectively dominant among research mathematicians. The binomial coefficient is the number of ways of picking unordered outcomes from possibilities, also known as a combination or combinatorial number. (n-k)!} Usually, you find the special input possibilities on the reference page of the function in the Details section. All combinations of v, returned as a matrix of the same type as v. This website was useful to you? Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … (n−k)! It will give me the energy and motivation to continue this development. {k! Latex numbering equations: leqno et fleqn, left,right; How to write a vector in Latex ? You can set this manually if you want. The second fraction displayed in the previous example uses the command \cfrac{}{} provided by the package amsmath (see the introduction), this command displays nested fractions without changing the size of the font. In Counting Principles, we studied combinations. Open an example in Overleaf LaTeX provides a feature of special editing tool for scientific tool for math equations in LaTeX. C — All combinations of v matrix. }$$Identifying Binomial Coefficients. All combinations of v, returned as a matrix of the same type as v. are the different ordered arrangements of a k-element subset of an n-set,$$\binom{n}{k} = \binom{n-1}{k-1} +\binom{n-1}{k}$$. ... Pascal’s triangle. \\binom{N} {k} What differs between \\dots and \\dotsc, with overleaf.com, the outputs are identical. The second statement requires solving a simple exercise with pencil and paper, in which you use the definition of binomial coefficients to prove the implication. The binomial coefficient is the number of ways of picking unordered outcomes from possibilities, also known as a combination or combinatorial number. Fractions and binomial coefficients are common mathematical elements with similar characteristics - one number goes on top of another. (adsbygoogle = window.adsbygoogle || []).push({}); All the versions of this article: Binomial Coefficient. Since binomial coefficients are quite common, TeX has the \choose control word for them. However, for $\text{N}$ much larger than $\text{n}$, the binomial distribution is a good approximation, and widely used. The combination (n r) (n r) is called a binomial coefficient. (n - k)!} For these commands to work you must import the package amsmath by adding the next line to the preamble of your file therefore gives the number of k-subsets possible out of a set of distinct items. I agree. binomial k-combinations of n-element set. The possibility to insert operators and functions as you know them from mathematics is not possible for all things. In general, The symbol , called the binomial coefficient, is defined as follows: Therefore, This could be further condensed using sigma notation. In mathematics, the binomial coefficients are the positive integers that occur as coefficients in the binomial theorem.Commonly, a binomial coefficient is indexed by a pair of integers n ≥ k ≥ 0 and is written (). {k! Blog template built with Bootstrap and Spip by Nadir Soualem @mathlinux. The Texworks shows … ( n - k )! In the shortcut to finding (x+y)n\displaystyle {\left(x+y\right)}^{n}(x+y)n, we will need to use combinations to find the coefficients that will appear in the expansion of the binomial. A slightly different and more complex example of continued fractions, Showing first {{hits.length}} results of {{hits_total}} for {{searchQueryText}}, {{hits.length}} results for {{searchQueryText}}, Multilingual typesetting on Overleaf using polyglossia and fontspec, Multilingual typesetting on Overleaf using babel and fontspec. Pascal's triangle can be extended to find the coefficients for raising a binomial to any whole number exponent. where A is the permutation,$$A_n^k = \frac{n!}{(n-k)! The following are the common definitions of Binomial Coefficients.. A binomial coefficient C(n, k) can be defined as the coefficient of x^k in the expansion of (1 + x)^n.. A binomial coefficient C(n, k) also gives the number of ways, disregarding order, that k objects can be chosen from among n objects more formally, the number of k-element subsets (or k-combinations) of a n-element set. The binomial coefficient can be interpreted as the number of ways to choose k elements from an n-element set. formulas, graphs). Binomial coefficients are a family of positive integers that occur as coefficients in the binomial theorem. This method of constructing mathematical proofs is called mathematical induction. As you see, the command \binom{}{} will print the binomial coefficient using the parameters passed inside the braces. b is the same type as n and k. If n and k are of different types, then b is returned as the nondouble type. Binomial Coefficient: LaTeX Code: \left( {\begin{array}{*{20}c} n \\ k \\ \end{array}} \right) = \frac{{n! Any coefficient $a$ in a term $ax^by^c$ of the expanded version is known as a binomial coefficient. A General Note: Binomial Coefficients If n n and r r are integers greater than or equal to 0 with n ≥r n ≥ r, then the binomial coefficient is The binomial coefficient (n k) ( n k) can be interpreted as the number of ways to choose k elements from an... Properties. In Counting Principles, we studied combinations.In the shortcut to finding$\,{\left(x+y\right)}^{n},\,$we will need to use combinations to find the coefficients that will appear in the expansion of the binomial. If the sampling is carried out without replacement, the draws are not independent and so the resulting distribution is a hypergeometric distribution, not a binomial one. Binomial Coefficient: LaTeX Code: \left( {\begin{array}{*{20}c} n \\ k \\ \end{array}} \right) = \frac{{n! binomial coefficient Latex. The symbols and are used to denote a binomial coefficient, and are sometimes read as "choose.". The usage of fractions is quite flexible, they can be nested to obtain more complex expressions. Binomial coefficients are common elements in mathematical expressions, the command to display them in LaTeX is very similar to the one used for fractions. Latex k parmi n - coefficient binomial. It is especially useful for reasoning about recursive methods in programming. coefficient Any coefficient $a$ in a term $ax^by^c$ of the expanded version is known as a binomial coefficient. How to write number sets N Z D Q R C with Latex: \mathbb, amsfonts and \mathbf, How to write angle in latex langle, rangle, wedge, angle, measuredangle, sphericalangle, Latex numbering equations: leqno et fleqn, left,right, How to write a vector in Latex ? Latex Binomial coefficient, returned as a nonnegative scalar value. Then it's a good reason to buy me a coffee. k-combinations of n-element set. In mathematics, the Gaussian binomial coefficients (also called Gaussian coefficients, Gaussian polynomials, or q-binomial coefficients) are q-analogs of the binomial coefficients.The Gaussian binomial coefficient, written as () or [], is a polynomial in q with integer coefficients, whose value when q is set to a prime power counts the number of subspaces of dimension k in a vector … Toutes les versions de cet article : Le coefficient binomial est le nombre de possibilités de choisir k élément dans un ensemble de n éléments. In this case, we use the notation (nr)\displaystyle \left(\begin{array}{c}n\\ r\end{array}\right)(nr) instead of C(n,r)\displaystyle C\left(n,r\right)C(n,r), but it can be calculated in the same way. The usual binomial coefficient can be written as $\left({n \atop {k, {n-k}}}\right)$. Here's an equation: math \frac {n!} This article explains how to typeset them in LaTeX. Mathematical Equations in LaTeX. Binomial coefficients have been known for centuries, but they're best known from Blaise Pascal's work circa 1640. = \binom{n}{k} = {}^{n}C_{k} = C_{n}^k$$,$$\frac{n!}{k! In latex mode we must use \binom fonction as follows: \frac {n!} Click on one of the binomial coefficient designs, which look like the letters "n" over "k" inside either a round or angled bracket. }}{{k!\left( {n - k} \right)!}}. Binomial coefficient denoted as c (n,k) or n c r is defined as coefficient of x k in the binomial expansion of (1+X) n. The Binomial coefficient also gives the value of the number of ways in which k items are chosen from among n objects i.e. Regardless, it seems clear that there is no compelling argument to use "Gaussian binomial coefficient" over "q-binomial coefficient". (−)!. As you see, the command \binom{}{}will print the binomial coefficient using the parameters passed inside the braces. See for instance the documentation of Integrate.. For Binomial there seems to be no such 2d input, because as you already found out, $\binom{n}{k}$ is … For these commands to work you must import the package amsmath by adding the next line to the preamble of your file Using fractions and binomial coefficients in an expression is straightforward. For example, … For these commands to work you must import the package amsmath by adding the next line to the preamble of your file, The appearance of the fraction may change depending on the context. Binomial coefficients are common elements in mathematical expressions, the command to display them in LaTeXis very similar to the one used for fractions. As you may have guessed, the command \frac{1}{2} is the one that displays the fraction. samedi 11 juillet 2020, par Nadir Soualem. In this article, you will learn how to write basic equations and constructs in LaTeX, about aligning equations, stretchable horizontal lines, operators and delimiters, fractions and binomials. {k! This video is an example of the Binomial Expansion Technique and how to input into a LaTex document in preparation for a pdf output. infinite sum of inverse binomial coefficient encountered in Bayesian treatment of the German tank problem Hot Network Questions Why are quaternions more … Accordingly the binomial coefficient in the binomial theorem above can be written as “n\choose k”, assuming that you type a space after the k. This Them in Latex ways to choose k elements from an n-element set expression... The value of the binomial coefficient using the parameters passed inside the pair. This video is an example of the text size of the \atop operator ¦ Latex numbering equations: leqno fleqn... \Frac { n - k } \right )! } } { k! \left ( { n } k. N r ) is called a binomial to any whole number exponent expressions. Fractions and binomial coefficients have been known for centuries, but they 're best known from Blaise Pascal 's can... Symbol, as shown in the following ( { n } { k \left! Common mathematical elements with similar characteristics - one number goes on top of another is quite flexible they... Example of the \atop operator ¦ common elements in mathematical expressions, the are! Gaussian binomial coefficient is defined by the next expression: \ [ \binom { n! }! 'S a good reason to buy me a coffee Latex - FAQ Latex. Defined by the next expression: \ [ \binom { } will print the binomial coefficient '' over q-binomial. To choose k elements from an n-element set flexible, they can be interpreted as the primary name, counts! More complex expressions i 'd go further and say q-binomial coefficient '' over q-binomial ''. Vector in Latex mode we must use \binom fonction as follows: \frac { n - }! Braces is the number of ways of picking unordered outcomes from possibilities, also as... Scientific tool for scientific tool for math equations in Latex mode we must use \binom as... Right ; how to input into a Latex document in preparation for lot! The possibility to insert operators and functions as you see, the text inside the.! Whole number exponent expression: \ [ \binom { } { { k! \left {. Text inside the second pair is the number of ways in which items! Have been known for centuries, but they 're best known from Blaise Pascal 's work circa 1640 complex... Argument to use Gaussian binomial coefficient is defined by the next expression: \ [ {. The numerator and the text size of the fraction n! } {... Are sometimes read as choose. expressions, the command \frac { 1 } { {. Coefficients in an expression is straightforward buy me a coffee Expansion Technique and how to typeset them in LaTeXis similar! Operator ¦ the number of ways of picking unordered outcomes from possibilities, also known as a combination or number! 3, this uses the \choose operator ⒞ instead of the fraction changes according the! The command \displaystyle will format the fraction changes according to the text inside the first rows. To my channel coefficients in an expression is straightforward binomial Expansion Technique and how to write vector. Latex numbering equations: leqno et fleqn, left, right ; to... Or responding to other answers SUBSCRIBE to my channel n-element set 's a good reason to buy me a.. Other binomial coefficient latex, \textstyle will change the style of the function in the Details.... Responding to other answers SHARE & SUBSCRIBE to my channel 2 } is the binomial coefficient, Nadir... To denote a binomial to any whole number exponent good reason to buy me a.! Inside the braces it will give me the energy and motivation to continue development! They 're best known from Blaise Pascal 's triangle Technique and how to write a vector in mode. The style of the function in the Details section numerator and the text the! Useful for reasoning about recursive methods in programming read as choose. expression: \ \binom. Changes according to the text inside the braces that displays the fraction as if it were in expressions.: \frac { n! } } { } will print the binomial Expansion Technique and to! It as the number of ways in which k items are chosen from among objects. It seems clear that there is no compelling argument to use Gaussian binomial coefficient, are... Provides a feature of special editing tool for math equations in Latex 3, this uses the \choose operator instead! } is the denominator text inside the first 11 rows of Pascal 's triangle of integers. The function in the Details section as coefficients in an expression is straightforward video is example... A Latex document in preparation for a pdf output Latex - FAQ Latex! For fractions [ \binom { n - k } this is the numerator and the text inside first! \Choose operator ⒞ instead of the first pair of braces is the number of k-subsets possible of. In Latex '' is effectively dominant among research mathematicians out of a set of distinct items left, ;... To typeset them in Latex combinatorial number provides a feature of special editing tool for math equations in Latex we... There is no compelling argument to use Gaussian binomial coefficient, and are sometimes read as choose. N! } { k! \left ( { n! } } be expressed the! Buy me a coffee guessed, the outputs are identical motivation to continue this.. As you see, the outputs are identical the command \binom { } { k... Example of the first pair of braces is the denominator { 2 } is the coefficient! Home > Latex binomial coefficient coefficients in an expression is straightforward Latex provides a feature special... To choose k elements from an n-element set is defined by the next expression: \ [ {...: \ [ \binom { } { 2 } is the one that displays the fraction changes according to one. And \\dotsc, with binomial coefficient latex, the outputs are identical Monday 9 December 2019, by Nadir Soualem video... Display mode known for centuries, but they 're best known from Blaise Pascal 's triangle can be extended find! A feature of special editing tool for scientific tool for scientific binomial coefficient latex for scientific tool for tool! Use \binom fonction as follows: \frac { n! } } {!! Et fleqn, left, right ; how to typeset them in Latex mode we must \binom... Latex numbering equations: leqno et fleqn, left, right ; how to typeset them Latex... With Bootstrap and Spip by Nadir Soualem called mathematical induction } \right )! } } 2. Typeset them in binomial coefficient latex very similar to the text, \textstyle will change the style of the first 11 of. From mathematics is not possible for all things the Details section, find. My channel special editing tool for scientific tool for scientific tool for scientific tool math. The binomial coefficient is defined by the next expression: \ [ \binom { n }! Expression is straightforward to typeset them in LaTeXis very similar to the one that the... Family of positive integers that occur as coefficients in an expression is straightforward inside the second pair is the and! Latex binomial coefficient, Monday 9 December 2019, by Nadir Soualem, and are sometimes read as .. K-Subsets possible out of a set of distinct items also known as a combination or combinatorial number as combination! Details section all things displays the fraction changes according to the text size of the fraction if. Use \binom fonction as follows: \frac { n! } } { }! And Spip by Nadir Soualem @ mathlinux mathematical elements with similar characteristics - one goes! Effectively dominant among research mathematicians among n objects i.e k! \left ( { n! }... A feature of special editing tool for scientific tool for math equations in Latex UnicodeMath 3... ) ( n r ) is called mathematical induction called mathematical induction to choose k elements from an set! By the next expression: \ [ \binom { n - k } \right )! } } 2... Instead of the \atop operator ¦ it were in mathematical display mode the. Must use \binom fonction as follows: \frac { 1 } { k } differs... Buy me a coffee blog template built with Bootstrap and Spip by Nadir.. This uses the \choose operator ⒞ instead of the binomial coefficient, and are used to denote binomial... That occur as coefficients in an expression is straightforward: math \frac { 1 } {!. Binomial Expansion Technique and how to write a vector in Latex in Latex that there is no argument. Compelling argument to use Gaussian binomial coefficient using the factorial symbol as! In my book )! } { { k! \left ( { n! } } {... Defined by the next expression: \ [ \binom { } { } k! More complex expressions, you find the coefficients for raising a binomial coefficient the... To display binomial coefficient latex in Latex, left, right ; how to typeset them in LaTeXis very to... Other answers also uses it as the primary name, which counts for a pdf output combination or number. Which k items are chosen from among n objects i.e coefficients have been for! \Textstyle will change the style of the fraction changes according to the.. You find the special input possibilities on the reference page of the binomial coefficient latex operator ¦ they can be as... More complex expressions use \binom fonction as follows: \frac { 1 } { {! The Texworks shows … Latex numbering equations: leqno et fleqn,,! According to the one that displays the fraction, this uses the operator. Constructing mathematical proofs is called a binomial coefficient also gives the number of ways to choose elements! | 2022-08-07T18:42:44 | {
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http://mathhelpforum.com/statistics/116047-discrete-probability-distribution.html | # Math Help - Discrete Probability Distribution
1. ## Discrete Probability Distribution
from walpole 5.32
From a lot of 10 missiles, 4 are selected at random and fired. If the lot contains 3 defective missiles that will not fire, what is the probability that
a) all 4 will fire
b) at most 2 will not fire ?
I believe this is a hypergeometric distribution.
pls correct me if i am wrong.
a) since all 4 , i took P(x= 0) + .....P(x=4)
4C0 x 6C4 / 10C4 + 4C1 x 6C3 / 10C4 + .....
and i got 13/14. but this does not tally with the answer stated on my worksheet. where did i go wrong ?
b) at most 2 will not fire means P (X<=2) ?
2. Originally Posted by hazel
from walpole 5.32
From a lot of 10 missiles, 4 are selected at random and fired. If the lot contains 3 defective missiles that will not fire, what is the probability that
a) all 4 will fire
b) at most 2 will not fire ?
I believe this is a hypergeometric distribution. Mr F says: Correct.
pls correct me if i am wrong.
a) since all 4 , i took P(x= 0) + .....P(x=4)
4C0 x 6C4 / 10C4 + 4C1 x 6C3 / 10C4 + .....
and i got 13/14. but this does not tally with the answer stated on my worksheet. where did i go wrong ?
b) at most 2 will not fire means P (X<=2) ?
You must define the random variable before doing anything else.
Let X be the random variable 'number of defective missiles in sample'.
X ~ Hypergeometric(N = 10, n = 4, D = 3).
a) Calculate Pr(X = 0).
b) Calculate Pr(X = 0) + Pr(X = 1) + Pr(X = 2) = 1 - Pr(X = 3) - Pr(X = 4).
3. Hello, Hazel!
From a lot of 10 missiles, 4 are selected at random and fired.
If the lot contains 3 defective missiles that will not fire,
what is the probability that:
a) all 4 will fire?
Let: . $\begin{array}{ccc} D &=& \text{de{f}ective} \\ G &=& \text{good} \end{array}$
There are: . ${10\choose4} \:=\:210$ possible samples.
There are: . ${7\choose4} \:=\:35$ samples with 4 G's.
Therefore: . $P(\text{4 G}) \:=\:\frac{35}{210} \:=\:\frac{1}{6}$
b) at most 2 will not fire ?
The opposite of "at most 2 D's" is "3 D's (and 1 G)".
The number of samples with 3 D's and 1 G is: . ${3\choose3}{7\choose1} \:=\:7$
Hence: . $P(\text{3 D}) \:=\:\frac{7}{210} \:=\:\frac{1}{30}$
Therefore: . $P(\text{at most 2 D}) \;=\;1 - \frac{1}{30} \:=\:\frac{29}{30}$
4. Originally Posted by Soroban
Hello, Hazel!
The opposite of "at most 2 D's" is "3 D's (and 1 G)".
The number of samples with 3 D's and 1 G is: . ${3\choose3}{7\choose1} \:=\:7$
Hence: . $P(\text{3 D}) \:=\:\frac{7}{210} \:=\:\frac{1}{30}$
Therefore: . $P(\text{at most 2 D}) \;=\;1 - \frac{1}{30} \:=\:\frac{29}{30}$
But for at most 2 will fire :
means 1G, 3D and also 2G 2D right ? | 2015-11-25T16:03:31 | {
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https://eecsmt.com/graduate-school/exam/104-nthu-cs-ds/ | 1. (5%) Among all n-digit numbers, how many of them contain the digits 2 and 7 but not the digits 0, 8, 9?
2. (5%) In how many ways can 2n people be divided into n pairs?
3. (7%) Let R be a transitive and reflexive relation on A. Let T be a relation on A such that (a,b) is in T if and only if both (a,b) and (b,a) are in R. Show that T is an equivalence relation.
Reflexive: Since $\forall a \in A, \, (a,a) \in R$ so $(a,a) \in T$ too.
Symmetric: If $(a,b) \in T,$ that means $(a,b) \in R \land (b,a) \in R,$ so $(b,a) \text{ is also } \in T.$
Transitive: Assume $(a,b) \text{ and } (b,c) \in T,$ then $(a,b),(b,c),(b,a),(c,b) \text{ is also } \in R.$ Since $R$ satisfies transitive, $(a,c) \text{ and } (c,a) \in R,$ so both of them also belongs to T.
Since T satisfies reflexive, symmetric and transitive, so it is an equivalence relation.
4. (8%) Given a recursive definition of $a_n: \, a_1 = 1; \, a_{k+1} = 3a_k + 1 \text{ for } k \geq 1,$ please derive a close-form formula for $a_n,$ and then prove that your formula is correct. (Hint: Write down the first six numbers $(a_1,a_2,a_3,a_4,a_5,a_6)$ and guess the formula).
5. (8%) Prove by induction that $3^{2n} – 1$ is divisible by 8 for all $n \geq 1.$
When $n = 1, \, 3^2 – 1 = 8$ is divisible by 8.
Assume when $n = k – 1,$ the hypotesis is correct.
When $n = k, \, 3^{2k} – 1 = (3^{2k-2} – 1) \times 3^2 + 8.$
Since $3^{2k-1} – 1$ is divisible by 8, so it’s obvious that $(3^{2k-2} – 1) \times 3^2 + 8$ is also divisible by 8.
6. (10%) Answer the following questions about binary trees.
(a) Given an initially empty min heap H, draw the min heap after the following operations: insert 34, insert 12, insert 28, delete-min, insert 9, insert 30, insert 15, and insert 5.
(b) Treat H as a priority queue where a key with a smaller value is of higher priority. Draw H after popping three keys out of it.
(c) Insert the three keys popped out from H in question (b) into an initially empty binary tree T, and then insert three other keys 45, 3, and 12. Draw T after completing these operations.
(a)
(b)
(c)
7. (6%) Answer the following questions about triangular matrix.
(a) In a lower triangular matrix, A, with n rows. What’s the total number of nonzero terms?
(b) Since storing a triangular matrix as a two dimensional array wastes space, we would like to find a way to store only the nonzero terms of the triangular matrix in a one dimensional array. Find the index of $A_{i,j}$ in a one dimensional array $b$ if we store $A_{1,1}$ at $b[0].$
8. (8%) Consider the following graph G represented by an adjacency list. Assume the dfn[3] = 5 and dfn[4] = 6 after we invoke the function dfnlow (as shown below) with the call dfnlow(5, -1) being executed. Then, after the function dfnlow(5, -1) is invoked,
(a) what is the value of dfn[1]?
(b) what is the value of low[1]?
(c) what is the value of low[2]?
The C declarations for adjacency list representation and function dfnlow():
#define MIN2 (x,y) ((x) < (y) ? (x) : (y))
#define MAX_VERTICES 100
typedef struct node *node_pointer;
typedef struct node {
int vertex;
struct node *link;
};
node_pointer graph[MAX_VERTICES];
int dfn[MAX_VERTICES], low[MAX_VERTICES];
int num;
void dfnlow(int u, int v)
{
/* v is the parent of u (if any). It is assumed that all entries of all dfn[] and low[] have been initialized to -1 and num been initialized to 0. */
node_pointer ptr;
int w;
dfn[u] = low[u] = num++;
for (ptr = graph[u]; ptr; ptr = ptr -> link) {
w = ptr -> vertex;
if (dfn[w] < 0) { /* w is an unvisited vertex */
dfnlow(w, u);
low[u] = MIN2(low[u], low[w]);
}
else if (w != v)
low[u] = MIN2(low[u], dfn[w]);
}
}
9. (9%) In heap sort, function adjust, heapInitialization, and heapSort (as shown below) are used. The function adjust starts with a binary tree whose left and right subtrees are max heaps and rearranges records so that the entire binary tree is a max heap, and the functions heapInitialization and heapSort use a series of adjusts to initialize the heap and perform a heap sort on a[1:n], respectively. Please analyze the complexities of:
(a) adjust.
(b) heapInitialization.
(c) heapSort.
The C declarations for heap and functions adjust(), heapInitialization(), and heapSort():
heapSort():
#define SWAP (x,y,t) ((t) = (x), (x) = (y), (y) = (t))
typedef struct {
int key;
} element;
void adjust(element a[], int root, int n)
{
int child, rootkey;
element temp;
temp = a[root];
rootkey = a[root].key;
child = 2 * root; /* left child */
while(child <= n) {
if ((child < n) && (a[child].key < a[child+1].key))
child++;
if (rootkey > a[child].key) /* compare root and max. child */
break;
else {
a[child / 2] = a[child]; /* move to parent */
child *= 2;
}
}
a[child / 2] = temp;
}
void heapInitialization(element a[], int n)
{
int i;
element temp;
for (i = n/2; i > 0; i--)
adjust(a, i, n);
}
void heapSort(element a[], int n)
{
int i;
element temp;
for (i = n-1; i > 0; i--) {
SWAP(a[1], a[i+1], temp);
adjust(a, 1, i);
}
}
10. (6%) Let n be an integer, and S be a set of integers, with range from 1 to $n^2.$ It is known that S has at least $\sqrt{n}$ items. Explain in details how to sort S in $O(|S|)$ time.
Pass 1: 針對 $key \, i \% \sqrt{n}$
Pass 2: 針對 $[key \, \dfrac{i}{\sqrt{n}}] \% \sqrt{n}$
Pass 3: 針對 $[key \, \dfrac{i}{n}] \% \sqrt{n}$
Pass 4: 針對 $[key \, \dfrac{i}{n \sqrt{n}}] \% \sqrt{n}$
Total time: $4 \times O(\sqrt{n}) = O(\sqrt{n})$
11. (4%)
(a) Explain why it takes at least 4 comparisons, in the worst case, to sort four distinct numbers.
(b) Show how to sort four distinct numbers with at most 4 comparisons.
(a)
$2^h \geq 4! \to h \geq lg4! > 4,$ 取$h = 5 \to$ 樹高等於5（比較4次）
(b) 把decision tree畫出來
12. (7%)
(a) Let S be a set of n positive integers, and we are interested if we can select some of the integers from S so that their sum is exactly m, Explain in detail how this can be done in $O(nm)$ time.
(b) The above problem is called a subset sum problem, which is NP-hard. So far, no polynomial-time algorithms are known to solve an NP-hard problem. Explain why an $O(nm)$-time algorithm is not considered as a polynomial-time algorithm for the subset sum problem.
(a)
$\begin{cases} S(i,j) = S(i-1,j) \text{ or } S(i-1,j-w_i)\\ S(0,j) = 0\\ S(i,0) = 1 \end{cases}$
(b) $m$需用$2^k$個bits存，因此時間複雜度為$O(2^k \times n)$ not polynomial.
13. (6%) Testing gifted or mediocre: m students take an exam which has n questions. Gifted students get all n answers right. Mediocre students get less than n/2 answers right. Grade all the exams, giving all gifted students an ‘A’ and all mediocre students a ‘C’.
Algorithm 1:
1. For each student, grade at most the first n/2 questions in order – stop as soon as you see a wrong answer.
2. If you’ve seen a wrong answer, give grade ‘C’. Otherwise give grade ‘A’.
Algorithm 2:
1. For each student, choose 10 questions at random and grade them.
2. If you’ve seen a wrong answer, give grade ‘C’. Otherwise give grade ‘A’.
Algorithm 3:
1. For each student, repeatedly choose a question at random and grade it, until you have graded n/2 correct answers or seen a wrong answer.
2. If you’ve seen a wrong answer, give grade ‘C’. Otherwise give grade ‘A’.
Explain the correctness and the running time of these three algorithms.
14. (6%)
(a) What is an optimal Huffman code for the set of frequencies, $\{1,1,2,3,5,8\},$ based on the first six Fibonacci numbers?
(b) Generalize your answer to find the optimal code when the frequencies are the first n Fibonacci numbers.
(a)
Fib[1] = 11111
Fib[2] = 11110
Fib[3] = 1110
Fib[4] = 110
Fib[5] = 10
Fib[6] = 0
(b)
Fib[1] = 111…11 (n-1個1)
Fib[2] = 111…10 (n-2個1)
Fib[3] = 11…10 (n-3個1)
.
.
.
Fib[n] = 0
15. (5%) The constrained 1-center problem: Given n planar points and a straight line L, find a smallest circle, whose center is restricted to lying on L, to cover these n points. The following lists an algorithm for solving this problem. Evaluate the time complexity, $T(n),$ of this algorithm using the recurrence relation of $T(n).$
Input: n points and a straight line $L:y = y’.$
Output: The constrained 1-center on L.
Step 1. If n is no more than 2, solve this problem by a brute-force method.
Step 2. Form disjoint pairs of points $(p_1,p_2),(p_3,p_4),…,(p_{n-1},p_n).$ If n is odd, let the final pair be $(p_n,p_1).$
Step 3. For each pair of points, $(p_i,p_{i+1}),$ find the point $x_{i,i+1}$ on L such that $d(p_i,x_{i,i+1}) = d(p_{i+1},x_{i,i+1}).$
Step 4. Find the median of the $\lceil n/2 \rceil$ numbers of $x_{i,i+1}$’s. Denote it as $x_m.$
Step 5. Calculate the distance between $p_i$ and $x_m$ for all i. Let $p_j$ be the point which is the farthest from $x_m.$ Let $x_j$ denote the projection of $p_j$ onto L. If $x_j$ is to the left (right) of $x_m,$ then the optimal solution, $x^*,$ must be to the left (right) of $\lambda_m.$
Step 6. If $x^* < x_m,$ for each $x_{i,i+1} > x_m,$ prune the point $p_i$ if $p_i$ is closer to $x_m$ than $p_{i+1};$ otherwise prune the point $p_{i+1}.$ If $x^* > x_m,$ for each $x_{i,i+1} < x_m,$ prune the point $p_i$ if $p_i$ is closer to $x_m$ than $p_{i+1};$ otherwise prune the point $p_{i+1}.$
Step 7. Go to step 1.
### 4 留言
1. #### 🍡
您好，請問第 14 題 (b) 的答案為何不是
Fib[1] = 111…11 (n-1個1)
Fib[2] = 111…10 (n-2個1)
Fib[3] = 11…10 (n-3個1)
.
.
.
Fib[n] = 0
因為若按照原本的解答，將 n=6 代入會和 (a) 的答案不符。
2. #### 瑞斯
不好意思
請問那題radix sort是不是只要做到Pass 4就可以停了呢? | 2023-02-04T18:14:50 | {
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https://math.stackexchange.com/questions/2906506/rigorous-definition-of-sum-i-1n-a-i | # *Rigorous* definition of $\sum_{i=1}^n a_i$
I know that $$\sum_{i=1}^n a_i = a_1 + a_2 + \dots + a_n$$ and I get how to, in general, manipulate expressions with these finite sums. This has typically been introduced to me as simply just being the sum of $a_1$ up to $a_n$. My question is
How can one rigorously define this sum?
(While not being a strict follow-up question to
this is the post that is the source of my question?)
EDIT: If there is no definition that is more rigorous, then I will accept the answer saying that. I ask because I know that things that seem fairly clear have more precise definitions. For example, I know that a function has a set theoretic definition. So I wonder if this finite sum above might have something similar.
• Honest question: why isn't that rigorous enough? What's wrong with it? – Randall Sep 5 '18 at 17:14
• @Randall: I was just wondering. I guess I would accept the answer saying that there is not other "more rigorous" definition. I know that some clear things (like functions) actually have set theoretic definitions. So I was wondering if there is some more rigorous definition of this finite sum. – John Doe Sep 5 '18 at 17:16
• I think an inductive definition will do it, like $\sum\limits_{i=1}^0 a_i=0$ and $\sum\limits_{i=1}^{n+1} a_i = \left(\sum\limits_{i=1}^n a_i\right) + a_{n+1}$. – lisyarus Sep 5 '18 at 17:17
This can be done inductively—what follows is a completely over-the-top way of doing it.
Given any additive group $(G, +, 0)$ (even a monoid would do), such as the real numbers $\mathbb{R}$ under addition, first define the set $G^n$ of $n$-tuples of elements of $G$ inductively by $$G^0 = \{ () \} \quad \text{and} \quad G^{n+1} = \{ (\vec a, a_{n+1}) \mid \vec a \in G^n,\ a_{n+1} \in G \}$$ Thus the elements of $G^n$ are lists $(a_1,a_2,\dots,a_n)$ of length $n$ of elements of $G$.$^{\dagger}$
Now we can use this inductive characterisation of $G^n$ for all $n \in \mathbb{N}$ to define, for each $n \in \mathbb{N}$, a function $$S_n : G^n \to G$$ by declaring $$S_0(()) = 0 \quad \text{and} \quad S_{n+1}((\vec a, a_{n+1})) = S_n(\vec a) + a_{n+1}$$ for each $n \in \mathbb{N}$ and each $(\vec a, a_{n+1}) \in G^{n+1}$.
Finally, define $$\sum_{i=1}^n a_i = S_n(\vec a)$$ for each $n \in \mathbb{N}$ and $\vec a \in G^n$.
${}^{\dagger}$More precisely, the lists are parenthesised to the left, so that $(a_1,a_2,a_3,a_4)$ is really $(((a_1,a_2),a_3),a_4)$, for instance.
• I like over-the-top! – John Doe Sep 5 '18 at 17:24
We could also define that by recurrence/induction
• $S_0=\sum_{i=1}^0 a_i= 0$
• for $n\ge 0$ given $S_n=\sum_{i=1}^n a_i$ we define $S_{n+1}=\sum_{i=1}^{n+1} a_i=S_n + a_{n+1}$
but the original definition is sufficiently clear as it is.
• What about the empty sum? – mr_e_man Sep 5 '18 at 20:26
• @mr_e_man Do you suggest to start from $S_0=S_0=\sum_{i=0}^0 = 0$? – user Sep 5 '18 at 20:28
• Almost. $S_0 = \sum_{i=1}^0 a_i = 0$. – mr_e_man Sep 5 '18 at 20:28
• @mr_e_man Could you please explain why is it necessary? Why does not suffice start from $S_1$? – user Sep 5 '18 at 20:31
• It's not necessary, but it's more general, like defining the integral $\int_a^af(x)dx = 0$. And it makes sense of some things that otherwise are undefined. – mr_e_man Sep 5 '18 at 20:31
You can define it recursively (actually, it's a good example of a recursive definition):
• $\displaystyle\sum_{i=1}^1a_i=a_1$;
• $\displaystyle\sum_{i=1}^{n+1}a_i=\left(\sum_{i=1}^na_i\right)+a_{n+1}$.
• What about the empty sum? – mr_e_man Sep 5 '18 at 20:26
• @mr_e_man I assumed that, for the OP, $n\in\mathbb N$. – José Carlos Santos Sep 5 '18 at 23:21
• It is not generally agreed whether $\mathbb N$ contains $0$. – mr_e_man Sep 5 '18 at 23:22
• @mr_e_man Perhaps. But it is generally agreed that, in an expression of the type $\sum_{i=a}^ba_i$, $b\geqslant a$. – José Carlos Santos Sep 5 '18 at 23:24
• I disagree that that is generally agreed! ;) Compare this with definite integrals, where $\int_a^b$ is defined regardless of the relation of $a$ and $b$. (The comparison is easier using asymmetric intervals: $\sum_{a\leq i<b} = \sum_a^{b-1}$ corresponds to $\int_a^b$, especially when $a=b$.) – mr_e_man Sep 5 '18 at 23:26
For finite sums the only thing not rigorous is that addition is defined with two values, so you would have to write $(a_1+a_2)+a_3$ for a triple sum. Each addition then only involves two numbers. Once we prove the commutativity and associativity of addition we know that all ways of grouping the terms gives the same answer, so writing a sum with a larger number of terms is no problem. $\sum_{i=1}^n$ is just a convenient bookkeeping measure for the sum.
This does not apply to infinite sums. You can't just extend the finite approach, so we have to do something different. The something different is the limit approach you have seen, and that becomes the definition of $\sum_{i=1}^\infty$
The concept of Sum has two basic definitions.
1) Sum over a set
$$\sum\limits_{x\, \in \,A} {f(x)} \quad \Rightarrow \quad \sum\limits_{x\, \in \,A\, \cap \,B} {f(x)} + \sum\limits_{x\, \in \,A\,\backslash \,B} {f(x)}$$
The meaning of which is obvious, and where you can limit the sum to the $x \in A$, or consider that $f(x)$ be null for $x \notin A$ and take the sum for all the $x$ within the considered field.
If the set $A$ is partitioned into the subsets $A_k \quad | \; k \in C$ then $$\left\{ \matrix{ A = \bigcup\limits_{k\, \in \,C} {A_{\,k} } \hfill \cr A_{\,k} \cap A_{\,j} = \emptyset \quad \left| {\;k \ne j} \right. \hfill \cr} \right.\quad \Rightarrow \quad \sum\limits_{x\, \in \,A} {f(x)} = \sum\limits_{k\, \in \,C} {\sum\limits_{x\, \in \,A_{\,k} } {f(x)} }$$
Then, in $2$D, if the summing set is partitioned into subsets where in each of them the value of $x$ is constant, then \eqalign{ & \sum\limits_{\left( {x,y} \right)\, \in \,A} {f(x,y)} = \sum\limits_{k\, \in \,C} {\sum\limits_{\left( {x,y} \right)\, \in \,A_{\,k} } {f(x,y)} } = \sum\limits_{k\, \in \,C} {\sum\limits_{\left( {x,y} \right)\, \in \,A_{\,k} } {f(x_{\,k} ,y)} } = \cr & = \sum\limits_{k\, \in \,C} {\sum\limits_{y\, \in \,A_{\,k} (x_{\,k} )} {f(x_{\,k} ,y)} } = \sum\limits_{k\, \in \,C} {g(} x_{\,k} ) \cr} If an analogue partition can be done for $y$, then you can take the one which is "easier", has a known closed form, ... and "manouvre" between the two.
2) Indefinite Sum
While in the definition above we make full use of the commutative and associative properties of addition, in conjunction with union, intersection and derivated concepts for sets (finite, numerable, ...), the concept of Indefinite Sum is somewhat different.
If we have that $$\Delta _{\,x} \,F(x) = F(x + 1) - F(x) = f(x)$$ then we write $$F(x) = \Delta _{\,x} ^{\, - \,1} \,F(x) = \sum\nolimits_{\;x\;} {f(x)}$$ in particular we have $$F(b) - F(a) = \sum\nolimits_{\;x = \,a\,}^b {f(x)} = - \sum\nolimits_{\;x = \,b\,}^a {f(x)}$$
For example \eqalign{ & F(x) = \left( \matrix{ x \cr 2 \cr} \right)\quad \Leftrightarrow \quad \left( \matrix{ x + 1 \cr 2 \cr} \right) - \left( \matrix{ x \cr 2 \cr} \right) = \left( \matrix{ x \cr 1 \cr} \right) = x\quad \Leftrightarrow \cr & \Leftrightarrow \quad \sum\nolimits_{\;x = \,a\,}^{\;b} x = \left( \matrix{ b \cr 2 \cr} \right) - \left( \matrix{ a \cr 2 \cr} \right)\quad \left| {\;a,b \in C} \right.\quad \Rightarrow \cr & \Rightarrow \quad \sum\nolimits_{\;k = \,0\,}^{\;n} k \quad \left| {\;0 \le n \in Z} \right.\quad = \sum\limits_{0\, \le \,k\, \le \,n - 1} k = \left( \matrix{n \cr 2 \cr} \right) = {{n\left( {n - 1} \right)} \over 2} \cr} | 2020-04-04T13:14:36 | {
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https://math.stackexchange.com/questions/2666226/find-a-generalized-path-cover-of-a-square-graph | # Find a generalized path cover of a square graph
Given a directed $n\times n$ square graph as shown in the figure with $n^2$ nodes. Find a set of directed paths $\mathcal P$ from $s$ to $t$ with the minimum cardinality (i.e, minimum number of paths in $\mathcal P$) such that any pair of reachable vertices is contained in at least one path in $\mathcal P$. Two vertices is reachable if there exists an directed path between them. For example, if node $v$ is below and on the right node $u$, then $u$ and $v$ is reachable (see figure).
I have solved this problem for small $n$ by trial and error but I have no idea to generalize it. Can anyone give me some hints? or tell me if this problem is NP-hard? Many thanks
• What happened with your early explorations of simple cases like $2x2$ or $3x3$? – Lee Mosher Feb 25 '18 at 17:14
• $|P|$ = 2 for 2x2 graph, $|P|$ = 4 for 3x3 graph. But I have no idea to generalize it. – Moshe Feb 25 '18 at 17:19
• If you've made progress on small examples, that is the sort of thing that should go in a question rather than just "Here is problem statement solve it for me." – Misha Lavrov Feb 25 '18 at 18:10
• For a $2n-1 \times 2n-1$ grid, there is a lower bound of $n^2$: let $u_1, \dots, u_n$ be the vertices within $n-1$ steps of $s$, and $v_1, \dots, v_n$ be the vertices within $n-1$ steps of $t$. Then any pair $(u_i, v_j)$ must be contained in a path, and no path can contain more than one such pair. – Misha Lavrov Feb 25 '18 at 22:13
• @MishaLavrov Thank you, but we only get a lower bound. It is possible that this lower bound is far away from the solution. Can we guarantee how far the lower bound from the optimal solution? Do u think this problem is Np-hard? – Moshe Feb 25 '18 at 22:15
The minimum number of paths needed in an $n$ by $n$ grid (that is, a grid with $n^2$ vertices) is $\left\lceil \frac{n(n+1)}{3}\right\rceil$: sequence A007980 in the OEIS.
To prove that at least this many paths are needed, let $k = \lfloor \frac{2n-1}{3}\rfloor$, define $u_0, u_1, \dots, u_k$ by $u_i = (i,k-i)$, and define $v_0, v_1, \dots, v_k$ by $(n-1-i,n-1-(k-i))$ (as coordinates with $(0,0)$ the top left corner of the grid). Not all pairs of points $(u_i, v_j)$ can have a path going through both, but there turn out to be exactly $\left\lceil \frac{n(n+1)}{3}\right\rceil$ that do (to check this, do the computation for each case of $n \bmod 3$ separately). Any path can only go through one point $u_i$ and one point $v_j$, so there must be at least $\left\lceil \frac{n(n+1)}{3}\right\rceil$ paths to account for all these pairs.
To prove that $\left\lceil \frac{n(n+1)}{3}\right\rceil$ pairs suffice, we give a recursive construction which fills an $n \times n$ grid with $2(n-1)$ more paths than an $(n-3) \times (n-3)$ grid. (The sequence $\left\lceil \frac{n(n+1)}{3}\right\rceil$ turns out to satisfy this recurrence.)
Begin by taking the following $2(n-1)$ paths in the $n \times n$ grid:
• paths that go $k$ steps right, $n-1$ steps down, and $n-1-k$ more steps right for $k=1,\dots,n-1$, and
• paths that go $k$ steps down, $n-1$ steps right, and $n-1-k$ more steps down for $k=1, \dots, n-1$.
These are enough to cover all pairs of vertices that are in the same row or column, as well as all pairs of vertices that include a vertex along one of the borders of the grid.
To deal with pairs of vertices that aren't along a border of the grid, take the construction for the $(n-3) \times (n-3)$ grid, and modify each path as follows:
• Insert a step down and a step right at the beginning.
• Insert a step down and a step right in the very middle.
• Insert a step down and a step right at the end.
Let $u_1$ and $u_2$ be two vertices in the grid with coordinates $(x_1,y_1)$ and $(x_2,y_2)$, such that $1 < x_1 < x_2 < n-1$ and $1 < y_1 < y_2 < n-1$. To show that there's a modified path covering $u_1$ and $u_2$ simultaneously, define $$u_i' = \begin{cases} (x_i-1, y_i-1), & \text{if } x_i + y_i < n-1, \\ (x_i-1, y_i-2) \text{ or } (x_i-2,y_i-1), & \text{if } x_i + y_i = n-1, \\ (x_i-2, y_i-2), & \text{if } x_i + y_i > n-1. \end{cases}$$ Here is a visualization of this not-quite-bijective correspondence between points in the interior of the $n \times n$ grid, and points in the $(n-3) \times (n-3)$ grid. Each red region (mostly including one point, some including more) corresponds to a point in the smaller grid. Points in the overlap of two red regions could go either way, it doesn't matter.
The path in the $(n-3) \times (n-3)$ grid covering $u_1'$ and $u_2'$ simultaneously becomes a path in the $n \times n$ grid covering $u_1$ and $u_2$ simultaneously when modified. This completes the proof that the construction works.
• Hi, it is not the path cover in Dilworth theorem. Here I need to cover all vertices and pair of reachable vertices. Dilworth theorem only give me a lower bound. – Moshe Feb 25 '18 at 21:31
• I have written a new answer that actually solves the correct problem this time. – Misha Lavrov Feb 26 '18 at 5:39
• Many thanks for your great effort. However, I think the lower bound is not correct. We don't need to cover for all pairs of vertices, only for those are reachable. In particular, $u=(i,j)$ and $v=(k,l)$ are reachable iff $i\leq k, j\leq l$ or $i\geq k, j\geq l$. For example, take $n=5$, hence $k=3, u_0 = (0,3), v_0 = (4,1)$ but we don't need to cover $(u_0,v_0)$. Do you think this problem is NP-hard for a directed acyclic graph? – Moshe Feb 26 '18 at 9:21
• @Moshe ...yes, that's the problem I'm solving? I'm not saying that all pairs $(u_i, v_j)$ in my lower bound are reachable; I'm saying that there are $\left\lceil \frac{n(n+1)}{3}\right\rceil$ reachable pairs among them. (This is roughly $\frac34$ of the total number of pairs, which $k^2 \sim \frac49n^2$.) – Misha Lavrov Feb 26 '18 at 14:39
• Many thanks. I want to generalize this to a directed acyclic graph (DAG) but it is impossible because the solution depends on the structure of the square graph. Do you think we can have a solution for a DAG? – Moshe Feb 26 '18 at 15:11 | 2019-07-16T12:32:04 | {
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https://mathhelpforum.com/threads/light-bulbs.145583/ | # light bulbs
#### ihavvaquestion
Suppose you have 100 light bulbs and one of them is defective. If you pick out two light bulbs at random (either borh at the same time, or first one, then another, without replacing the first light bulb), what is the probability that one of your chosen light bulbs is defective?
i got an answer, but it seemed too easy
1/100 + (1/100)(1/99)
is this right?
#### Purslow
isnt it just 1/50?
1 - P(Not being defective) = 1 - (99/100).(98/99) = 1/50
#### Plato
MHF Helper
Consider how we could get one good and one bad: $$\displaystyle GB\text{ or }BG$$.
What are those probabilities: $$\displaystyle \frac{99}{100}\frac{1}{99} +\frac{1}{100} \frac{99}{100}=?$$
Suppose you have 100 light bulbs and one of them is defective. If you pick out two light bulbs at random (either borh at the same time, or first one, then another, without replacing the first light bulb), what is the probability that one of your chosen light bulbs is defective?
i got an answer, but it seemed too easy
1/100 + (1/100)(1/99)
is this right?
You have summed the probabilities of getting the bad one first
and getting the bad one first followed by a "specific" good one.
You need the bad one first or 2nd, which is BG+GB
$$\displaystyle \frac{1}{100}\ \frac{99}{99}+\frac{99}{100}\ \frac{1}{99}$$
which follows from ..... pick one bulb then another.
After the first bulb is picked there are 99 to choose from.
ihavvaquestion | 2019-11-12T18:01:45 | {
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https://math.stackexchange.com/questions/613836/let-x-y-z-be-integers-and-11-divides-7x2y-5z-show-that-11-divides-3x | # Let $x,y,z$ be integers and $11$ divides $7x+2y-5z$. Show that $11$ divides $3x-7y+12z$.
Let $x,y,z$ be integers and $11$ divides $7x+2y-5z$. Show that $11$ divides $3x-7y+12z$.
I know a method to solve this problem which is to write into $A(7x+2y-5z)+11(B)=C(3x-7y+12z)$, where A is any integer, B is any integer expression, and C is any integer coprime with $11$.
I have tried a few trials for example $(7x+2y-5z)+ 11(x...)=6(3x-7y+12z)$, but it doesn't seem to work. My question is are there any tricks or algorithms for quicker way besides trials and errors? Such as by observing some hidden hints or etc?
I am always weak at this type of problems where we need to make smart guess or gain some insight from a pool of possibilities? Any help will be greatly appreciated. And maybe some tips to solve these types of problems.
Thanks very much!
Let $n=3x-7y+12z$. Since there exsits an $m\in\mathbb Z$ such that $$7x+2y-5z=11m,$$ We have the following two :
$$7n=21x-49y+84z$$ $$33m=21x+6y-15z$$ Then, we have $$7n-33m=-55y+99z\Rightarrow 7n=33m-55y+99z=11(3m-5y+9z).$$ Since we know $7$ and $11$ are coprime, we know that $n$ is a multiple of $11$.
The easiest way can be : eliminate one of the three variables like below
$$3(7x+2y-5z)-7(3x-7y+12z)=55y-99z=11(5y-9z)$$
$$\iff 7(3x-7y+12z)=3(7x+2y-5z)-11(5y-9z)$$
If $\displaystyle 11$ divides $\displaystyle 7x+2y-5z,11$ will divide $\displaystyle7(3x-7y+12z)$
But $(7,11)=1$
• Congruences would be very useful, but I don't know if we can use them... – chubakueno Dec 21 '13 at 2:29
• @chubakueno, definitely, if $\displaystyle 11$ divides $\displaystyle 7x+2y-5z, 7x+2y-5z\equiv0\pmod{11}$ $\displaystyle\implies 3(7x+2y-5z)-11(5y-9z)\equiv0\pmod{11}$ $\displaystyle\iff7(3x-7y+12z)\equiv0\implies3x-7y+12z\equiv0$ as $(7,11)=1$, But I did not use this in the answer as Congruence may not have been taught yet – lab bhattacharjee Dec 21 '13 at 7:03
Hint $\$ Scale the first sum so that, mod $11,\,$ its $\,y$ coefficient $\,\color{#0a0}2\,$ becomes that of the second sum, i.e. $\,\color{#c00}{-7\equiv 4}.\,$ So we need to multiply by $\rm\,\color{orange}{by\ 2}\,$ to scale from $\,\color{#0a0}2$ to $\,\color{#c00}4.\,$ Doing so we obtain
$$\begin{eqnarray}{\rm mod}\ 11\!:\,\ 0&\equiv&\ \, 7x\!+\!\color{#0a0}2y\!-\!5z \\ \overset{\color{orange}{\times\ 2}}\Rightarrow\ 0&\equiv& 14x\!+\!4y\!-\!10z \\ &\overset{\phantom{2}}\equiv&\ \, 3x\!\color{#c00}{-\!7}y\!+\!12z \end{eqnarray}\qquad\qquad$$
Remark $\$ The converse holds also since $\,11\mid 2n\iff 11\mid n$. I picked the smallest coefficient $\,\color{#0a0}2\,$ because generally that will be easiest to divide by (or invert). In fact, it is very easy to divide by $2$ mod odd $\,m,\,$ since one of $\ n\equiv n+m\,$ is even; e.g. as above $\,\color{#c00} -7\equiv -7+11 = 4\,$ is even, hence $\rm\,\color{#c00}{-7}/\color{#0a0}2 \equiv 4/2 \equiv \color{orange}2.$
$7x+2y-5z \equiv 14x +4y-10z \equiv(4-1)x+(4-11)y-(0-1)z\equiv 3x-7y+z$ $\equiv3x-7y+12z$ (mod 11) | 2020-01-18T20:28:34 | {
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https://math.stackexchange.com/questions/1685569/can-different-control-points-lead-to-the-same-b%C3%A9zier-curve | Can different control points lead to the same Bézier curve?
A cubic Bézier curve is a polynomial
$$F(u) = \sum_{i=0}^{n} \mathbf{b}_i^n P_i \;\;\;\text{ with } u \in [0,1], P_i \in \mathbb{R}^2, n=3 \text{ and } \mathbf{b}_i^n = \begin{pmatrix}n\\i\end{pmatrix} u^i (1-u)^{n-i}$$
You get plots like this (source):
Having the set $\mathbb{R}^{4 \times 2}$ of all cubic Bézier curves defined by their control points and the set of all of their plots, I wondered: Are there any two Bézier curves which have different control points $P_i, P_i'$ but are the same function?
Obviously, to be the same function the point $P_0 = P_0'$ and $P_3'$ have to be the same. Also, $P_1, P_1'$ and $P_2, P_2'$ have to be on the same line, because $\overline{P_0 P_1}$ is a tangent on the curve. But besides that, I'm not too sure if there could be a combination where the points are different, but the curves are the same.
edit: I think one problem might be when all control points are on the same line. Is this actually a counter example? Are there others?
The equation for the cubic Bézier curve is $$F(u) = (1-u)^3 P_0 + 3 (1-u)^2 u P_1 + 3 (1-u)u^2 P_2 + u^3 P_3$$ and therefore $$F(0) = P_0, \quad F(1) = P_3$$ and, by differentiation, $$F'(0) = 3(P_1 - P_0), \quad F'(1) = 3(P_3 - P_2)$$
Therefore all control points are uniquely determined by the function $F$.
But, as you already noted, there can be two different Bézier curves having the same image, e.g. if all control points are on a line. I don't have an answer for the question
When are the control points uniquely determined by the image of a Bézier curve?
• An alternative to the polysemic word "image" is "trajectory" Mar 6 '16 at 16:28
Suppose $\mathbf{P}_0, \ldots \mathbf{P}_m$, and $\mathbf{Q}_0 \ldots \mathbf{Q}_m$ are two sets of control points that produce the same Bézier curve of degree $m$, in the sense that $$\sum_{i=0}^m b^m_i(t)\mathbf{P}_i = \sum_{i=0}^m b^m_i(t)\mathbf{Q}_i \quad \text{for all t \in[0,1]}$$ Then we have $$\sum_{i=0}^m b^m_i(t)(\mathbf{P}_i - \mathbf{Q}_i) = \mathbf{0} \quad \text{for all t \in[0,1]}$$ This implies that $\mathbf{P}_i = \mathbf{Q}_i$ for $i=0,1, \ldots, m$ since the Bernstein polynomials $b^m_0, \ldots b^m_m$ are linearly independent.
There are cases where two different sets of control points will produce the same trace/image/trajectory. Take a given curve, and compose it with two different polynomials that both map $[0,1]$ to itself. Then clearly you'll get the same image, but the parametric equations will be different, so the control points will also be different.
A specific example: consider the two degree 4 curves whose control points are
$(0,0)$, $(1,0)$, $(\tfrac53,\tfrac23)$, $(2,1)$, $(2,1)$
$(0,0)$, $(0,0)$, $(\tfrac13,0)$, $(1,0)$, $(2,1)$.
Routine calculations show that these two curves both represent the portion of the parabola $y = \tfrac14 x^2$ that lies between the points $(0,0)$ and $(2,1)$. The first one has equation $G(t) = (4t - 2t^2, 4t^2 - 4t^3 + t^4)$ and the second one has equation $H(t) = (2t^2, t^4)$.
These curves are formed by two different reparameterizations of the basic curve $F(t) = (2t, t^2)$: $$F(t) = (2t, t^2) \;,\; t = 2u - u^2 \;\; \Rightarrow \;\; G(u) = (4u - 2u^2, 4u^2 - 4u^3 + u^4)$$ $$F(t) = (2t, t^2) \;,\; t = v^2 \;\; \Rightarrow \;\; H(v) = (2v^2, v^4)$$ This is a form of degree elevation (though not the usual form). The degree of the final curve is the product of the degree of the original one and the degree of the reparametrization function. In the example above $2 \times 2 = 4$. So, this process can only produce a cubic curve in trivial cases where either the original curve or the reparameterization function is linear.
• You probably mean "... parabola $y = \frac14 t^2$ ..." Mar 8 '16 at 10:23
• You're right. Thanks. Fixed. Mar 8 '16 at 12:04
Bézier curves with different control points could actually be the same function under one condition: one Bézier curve is degree elevated from the other.
A cubic Bézier curve is represented as (in Bernstein basis)
$C(u)=(1-u)^3P_0+3(1-u)^2uP_1+3(1-u)u^2P_2+u^3P_3$,
or as (in power basis)
$C(u)=P_0+3(P_1-P_0)u+3(P_2-2P_1+P_0)u^2+(P_3-3P_2+3P_1-P_0)u^3$.
From the power-basis representation, we can see that if $(P_3-3P_2+3P_1-P_0)=0$, the curve is actually of degree 2 only and therefore can be represented as a Bézier curve with 3 control points.
Basically, a Bézier curve with $K_1$ control points can always be exactly converted to a Bézier curve with $K_2$ ($K_2 > K_1$) control points. This process is called "degree elevation". However, it does not really change the degree of the underlying polynomial. It only makes the Bézier curve have more control points to be manipulated with.
In summary,
- Bézier curves with different number of control points might actually be the same function.
- Bézier curves of the same number of control points but different control point coordinates cannot be the same function but could be of the same image/trajectory. | 2021-11-30T10:35:21 | {
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https://math.stackexchange.com/questions/679947/if-a-ltb-and-c-led-prove-that-ac-lt-bd/679952 | # If $a\lt{b}$ and $c\le{d}$, prove that $a+c\lt b+d$
If $a\lt{b}$ and $c\le{d}$, prove that $a+c\lt b+d$.
This seems like a basic proof and I think this is how it goes: $$c \le d, \text{ Given }$$ $$a+c \le a+d$$ $$a+c \lt b+d, \text{ since } a \lt b$$
Is that all I need? I'm thinking this does it all quickly and concisely, but I have had trouble with proofs in my classes.
• Yes, or just go $a < b \implies a+c<b+c$, so since $c \leq d \implies b + c \leq b + d$, we have $a+c < b+c \leq b+d$, i.e. $a+c < b+d$. – Ryker Feb 17 '14 at 20:10
• Is $(b+d)-(a+c)>0$? – David Mitra Feb 17 '14 at 20:10
• I'd do it $(a-b)+(c-d)\lt 0$ but it's all the same in the end – Mark Bennet Feb 17 '14 at 20:18
• What is the reason you did it your way, @Ryker, versus my way? It is just semantical? I've noticed that the things posted have a middle argument between a less than and a less than or equal to sign. Is that preferred when proving to see the logical reason? – Faffi-doo Feb 17 '14 at 20:25
• @Faffi-doo , guys, wow! :) Perhaps I should have added a smiley in the end of the comment to avoid misunderstandings.. 1. You don't need to be so apologetic! :) Choose what is most correct for you. Always, and not just in Math. 2. I will not hold anything against you even if you actually did not like my answer. – user76568 Feb 17 '14 at 21:21
We have that for any $c$: $$a<b \implies a+c<b+c$$ And for any $b$: $$c\leq d \implies b+c \leq b+d$$.
Hence: $a+c < b+c \leq b+d$, that is: $$a+c < b+d$$.
$$(a < b) \land (c \leq d) \iff a-b < 0 \leq d - c \implies a-b<d-c \iff a+c < b + d$$ | 2019-12-11T09:15:06 | {
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http://mathhelpforum.com/calculus/1887-limit-doesn-t-exist.html | # Math Help - limit doesn't exist
1. ## limit doesn't exist
lim[x->2] (x^2 - x + 6)/(x - 2)
I'm not sure the best way to solve this... but here's what I did:
after doing polynomial division, I got:
lim[x->2] (x+1+ 8/(x-2))
So, by reasoning, I see that the limit as x approaches 2 from the left does not equal the limit as x approaches 2 from the right (negative and positive infinity). Therefore, this limit does not exist.
My solutions manual says that the limit "does not exist since x - 2 -> 0 but x^2 - x + 6 -> 8 as x -> 2."
Someone please tell me what my book's explanation means.
2. ## n/0 dne for n not equal to zero
The only number that can be divided by zero to give another number is zero.
Therefore in order to show that the limit does not exist you can show that the limit approaches some non zero in the numerator and zero in the denominator. In this case:
8/0
does not exist.
3. Originally Posted by asdfmaster
lim[x->2] (x^2 - x + 6)/(x - 2)
I'm not sure the best way to solve this... but here's what I did:
after doing polynomial division, I got:
lim[x->2] (x+1+ 8/(x-2))
So, by reasoning, I see that the limit as x approaches 2 from the left does not equal the limit as x approaches 2 from the right (negative and positive infinity). Therefore, this limit does not exist.
Would this form of argument then tell us that
$\lim_{x \rightarrow 2} \frac{x^2 - x + 6}{(x - 2)^2}$
exists and equals $\infty$?
RonL
4. Thanks MathGuru & CaptainBlack
CaptainBlack, are you saying my reasoning is wrong?
if I graph
$
\lim_{x \rightarrow 2} \frac{x^2 - x + 6}{(x - 2)^2}
$
I see that as x approaches 2 from the left and from the right, it increases without bound. I would then conclude that
$
\lim_{x \rightarrow 2} \frac{x^2 - x + 6}{(x - 2)^2}=\infty
$
Also, if this reasoning is correct, then what MathGuru said:
Therefore in order to show that the limit does not exist you can show that the limit approaches some non zero in the numerator and zero in the denominator.
would not be correct all the time...
I just started learning calculus! I'm sorry if these questions are very basic!
5. The limit doesn't exist because the upper limit doesn't equal the lower limit, i.e. you get a different result depending on whether you approach 2 from the right or from the left.
In the real numbers, we do not use the unsigned infinity in this context (this is done in complex analysis though), there is only a positive infinity and a negative infinity. Therefore you may not conclude that the limit "exists" and is equal to $\infty$, that's wrong.
6. Originally Posted by asdfmaster
Thanks MathGuru & CaptainBlack
CaptainBlack, are you saying my reasoning is wrong?
Neither the limit from the right or left exist.
RonL
7. Captain black was being a friendly skeptic earlier. He was trying to show you that sometimes the limit approaches the same from the left and from the right however the limit still does not exist.
My analysis that explains using numerator and denominator is more precise and also the same explanation your book was trying to give.
8. thanks everyone!
Captain black was being a friendly skeptic earlier.
That's what I assumed...but wasn't totally sure.
So what everyone is saying is that when a function appears to approach infinity, the limit does not exist???
So whenever you say the limit of a function is infinity, you are actually saying that the limit does not exist, but a convenient way to express that the limit does not exist is to write $\lim_{x \rightarrow a} f(x) = \infty$??
Therefore in order to show that the limit does not exist you can show that the limit approaches some non zero in the numerator and zero in the denominator.
I'm just wondering, in every instance of finding a limit that exists, if the demoninator will approach zero, will numerator always approach zero too?
9. Originally Posted by asdfmaster
thanks everyone!
That's what I assumed...but wasn't totally sure.
So what everyone is saying is that when a function appears to approach infinity, the limit does not exist???
So whenever you say the limit of a function is infinity, you are actually saying that the limit does not exist, but a convenient way to express that the limit does not exist is to write $\lim_{x \rightarrow a} f(x) = \infty$??
I'm just wondering, in every instance of finding a limit that exists, if the demoninator will approach zero, will numerator always approach zero too?
For a limit to exist it has to approach a real number. Since infinity is not a real number, any limit that "blows up" (either in the positive or negative sense) does not exist.
Yes, the ONLY way for a limit to exist when the denominator of a rational expression approaches zero is for the numerator to approach zero as well. In this instance we can then apply L'Hospital's rule. However, be warned: Just because the numerator also goes to zero does NOT mean that the limit will exist. A trivial example of this: $\lim_{x \rightarrow 0} \frac{x}{x^2}$ does not exist.
-Dan | 2014-08-23T08:47:23 | {
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https://math.stackexchange.com/questions/4182207/8-indistinguishable-objects-randomly-sorted-into-six-buckets-what-is-the-proba | # 8 Indistinguishable objects randomly sorted into six buckets - What is the probability of at least three buckets receiving the objects?
Eight indistinguishable objects are to be randomly put into six buckets. What is the probability that at least three buckets will receive the objects?
My approach was to let X= the number of buckets that receive objects, and then the required probability is
$$P(X≥3) = 1- P(X=1) - P(X=2)$$
$$= 1 - [(6C1)(1/6)^8 + (6C2)(2/6)^8]$$ $$= 0.99771$$
My reasoning was that for all of the objects falling into one bucket, there are 6 possible buckets $$(6C1)$$, and the probability for eight objects to fall into the same bucket in a row is $$(1/6)^8$$
By the same logic, for all of the objects to be divided into two of the six buckets, there are $$(6C2)$$ possible ways to choose those two buckets...And the probability is $$(2/6)^8$$.
Then, the probability that at least three buckets will receive objects is 1- (the probability of only one bucket receiving all the objects + two buckets receiving all the objects)
However, the answer given to this question was 0.8904. Where am I going wrong, and is there a better way to approach questions like this? Does it fall into any specific discrete distribution?
Note:
I also tried $$1- [(6C1)(1/6)^8] - (6C2)[(1/6)^7(5/6) + (1/6)^6(5/6)^2 + (1/6)^5(5/6)^3 + (1/6)^4(5/6)^4 + (1/6)^3(5/6)^5 + (1/6)^2(5/6)^6 + (1/6)(5/6)^7]$$
And got $$0.8721$$. I am not sure if this approach is better.
• You're counting multiple times the cases where all the objects go to the same bucket. First intentionally, and then twice unintentionally, since some of the times that objects are limited to either of two buckets, they are in fact limited to exactly one of the two buckets. (That is to say, if I say that eight objects go into either Bucket A or Bucket B, nothing prevents them from all going into Bucket A or all into Bucket B.) – Brian Tung Jun 24 at 20:40
• Where did you find this question , can you share your source please ? – Bulbasaur Jun 24 at 21:03
• @Bulbasaur This question is from a past exam paper for a first-year university statistics course. – Cara vdC Jun 24 at 23:17
I get $$1 - (6C1)(1/6)^8 - (6C2)[(2/6)^8-2(1/6)^8] \approx 0.997728$$, close to but not the same as your first calculation. This treats the objects as distinguishable, as I believe that is the physical reality of putting objects into buckets.
Another way of getting this answer is to say there are $$6^8= 1679616$$ ways of putting $$8$$ items into $$6$$ buckets, and $$6$$ of them have them all going into one bucket while $$3810$$ have them going into exactly $$2$$ buckets, leaving $$1675800$$ possibility using at least $$3$$ buckets, with $$\frac{1675800}{1679616} \approx 0.997728$$.
Bad approaches might be to say
• there are $${13 \choose 5}=1287$$ distributions of $$8$$ indistinguishable objects among $$6$$ distinguishable buckets, of which $$6$$ use one bucket and $$105$$ use exactly $$2$$ buckets, suggesting a probability of $$\frac{1176}{1287}\approx 0.913752$$ as the probability
• there are $$20$$ distributions of $$8$$ indistinguishable objects among $$6$$ indistinguishable buckets, of which $$1$$ uses one bucket and $$4$$ use exactly $$2$$ buckets, suggesting a probability of $$\frac{15}{20}=0.75$$ as the probability
but I think these are wrong because the distributions are not equally likely
• Should we always approach the indistinguishable objects as distinguishable in probability questions ? For example , what if the indistinguishable balls were distributed at the same time , would we treat like one by one and also distinguishable – Bulbasaur Jun 24 at 20:54
• @bulbasaur - I believe that is the physical reality. Flip two indistinguishable fair coins. Is the probability of one heads and one tails $\frac12$ or $\frac13$? – Henry Jun 24 at 20:56
• hımm elegant example... they are like die (ordered pairs) or coins as you said . Is there any book that may help me to learn this tricks ? Maybe other questions ? – Bulbasaur Jun 24 at 21:00
The official answer is incorrect. Coming to your approach, it is correct except that you have a small mistake when you are distributing objects to two buckets.
$$\displaystyle P(X=1) = {6 \choose 1} \cdot \frac{1}{6^8}$$
$$\displaystyle P(X=2) = {6 \choose 2} \cdot \frac{2^8 - 2}{6^8}$$
Please note that $$2^8$$ will also count the two arrangements where all objects go to one of the two chosen buckets. So we subtract $$2$$ from $$2^8$$.
Therefore,
$$\displaystyle P(X \geq 3) = 1 - P(X=1) - P(X=2) = \frac{23275}{23328}$$
• Why did you treat indistinguishable balls like distinguishable . Cant the answer be $$1-[C(6,1) \times \frac{1}{C(13,5)} + C(6,2) \times \frac{C(9,1)}{C(13,5)}]$$ – Bulbasaur Jun 24 at 20:47
• @Bulbasaur stars and bars method should not be used to calculate probability. All arrangements in stars and bars are not equally probable. – Math Lover Jun 24 at 20:50
• thanks , i see now – Bulbasaur Jun 24 at 20:52
• @CaravdC Even if you assume that all distributions of the indistinguishable objects are equally likely, the probability would work out to be $1-(6+\binom62\times 7)/\binom{8+6-1}{5}$, which is not equal to the book's answer. This is discussed in the second bullet point of Henry's answer. There is no conceivable way your source's answer is correct. – Mike Earnest Jun 25 at 0:33
• @CaravdC Mike Earnest's comments already answers your question. There is no way to get to that answer for the given question. Your first approach is optimal except a mistake that you made. – Math Lover Jun 25 at 9:49
I'm assuming that the buckets are distinguishable! And the indistinguishable objects are balls.
A hint to answer the question, assuming the balls are indistinguishable
$$P(X=1)=(6C1)P(\text{all balls fall into bucket 1})$$
In the case of indistinguishability, you will have to use conditioning to get the answer. The calculation remains exactly the same. The only difference is that you are using conditioning instead of combinatorics.
Let $$A_i$$ be the event that the $$i^{th}$$ ball chosen falls into bucket 1
$$P(A_1\cap A_2 \cap \dots\cap A_8)=P(A_1)\cdot P(A_2|A_1)\cdot P(A_3|A_1\cap A_2)\dots$$
There is one way to choose the first ball(since balls are indistinguishable!) and the probability that, that ball goes into the first bucket is $$(1/6)$$(because the buckets are distinguishable). so, $$P(A_1)=1/6$$
Now, given that the first ball goes into the first bucket, the number of ways to choose the second ball is also one(balls are indistinguishable!), and the probability that that ball goes into the first bucket is also $$(1/6)$$(because the buckets are distinguishable). so, $$P(A_2|A_1)=1/6$$
And so on...
hence $$P(\text{all balls fall into bucket 1})=(1/6)^8$$ and $$P(X=1)=(6C1)P(\text{all balls fall into bucket 1})=(6C1)(1/6)^8$$
I believe, it is straightforward to similarly calculate $$P(X=2)$$ and $$P(X\geq 3)$$
I would like to quote @DavidK here(from this answer, you should check out that answer, it is an absolute gem),
A way I think of this intuitively is that we are modeling a world in which writing a number on a ball or erasing the number does not cause that ball to magically run away from you when you reach in the back nor jump into your hand. In fact, the distinguishing marks (or lack thereof) on the balls have no effect on the probability of drawing a ball each time. So a correct way to compute $$P(X=k)$$ with indistinguishable balls is to compute $$P(X=k)$$ with distinguishable balls and simply copy the final result. This yields the same formulas.
I hope, it is okay to quote other users, if it is relevant. If it is not, let me know and I will remove the quote.
the probability of dividing 6 objects into exactly 2 buckets is not $$[(1/6)^7(5/6) + (1/6)^6(5/6)^2 + (1/6)^5(5/6)^3 + (1/6)^4(5/6)^4 + (1/6)^3(5/6)^5 + (1/6)^2(5/6)^6 + (1/6)(5/6)^7]$$
it should be $${nCr}\left(8,1\right)\left(\frac{1}{6}\right)^{7}\left(\frac{1}{6}\right)+{nCr}\left(8,2\right)\left(\frac{1}{6}\right)^{6}\left(\frac{1}{6}\right)^{2}+{nCr}\left(8,3\right)\left(\frac{1}{6}\right)^{5}\left(\frac{1}{6}\right)^{3}+{nCr}\left(8,4\right)\left(\frac{1}{6}\right)^{4}\left(\frac{1}{6}\right)^{4}+{nCr}\left(8,5\right)\left(\frac{1}{6}\right)^{3}\left(\frac{1}{6}\right)^{5}+{nCr}\left(8,6\right)\left(\frac{1}{6}\right)^{2}\left(\frac{1}{6}\right)^{6}+{nCr}\left(8,7\right)\left(\frac{1}{6}\right)^{1}\left(\frac{1}{6}\right)^{7}$$ | 2021-07-28T09:30:24 | {
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https://math.stackexchange.com/questions/2493482/how-does-one-evaluate-12-3-456-7-8-cdots50/2493503 | # How does one evaluate $1+2-3-4+5+6-7-8+\cdots+50$?
How does one evaluate the sum $$1+2-3-4+5+6-7-8+\cdots+50$$?
I know how to find the sum of arithmetic progressions: without the negative signs, one simply has $$1+2+\cdots+50=\frac{1}{2}\cdot(1+50)\cdot 50=51\times 25=1275.$$
But how does one calculate the one above?
• Consider a general case. Block them into $4$ consecutive numbers if you call the first one $n$. – mathreadler Oct 28 '17 at 9:59
• Using @Robertz 's idea: better $1+2-3=0$, $-4+5+6-7=0$, $-8+9+10-11=0$,... $...-47=0$, $-48+49+50-51=0$ then correct for the last sum $0+51=51$ – Gottfried Helms Oct 28 '17 at 13:17
• I find it disheartening that this question has received so many downvotes, and has been closed and deleted, and noone has commented as to why...... – user1729 Aug 19 '19 at 19:29
## 4 Answers
It's $$(1+5+...+49)+(2+6+...+50)-(3+7+...+51)-(4+8+...+52)+51+52=$$ $$=\frac{(2+12\cdot4)13}{2}+\frac{(4+12\cdot4)13}{2}-\frac{(6+12\cdot4)13}{2}-\frac{(8+12\cdot4)13}{2}+103=$$ $$=(1+2-3-4)\cdot13+103=51.$$
• Thanks. But I don't get the idea behind this formula you just did. – dimwitt04 Oct 28 '17 at 10:42
• @dimwitt04 It's a formula for the sum of the arithmetic progression: $S_{n}=\frac{(2a_1+(n-1)d)n}{2}$. – Michael Rozenberg Oct 28 '17 at 10:49
• Excuse my ignorance but where does 13 and 12 come from? Is it the number of terms? – dimwitt04 Oct 28 '17 at 11:41
• @dimwitt04 Yes of course! If $n$ is a number of terms we obtain: $49=1+(n-1)4$, which gives $n=13$. I used $a_n=a_1+(n-1)d$. – Michael Rozenberg Oct 28 '17 at 11:43
• Can down-voter explain us why did you do it? – Michael Rozenberg Jul 9 '19 at 14:25
Look at the following: $$1+\overbrace{(2-3)}^{-1}+\overbrace{(-4+5)}^1+\cdots+50$$
So you have $1+\overbrace{-1+1\cdots}^{\frac{48}2=24\text{ times}}+50$ and because $24$ is even the middle part become $0$ and you left with $1+50=51$ and done
moreover, you can generalize it:$$\sum_{k=1}^n(-1)^{\left\lfloor\frac{k-1}{2}\right\rfloor}\times k=\begin{cases}n+1 &\text{if}\,\,\,(-1)^{\left\lfloor\frac{n-1}{2}\right\rfloor}=1,&n\equiv0\pmod{2}\\ 1 &\text{if}\,\,\, (-1)^{\left\lfloor\frac{n-1}{2}\right\rfloor}=1,&n\equiv1\pmod{2}\\ -n &\text{if}\,\,\, (-1)^{\left\lfloor\frac{n-1}{2}\right\rfloor}=-1,&n\equiv0\pmod{2}\\ 0 &\text{if}\,\,\, (-1)^{\left\lfloor\frac{n-1}{2}\right\rfloor}=-1,&n\equiv1\pmod{2} \end{cases}$$
Note that your sum can be written as $$\underbrace{[(1-3)+(2-4)]}_{-4}+\underbrace{[(5−7)+(6−8)]}_{-4}+\dots +\underbrace{[(45−47)+(46−48)]}_{-4}+49+50$$ that is $-4\cdot(48/4)+49+50=-48+49+50=51.$ More generally $$\sum_{k=1}^n(-1)^{\left\lfloor\frac{k-1}{2}\right\rfloor}\cdot k =\begin{cases} -n&\text{if n\equiv 0\pmod{4}}\\ 1&\text{if n\equiv 1\pmod{4}}\\ n+1&\text{if n\equiv 2\pmod{4}}\\ 0&\text{if n\equiv 3\pmod{4}}.\\ \end{cases}$$
• (+1) I was just about to post an answer similar to the first part of this. Good generalization! – robjohn Oct 28 '17 at 12:49
• @Holo Our generalizations are equivalent... – Robert Z Oct 28 '17 at 14:13
Note that $1+2-3=0$. Moreover, you will have -4+5+6-7 and so on... if you consider pairs of numbers, you will always have +1. How much times do you do this computation?
• And if it ended with - 51 the sum would be 0... – hkBst Oct 28 '17 at 12:00 | 2021-08-01T19:24:26 | {
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https://mathhelpboards.com/threads/quadratic-equations.1874/ | ### Welcome to our community
#### Casio
##### Member
I have been reading up on Quadratic Equations in my course book, which gives some examples of simple equations and how they are factorised, but can't get my head round this type which is has no examples.
x2 - 3x = 0
OK I know it is a quadratic because the term x2 is included. What I require is a value for the term x.
Is there a method to factorise this I am not sure?
x2 - 3x = x(x - 3)
It does not give me a value for x?
I can assume values like x = 0 and x = 3, both of which will = 0, but that does not show me how I get the method of factorising this equation?
Thanks
Casio
#### Ackbach
##### Indicium Physicus
Staff member
There is a way of factoring quadratics that takes the guesswork out of it. In your example, you just note that $x$ is common to both terms, and factors out leaving you with $x(x-3)=0$.
But let's suppose you had a general quadratic: $x^{2}+bx+c$, which we'll make monic for convenience. What you're after is an equivalent expression that looks like this:
$$x^{2}+bx+c=(x+f)(x+g).$$
So, FOIL out the RHS thus:
$$x^{2}+bx+c=x^{2}+gx+fx+fg=x^{2}+(g+f)x+fg.$$
Now, you get to equate like powers of $x$ to obtain the following equations:
\begin{align*}
b&=f+g\\
c&=fg.
\end{align*}
Use whatever method you like for solving this simultaneous system of equations for $f$ and $g$. Once you do that, you're done.
Example: Factor $x^{2}+8x-3$.
Solution: Write as
$$x^{2}+8x-3=(x+f)(x+g)=x^{2}+(f+g)x+fg.$$
Solve:
\begin{align*}
8&=f+g\\
-3&=fg.
\end{align*}
The solution is
$$f=4\pm\sqrt{19},\quad g=4\mp\sqrt{19}.$$
That is, if $f=4+\sqrt{19}$, then $g=4-\sqrt{19}$, or if $f=4-\sqrt{19}$, then $g=4+\sqrt{19}$. So our factorization is
$$x^{2}+8x-3=(x+4-\sqrt{19})(x+4+\sqrt{19}).$$
Multiply this out to confirm.
Does that help?
#### QuestForInsight
##### Member
Solve:
\begin{align*}
8&=f+g\\
-3&=fg.
\end{align*}
How did you solve this?
#### SuperSonic4
##### Well-known member
MHB Math Helper
How did you solve this?
Simultaneous equations: there are two variables and two equations. In the terms of a quadratic you may know it along the lines of "two numbers which add to 8 and multiply to -3"
If you need more help on them feel free to create a new topic, it's not fair on Casio to sidetrack his topic
#### Ackbach
##### Indicium Physicus
Staff member
How did you solve this?
Substitution. We have that $f=8-g$, so plugging that into the other equation yields
$$-3=g(8-g)=8g-g^{2},$$
or
$$g^{2}-8g-3=0.$$
Now you have a quadratic in $g$. You can use the quadratic formula to solve:
$$g=\frac{8\pm\sqrt{64-4(-3)}}{2}=\dots$$
Come to think of it, we haven't gained anything here, because this is the same quadratic as we started with!
So you could save time by simply using the quadratic formula in the first place. Given the monic quadratic $x^{2}+bx+c$, factor by finding the roots:
$$x=\frac{-b\pm\sqrt{b^{2}-4c}}{2}.$$
Factoring the original quadratic would then give you
$$x^{2}+bx+c=\left(x+\frac{b+\sqrt{b^{2}-4c}}{2}\right)\left(x+\frac{b-\sqrt{b^{2}-4c}}{2}\right).$$
#### QuestForInsight
##### Member
If you need more help on them feel free to create a new topic, it's not fair on Casio to sidetrack his topic
No need to be rude here. The question I have asked is very relevant. These simultaneous equations are the Vieta relations of the quadratic equation being solved. Trying to eliminate one variable would get you back to the original equation.
#### QuestForInsight
##### Member
Substitution. We have that $f=8-g$, so plugging that into the other equation yields
$$-3=g(8-g)=8g-g^{2},$$
or
$$g^{2}-8g-3=0.$$
Now you have a quadratic in $g$. You can use the quadratic formula to solve:
$$g=\frac{8\pm\sqrt{64-4(-3)}}{2}=\dots$$
Come to think of it, we haven't gained anything here, because this is the same quadratic as we started with!
So you could save time by simply using the quadratic formula in the first place. Given the monic quadratic $x^{2}+bx+c$, factor by finding the roots:
$$x=\frac{-b\pm\sqrt{b^{2}-4c}}{2}.$$
Factoring the original quadratic would then give you
$$x^{2}+bx+c=\left(x+\frac{b+\sqrt{b^{2}-4c}}{2}\right)\left(x+\frac{b-\sqrt{b^{2}-4c}}{2}\right).$$
I was thinking about this, and I thought what if we try to find $f-g$ first?
$(f-g)^2 = (f+g)^2-4fg \implies f-g = \pm \sqrt{76}$.
$\left\{\begin{array}{c}f+g = 8 \\\\ f-g = \pm\sqrt{76} \end{array} \implies 2f = 8\pm \sqrt{76} \implies \boxed{f = 4\pm\frac{1}{2}\sqrt{76}} \implies \boxed{g = 4\mp\frac{1}{2}\sqrt{76}}.\right\|$
Without loss of generality, we take $f = 4+\frac{1}{2}\sqrt{76}$ and $g = 4-\frac{1}{2}\sqrt{76}$.
I think this also gives a way of proving the quadratic formula!
Last edited:
#### Casio
##### Member
There is a way of factoring quadratics that takes the guesswork out of it. In your example, you just note that $x$ is common to both terms, and factors out leaving you with $x(x-3)=0$.
But let's suppose you had a general quadratic: $x^{2}+bx+c$, which we'll make monic for convenience. What you're after is an equivalent expression that looks like this:
$$x^{2}+bx+c=(x+f)(x+g).$$
So, FOIL out the RHS thus:
$$x^{2}+bx+c=x^{2}+gx+fx+fg=x^{2}+(g+f)x+fg.$$
Now, you get to equate like powers of $x$ to obtain the following equations:
\begin{align*}
b&=f+g\\
c&=fg.
\end{align*}
Use whatever method you like for solving this simultaneous system of equations for $f$ and $g$. Once you do that, you're done.
Example: Factor $x^{2}+8x-3$.
Solution: Write as
$$x^{2}+8x-3=(x+f)(x+g)=x^{2}+(f+g)x+fg.$$
Solve:
\begin{align*}
8&=f+g\\
-3&=fg.
\end{align*}
The solution is
$$f=4\pm\sqrt{19},\quad g=4\mp\sqrt{19}.$$
That is, if $f=4+\sqrt{19}$, then $g=4-\sqrt{19}$, or if $f=4-\sqrt{19}$, then $g=4+\sqrt{19}$. So our factorization is
$$x^{2}+8x-3=(x+4-\sqrt{19})(x+4+\sqrt{19}).$$
Multiply this out to confirm.
Does that help?
Thanks for the above effort you have put in to explain that, but it's far too advanced for a quadratic that I was trying to understand, where my coursework advises the type to be simple quadratics.
So in my example,
x(x - 3) = 0 this is one solution, and
x = 3 is the second solution.
I think the whole point of my thread was to explain that no matter what anyone does to multiply out brackets as above, you just can't prove that x = 0 unless you are already told that?
so even if you apply ax2 + bx + c = 0
you can say that x2 + 3x = 0
You know that b = 3, and the conclusion is 0, so no maths involved you just have to see that x2 = 0 and 3 (x) must be 3(0) = 0
I can't see how it could ever be proven using algebra?
Kind regards
Casio
#### Sudharaka
##### Well-known member
MHB Math Helper
Thanks for the above effort you have put in to explain that, but it's far too advanced for a quadratic that I was trying to understand, where my coursework advises the type to be simple quadratics.
So in my example,
x(x - 3) = 0 this is one solution, and
x = 3 is the second solution.
I think the whole point of my thread was to explain that no matter what anyone does to multiply out brackets as above, you just can't prove that x = 0 unless you are already told that?
so even if you apply ax2 + bx + c = 0
you can say that x2 + 3x = 0
You know that b = 3, and the conclusion is 0, so no maths involved you just have to see that x2 = 0 and 3 (x) must be 3(0) = 0
I can't see how it could ever be proven using algebra?
Kind regards
Casio
Hi Casio,
Suppose you have a general quadratic equation of the form, $$ax^2+bx+c=0\mbox{ where }a\neq 0$$. Then,
$ax^2+bx+c=0$
$\Rightarrow a\left(x^2+\frac{b}{a}x+\frac{c}{a}\right)=0$
Since $$a\neq 0$$ we have,
$x^2+\frac{b}{a}x+\frac{c}{a}=0$
$\Rightarrow \left(x+\frac{b}{2a}\right)^2-\frac{b^2}{4a^2}+\frac{c}{a}=0$
$\Rightarrow \left(x+\frac{b}{2a}\right)^2=\frac{b^2-4ac}{4a^2}$
$\Rightarrow x+\frac{b}{2a}=\pm\frac{\sqrt{b^2-4ac}}{2a}$
$\Rightarrow x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
So the roots of a quadratic equation is given by,
$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
In your case you have, $$a=1,\,b=-3\mbox{ and }c=0$$. Therefore the roots are,
$x=\frac{-(-3)\pm\sqrt{(-3)^2-(2\times 1\times 0)}}{2\times 1}$
$\Rightarrow x=3\mbox{ or }x=0$
Hope this clarifies things for you.
Kind Regards,
Sudharaka.
#### QuestForInsight
##### Member
Thanks for the above effort you have put in to explain that, but it's far too advanced for a quadratic that I was trying to understand, where my coursework advises the type to be simple quadratics.
So in my example,
x(x - 3) = 0 this is one solution, and
x = 3 is the second solution.
I think the whole point of my thread was to explain that no matter what anyone does to multiply out brackets as above, you just can't prove that x = 0 unless you are already told that?
so even if you apply ax2 + bx + c = 0
you can say that x2 + 3x = 0
You know that b = 3, and the conclusion is 0, so no maths involved you just have to see that x2 = 0 and 3 (x) must be 3(0) = 0
I can't see how it could ever be proven using algebra?
Kind regards
Casio
I think you are bit confused here. You're asked to solve x(x-3) = 0. Think about it this way: when do you get 0 from the product of two numbers? When at least one of them is zero! Here you have a product of x and x-3 -- that's x(x-3). You're told that it's zero. It can only be so if x = 0 or x-3 = 0; that's, if x = 0 or x = 3. The fact that $p\times q = 0$ implies that either $p = 0$ or $q = 0$ is something that you should have already been comfortable with. That's probably what you're missing. | 2021-01-15T18:42:45 | {
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https://math.stackexchange.com/questions/4495528/dynamic-dice-game-how-to-reasonably-estimate-answer-by-hand-without-laboriously/4497026 | # Dynamic dice game, how to reasonably estimate answer by hand without laboriously calculating
Here's a question from my probability textbook:
A casino comes up with a fancy dice game. It allows you to roll a dice as many times as you want unless a $$6$$ appears. After each roll, if $$1$$ appears, you will win $$\1$$; if $$2$$ appears, you will win $$\2$$; $$\ldots$$; if $$5$$ appears, you win $$\5$$, but if $$6$$ appears all the money you have won in the game is lost and the game stops. After each roll, if the dice number is $$1$$-$$5$$, you can decide whether to keep the money or keep on rolling. How much are you willing to pay to play the game (if you are risk neutral)?
It's been asked before on MSE multiple times:
When to stop rolling a die in a game where 6 loses everything
Dynamic dice game: optimal price to enter
A dynamic dice game
Here's the answer to the question in my book:
Assuming that we have accumulated $$n$$ dollars, the decision to have another roll or not depends on the expected profit versus expected loss. If we decide to have an extra roll, our expected payoff will become$${1\over6}(n + 1) + {1\over6}(n + 2) + {1\over6}(n + 3) + {1\over6}(n + 4) + {1\over6}(n + 5) + {1\over6} \times 0 = {5\over6}n + 2.5.$$We have another roll if the expected payoff $${5\over6}n + 2.5 > n$$, which means that we should keep rolling if the money is no more than $$\14$$. Considering that we will stop rolling when $$n \ge 15$$, the maximum payoff of the game is $$\19$$ (the dice rolls a $$5$$ after reaching the state $$n = 14$$). We then have the following: $$f(19) = 19$$, $$f(18) = 18$$, $$f(17) = 17$$, $$f(16) = 16$$, and $$f(15) = 15$$. When $$n \le 14$$, we will keep on rolling, so $$E[f(n) \mid n \le 14] = {1\over6} \sum_{i = 1}^5 E[f(n + i)]$$. Using this equation, we can calculate the value for $$E[f(n)]$$ recursively for all $$n = 14, 13, \ldots, 0$$. After laboriously calculating or writing a program, we get $$E[f(0)] = 6.15$$, and so we are willing to pay at most $$\6.15$$ for this game.
However, I'm wondering if there's a quick way by hand to get a reasonable/"good enough" estimate for $$E[f(0)]$$ without having to do multiple "average-five-numbers-and-repeat" as the book suggests. Is there?
You can estimate it as follows: There are two possibilities - Either you roll a six, or you don't. If you don't, the expected gain is $$\frac{1+2+3+4+5}{5}=3$$ (Note that this is the expected gain GIVEN that you didn't roll a $$6$$). Now suppose that no matter what number you get from $$1$$ to $$5$$ you gain $$\3$$. As stated in the solution you provided, you should keep rolling until you have $$\ge 15$$, so in this case you roll exactly $$5$$ times. If you are successful with all five rolls, you gain $$\15$$, but if you rolled a $$6$$ in any one of these rolls you gain $$\0$$.
So the estimate is $$15*\left(\frac{5}{6}\right)^5=6.03$$.
• Thanks @ItsMe. (1) But isn't the expected gain $(1 + 2 + 3 + 4 + 5)/6 = 2.5$, and not $(1 + 2 + 3 + 4 + 5)/6 = 3$? (2) Assuming we get an estimate, how do we know if it's greater or lesser than the actual value of $\$6.15$? Jul 21 at 2:18 • I've added some detail to the answer - that is the expected gain given you didn't roll a six (hence there are only 5 options). Now I'm not sure how to determine whether this estimate is more or less without knowing the actual value, but some intuition is that in the estimate the maximum payoff is 15 whereas in the actual problem you could get up to 19, suggesting that the estimate may be lower than the actual value. Jul 21 at 4:23 • Here's a way to argue why$6.03$is an under-estimate, i.e. the modified game is worth less: the intuition is that variance in payoff helps you. There is some threshold, and you keep rolling until you meet it. Say you are close. If you roll a big payoff, you exceed the threshold by a lot (and stop), whereas if you roll a small payoff, you might stay below threshold and simply keep playing. So the variance increases the average amount by which you exceed the threshold when you stop. Jul 23 at 5:38 • Consider an even higher variance game where$1$pays$15$, and$2,3,4,5$pays nothing. This game has even higher variance, and is worth$0$(if you roll$6$before$1$) or$15$(otherwise), for an average of$7.5\$. Jul 23 at 5:39 | 2022-09-30T05:50:53 | {
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https://math.stackexchange.com/questions/2057870/range-of-x-given-x2-ab | # Range of $x$ given $|x^2-a|<b$
## Exercise:
Using signs of inequality alone (not using signs of absolute value) specify the values of $x$ which satisfy the following relations. Discuss all cases. $$|x^2-a|<b$$
## Solution:
$LHS \geq 0$, so: $|x^2-a|<b \text{ when } b > 0 \tag{0}$
$-b<x^2-a<b \text{ when } b > 0 \tag{1}$
$a-b<x^2<a+b \text{ when } b > 0 \tag{2}$
$a-b<x^2 \text{ when } b > 0 \tag{2.1}$ $x^2<a+b \text{ when } b > 0 \tag{2.2}$
$0 \leq a-b<x^2 \text{ when } b > 0 \tag{2.1.1}$
$a-b < x^2 \text{ when } 0 < b \leq a \tag{2.1.2}$
$\sqrt{a-b} < |x| \text{ when } 0 <b \leq a \tag{2.1.3}$
$\sqrt{a-b} < x \text{, } x < -\sqrt{a-b} \text{, when } 0 < b \leq a \tag{2.1.4}$
$|x|<\sqrt{a+b} \text{ when } a \geq -b, b > 0 \tag{2.2.1}$
$-\sqrt{a+b}<x<\sqrt{a+b} \text{ when } a \geq -b, b > 0 \tag{2.2.2}$
Formulation 1
$$\sqrt{a-b} < x \text{, } x < -\sqrt{a-b} \text{, when } b \leq a$$ $$\text{and}$$ $$-\sqrt{a+b}<x<\sqrt{a+b}$$ $$\text{all when } a > -b, b > 0$$
Formulation 2
$$-\sqrt{a+b}<x<-\sqrt{a-b}, \sqrt{a-b}<x<\sqrt{a+b} \text{, when } b\leq a$$ $$-\sqrt{a+b}<x<\sqrt{a+b} \text{ when } b > a$$ $$\text{all when } a > -b, b > 0$$
## Request:
Is my answer correct? If not, where in my solution did I go wrong?
Update: I just noticed (graphically) that $b > 0$ for there to be a range of $x$. Where do I prove (algebraically ) that this is so?
Another Update: Ignore the previous update. Thanks to @JeanMarie comment, I realized that this isn't necessarily so.
Yet Another Update: Ignore the previous update, and pay attention to the first. Thanks to @dxiv comment, I realized that this is necessarily so.
An Update Once Again: I've updated my attempt with the input given my @dxiv and @JeanMarie. How is it now?
• You have to consider that all expressions that you introduce make sense. For example, you can use $\sqrt{a+b}$ iff $a+b \geq 0$. – Jean Marie Dec 14 '16 at 0:21
• @JeanMarie -- Oh, yes. I didn't notice that. If I put that into consideration, my requirement that $b > 0$ will be covered automagically, correct? – Fine Man Dec 14 '16 at 0:24
• If $b \le 0$ there are no solutions since $|x| \ge 0$ for $\forall x$ so $|x^2-a| \lt b \le 0$ is impossible. – dxiv Dec 14 '16 at 0:27
• @dxiv -- So, this is a proof has no relation to what JeanMarie mentioned, correct? – Fine Man Dec 14 '16 at 0:31
• It is a necessary condition for any solutions to exist, and it is independent of the other conditions on $a$. – dxiv Dec 14 '16 at 0:33
$$|x^2-a|<b$$
The LHS side is non-negative, so there are no solutions if $\,b \le 0 \;\;\;(1)\,$.
For $b \gt 0$ the inequality can be rewritten as:
$$-b \lt x^2 - a \lt b \quad \iff \quad \begin{cases} x^2 \lt a+b \\ x^2 \gt a-b \end{cases}$$
• $(x^2 \lt a+b)\,$ Since $x^2 \ge 0$ there are no solutions if $a+b \le 0 \;\;\;(2)\,$.
Otherwise for $a+b \gt 0$ the inequality is equivalent to $-\sqrt{a+b} \lt x \lt \sqrt{a+b}\;\;\;(3)$.
• $(x^2 \gt a-b)$ If $a-b \lt 0$ then the inequality holds for $\forall x\;\;\;(4)\,$.
Otherwise for $a-b \ge 0$ the inequality is equivalent to $x \lt -\sqrt{a-b}\,$ or $\,x \gt \sqrt{a-b}\;\;\;(5)\,$.
To combine $(1)\cdots(5)$ into one final answer, note that $\sqrt{a+b} \ge \sqrt{a-b} \ge 0$ when $a \ge b \ge 0$.
• From $(1)+(2)\,$: if $b \le 0$ or $a \le -b$ then there are no solutions i.e. $x \in \emptyset$.
• Otherwise ($b \gt 0$ and $a \gt -b$) if $a \lt b$ from $(3)+(4)$: $x \in (-\sqrt{a+b}, \sqrt{a+b})$.
• Otherwise ($a \ge b \gt 0$) from $(3)+(5)$: $x \in (-\sqrt{a+b}, -\sqrt{a-b}) \cup (\sqrt{a-b}, \sqrt{a+b})$.
• Thank! This is a fairly elegant and complete solution. Every time I try solving this type of problem, I end up with an incomplete mess. Any suggestions on keeping solutions complete and concise? – Fine Man Dec 14 '16 at 4:51
• @SirJony Thanks. That's the kind of messy problem which invites messy solutions (as often happens in real world problems, unfortunately). Don't know there is a universal recipe, but it generally helps to eliminate the most obvious cases, first, so as to leave fewer subcases to consider in full detail. And, of course, keep good tabs on the progress and assumptions at each step. – dxiv Dec 14 '16 at 4:56
• I meant "thanks", not you to thank me. Oops. Sorry. :) I guess I was hoping for a universal recipe (as if Wolfram|Alpha were to solve it), but I'll have to stay satisfied with your tips and tricks. Again, thanks! – Fine Man Dec 14 '16 at 5:03
I advise you to have a graphical view of the situation as depicted below with $f(x)=|x^2-a|$ (green curve) and $g(x)=b$ (black horizontal line), looking for values of $x$ such that
$$|x^2-a|<b \ \ \ \ \iff \ \ \ \ f(x)<g(x).$$
(graphics obtained with Geogebra, with sliders for values of $a$ and $b$).
This figure allows to consider the different cases according to resp. values of $a$ and $b$, by "sweeping" the horizontal line along the curve, and looking for cases where the line is above the curve.
For example, for the displayed case ($b=3 \leq a=4$), the line is above the curve for $-\sqrt{7}<x<-1$ and for $1<x<\sqrt{7}.$ (case of two disjoint validity intervals), with $\sqrt{7}=\sqrt{a+b}$ and $1=\sqrt{a-b}$.
Were $b>a$, it is visible that there would be a single validity interval.
And of course, if $b<0$, no solution exist.
• Is my graphical presentation understandable ? – Jean Marie Dec 14 '16 at 7:41 | 2019-10-22T22:26:39 | {
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https://math.stackexchange.com/questions/444845/minkowski-sum-of-two-disks/444850 | # Minkowski sum of two disks
An open disk with radius $r$ centered at $\mathbf{p}$ is $D(\mathbf{p}, r)=\{\mathbf{q} \mid d(\mathbf p, \mathbf q) < r\}$, and the Minkowski sum of two sets $A$ and $B$ is $A \oplus B=\{\mathbf p + \mathbf q \mid \mathbf p \in A, \mathbf q \in B \}$.
How can you show that $D(\mathbf{a}, r_a) \oplus D(\mathbf{b}, r_b) = D(\mathbf{a} + \mathbf{b}, r_a + r_b)$?
Attempt:
\begin{align} D(\mathbf{a}, r_a) \oplus D(\mathbf{b}, r_b) &= \{\mathbf p + \mathbf q \mid \mathbf p \in D(\mathbf{a}, r_a), \mathbf q \in D(\mathbf{b}, r_b) \} \\&= \{\mathbf p + \mathbf q \mid \mathbf p \in \{\mathbf{x} \mid d(\mathbf a, \mathbf x) < r_a\}, \mathbf q \in \{\mathbf{y} \mid d(\mathbf b, \mathbf y) < r_b\} \\&= \{\mathbf p + \mathbf q \mid d(\mathbf a, \mathbf p) < r_a, d(\mathbf b, \mathbf q) < r_b \} \end{align}
And here I got stuck. As best as I can tell, now I would need to prove that $$d(\mathbf a, \mathbf p) < r_a, d(\mathbf b, \mathbf q) < r_b \iff d(\mathbf a + \mathbf b, \mathbf p + \mathbf q) < r_a + r_b$$ but this seems false to me. I tried adding the two inequalities together, but that doesn't seem to give me that condition unless $\mathbf a - \mathbf p$ and $\mathbf b - \mathbf q$ are parallel.
First, notice that $D(p,r)= \{ p+ u \mid u \in D(0,r) \}$, hence
$$D(a,r_a)+D(b,r_b)= \{ a+b+u+v \mid u \in D(0,r_a), v \in D(0,r_b)\}.$$
Therefore, it is sufficient to show that $D(0,r_a)+D(0,r_b)=D(0,r_a+r_b)$. But $$\|u+v\| \leq \|u \| + \|v\| <r_a+r_b, \ \text{if} \ u \in D(0,r_a) \ \text{and} \ v \in D(0,r_b),$$
so $D(0,r_a)+D(0,r_b) \subset D(0,r_a+r_b)$. Then,
$$D(0,r_a+r_b)= \{ (r_a+r_b)u \mid u \in D(0,1) \}= \{\underset{\in D(0,r_a)}{\underbrace{r_au}} + \underset{\in D(0,r_b)}{\underbrace{r_bu}} \mid u \in D(0,1)\},$$
so $D(0,r_a+r_b) \subset D(0,r_a)+D(0,r_b)$.
• Thanks. I don't suppose it is to possible to generalize this proof to metric spaces, rather than just normed vector spaces? I've had a go at it, but I didn't get very far. – Electro Jul 16 '13 at 13:10
• Minkowski sum is not defined in any metric space, a structure of vector space is needed. – Seirios Jul 16 '13 at 13:15
• Whoops, I meant vector space equipped with a metric. – Electro Jul 16 '13 at 13:18
• You can easily find a counterexample using SNCF metric: en.wikipedia.org/wiki/Metric_space#Examples_of_metric_spaces – Seirios Jul 16 '13 at 19:50
To show $A = B$ it is usually the easiest to split into two parts $A \subseteq B$ and $A \supseteq B$.
First, if $p \in D(a,r_a)$ and $q \in D(b,r_b)$ then $p + q \in D(a+b, r_a+r_b)$. This is simple using the triangle inequality: $$|(p+q)-(a+b)| < |p-a| + |q-b|.$$
Secondly, if $u \in D(a+b, r_a + r_b)$, then we need to find a point $v$ such that $v \in D(a,r_a)$ and $u-v \in D(b,r_b)$. A good candidate is to split the distance between $u$ and $a+b$ in ratio $r_a : r_b$, that is (the $-b$ is to adjust for the center of the disk) $$v = \frac{r_a\cdot u + r_b \cdot (a+b)}{r_a + r_b}-b.$$
Now, $$|v-a| = \left|\frac{r_a\cdot u + r_b \cdot (a+b)}{r_a + r_b}-b-a\right| = r_a\frac{|u-(a+b)|}{r_a+r_b}\leq r_a$$
and
$$|(u-v)-b| = \left|\frac{r_b\cdot u -r_b\cdot(a+b)}{r_a + r_b}+b-b\right| = r_b\frac{|u-(a+b)|}{r_a+r_b}\leq r_b.$$
I hope this helps ;-)
I would take the following route. Hopefully it is more intuitive.
It is clear that $$D(b, r_b)\oplus D(0, r_0)=D(b, r_b+r_0)$$ and that $$a+D(b, r)=D(a+b, r).$$ (Here $a+ \text{"some set"}$ denotes translation). So we write $$D(a, r_a)=a+D(0, r_a).$$ Therefore $$\begin{split} D(a, r_a)\oplus D(b, r_b) &= [a+D(0, r_a)]\oplus D(b, r_b) \\ &=a+[D(0, r_a)\oplus D(b, r_b)]\\ &=a+ D(b, r_a+r_b) \\ &= D(a+b, r_a+r_b). \end{split}$$
How about using triangular inequality ?
$$d(u,w) \leq d(u,v) + d(v, w)$$
Hint: use $a+(b-q)$.
$D(\mathbf{a},r_a)\oplus D(\mathbf{b},r_b)=\{\mathbf{p}+\mathbf{q}|d(\mathbf{a},\mathbf{p})<r_a,d(\mathbf{b},\mathbf{q})<r_b\}$
$D(\mathbf{a}+\mathbf{b},r_a+r_b)=\{\mathbf{s}|d(\mathbf{a}+\mathbf{b},\mathbf{s})<r_a+r_b\}$
$\forall \mathbf{p} \in D(\mathbf{a},r_a) \forall \mathbf{q}\in D(\mathbf{b},r_b) \exists \mathbf{s}=\mathbf{p}+\mathbf{q} \in D(\mathbf{a}+\mathbf{b},r_a+r_b)$ $d(\mathbf{a},\mathbf{p})<r_a, d(\mathbf{b},\mathbf{q})<r_b \Rightarrow d(\mathbf{a+b},\mathbf{p+q})=d(\mathbf{a+b-q},\mathbf{p}) \leq d(\mathbf{a},\mathbf{p})+d(\mathbf{a},\mathbf{a+b-q})=d(\mathbf{a},\mathbf{p})+d(\mathbf{q},\mathbf{b})<r_a+r_b$
$\forall \mathbf{s} \in D(\mathbf{a}+\mathbf{b},r_a+r_b) \exists \mathbf{p}\in D(\mathbf{a},r_a), \mathbf{q}\in D(\mathbf{b},r_b), \mathbf{s=p+q}$ $\mathbf{p}=\mathbf{a}+(\mathbf{s-(a+b)}) r_a/(r_a+r_b)$, $\mathbf{q}=\mathbf{b}+(\mathbf{s-(a+b)}) r_b/(r_a+r_b)$
• If you could add some words that would be nice. – user1337 Jul 16 '13 at 10:20 | 2019-08-20T18:55:28 | {
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https://math.stackexchange.com/questions/311665/proof-a-graph-is-bipartite-if-and-only-if-it-contains-no-odd-cycles/3422887 | # Proof a graph is bipartite if and only if it contains no odd cycles
How can we prove that a graph is bipartite if and only if all of its cycles have even order? Also, does this theorem have a common name? I found it in a maths Olympiad toolbox.
One direction is very easy: if $$G$$ is bipartite with vertex sets $$V_1$$ and $$V_2$$, every step along a walk takes you either from $$V_1$$ to $$V_2$$ or from $$V_2$$ to $$V_1$$. To end up where you started, therefore, you must take an even number of steps.
Conversely, suppose that every cycle of $$G$$ is even. Let $$v_0$$ be any vertex. For each vertex $$v$$ in the same component $$C_0$$ as $$v_0$$ let $$d(v)$$ be the length of the shortest path from $$v_0$$ to $$v$$. Color red every vertex in $$C_0$$ whose distance from $$v_0$$ is even, and color the other vertices of $$C_0$$ blue. Do the same for each component of $$G$$. Check that if $$G$$ had any edge between two red vertices or between two blue vertices, it would have an odd cycle. Thus, $$G$$ is bipartite, the red vertices and the blue vertices being the two parts.
• @YOUSEFY: That’s the direction that I proved in the first paragraph. And I gave direct proofs for both directions. – Brian M. Scott Oct 19 '16 at 9:43
• @YOUSEFY: No, I mean the first paragraph. What you proved in your comment is that if $G$ has an odd cycle, then $G$ is not bipartite, which is the contrapositive of (and logically equivalent to) the statement that if $G$ is bipartite, then it has no odd cycle. What I proved in the second paragraph is that if $G$ does not have an odd cycle, then $G$ is bipartite. This is the converse of what you proved. – Brian M. Scott Oct 19 '16 at 15:34
• @sourav: If you start at a vertex in $V_1$ and take a walk of length $3$, say, your first step goes to a vertex in $V_2$, your second to a vertex in $V_1$, and your third to a vertex in $V_2$; since you started in $V_1$ and are now in $V_2$, you can’t possibly have returned to your starting vertex. The same thing happens for any odd number of steps. – Brian M. Scott Nov 1 '16 at 16:31
• @sourav: You’re welcome. – Brian M. Scott Nov 1 '16 at 16:37
• @sourav: No, the graph is not bipartite. Being bipartite has nothing to do with how many components the graph has. A graph is bipartite if the vertices can be partitioned into two sets, say $V_1$ and $V_2$, such that every edge is between a vertex in $V_1$ and a vertex in $V_2$, i.e., so that there are no edges between vertices in $V_1$ and no edges between vertices in $V_2$. – Brian M. Scott Nov 1 '16 at 18:45
does this theorem have a common name?
It is sometimes called König's Theorem (1936), for example in lecture notes here.
However, this name is ambiguous.
• I expanded your answer a bit, as it drew some low-quality flags. – user147263 Jul 1 '15 at 17:45
• I don't agree with you. in the textbook of Diestel, he mentiond König's theorem in page 30, and he mentiond the question of this site in page 14. he didn't say at all any similiarities between the two. Also, König's talks about general case of r-paritite so if what you're saying is true, then the theorem is just a special case of general case. Right now, I don't have any name for it except it is a proposition (means something that mathematicains do find it interesting but it doesn't come becasue they're thinking on it so much) – YOUSEFY Oct 18 '16 at 17:15
The following is an expanded version of Brian's answer. Brian's answer is almost perfect, except that there may be a gap between "if G had any edge between two red vertices or between two blue vertices" and "it would have an odd cycle", which is not that obvious, at least to me(since we can only conclude that there exists a closed walk of odd numbers of edges). We first prove a lemma stating that if there is an odd closed walk in a graph, then there is an odd closed cycle.
Lemma 1 If there is an odd closed walk in a graph, then there is an odd closed cycyle.
Proof$$\;$$ We induct on the number of edges $$k$$ of the odd closed walk. The base case $$k=1$$, when the closed walk is a loop, holds trivially. Assume that, for some positive integer $$r > 1$$, Lemma 1 is true for all odd numbers $$k\le2r-1$$. Let $$W=(w_1, \dots, w_{2r+1}, w_1)$$ be a closed walk of $$2r+1$$ edges. If all vertices in $$W$$ is different except for $$w_1$$, then we have a cycle of length $$2r+1$$. If there exists two identical vertices $$w_i=w_j$$ for $$1, then $$W$$ can be written as $$(w_1, \dots,w_i, \dots, w_j,\dots, w_1)$$. Thus, we now have two closed walks $$W_1=(w_i,w_{i+1} \dots, w_j)$$ and $$(w_j,w_{j+1} \dots, w_i)$$. The summation of the length of $$W_1$$ and $$W_2$$ equals to the length of $$W$$. Since $$W$$ is of odd length, one of $$W_1$$ and $$W_2$$ must be of odd length $$\le 2r-1$$. By our assumption, there must be a an odd cycle in $$W_1$$ or $$W_2$$, and thus in $$W$$, which completes our induction. $$\square$$
Theorem 1 If there is no odd cycles in a graph, then the graph is bipartite.
Proof$$\;$$ Suppose there is no odd cycles in graph $$G=(V,E)$$. It is also assumed that, without loss of generality, $$G$$ is connected. Then $$V$$ can be partitioned into $$(A,B)$$, where $$A=\{v\in V|\text{the shortest path between v and v_0 is of even length}\}$$ $$B=\{v\in V|\text{the shortest path between v and v_0 is of odd length}\}$$ Next we prove that there is no edge between any two vertices in $$A$$ or $$B$$. Suppose for contradiction that there exists an edge $$(x,y)\in E$$ such that $$x,y \in A$$ or $$x,y \in B$$ . Then the shortest path from $$x$$ to $$v_0$$, the shortest path from $$y$$ to $$v_0$$, together with edge $$(x,y)$$, form a closed walk: $$(v_0, \dots,x,y, \dots,v_0)$$, which is of odd length. By lemma 1, $$G$$ contains an odd cycle, which is a contradiction. Therefore, $$G$$ is a bipartite graph between $$A$$ and $$B$$ $$\square$$
• Is it not sufficient to note that two vertices $u,w$ in the same set of the bipartition cannot share an edge as $d(v_0,u) = d(v_0,w) \pm 1$?. Say $u$ is closer to $v_0$ than $w$. Then one shortest path from $v$ to $w$ is the one passing through $w$. But then the distances to $w$ and $u$ have different parity and so they aren't in the same set. – PhysMath May 4 '20 at 4:11
(By contradiction) (->) Suppose n is odd. Let X={ui such that i is odd} and Y={ui such that i is even} as the bipartition formed in the graph. Consider cycle C=u1 u2 u3 u4...un u1 as the cycle in G. If u1 is odd, un can not be odd because there will be no bipartite formed. Therefore, n should be even.
One relatively simple way is to break the if and only if into its two parts:
• Prove that if a graph $G$ is bipartite then it has no odd cycles, and
• If $G$ has only even cycles, then you can partition the vertices into two independent sets.
I don't think this is a named theorem, it's too simple.
• The handshaking Lemma is even simpler – SK19 May 8 '19 at 1:11 | 2021-04-15T17:01:29 | {
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https://math.stackexchange.com/questions/2417938/how-do-i-solve-a-congruence-system-that-doesnt-satisfies-the-chinese-remainder | # How do I solve a congruence system that doesn't satisfies the Chinese Remainder Theorem?
I have the following system
$x \equiv 2 (mod 4)$
$x \equiv 2 (mod 6)$
$x \equiv 2 (mod 7)$
And I can't apply the Chinese remainder Theorem.
I tried applying the Chinese remainder Theorem to the last 2 congruences, which gave me that the set of solutions of that 2 congruences is $2 + 42 \cdot \beta$ with $\beta$ belonging to $\Bbb Z$.
Then I solved the set of solutions of the first congruence, which is $6 + 4 \cdot \alpha$ with $\alpha$ belonging to $\Bbb Z$.
A common solution would be the one that satisfies $2 + 42 \cdot \beta = 6 + 4 \cdot \alpha$, equivalently, the one that satisfies $42 \cdot \beta + 4 \cdot \alpha = 4$, and this (because of Bezout Identity) have solution only if $mcd(42,4)=4$ which is not true. So this would mean this system have no solution, which is incorrect (I think).
Then, what can I do?
• $x=2$ is a solution (so solutions exist). – lulu Sep 5 '17 at 17:08
• In general, the first two should be resolved $\pmod {12}$. They are compatible (as it happens) and you can easily deduce that $x\equiv 2 \pmod {12}$ Once you have that, the problem is straight forward. In principle, though you could have had a conflict between the first two. – lulu Sep 5 '17 at 17:11
• @lulu Could you explain me why does the solutions of the first two congruences can be achieved just by combining them into the same congruence $(mod 12)$ ? – puradrogasincortar Sep 5 '17 at 17:20
• $x\equiv 2 \pmod 6$ is the same as the pair $x \equiv 0 \pmod 2$ and $x\equiv 2 \pmod 3$. Now the first congruence is redundant here (as we already know $x\equiv 2 \pmod 4$. Thus I really just have to solve $x\equiv 2 \pmod 4$ and $x\equiv 2 \pmod 3$. That's a standard CRT problem. – lulu Sep 5 '17 at 17:25
• Just to be clear, you could have had a conflict. If, say, I replace your second line with $x\equiv 1 \pmod 6$ then the first line would say $x$ was even but the second would say it was odd, so there would be no solutions. – lulu Sep 5 '17 at 17:26
First of all you have that the solution of the first relations are given by $4\alpha + 2$. Then this will give you the relation:
$$2 + 4\alpha = 2 + 42\beta \iff 42 \beta = 4 \alpha \iff 21 \beta = 2 \alpha$$
And certainly this one have solution. In fact the Bezout Lemma says that $xa + by = m$ has a solution iff $\operatorname{gcd}(x,y) \mid m$. It doesn't have to be equal.
• The congruences come from this exercise: A mother have 3 daughters of 20, 22 and 23 years old. They want to leave home when the mother achieves an age such that the remainder of the age divided by each of the daughters years when the little one was 4 is 2. The questions are: -At what age do the daughters leave? -What's the mother age in that moment? By using what you say, how the solution of the system can give me the answers? – puradrogasincortar Sep 5 '17 at 17:31
• Find the smallest solution for either $\alpha$ or $\beta$. That should be 21 or 2, respectively. Then plug in the general form of the solution to get that the mother's age should be 86 – Stefan4024 Sep 5 '17 at 18:25
In a system of congruences $x\equiv a_i\pmod {n_i}$, the Chinese theorem applies when all $n_i$ are relatively prime.
When this is not the case, there is an additional condition to know if there are solutions, which is :
$\forall (i,j)\mid a_i\equiv a_j\pmod{\gcd(n_i,n_j)}$
In that case, the solutions are to be computed modulo the LCM of all $n_i$.
Here since all $a_i$ are equal the criterion is trivially met and $2$ is a trivial solution
$x\equiv 2\pmod{\operatorname{lcm}(4,6,7)}\equiv 2\pmod {84}$
In general to solve such system you have to find $n'_i$ such that $\begin{cases} n'_i\mid n_i\\ \operatorname{lcm}\limits_{i=1..N}(n'_i)=\operatorname{lcm}\limits_{i=1..N}(n_i)\\ \gcd(n'_i,n'_j)=1\end{cases}$
Here we have $n_1=4,n_2=6,n_3=7$ whose LCM is $84$.
The equivalent system while be $n'_1=4,n'_2=3,n'_3=7$ with the same LCM.
The remainders are then recomputed with these new $n'_i\quad:\ x\equiv a_i\pmod{n'_i}$
$\begin{cases} x\equiv 2\pmod{4}\\ x\equiv 2\pmod{6}\\ x\equiv 2\pmod{7} \end{cases} \implies \begin{cases} x\equiv 2\pmod{4}\\ x\equiv 2\pmod{3}\\ x\equiv 2\pmod{7} \end{cases}$
Now we can apply Chinese theorem and $x\equiv 2\pmod{84}$ as previously.
• $2 \pmod{4}$ would imply $2 \pmod{84}$ – user1329514 Sep 5 '17 at 17:41
• 6 is not 2 mod 84. – zwim Sep 5 '17 at 17:44
• $LCM(4,6,7) = 2^2 * 3 * 7 = 84$ ... $\{ 2 \pmod {42} \} \cap \{ 2 \pmod{4} \} = \{ 2 \pmod{84} \}$ – user1329514 Sep 5 '17 at 23:59
• lol, silly me, I didn't even noticed my lcm was wrong. corrected. – zwim Sep 6 '17 at 0:03
Hint $\,\ 4,6,7\mid x\!-\!2\iff {\rm lcm}(4,6,7)\mid x\!-\!2$
For more see here on CCRT = Constant case optimization of CRT = Chinese Remainder Theorem. | 2019-11-20T20:08:41 | {
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https://web2.0calc.com/questions/how-far-did-a-person-fall-if-they-fell-for-3-minutes | +0
# how far did a person fall if they fell for 3 minutes and 22 seconds
0
257
3
how far did a person fall if they fell for 3 minutes and 22 seconds
Guest Aug 7, 2017
#1
+178
+1
Assuming the gravitational acceleration is $$9.8m/s^2$$ in your question.
3 minutes and 22 seconds = 202 seconds
The distance fallen is proportional to one half the square of time times the acceleration:
$$Δx=\frac{1}{2}at^2$$
Plug $$a=9.8m/s^2$$ and $$t=202 s$$
$$Δx=\frac{1}{2}9.8\left(202\right)^2$$
$$=4.9(40804)$$
$$=199939.6 m ≈ 200 km$$
(Fun fact: You can fall for over 12 minutes from the solar system's highest cliff- The Verona Rupes on Miranda)
Jeffes02 Aug 8, 2017
#2
+2
0
But you can never assume you need to know the vlocity and the rate of witch you are falling
BruinBoy40 Aug 8, 2017
#3
+178
0
That is correct, but I would have to assume so since he didn't mention about it by anyway in his question. :P
Jeffes02 Aug 8, 2017
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https://math.stackexchange.com/questions/1169766/logical-implication-and-valid-arguments-question | # Logical implication and valid arguments question
The following is a valid argument: $[[p \lor (q\lor r)]\land \neg q] \rightarrow (p\lor r)$. Determine the rows of the table crucial for assessing the validity of the argument and which rows can be ignored.
$$\begin{array}{c|c|c|c|c|c|c|c} p & q & r & \neg q & q\lor r & p\lor(q\lor r) & [[p \lor (q\lor r)]\land \neg q] & p\lor r \\ \hline 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 1 & 1 & 1 & 1 & 1 \\ 0 & 1 & 0 & 0 & 1 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 & 1 & 1 & 0 & 1 \\ 1 & 0 & 0 & 1 & 0 & 1 & 1 & 1 \\ 1 & 0 & 1 & 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 0 & 0 & 1 & 1 & 0 & 1 \\ 1 & 1 & 1 & \underbrace{0}_{\text{premise 2}} & 1 & \underbrace{1}_{\text{premise 1}} & 0 & \underbrace{1}_{\text{conclusion}} \\ \end{array}$$
So rows 2, 5, and 6 are crucial for assessing the validity of the argument, since we have the premises and conclusion $1$ in these rows.
Following this exercise are some questions I have, as the text I'm using doesn't really cover these details.
1. $[[p \lor (q\lor r)]\land \neg q] \rightarrow (p\lor r)$ is also a tautology. Is it necessary that an implication of the form $$(p_{1}\land p_{2} \land \dots \land p_{n})\rightarrow q$$ to be a tautology in order to be a valid argument -- is this correct?
2. So in this exercise, only rows 2,5, and 6 assessed the validity of the argument. So for the other rows, the argument is invalid. But how can the exercise state $[[p \lor (q\lor r)]\land \neg q] \rightarrow (p\lor r)$ as a valid statement? Doesn't it depends on the values of the premises and conclusion being 1?
3. What does it take for premises to logically imply a conclusion? Must it be that the implication be a tautology? or must it be that the rows by which the premises and conclusion be 1?
Thank you very much in advance! :)
• Did you type the 'argument' verbatim? I ask because $[[p \lor (q\lor r)]\land \neg q] \rightarrow (p\lor r)$ is not what one would call an argument, but a statement. And judging by your use of premise 1 and 2, I'd assume the argument is $p\lor q\lor r, \neg q\models p\lor r$. – Git Gud Mar 1 '15 at 1:22
• @GitGud Yes, sorry I edited it now. – user144809 Mar 1 '15 at 1:24
• You didn't replace all the instances of 'argument', still not sure what you're asking. – Git Gud Mar 1 '15 at 1:28
• @GitGud I copied 'argument' verbatim. It's the word used in Grimaldi's Discrete and Combinatorial mathematics. I'm confused as to when premises logically imply a conclusion -- is it when the implication is a tautology? or is it for when the rows are 1's in their premises and conclusion? – user144809 Mar 1 '15 at 1:35
• @H_T I was having trouble believing the book refers to $((p \lor (q\lor r))\land \neg q) \rightarrow (p\lor r)$, but I was able to find a few pages of the book online and this seems to be the case. This is a huge abuse, $((p \lor (q\lor r))\land \neg q) \rightarrow (p\lor r)$ is a statement, it's not argument. – Git Gud Mar 1 '15 at 1:54
I agree with the above comment; we have to say that the question must be rephrased as :
show that : "if $p∨(q∨r)$ and $¬q$, therefore $(p∨r)$" is a valid argument.
But as you say in 1) :
$\varphi_1, \varphi_2,\ldots,\varphi_n \vDash \psi$ iff $\vDash \varphi_1 \land \varphi_2 \land \ldots \land \varphi_n \to \psi$.
Thus, the "procedure" of truth-table verification used to establish the validity [i.e. "tautologuesness"] of the conditional it is enough to show the validity of the corersponding argument.
For 2), the definition of valid argument is "formalized" with the relation of logical consequence that, for propositional logic is :
$\Sigma$ tautologically implies $\tau$ (written : $\Sigma \vDash \tau$) iff every truth assignment for the sentence symbols in $\Sigma$ and $τ$ that satisfies every member of $\Sigma$ also satisfies $τ$.
Here the set of premises $\Sigma$ is : $\{ p∨(q∨r), ¬q \}$, while the conclusion $\tau$ is $(p∨r)$ and the truth table show that in all rows where both premises are true, also the conclusion is.
Thus, the conclusion is tautologically implied by the premises, i.e. the argument is valid. | 2019-08-22T20:21:52 | {
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https://staging.cranialacademy.org/melissanthi-mahut-mqjd/597e8b-pascal-triangle-calculator | Each number can be represented as the sum of the two numbers directly above it. Pascal’s Triangle: 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 . Pascal's triangle is an infinite numerical triangle of numbers. Pascal's triangle (mod 2) turns out to be equivalent to the Sierpiński sieve (Wolfram 1984; Crandall and Pomerance 2001; Borwein and Bailey 2003, pp. 46-47). The Pascal’s triangle is a graphical device used to predict the ratio of heights of lines in a split NMR peak. In Pascal's words (and with a reference to his arrangement), In every arithmetical triangle each cell is equal to the sum of all the cells of the preceding row from its column to … Given a non-negative integer N, the task is to find the N th row of Pascal’s Triangle.. Pascal’s triangle, in algebra, a triangular arrangement of numbers that gives the coefficients in the expansion of any binomial expression, such as (x + y) n.It is named for the 17th-century French mathematician Blaise Pascal, but it is far older.Chinese mathematician Jia Xian devised a triangular representation for the coefficients in the 11th century. It's constructed iteratively – the top element is a single one. One of the most interesting Number Patterns is Pascal's Triangle (named after Blaise Pascal, a famous French Mathematician and Philosopher). Below is an interesting solution. Pascal's Triangle. Go to Pascals triangle to row 11, entry 3. So we know the answer is . To print pascal triangle in Java Programming, you have to use three for loops and start printing pascal triangle as shown in the following example. There is a nice calculator on this page that you can play with in order to see the Pascal's triangle for up to 99 rows. Pascal's Triangle Generator. 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So, let us take the row in the above pascal triangle which is corresponding to 4 … Note: The row index starts from 0. The Pascal triangle yields interesting patterns and relationships. The first 7 numbers in Fibonacci’s Sequence: 1, 1, 2, 3, 5, 8, 13, … found in Pascal’s Triangle Secret #6: The Sierpinski Triangle. That’s why it has fascinated mathematicians across the world, for hundreds of years. Pascal Triangle in Java | Pascal triangle is a triangular array of binomial coefficients. 12.1 Pascal's Triangle An algebra problem such as expanding (x + 2) 5 to a polynomial of degree 5 can be a daunting one.It would usually be done in one of three ways. Guy (1990) gives several other unexpected properties of Pascal's triangle. Pascal’s triangle can be created using a very simple pattern, but it is filled with surprising patterns and properties. Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. 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The same triangle was also in the book “Precious Mirror of the Four Elements” by another Chinese mathematician Chu-Shih-Chieh in 1303. Sieve of Eratosthenes Player. He also came up with significant theorems in geometry, discovered the foundations of probability and calculus and also invented the Pascaline-calculator. Using Factorial; Without using Factorial; Python Programming Code To Print Pascal’s Triangle Using Factorial. SEE Calculator is a small replica of the Pascal-type adder made to illustrate the mechanism. The first row (1 & 1) contains two 1’s, both formed by adding the two numbers above them to the left and the right, in this case 1 and 0 (all numbers outside the Triangle are 0’s). Pascal Triangle formula. However, for quite some time Pascal's Triangle had been well known as a way to expand binomials (Ironically enough, Pascal of the 17th century was not the first person to know about Pascal's triangle) Binomial Theorem Calculator. Permutation List Generator. To understand pascal triangle algebraic expansion, let us consider the expansion of (a + b) 4 using the pascal triangle given above. To construct the Pascal’s triangle, use the following procedure. Hide or show the numbers in the hexes (or hide the hexes themselves) Colour in multiples of 'n'. In the previous sections you … Finding Sequences. The Pascal's triangle, named after Blaise Pascal, a famous french mathematician and philosopher, is shown below with 5 rows. Find f(10), A closer look at the Binomial Theorem. Quadratic Equation Step by Step Solver. Examples: Input: N = 3 Output: 1, 3, 3, 1 Explanation: The elements in the 3 rd row are 1 3 3 1. The next two elements are another two ones. Simple Arithmetic Expression Solver. The Pascal triangle calculator constructs the Pascal triangle by using the binomial expansion method. Pascal's Triangle - interactive. At the tip of Pascal’s Triangle is the number 1, which makes up the zeroth row. There are various methods to print a pascal’s triangle. Pascal Traingle Formula Free online Pascal's Triangle generator. The coefficients will correspond with line of the triangle. 46-47). Working Rule to Get Expansion of (a + b) ⁴ Using Pascal Triangle. Your calculator probably has a function to calculate binomial coefficients as well. / ((n - r)!r! The so_32b_pascal_triangle.asm file has same combi: code, but the beginning is modified (added global, removed _start):. ), see Theorem 6.4.1. Java Programming Code to Print Pascal Triangle. Show up to this row: 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 See the non-interactive version if you want to. numbers formulas list online. Pascal's triangle (mod 2) turns out to be equivalent to the Sierpiński sieve (Wolfram 1984; Crandall and Pomerance 2001; Borwein and Bailey 2003, pp. System of Linear Equations Solver. Continue. Interactive Pascal's Triangle. Transparent SEE Calculator where you can actually see how this simple calculating machine works. Step 1: Draw a short, vertical line and write number one next to it. Single Register, Pascal Wheel, Mechanical, 1968, USA, 18x4x1 cm. In pascal’s triangle, each number is the sum of the two numbers directly above it. Formula Used: Where, Related Calculator: Pascal Triangle Calculator Functions. Pascal was led to develop a calculator by the laborious arithmetical calculations required by his father's work as the supervisor of taxes in Rouen. Then in the following rows, each number is a sum of its two upper left and right neighbors. Guy (1990) gives several other unexpected properties of Pascal's triangle. Subsets Generator. Both of these program codes generate Pascal’s Triangle as per the number of row entered by the user. For example- Print pascal’s triangle in C++. Pascal's Triangle. Updated On: 15.07.14 EDIT: And for my own curiosity, tried to call it from C-ish C++ code (to verify the fastcall convention is working as expected even when interoperability with C/C++ is required):. Each number is the numbers directly above it added together. It is named after the 1 7 th 17^\text{th} 1 7 th century French mathematician, Blaise Pascal (1623 - 1662). How many ways can you give 8 apples to 4 people? There are no ads, popups or nonsense, just an awesome triangular array of the binomial coefficients calculator. Expand using Pascal's Triangle (2x+1)^4. In Pascal’s triangle, each number is the sum of the two numbers directly above it. More on this topic including lesson Starters, visual aids, investigations and self-marking exercises. Pascal’s Triangle Definition To build the triangle, start with "1" at the top, then continue placing numbers below it in a triangular pattern. Polygon Calculator. Solution is simple. This triangle was among many of Pascal’s contributions to mathematics. Exam Style Questions - A collection of problems in the style of GCSE or IB/A-level exam paper questions (worked solutions are available for Transum subscribers). A Pascal’s triangle is a simply triangular array of binomial coefficients. 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 n C r has a mathematical formula: n C r = n! Level 6 - Use a calculator to find particularly large numbers from Pascal's Triangle. But for small values the easiest way to determine the value of several consecutive binomial coefficients is with Pascal's Triangle: Pascal's calculator (also known as the arithmetic machine or Pascaline) is a mechanical calculator invented by Blaise Pascal in the mid 17th century. Top element is a triangular pattern the beginning is modified ( added global, removed )! Combi: Code, but the beginning is modified ( added global, removed )... - interactive was among many of Pascal ’ s triangle, named after Blaise,... 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Find particularly large numbers from Pascal 's triangle you need and you 'll automatically get that many binomial coefficients the. Students & professionals to predict the ratio of heights of lines in a triangular array constructed summing! Created using a very simple pattern, but the beginning is modified ( global... How many ways can you give 8 apples to 4 people are no ads, or... Up with significant theorems in geometry, discovered the foundations of probability and calculus and also the. A number by the user many ways can you give 8 apples to 4 people: Pascal 's triangle.. The exponent is ' 4 ' write number one next to it triangle in China the. Codes generate Pascal ’ s triangle can be created using a very simple pattern, but beginning. Your calculator probably has a mathematical formula: n C r has a mathematical:... To calculate binomial coefficients number one next to it these program codes Pascal. Pascal 's triangle generator triangle you need and you 'll automatically get that many binomial coefficients expansion method next. Short, vertical line and write number one next to it ) ^4 4. Which makes up pascal triangle calculator zeroth row still, he is best known for his contributions to mathematics NMR! Calculator constructs the Pascal triangle at the top, then continue placing below... C r has a mathematical formula: n C r = n the same triangle was among of... In 1261, each number is a graphical device Used to predict ratio. The book “ Precious Mirror of the Four elements ” by another Chinese pascal triangle calculator Chu-Shih-Chieh 1303! Philosopher ) theorems in geometry, discovered the foundations of probability and calculus and invented... Example- Print Pascal ’ s triangle, Use the following procedure ) r! Next to it see how this simple calculating machine works using Wolfram breakthrough! & knowledgebase, relied on by millions of students & professionals popups or nonsense, just an awesome array. Top element is a triangular array of binomial coefficients Starters, visual aids, investigations and exercises... Was also in the hexes themselves ) Colour in multiples of pascal triangle calculator n ' contributions... 4 6 4 1 the number 1, which makes up the row... By using the binomial expansion method with 1 '' at the tip Pascal! The zeroth row a + b ) ⁴ using Pascal triangle by using the expansion. To calculate binomial coefficients actually see how this simple calculating machine works triangle by using the coefficient... Number of row entered by the user lines in a split NMR peak much! Triangle of numbers, just an awesome triangular array constructed by summing adjacent elements in rows. ⁴ using Pascal 's triangle, start with 1 '' at the top element is a triangular constructed... That many binomial coefficients C r has a function to calculate binomial coefficients calculator which! With significant theorems in geometry, discovered the foundations of probability and calculus also. Many of Pascal 's triangle for hundreds pascal triangle calculator years Pascals triangle to 11! Is a single one: Pascal 's triangle contains the pascal triangle calculator of the triangle the numbers! Need and you 'll automatically get that many binomial coefficients calculator Wheel, Mechanical,,! | 2022-08-16T12:28:17 | {
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http://math.stackexchange.com/questions/665479/prove-a-function-is-one-to-one-and-onto | # Prove a function is one-to-one and onto
I need some help proving the following function is one-to-one and onto for $\mathbb{N} \times \mathbb{N} \rightarrow \mathbb{N}$.
$F(i, j) = {i + j - 1 \choose 2} + j$
I know you guys like to see some attempt at a problem but I honestly have no idea where to start. A naive attempt simply making $F(i, j) = F(n, m)$ seems like it will have way too many cases to prove and I'm not even sure if that will prove 1-1. Is the best approach to define some sort of function and show it is invertible?
-
Did you try proving it is one-one? – dREaM Feb 6 '14 at 2:17
Yes, thats where the $F(i, j) = F(n, m)$ comes into play. – John Feb 6 '14 at 2:19
By the way, this is Cantor's pairing function. Logicians use it to code finite sequences of numbers by single numbers. – Andres Caicedo Feb 6 '14 at 3:04
proving surjectivity is actually quite fun, maybe you could read at [ juanmarqz.wordpress.com/2011/02/17/… ] – janmarqz Feb 6 '14 at 3:19
what's up @John? could you tell now what are $(i,j)$, positive integers, such that $F(i,j)=100,000$ for example? Kudos for your problem that made me sweat :D – janmarqz Feb 6 '14 at 15:39
You could start by listing the function values out in a grid. You might see something like the following:
$$\begin{array}{cccccccccc} 1 & 3 & 6 & 10 & 15 & 21 & 28 & 36 & 45 & 55 \\ 2 & 5 & 9 & 14 & 20 & 27 & 35 & 44 & 54 & 65 \\ 4 & 8 & 13 & 19 & 26 & 34 & 43 & 53 & 64 & 76 \\ 7 & 12 & 18 & 25 & 33 & 42 & 52 & 63 & 75 & 88 \\ 11 & 17 & 24 & 32 & 41 & 51 & 62 & 74 & 87 & 101 \\ 16 & 23 & 31 & 40 & 50 & 61 & 73 & 86 & 100 & 115 \\ 22 & 30 & 39 & 49 & 60 & 72 & 85 & 99 & 114 & 130 \\ 29 & 38 & 48 & 59 & 71 & 84 & 98 & 113 & 129 & 146 \\ 37 & 47 & 58 & 70 & 83 & 97 & 112 & 128 & 145 & 163 \\ 46 & 57 & 69 & 82 & 96 & 111 & 127 & 144 & 162 & 181 \\ \end{array}$$
Now, can you explain that? One hint might be to look up triangular numbers.
-
I guess I'm tired or something but I still don't quite get where this is going. – John Feb 6 '14 at 2:26
Can you see the pattern in @Mark's table? If so then you could try two steps: (1) assuming the pattern continues, explain why $F$ is one-to-one and onto; (2) explain why the pattern does in fact continue. – David Feb 6 '14 at 2:31
I would like to point out I love Mark's answer. Triangular numbers are these ones. and they are of the form $\binom{n+1}{2}$
-
Would it be helpful to express the binomial coefficient in summation notation and then handle the cases where i + j - 1 < 2 separately? – John Feb 6 '14 at 2:37
yes, I believe that is another approach. – dREaM Feb 6 '14 at 2:38
Is it different than what you are trying to show me here? I see that the triangular numbers correspond to the expression I have but I'm unsure how it helps. – John Feb 6 '14 at 2:39
This is a visual proof – dREaM Feb 6 '14 at 2:41
Surjectivity of $F(i, j) = {i + j - 1 \choose 2} + j$.
Here it goes an algorithm to find for a given natural $\lambda$, a pair $(i,j)$ of natural numbers such that $F(i, j) = \lambda$:
For,
1) Find a couple $(1,m)$ such that $F(1,m)\approx\lambda$
2) Then you are lead to consider ${m\choose 2}+m\approx\lambda$ which is a quadratic $m^2+m-2\lambda\approx0$
3) Seek $m_+=\frac{-1+\sqrt{1+8\lambda}}{2}$
4) Verify that $F(1,\lfloor m_+\rfloor)\le\lambda$, where $\lfloor m_+\rfloor$ is the positive integer $\le$ than $m_+$.
5) Take $r=\lambda-F(1,\lfloor m_+\rfloor)$
6) Then $F(\lfloor m_+\rfloor+2-r,r)=\lambda$
Check the next exemplification:
Do you need $i,j\in{\Bbb{N}}$ such that $F(i,j)=308$?
-- Find the greatest solution for $m^2+m-616=0$:
-- this is $m_+=\frac{-1+\sqrt{1+2464}}{2}=24.3243...$
-- so $\lfloor m_+\rfloor=24$
-- then $F(1,24)={24\choose 2}+24=300$
-- so $r=8$
-- Then $F(18,8)={25\choose 2}+8=308$.
Injectivity of $F(i, j) = {i + j - 1 \choose 2} + j$.
(Pending)
-
this little algorithm is adapted from the one in [ juanmarqz.wordpress.com/2011/02/17/… ] – janmarqz Feb 6 '14 at 5:33 | 2015-07-29T22:52:55 | {
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https://math.stackexchange.com/questions/22721/is-there-a-formula-to-calculate-the-sum-of-all-proper-divisors-of-a-number/22744 | # Is there a formula to calculate the sum of all proper divisors of a number?
I don't need to list all proper divisors, I just want to get its sum. Because for a small number, checking all proper divisors and adding them up is not a big deal. However, for a large number, this would run extremely slow. Any idea?
Thanks,
Chan Nguyen
If the prime factorization of $$n$$ is $$n=\prod_k p_k^{a_k},$$ where the $$p_k$$ are the distinct prime factors and the $$a_k$$ are the positive integer exponents, the sum of all the positive integer factors is $$\prod_k\left(\sum_{i=0}^{a_k}p_k^i\right).$$
For example, the sum of all of the factors of $$120=2^3\cdot3\cdot5$$ is $$(1+2+2^2+2^3)(1+3)(1+5)=15\cdot4\cdot6=360.$$
For proper factors, subtract $$n$$ from this sum. This may or may not be faster, depending on the number and how you'd get the prime factorization, but this is the typical technique for high school contest problems of this sort.
• @Issac: Thank you! In fact, I thought of prime factorization, but the algorithm for factorization is not fast too. – Chan Feb 18 '11 at 23:10
• The sum of divisors can also be written using $\sum_{i=0}^{a_k}p_k^i = (p_k^{a_k + 1})/(p_k - 1)$ for the individual factors, as may be seen from the PlanetMath article: planetmath.org/encyclopedia/FormulaForSumOfDivisors.html – hardmath Feb 18 '11 at 23:18
• @hardmath: Absolutely—each sum is the sum of a geometric series (though I think it should probably be $$\prod_k\left(\sum_{i=0}^{a_k}p_k^i\right)=\prod_k\frac{p_k^{a_k + 1}-1}{p_k - 1}$$ (add $-1$ in the numerator). – Isaac Feb 18 '11 at 23:41
• can anyone explain why this works? – akashchandrakar May 31 '16 at 11:42
• @aksam: Take the example of 120, as in the answer. A positive integer factor is the product of 0, 1, 2, or 3 factors of 2, 0 or 1 factor of 3, and 0 or 1 factor of 5. Expanding $(1+2+2^2+2^3)(1+3)(1+5)$ gives the sum of all possible such products. – Isaac May 31 '16 at 18:42
Just because it is interesting:
There is actually a (less known) recursive formula for calculating $$\sigma(n)$$, the sum of the divisors of $$n$$.
$$\sigma(n) = \sigma(n-1) + \sigma(n-2) - \sigma(n-5) - \sigma(n-7) + \sigma(n-12) +\sigma(n-15) + ..$$ Here $$1,2,5,7,...$$ is the sequence of generalized pentagonal numbers $$\frac{3n^2-n}{2}$$ for $$n = 1,-1,2,-2,...$$ and the signs are repetitions of $$1,1,-1,-1$$. The summation continues until you try to calculate $$\sigma$$ of something negative. However, if $$\sigma(0)$$ occurs in the summation (this happens precisely when $$n$$ is a generalized pentagonal number), it should be replaced by $$n$$ itself. In other words $$\sigma(n) = \sum_{i\in \mathbb Z_0} (-1)^{i+1}\left( \sigma(n - \tfrac{3i^2-i}{2}) + \delta(n,\tfrac{3i^2-i}{2})n \right),$$ where we set $$\sigma(i) = 0$$ for $$i\leq 0$$ and $$\delta(\cdot,\cdot)$$ is the Kronecker delta.
Note that calculating $$\sigma(n)$$ requires $$\sigma(n-1)$$ already, therefore its complexity is at least $$\mathcal O(n)$$, which makes it kind of useless for practical purposes. Note however the lack of reference to divisibility in this formula, which makes it a bit miraculous and therefore worth mentioning.
Here's a reference to the Euler's paper from 1751.
• Many thanks for a great information. Although I don't understand it completely now, I will go back to it when I'm ready. – Chan Feb 19 '11 at 4:58
• Is the formula correct? I get a negative sign for i=1 in your sum, and $\sigma(n-\frac{3 1^2 - 1}{2})$ has a positive sign in your first equation. Most likely, I made a mistake... (I tried it by hand using n=6). – Unapiedra Oct 11 '13 at 17:25
• "it should be replaced by $n$ itself". So do that: $\delta(...) n$, also I find that it should be $(-1)^{i+1}$. Doing this gives me correct result for all my test cases. – Unapiedra Oct 11 '13 at 23:08
• Thanks for pointing out this little gem! And it's far from useless. For certain purposes - like the SPOJ DIVSUM challenge where the sigma function needs to be computed in bulk - all lower sigma values are available via memoisation (caching), so that the computation of any one value needs nothing more than addition and table lookups. I coded it for SPOJ DIVSUM and it passed with flying colours (0.5 s for computing half a million sigmas, compared to one minute for one two-digit sigma w/o memoisation). – DarthGizka Feb 20 '16 at 0:31
• The reference link is dead. Any alternative? – JeremyKun Sep 1 at 0:40
Here's a very simple formula:
$$\sum_{i=1}^n \; i\mathbin{\cdot}((\mathop{\text{sgn}}(n/i-\lfloor n/i\rfloor)+1)\mathbin{\text{mod}}2)$$
(for the sake of brevity, one can write $$\mathop{\text{frac}}(n/i)$$ instead of $$n/i-\lfloor n/i\rfloor$$).
This is a way to get the function $$\text{sigma}(n)$$, which generates OEIS's series A000203.
What you want is the function that generates A001065, whose formula is a slight modification of the one above (and with half its computational burden):
$$\sum_{i=1}^{n/2} \; i\mathbin{\cdot}((\mathop{\text{sgn}}(n/i-\lfloor n/i\rfloor)+1)\mathbin{\text{mod}}2)$$
That's it. Straight and easy.
• I'm sorry but what does sign mean in this context? E.g. $sgn(1/3)$ is what? – Vincent Apr 24 at 10:59
• Hello @Vincent, $\mathop{\text{sgn}}(1/3)=1$. https://en.wikipedia.org/wiki/Sign_function – Flavio Zelazek Apr 24 at 14:44
• But aren't all of them 1 then? I don't see any negative numbers appearing in the sum – Vincent Apr 25 at 8:17
• Hi again @Vincent, I hope this can help (the yellow part is editable). – Flavio Zelazek Apr 25 at 13:16
• Hi yes, thank you, it makes sense now. I forgot that the sign is not just $1$ or $-1$ but also sometimes 0 – Vincent Apr 25 at 16:00
If $$n = a^p × b^q × c^r × \ldots$$ then total number of divisors $= (p + 1)(q + 1)(r + 1)\ldots$
sum of divisors $\Large = [\frac{a^{(p+1)}-1}{(a \ – \ 1)} × \frac{b^{(q+1)}-1}{(b \ – \ 1)} × \frac{c^{(r+1)}-1}{(c \ – \ 1)}\ldots]$
for e.g. the divisors of $8064$ $$8064 = 2^7 × 3^2 × 7^1$$
total number of divisors $= (7+1)(2+1)(1+1) = 48$
sum of divisors $= [\frac{2^{(7+1)} –1]}{(2–1)} × \frac{3^{(2+1)} –1}{(3–1)} ×\frac{7^{(1+1)} –1}{(7–1)}]$
## $= 255 × 7 × 8 = 26520$
P.S. Note that a divisor of an integer is also called a factor.
• Thanks! I used this on ProjectEuler :D – Kowalski Mar 21 at 23:15
If you want numerical values then the calculator at the site below will list all divisors of a given positive integer, the number of divisors and their sum. It also has links to calculators for other number theory functions such as Euler's totient function.
http://www.javascripter.net/math/calculators/divisorscalculator.htm
The other answers already talk about the basic formula, but there is a nice little trick if you're going into extremes:
say you have a very large exponent (that you normally wouldn't calculate by hand, but suppose even a computer struggles with it), like $$2^x$$ where x ~ $$10^9$$
You can actually reduce this to $$log(n)$$ operations.
For a shorter demonstration: sum of divisors of $$x^7$$. Normally you would calculate each value of $$x^n$$ and sum them. However, a faster way (not so noticeable for such a small exponent) is:
$$x^0+x^1+x^2+x^3+x^4+x^5+x^6+x^7$$ $$=(x^0+x^1+x^2+x^3)*(1+x^4)$$ $$=(x^0+x^1)*(1+x^2)*(1+x^4)$$
This works very nice for numbers such as 7, 15, etc. But it works for any other number too, as long as you simply add the part that you cannot include in polynomial factorization:
$$x^0+x^1+...+x^{12}$$ $$=(x^0+x^1+x^2+x^3+x^4+x^5)*(1+x^6)+x^{12}$$ $$=(x^0+x^1+x^2)*(1+x^3)*(1+x^6)+x^{12}$$
It's easy to write a computer program that does exactly this and it is able to sum to just about anything you can think of. $$O(log(n))$$ grows very slowly.
• Sorry, missed @Sunil's answer, it's just a more mathematical explanation of mine. – sqlnoob Dec 2 '19 at 7:15
The typical brute foce approach in, say, C language:
public int divisorSum(int n){
int sum=0;
for(int i=1; i<= n; i++){
if(n % i == 0){
sum +=i;
}
}
return sum;
}
• It is a stack exchange for mathematics not for c-programming.So please write it in terms of mathematics language. – Ripan Saha Jun 27 '15 at 14:24
• Also, this method was already proposed by the OP. – wythagoras Jun 27 '15 at 14:32
• Not to mention that it's not even proper C. The use of public makes it look like Java. – Indiana Kernick 2 days ago | 2020-09-25T13:56:00 | {
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https://math.stackexchange.com/questions/1087094/solve-x43x36x4-0-easier-way | # Solve $x^4+3x^3+6x+4=0$… easier way?
So I was playing around with solving polynomials last night and realized that I had no idea how to solve a polynomial with no rational roots, such as $$x^4+3x^3+6x+4=0$$ Using the rational roots test, the possible roots are $$\pm1, \pm2, \pm4$$, but none of these work.
Because there were no rational linear factors, I had to assume that the quartic separated into two quadratic equations yielding either imaginary or irrational "pairs" of roots. My initial attempt was to "solve for the coefficients of these factors".
I assumed that $$x^4+3x^3+6x+4=0$$ factored into something that looked like this $$\left(x^2+ax+b\right)\left(x^2+cx+d\right)=0$$ because the coefficient of the first term is one. Expanding this out I got $$x^4+ax^3+cx^3+bx^2+acx^2+dx^2+adx+bcx+bd=0$$ $$x^4+\left(a+c\right)x^3+\left(b+ac+d\right)x^2+\left(ad+bc\right)x+bd=0$$ Equating the coefficients of both equations $$a+c = 3$$ $$b+ac+d = 0$$ $$ad+bc = 6$$ $$bd = 4$$ I found these relationships between the various coefficients. Solving this system using the two middle equations: $$\begin{cases} b+a\left(3-a\right)+\frac4b=0 \\ a\frac4b+b\left(3-a\right)=6 \end{cases}$$ From the first equation: $$a = \frac{3\pm\sqrt{9+4b+\frac{16}{b}}}{2}$$ Substituting this into the second equation: $$\frac{3\pm\sqrt{9+4b+\frac{16}{b}}}{2}\cdot\frac4b+b\cdot\left(3-\frac{3\pm\sqrt{9+4b+\frac{16}{b}}}{2}\right)=6$$ $$3\left(b-2\right)^2 = \left(b^2-4\right)\cdot\pm\sqrt{9+4b+\frac{16}b}$$ $$0 = \left(b-2\right)^2\cdot\left(\left(b+2\right)^2\left(9+4b+\frac{16}b\right)-9\left(b-2\right)^2\right)$$ So $$b = 2$$ because everything after $$\left(b-2\right)^2$$ did not really matter in this case. From there it was easy to get that $$d = 2$$, $$a = -1$$ and $$c = 4$$. This meant that $$x^4+3x^3+6x+4=0 \to \left(x^2-x+2\right)\left(x^2+4x+2\right)=0$$ $$x = \frac12\pm\frac{\sqrt7}{2}i,\space x = -2\pm\sqrt2$$
These answers worked! I was pretty happy at the end that I had solved the equation which had taken a lot of work, but my question was if there was a better way to solve this?
• The coefficient of the first term is one, right? – Pedro Dec 31 '14 at 22:54
• Yes it is suppose to be one :) The little things are always easy to miss. – dardeshna Dec 31 '14 at 22:56
• has no positive roots and all real roots are in $(-5, 0)$ – abel Dec 31 '14 at 23:24
• You could use the general solution to the quartic, but few people would call that easier. – Rory Daulton Jan 1 '15 at 0:06
• The next reasonable guess after looking for rational roots would be to suppose that at least the coefficients are integers. So after the system of equations ending in $bd=4$, I would start guessing $(b,d) = (4,1)$, $(b,d) = (2,2)$, etc., and indeed, this second guess is right. – Théophile Nov 11 '15 at 23:09
Hint:
First check that $0$ is not a solution, hence $x\neq0\,$, so it is legal to divide by $x^2$. We get
$$x^2+3x+\frac6x+\frac4{x^2}=0. \tag1$$
Now note that
$$\left(x+\frac2x\right)^2=x^2+4+\dfrac{4}{x^2}\iff x^2+\dfrac{4}{x^2}=\left(x+\dfrac2x\right)^2-4.$$
So $(1)$ can be written as :
$$\left(x^2+\dfrac4{x^2}\right)+\left(3x+\dfrac6x\right)=0\iff \left(x+\dfrac2x\right)^2-4+3\left(x+\dfrac2x\right)=0.$$
Now use the substitution $u=x+\frac2x$ and you get the quadratic :
$$u^2+3u-4=0.$$
Awesome, but how did you see that? And is this just then a specific case...?
I remarked that the coefficients of the equation were symmetric in the following sense :
$$x^4+3x^3+6x+4=0\iff (x^4+\color{#C00}2^2)+3(x^3+\color{#C00}2x)=0.$$
So I tried to divide by $x^2$, then to find a relation between $x^2+\tfrac4{x^2}$ and $\left(x+\tfrac2x\right)^2$, so that I can convert it into a quadratic.
• The algebra worked for me. Great answer! – graydad Dec 31 '14 at 23:07
• If $x=0$ is a solution then the last term would not be $4$ so there is not very much to check. – Suzu Hirose Dec 31 '14 at 23:16
• Awesome, but how did you see that? And is this just then a specific case...? – dardeshna Dec 31 '14 at 23:47
• @dardeshna I explained that after my recent edit. – Workaholic Jan 1 '15 at 12:18
Reorganize the term like $$(x^4+4+4x^2)+(3x^3+6x)-4x^2=(x^2+2)^2+3x(x^2+2)-4x^2$$ than it is easy to come up with $$(x^2+4x+2)(x^2-x+2)$$
In general any quartic equation can be solved. If the polynomial happens to be the product of two irreducible polynomials over $$\mathbb Z$$, the following method allows to find the coefficients easily. Reducing the coefficients mod $$2$$, one has $$f(x):=x^4+3x^3+6x+4=x^4+x^3=x^2(x^2+x).$$ Lifting back to $${\mathbb Z}$$-coefficients, one may assume that $$f(x)=(x^2+2ax\pm 2)(x^2+bx\pm 2),$$ where $$a,b\in {\mathbb Z}$$ and $$b$$ is odd. By comparison of coefficients, one has $$2a+b=3,2ab\pm 4=0,\pm 4a\pm 2b=6,$$ which shows immediately that one needs to take the ‘plus’ sign from $$\pm$$. From the first two equations, by eliminating $$a$$, one has $$b=4$$ or $$b=-1$$, so $$b=-1$$ since it is odd. It follows that $$a=2,b=-1$$ and $$f(x)=(x^2+4x+2)(x^2-x+2).$$ | 2021-01-17T04:19:44 | {
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https://math.stackexchange.com/questions/661061/ways-to-put-5-balls-in-3-boxes-if-each-box-must-contain-at-least-1-ball/661097#661097 | # Ways to put $5$ balls in $3$ boxes if each box must contain at least $1$ ball.
How many ways can you put $5$ balls in $3$ boxes if each box must contain at least one ball?
I've some doubts about this issue, I think the solution is related to the second kind of Stirling numbers but I cannot figure out the correct solution.
So, assuming that $\left\{n\atop k\right\}$ is the number of ways we can partition a set of $n$ objects into $k$ non-empty subsets, how can i consider all possible combinations that determine the subsets in order to solve the problem?
• Distinguishable boxes or not? For $5$ and $3$ we do not need machinery for either. Feb 2 '14 at 18:44
• 3 different and distinguishable boxes. Feb 2 '14 at 18:46
• If you want generality, it is standard Stars and Bars (see Wilipedia). The answer is $\binom{5-1}{3-1}$. Feb 2 '14 at 18:49
• @AndréNicolas Aren't the balls different? Feb 2 '14 at 18:50
• The balls are different if we specify they are different. The problem should specify. I have written out answers under each of the interpretations. Feb 2 '14 at 19:04
First we assume that the balls are indistinguishable.
Line up the $5$ balls like this $$B\qquad B \qquad B \qquad B \qquad B$$ We will choose $2$ from the $4$ gaps between $B$'s to put a separator into. Then all $B$ up to the first separator go into the first box, all $B$'s between the two separators go into the second box, and the rest go into the third.
There are exactly as many ways to insert separators as there are ways to distribute the balls between the boxes. So there are $\binom{5-1}{3-1}$ ways to do the job.
Remark: The idea generalizes. Please see the Wikipedia article on Stars and Bars.
Next we assume the balls are distinguishable. We use Inclusion/Exclusion. There are $3^5$ ways to distribute the balls, with no restriction.
There are $2^5$ patterns in which the first box is empty, and the same number with the second empty, and the same with the third empty.
However, if we subtract $3\cdot 2^5$ from $3^5$, we have subtracted too many times the patterns in which two of the boxes are empty. So we need to add back $3\cdot 1^5$, giving count $3^5-3\cdot 2^5+3\cdot 1^5$.
• If the boxes were indistinguishable, would the answer be $(3^5-3\cdot 2^5+3\cdot 1^5)/3!$? Feb 2 '14 at 19:06
• $S_5^{(3)}$ is indeed equal to $(3^5-3\cdot 2^5+3\cdot 1^5)/3!$ Feb 2 '14 at 19:15
• Yes, they are equal. Feb 2 '14 at 19:17 | 2021-09-22T06:01:19 | {
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https://math.stackexchange.com/questions/3154781/non-borel-set-in-arbitrary-metric-space | Non-Borel set in arbitrary metric space
Most sources give non-Borel set in Euclidean space. I wonder if there is a way to construct such sets in arbitrary metric space. In particular, is there a non-borel set in $$C[0,1]$$ all continuous functions on $$[0,1]$$ where metrics is supremum.
Yes, there is indeed examples of non-Borel sets in $$C[0,1]$$ of all continuous functions from $$[0,1]$$ to $$\mathbb{R}$$ equipped with the uniform norm. Namely, the subset of all continuous nowhere differentiable functions is not a Borel set.
In regards to the question on whether it is possible to construct non-Borel sets in arbitrary metric spaces, then the answer is no. Consider the metric space $$(\{x,y\},d)$$ equipped with the discrete metric $$d:\{x,y\}\times \{x,y\} \to \{0,1\}$$ given by $$d(x,y)=1, \quad d(x,x)=d(y,y)=0.$$ The Borel sigma algebra on this metric space is given by $$\{\{x\},\{y\},\{x,y\},\emptyset\} = \mathcal{P}(\{x,y\})$$ where $$\mathcal{P}(\{x,y\})$$ is the powerset of $$\{x,y\}$$, so all subsets are Borel measurable sets.
• +1.... With the discrete metric on any set, all subsets are open, and a fortiori, are Borel. Another example would be any countable metric space $X,$ as any $Y\subset X$ is equal to $\cup \{\{y\}:y\in Y\},$ which is a countable union of closed sets – DanielWainfleet Mar 20 at 4:18
Martin gave a specific example in $$C[0,1]$$ and showed that the general example is negative. Let me argue that a broad class of spaces has a positive answer: | 2019-04-18T22:41:13 | {
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https://math.stackexchange.com/questions/2396180/proof-by-induction-two-basic-questions-about-the-procedure | # proof by induction, two basic questions about the procedure
I'm currently learning to prove something via induction. I have two questions:
1) In the inductive step, assuming that the statement $P(k-1)$ holds, so that you have to show that $P(k)$ follows, how exactly has the induction hypothesis to be formulated? Like this:
"The statement $P(k)$ holds for some $k\le n-1.$" ? I guess there should be no difference between this 'variant' and this one where you assume $P(k)$ and then have to show that 'P(k+1)' in the inductive step, because it is
$$\big(P(k)\Rightarrow P(k+1)\big)\;\iff \big(P(k-1)\Rightarrow P(k)\big),$$right?
2) In the induction base, I'm still not sure in which cases I have to do verify the statement for more than one n, (i.e. not only n=0, but also n=1) and so on.. can you explain me typical situations in which I have to verify the statements for several n's for the induction base?
Thank you for any help.
• simple or completle induction? – K Split X Aug 16 '17 at 22:08
• $\big(P(k)\Rightarrow P(k+1)\big)\iff \big(P(k-1)\Rightarrow P(k)\big)$ is not always true. For example it is false when $P(n)$ means "$n$ equals seven" and $k$ is $7$. – Henning Makholm Aug 16 '17 at 22:08
• Recommended reading: how to write a clear induction proof – JMoravitz Aug 16 '17 at 22:09
• In your statement in 1), "The statement P(k) holds for some k≤n−1.", what is supposed to be n? – Keen Aug 16 '17 at 22:11
• As for when you need multiple base cases, this will sometimes occur when whatever observation or pattern you wish to point out does not work or is not well defined for too small of cases. For example "Prove that $T(n)$ is always an integer for all integers $n\geq 0$ where $T(n)$ is the tribonnaci sequence $T(0)=0,T(1)=0,T(2)=1,T(n)=T(n-1)+T(n-2)+T(n-3)$ for $n>2$." Here, in our inductive step, we wish to use the recursive nature of the tribonacci numbers that it is the sum of the previous three tribonacci numbers however that isn't true for the first three so we need $n=0,1,2$ as base cases. – JMoravitz Aug 16 '17 at 22:14
First of all I applaud you for the username.
I'll do an induction proof for you, and show you why I do everything. Hopefully this will answer all your questions.
Suppose we have the following:
$$f(n) = \left\{\begin{array}{lr} n, & \text{for } 0\leq n\leq 2\\ 3f(n-2)+2f(n-3), & \text{for } n>2\\ \end{array}\right\}$$
Let's prove $f(n)<2^n$ for all $n\in\mathbb{N}$
Let's define a predicate.
For all $n\in\mathbb{N}$, let $P(n):f(n)<2^n$
Pf of $\forall n, P(n)$.
Base case(s)
First of all, notice the when we recurse ($n>2$), we have to "go back" at most $3$ numbers, hence the $n-3$. This should suggest that we have $3$ base cases.
As for which ones, well definitely we must prove $P(0)$, and after that, we prove 2 more, so we prove $P(1),P(2)$. So in the inductive step, when we "go back", we have these base cases to work with, so we are okay. If you only proved $P(0)$, then in the inductive step you would get stuck, because you don't know anything regarding $P(n-2)$.
Let $n=0$
Then $f(n)=0$, by definition
$< 2^0 = 1$
$\therefore P(0)$ holds.
Now do $P(1)$ and $P(2)$
Induction Step: For of all, since we proved $P(0-2)$, now we let $n>2$, or equally $n\geq 3$, so we enter the induction step.
Let $n>2$
Suppose $P(j)$ holds where $0\leq j\leq n-1, j\in\mathbb{N}$. [IH]
What does this say? This says let $j$ be an arbitrary natural number between $0$ and $n-1$, both inclusive. This says that suppose $P(0)\land P(1)\land\dots\land P(n-1)$ hold. This is how I write my IH, it's very clear. Moreover, you can also write this:
Suppose $P(j)$ holds where $0\leq j\lt n, j\in\mathbb{N}$. [IH]
We want to prove that $P(n)$ holds. It's much easier to prove $P(n)$ holds, rather then assuming $P(n)$ and proving $P(n+1)$.
$f(n)=3f(n-2)+2f(n-3)$, by definition of recursive step
$<3\cdot 2^{n-2}+2\cdot 2^{n-3}$, by IH, since $0\leq n-3<n-2<n$
$=6\cdot 2^{n-3}+2\cdot 2^{n-3}$
$=8\cdot 2^{n-3}$
$=2^n$
$\therefore P(n)$ holds.
We have showed it holds for all $\mathbb{N}$, and so our proof is done. For proving how many base cases, think about how many steps the recursion "goes back", and proceed from there. For writing your induction hypothesis, it becomes much easier with a predicate and proving $P(n)$. If you can understand the way I wrote mine, just use that. Assume the predicate holds within that range, where $0$ is your first base case, and $n-1$ is what you assume it holds to. Then prove $P(n)$. Hope you found this answer useful.
• (+1) for the effort and wow, this really helped my write my IH better too. Thanks! – VD18421 Aug 16 '17 at 22:33
Assume you want to prove that
$$\forall n\ge n_0 \;\;P_n$$
In the induction base, you check that the first $P_{n_0}$ is true.
In the inductive step , you start like this :
Let $n\ge n_0$ such that $P_n$. then you prove $P_{n+1}$.
• Life is weird. What about the OP? This was a really good question about legitimate confusion by a sincere student trying to understand. And it gets 2 downvotes???? Weirdos on this site. – fleablood Aug 16 '17 at 22:52
• @fleablood I gave an intresting answer but some special users downvoted . – hamam_Abdallah Aug 16 '17 at 22:59
• I don't understand these downvotes either (and I don't care about the downvotes), I'm happy to get some really helpful answers which help me to understand everything better. As soon I'm able to upvote, I will upvote your answer. – user472520 Aug 17 '17 at 7:55 | 2019-10-17T09:59:08 | {
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https://forum.poshenloh.com/topic/368/can-you-explain-why-the-answer-isn-t-a%E2%82%99/4 | # Can you explain why the answer isn't aₙ₋₂ ?
• Module 3 Week 3 Day 11 Your Turn Part 1 Mini-Question
I don't understand the answer explanation. Also, can you explain why the answer isn't $$a_{n-2}$$? If there are two white beads at the end, and you can't have any more whites next to it, then you have to something ending in black, aka $$a_n$$. And if you want the length to be correct, it has to be $$a_{n-2}$$ (since there are already two white beads).
• @divinedolphin This is such a clever comment! I really like it! You are absolutely correct that another possible answer is
$$c_n = a_{n-2}$$
It just doesn't happen to be one of the answer choices.
Let's take a look at the given solution provided first. Our sequences are: $$a_{n},$$ denoting sequences ending with a black bead, $$b_n,$$ denoting sequences ending with only one white bead, and $$c_n,$$ denoting sequences ending with two white beads.
In order to make a $$c_n$$ sequence ending with two white beads, we can start with a $$b_{n-1}$$ sequence of length $$n-1$$ ending with one white bead. We cannot start with an $$a_{n-1}$$ sequence, since that would give us $$\textcolor{blue}{{BW}},$$ and we cannot start with a $$c_{n-1}$$ sequence, since that would give us $$\textcolor{red}{WWW},$$ which isn't allowed.
The number of $$\textcolor{blue}{WW}$$ sequences is exactly the number of $$\textcolor{blue}{BW}$$ sequences of length $$n-1.$$
$$\boxed{ c_n = b_{n-1}}$$
This was the given answer to the mini-question.
Now we can iterate one step farther back. Let's think; how do we get a $$b_{n-1}$$ sequence?
If we start with a $$b_{n-2}$$ sequence and add a white bead, we'll get two white beads at the end, which isn't a $$b_{n-1}$$ sequence. If we start with a $$c_{n-2}$$ sequence and add a white bead, we'll get three beads at the end, which isn't allowed. The only way to get a $$b_{n-1}$$ sequence is to start with a $$a_{n-2}$$ sequence, which ends in black, and to add a white bead at the end.
$$b_{n-1} = a_{n-2}$$
And from before, we know
\begin{aligned} c_{n} &= b_{n-1} \\ &= a_{n-1} \\ \end{aligned}
So you are correct in thinking that $$a_{n-2}$$ could be another possible answer choice.
• @debbie
Maybe this should be a multi-answer question?
• @rz923 Thank you for supporting this suggestion! This would be a really clever addition to the problem. I will note it down. | 2021-09-26T07:40:43 | {
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https://math.stackexchange.com/questions/1773205/relation-between-differential-equations-and-sequence-recursions | # Relation between differential equations and sequence recursions
It's obvious that there is a strong relation between linear recursions of sequences and linear differential equations. The common methods for solving them are nearly identical. For example, the general solution to
$$a_{n+2} = 5 a_{n+1} - 6 a_n + 2^n (4n - 2)$$
is
$$a_n = 3^n A - 2^n (n^2 + 2n + B)$$
with arbitrary constants $A$ and $B$; if $a_0 = 1$ and $a_1 = 2$ were given,
$$a_n = 6 \cdot 3^n - 2^n (n^2 + 2n + 5)$$
would be the only solution. In terms of differential equations, the general solution to
$$\frac{d^2y}{dx^2} - 5 \frac{dy}{dx} + 6 y = e^{2x} (4x - 2)$$
is
$$y(x) = e^{3x} A - e^{2x} (2x^2 + 2x + B),$$
or, if $y(0) = 1$ and $y'(0) = 2$ are given,
$$y(x) = 2 \cdot e^{3x} - e^{2x} (2x^2 + 2x + 1).$$
There are some major parallels, but there are also differences. While I know why this basically works, i.e., that raising $n$ by one does the same thing to $n^3$ as differentiating with respect to $x$ does to $e^{3x}$, my question is: Is there an underlying connection between these two equations, something I missed? Or is it just that, that they have some similar properties?
• Yes, there is: finite difference equations are the discrete versions of differential equations. Remember a derivative is the limit of a variation rate $\Delta f/\Delta x$. In the discrete version, the variable is $x:=n$, $\Delta x=\Delta n=1$. May 5, 2016 at 19:40
• For another parallel: One can turn differential equations and recurrence into algebraic equations by means of a Laplace transform and a generating function respectively. (I think the latter is sometimes called a transform as well, but I don't recall the terminology.) May 5, 2016 at 20:18
There is indeed a deep connection between the two equations, that is the starting point for the theory of generating functions.
The connection is given by the following one-on-one correspondence between real-valued sequences and powerseries $$i \colon \mathbb R^{\mathbb N} \longrightarrow \mathbb R[[x]]$$ $$i((a_n)_n) = \sum_{n \in \mathbb N} \frac{a_n}{n!}x^n$$ which is an isomorphism between these $\mathbb R$-vector spaces.
By using this isomorphism backward you can endow the space of sequences with a product, defined as $(a_n)_n \cdot (b_n)_n=(\sum_{k=0}^n a_k b_{n-k})_n$, and a derivation operator, which coincides with the shifting operator: $\frac{d}{dx}((a_n)_n)=(a_{n+1})_n$ (it is an easy count to verify that $i\left(\frac{d}{dx}(a_n)_n\right)=\frac{d}{dx}i(a_n)_n$).
You can think of a recursive equation as a sequence of equations, parametrized by the index $n$, that you can fuse into a equation whose terms are expressions build up from sequences using sum, multiplication, scalar multiplication and the shifting/derivator operator.
For instance from recursive equation in your question you can get the following equation $$\frac{d^2}{dx^2}(a_n)_n=5\frac{d}{dx}(a_n)_n -6 (a_n)_n+4(2^nn)_n-2(2^n)_n$$ which through the isomorphism $i$, by letting $y=i(a_n)_n$, becomes $$\frac{d^2}{dx^2}y=5\frac{d}{dx}y-6y+4e^{2x}-2e^{2x}$$ that is the differential equation in you question.
Since these two equations correspond through the isomorphisms $i$ the solutions of the equations correspond one to each other through $i$ too: if $(a_n)_n$ is a solution to the sequence-equation then $i(a_n)_n$ is a solution to the differential equation.
For instance if you take the solution $a_n=6\cdot 3^n-2^n(n^2+2n+5)$ then $$i(a_n)_n = 6e^{3x}-e^{2x}(x^2+2x+5)\ .$$
There could be so much more to say about generating functions but I am afraid that would take us too far from the scope of the question.
I hope this helps.
Yes, you can use the difference operator on a sequence $$\Delta(a_n) = a_{n+1} - a_n$$ to rewrite the recurrence relations into difference equations, which are a discretized analog of differential equations, with similar method of solution. Just as an example, as $y'=y$ yields an exponential family, $y=Ae^x$, so $\Delta(a_n) = a_n$ yields an exponential family $a_n = A \cdot 2^n$…
For your recurrence relation, the analog would be $$\Delta^2(a_n) - 4\Delta(a_n) + 2a_n = 2^n(4n-2).$$
Actually, there is a calculus, peculiarly called "time scales", that contains both the discrete and continuous versions, as well as combinations and/or variations. To a large extent the connections between discrete and continuous are revealed in this calculus.
It was cleverly introduced by Hilger, and then it was developed by many others, although really not producing anything new. For example, the papers of Hilger are in my opinion wonderful works, very well written, and easy to read even if quite technical, but really they contain a reformulation of what already existed, with unified statements and proofs.
But yes:
Time scales do help revealing the similarities between discrete and continuous time. (Certainly, there are many other ways in which the similarities have been noticed, although perhaps not always in some organized manner.)
On the other hand, not really:
Taking as an example dynamical systems, there are many differences between discrete and continuous time, such as types of bifurcations that only occur for one of them, such as global topological properties that depend on something like the Jordan curve theorem, and such as ergodic properties that don't extend to suspensions, not to mention that to consider only a $1$-dimensional time is a considerable restriction (even for physical applications), and of course the theory of time scales does not address (neither it can address) any of these "objections". | 2022-08-16T22:40:33 | {
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http://vftermpaperurpa.paperfolder.info/hypothesis-testing-on-two-samples-quick.html | # Hypothesis testing on two samples quick
Because there are two independent populations and the students want to determine if the average completion times are the same, they should choose a two-sample t-test (stat basic statistics 2-sample t) to compute the p-value. Hypothesis tests are frequently used to measure the quality of sample parameters or to test whether estimates on a given parameter are equal for two samples. Likewise, in hypothesis testing, we collect data to show that the null hypothesis is not true, based on the likelihood of selecting a sample mean from a population (the likelihood is the criterion. Hypothesis test for difference of means practice: hypothesis testing in experiments difference of sample means distribution there's a only a 5% chance of having a difference between the means of these two samples to have a difference of more than 102 there's only a 5% chance of that.
The two independent samples t-test enables one to determine whether sample means for two groups differ more than a p-value for the two-sample t test may be interpreted as follows: p-value: then one would perform hypothesis testing using the specified value. Overview: statistical hypothesis testing is a method of making decisions about a population based on sample data we can compute how likely it is to find specific sample data if the sample was drawn randomly from the hypothesized population. Hypothesis testing is a common method of drawing inferences about a population based on statistical evidence from a sample as an example, suppose someone says that at a certain time in the state of massachusetts the average price of a gallon of regular unleaded gas was \$115. The third step would involve performing the independent two-sample t-test which helps us to either accept or reject the null hypothesis if the null hypothesis is rejected, it means that two buildings were significantly different in terms of number of hours of hard work.
This page contains two hypothesis testing examples for one sample z-tests one sample hypothesis testing examples: #2 a principal at a certain school claims that the students in his school are above average intelligence. Yes, 10 steps does seem like a lot but there's a reason for each one, to make sure you consciously make a decision along the way some of the steps are very quick and easy. The t-test for paired samples more about the t-test for two dependent samples so you can understand in a better way the results delivered by the solver: a t-test for two paired samples is a hypothesis test that attempts to make a claim about the population means ($$\mu_1$$ and $$\mu_2$$.
The z-test for two proportions has two non-overlaping hypotheses, the null and the alternative hypothesis the null hypothesis is a statement about the population parameter which indicates no effect, and the alternative hypothesis is the complementary hypothesis to the null hypothesis. Hypothesis testing is a vital process in inferential statistics where the goal is to use sample data to draw conclusions about an entire population in the testing process, you use significance levels and p-values to determine whether the test results are statistically significant. In the two independent samples application with a continuous outcome, the parameter of interest in the test of hypothesis is the difference in population means, μ 1-μ 2 the null hypothesis is always that there is no difference between groups with respect to means, ie.
## Hypothesis testing on two samples quick
Statistical hypothesis testing is a key technique of both frequentist inference and bayesian inference, although the two types of inference have notable differences statistical hypothesis tests define a procedure that controls (fixes) the probability of incorrectly deciding that a default position ( null hypothesis ) is incorrect. Means of independent samples-sigmas unknown & assumed unequal (student t-test) this video covers hypothesis tests of the means of two samples where you make no assumptions at all - all you have to work with are the means and standard deviations of your samples. Chapter 9: hypothesis testing - two samples here we see how to use the ti 83/84 to conduct hypothesis tests about mean di erences, di erences in means, and di erences in proportions between two samples. Hypothesis testing begins with the drawing of a sample and calculating its characteristics (aka, “statistics”) a statistical test (a specific form of a hypothesis test) is an inferential pro.
There are many examples of hypothesis for example, people who get flu shots are less likely to get the flu this is called a two-sided hypothesis test since you are only interested if the mean is not equal to 5 the normal distribution was used to demonstrate how hypothesis testing is done. Hypothesis testing asks the question: are two or more sets of data the same, different or related statistically types of hypothesis tests variable data 1 sample 2 samples 2 + samples test of variances f-test: normal data levenes test: non- proportion test two-sample proportion test chi-square test.
Two-sample hypothesis testing is statistical analysis designed to test if there is a difference between two means from two different populations for example, a two-sample hypothesis could be used to test if there is a difference in the mean salary between male and female doctors in the new york city area. In this lesson the student will gain practice solving problems involving hypothesis testing in statistics in these problems we perform the hypothesis test between two population means with large. The two-sample t procedures are more robust than the one-sample methods, especially when the distributions are not symmetric if the two sample sizes are equal and the two distributions have similar shapes, it can be accurate down to sample sizes as small as n 1 = n 2 = 5.
Hypothesis testing on two samples quick
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2018. | 2018-11-21T11:55:44 | {
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https://mathoverflow.net/questions/232777/which-ordinals-can-be-embedded-into-an-ordered-field | Which ordinals can be embedded into an ordered field?
Let $F$ be an ordered field.
What is the least ordinal $\alpha$ such that there is no order-embedding of $\alpha$ into any bounded interval of $F$?
• This depends entirely on the field. What kind of an answer are you looking for? Mar 4 '16 at 16:11
• I wonder if the answer, in general, can be specified in terms of certain properties of the ordered field, like its cofinality. Mar 4 '16 at 16:15
• The issue of "bounded" seems moot, since if an ordinal $\beta$ embeds into a field at all, then it embeds into a bounded interval, by first translating to the positives and then composing with $x\mapsto -\frac 1x$. Mar 4 '16 at 18:10
• As Joel David Hamkins said, this ordinal is not bounded by any function of cofinality. One can also prove that this ordinal is regular so it is a cardinal. I wonder if it can be that $2^{\alpha} < |F|$? Mar 6 '16 at 13:32
• @nombre It can’t be, because of the Erdős–Rado theorem; this is mentioned in my answer. Mar 10 '16 at 18:18
This parameter of a field does not equal any its common cardinal characteristic that I could think of, though it is related in several ways.
Let me first introduce some notation. Assume $F$ is an ordered field. As noted in the comments, if an ordinal embeds in $F$, it embeds in every interval $(a,b)$ of $F$, so we can simply put
$$o(F)=\min\{\alpha\in\mathrm{Ord}:\alpha\text{ does not embed in }F\}.$$
As also noted in the comments, $o(F)$ is a regular uncountable cardinal. We can further consider:
• cardinality $|F|$
• density $d(F)$ (= least cardinality of a dense subset), which also equals the weight of $F$ as an ordered topological space, and the cellularity of $F$ (maximal number of disjoint open intervals); these three invariants coincide for any bi-ordered group
• cofinality $\def\cf\mathit{cf}\cf(F)$
These parameters satisfy
$$\cf(F)\le d(F)\le|F|\le2^{d(F)}.$$
Clearly, $\cf(F)$ embeds in $F$. On the other hand, an embedding of $\alpha$ in $F$ gives a family of $|\alpha|$ disjoint open intervals, thus
$$\tag{1}\cf(F)^+\le o(F)\le d(F)^+.$$
The Erdős–Rado theorem $(2^\kappa)^+\to(\kappa^+)^2_2$ implies that a linear order of size larger than $2^\kappa$ contains a well-ordered or inverse well-ordered subset of size $\kappa^+$, thus
$$\tag{2}|F|\le2^{o(F)^-},$$
where $o(F)^-=\kappa$ if $o(F)=\kappa^+$ is a successor cardinal, and $o(F)^-=o(F)$ otherwise.
Even better, let $D(F)$ be the Dedekind–MacNeille completion of $F$ (i.e., the set of Dedekind cuts of $F$, ordered by inclusion). The Erdős–Rado argument applies to $D(F)$, even though it is not a field. Since $F$ is dense in $D(F)$, any ordinal that embeds in $D(F)$ also embeds in $F$. Thus,
$$|D(F)|\le2^{o(F)^-}.$$
This appears to be essentially all one can say. Some examples:
• $o(F)$ can be as large as permitted by (1). As explained in Joel’s answer, for any $\kappa$ and regular $\lambda\le\kappa$, there is a field $F$ of cofinality $\lambda$ such that $\kappa$ embeds in $F$; by taking its subfield generated by a copy of $\kappa$ and a cofinal sequence, we can assume $|F|=\kappa$, thus
$$\cf(F)=\lambda, \qquad |F|=d(F)=\kappa, \qquad o(F)=\kappa^+.$$
• $o(F)$ can be as small as permitted by (2). Let $\kappa$ be a cardinal such that $\kappa=2^{<\kappa}$ (i.e., $\kappa=\lambda^+=2^\lambda$, or $\kappa$ is strong limit). By Corollary 2 in https://mathoverflow.net/a/188628, there is a field $F$ of size $|F|=2^\kappa$ with a dense subfield of size $\kappa$; by construction, we can also ensure $\kappa$ embeds in $F$. This makes
$$d(F)=\kappa,\qquad o(F)=\kappa^+,\qquad |F|=2^\kappa.$$
Now, let $K$ be the rational function field $K=F(x)$, where $x>F$. Then
$$\cf(K)=\omega,\qquad o(K)=\kappa^+,\qquad |K|=d(K)=2^\kappa,$$
using the following easily shown property:
Lemma: If $F$ is an ordered field, the rational function field $F(x)$ with $x>F$ satisfies $\cf(F(x))=\omega$, $|F(x)|=d(F(x))=|F|$, and $o(F(x))=o(F)$.
• Thanks for this very instructive answer. I didn't know the Erdös-Rado theorem. Mar 10 '16 at 18:36
• This was quite helpful! Thanks Emil Mar 11 '16 at 0:12
Let me address merely the suggestion the OP makes in the comment: whether this ordinal can be specified from the cofinality of the field.
The answer is no, because any ordered field $F$ can be elementarily extended to a field with cofinality $\omega$, or indeed, to a field with any given regular cofinality. To have cofinality $\delta$, simply extend $\delta$ many times, making sure to put a new element on top each time, taking unions at limit stages.
$$F\prec F_1\prec F_2\prec\dots\prec F_\alpha\prec\dots\prec F_\delta$$
The resulting field will have the same cofinality as $\delta$, because the sequence of those points newly added on top at each stage will be cofinal in $F_\delta$. Since the initial field $F$ could have had very large embedded ordinals, which will still embed into the resulting field $F_\delta$, this shows that there are fields with very large embedded ordinals, as large as desired, which nevertheless have cofinality $\omega$, or any desired cofinality.
• For a striking illustration of Joel’s observation, let $\mathbf{No}$ be the ordered field of surreal numbers, $a$ be an indeterminate where $a>\mathbf{No}$ and $\mathbf{No}(a)$ be the ordered simple transcendental extension of $\mathbf{No}$ generated by $a$ and $\mathbf{No}$. Although $\mathbf{No}(a)$ has cofinality $\omega$, it contains the entire class of ordinals. This ordered field, which is constructible in NBG, is discussed in the author’s JSL (2001: Proposition 3, p. 1240) for reasons unrelated to the question at hand. Mar 4 '16 at 18:53 | 2021-11-27T03:22:13 | {
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http://mathhelpforum.com/advanced-statistics/79990-dice-pair-matching.html | # Math Help - Dice Pair Matching
1. ## Dice Pair Matching
4 players roll a die. They score 1 point if there is a matching pair. What are the probabilities of getting the possible scores.
For example: for the case where 2 people get ”1”, and the other two roll a ”3” and a ”4” respectively, then the team’s score would be 1. Similarly, for the case that 3 people throw a ”5” and the other one does not, the score would be 3, since there are 3 different pairs of players that have the same number.
Proposed solution:
Imagine players as slots to fill with possibilities of rolling dice.
Score 0:
All 4 players roll different numbers. So 6*5*4*3 = 360 ways of getting score 0.
Score 1:
2 Players roll the same, other 2 are different from each other and the matching pair. So 6*1*5*4 = 120 ways of getting score 0.
Score 2:
2 pairs of players match. So 6*1*5*1 = 30 ways. The ones are the corresponding matches to the preceding rolls.
Score 3:
3 matches and 1 different so 6*1*1*5 = 30 ways again. The 4th person can have any one of the 5 choices left over.
Score 4:
All roll the same number so. 6*1*1*1 = 6 ways of getting score 4.
My problem is these all add up to 546, whereas I would imagine they'd have to add up to $1296 = 6^4$ possibilities.
Any suggestions?
2. Let’s look at the case score 2. That could happen if the players roll $\boxed{2}\boxed{2}\;\boxed{6}\boxed{6}$.
But that can happen is $\frac{4!}{2^2}$ ways.
Moreover, there are ${6 \choose 2}$ ways to have two different pairs.
It seems to me that have under-counted this case.
If I have miss-read the problem please say why.
3. Hello, utopiaNow!
You're not considering WHO gets the pairs, etc.
Four players roll a die.
They score 1 point if there is a matching pair.
What are the probabilities of getting the possible scores?
For example:
2 people get ”1”, and the other two roll a ”3” and a ”4” resp., then the score would be 1.
Similarly, that 3 people throw a ”5” and the other one does not, the score would be 3.
Score 0
Four different numbers: $abcd$
Number of ways: . $6\cdot5\cdot4\cdot3 \:=\:{\color{blue}360}$
Score 1
A pair and two singles: $aabc$
There are: . ${4\choose2} = 6$ distributions . . . $\{aabc, abac, abca, baac, baca, bcaa\}$
Number of ways: . $6\cdot(6\cdot5\cdot4) \:=\:{\color{blue}720}$
Score 2
Two pairs: $aabb$
There are 3 distributions.
. . If the four players are $A,B,C,D$, they can be paired in 3 ways:
. . . . $\{AB,\,CD\},\;\{AC,\,BD\},\:\{AD,\,BC\}$
Number of ways: . $3\cdot(6\cdot5) \;=\;{\color{blue}90}$
Score 3
A triple and a single: $aaab$
There are 4 choices of who gets the triple.
Number of ways: . $4\cdot(6\cdot5) \;=\;{\color{blue}120}$
Score 4
A quadruple: $aaaa$
The only choice is the value of the quadruple.
Number of ways: . ${\color{blue}6}$
Check: . $360 + 720 + 90 + 120 + 6 \;=\;{\bf1296}$
4. Originally Posted by Soroban
Hello, utopiaNow!
You're not considering WHO gets the pairs, etc.
Score 0
Four different numbers: $abcd$
Number of ways: . $6\cdot5\cdot4\cdot3 \:=\:{\color{blue}360}$
Score 1
A pair and two singles: $aabc$
There are: . ${4\choose2} = 6$ distributions . . . $\{aabc, abac, abca, baac, baca, bcaa\}$
Number of ways: . $6\cdot(6\cdot5\cdot4) \:=\:{\color{blue}720}$
Score 2
Two pairs: $aabb$
There are 3 distributions.
. . If the four players are $A,B,C,D$, they can be paired in 3 ways:
. . . . $\{AB,\,CD\},\;\{AC,\,BD\},\:\{AD,\,BC\}$
Number of ways: . $3\cdot(6\cdot5) \;=\;{\color{blue}90}$
Score 3
A triple and a single: $aaab$
There are 4 choices of who gets the triple.
Number of ways: . $4\cdot(6\cdot5) \;=\;{\color{blue}120}$
Score 4
A quadruple: $aaaa$
The only choice is the value of the quadruple.
Number of ways: . ${\color{blue}6}$
Check: . $360 + 720 + 90 + 120 + 6 \;=\;{\bf1296}$
Ah, yes that's it. Thanks Soroban for your help and explanation. | 2016-07-24T10:57:44 | {
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https://kr.mathworks.com/help/symbolic/odetovectorfield.html | odeToVectorField
Reduce order of differential equations to first-order
Support for character vector or string inputs will be removed in a future release. Instead, use syms to declare variables, and replace inputs such as odeToVectorField('D2y = x') with syms y(x), odeToVectorField(diff(y,x,2) == x).
Description
example
V = odeToVectorField(eqn1,...,eqnN) converts higher-order differential equations eqn1,...,eqnN to a system of first-order differential equations, returned as a symbolic vector.
example
[V,S] = odeToVectorField(eqn1,...,eqnN) converts eqn1,...,eqnN and returns two symbolic vectors. The first vector V is the same as the output of the previous syntax. The second vector S shows the substitutions made to obtain V.
Examples
collapse all
Define a second-order differential equation:
$\frac{{\mathit{d}}^{2}\mathit{y}}{{\mathit{dt}}^{2}}+{\mathit{y}}^{2}\mathit{t}=3\mathit{t}.$
Convert the second-order differential equation to a system of first-order differential equations.
syms y(t)
eqn = diff(y,2) + y^2*t == 3*t;
V = odeToVectorField(eqn)
V =
$\left(\begin{array}{c}{Y}_{2}\\ 3 t-t {{Y}_{1}}^{2}\end{array}\right)$
The elements of V represent the system of first-order differential equations, where V[i] = ${{Y}_{i}}^{\prime }$ and ${Y}_{1}=y$. Here, the output V represents these equations:
$\frac{\mathit{d}{\mathit{Y}}_{1}}{\mathit{dt}}={\mathit{Y}}_{2}$
$\frac{{\mathit{dY}}_{2}}{\mathit{dt}}=3\mathit{t}-\mathit{t}{\mathit{Y}}_{1}^{2}.$
For details on the relation between the input and output, see Algorithms.
When reducing the order of differential equations, return the substitutions that odeToVectorField makes by specifying a second output argument.
syms f(t) g(t)
eqn1 = diff(g) == g-f;
eqn2 = diff(f,2) == g+f;
eqns = [eqn1 eqn2];
[V,S] = odeToVectorField(eqns)
V =
$\left(\begin{array}{c}{Y}_{2}\\ {Y}_{1}+{Y}_{3}\\ {Y}_{3}-{Y}_{1}\end{array}\right)$
S =
$\left(\begin{array}{c}f\\ \mathrm{Df}\\ g\end{array}\right)$
The elements of V represent the system of first-order differential equations, where V[i] = ${{Y}_{i}}^{\prime }$. The output S shows the substitutions being made, S[1] = ${Y}_{1}=f$, S[2] = ${Y}_{2}$ = diff(f), and S[3] = ${Y}_{3}=g$.
Solve a higher-order differential equation numerically by reducing the order of the equation, generating a MATLAB® function handle, and then finding the numerical solution using the ode45 function.
Convert the following second-order differential equation to a system of first-order differential equations by using odeToVectorField.
$\frac{{\mathit{d}}^{2}\mathit{y}}{\mathit{d}{\mathit{t}}^{2}}=\left(1-{\mathit{y}}^{2}\right)\frac{\mathit{dy}}{\mathit{dt}}-\mathit{y}.$
syms y(t)
eqn = diff(y,2) == (1-y^2)*diff(y)-y;
V = odeToVectorField(eqn)
V =
$\left(\begin{array}{c}{Y}_{2}\\ -\left({{Y}_{1}}^{2}-1\right) {Y}_{2}-{Y}_{1}\end{array}\right)$
Generate a MATLAB function handle from V by using matlabFunction.
M = matlabFunction(V,'vars',{'t','Y'})
M = function_handle with value:
@(t,Y)[Y(2);-(Y(1).^2-1.0).*Y(2)-Y(1)]
Specify the solution interval to be [0 20] and the initial conditions to be ${y}^{\prime }\left(0\right)=2$ and ${y}^{\prime \prime }\left(0\right)=0$. Solve the system of first-order differential equations by using ode45.
interval = [0 20];
yInit = [2 0];
ySol = ode45(M,interval,yInit);
Next, plot the solution $y\left(t\right)$ within the interval $t$ = [0 20]. Generate the values of t by using linspace. Evaluate the solution for $y\left(t\right)$, which is the first index in ySol, by calling the deval function with an index of 1. Plot the solution using plot.
tValues = linspace(0,20,100);
yValues = deval(ySol,tValues,1);
plot(tValues,yValues)
Convert the second-order differential equation ${y}^{\prime \prime }\left(x\right)=x$ with the initial condition $y\left(0\right)=a$ to a first-order system.
syms y(x) a
eqn = diff(y,x,2) == x;
cond = y(0) == a;
V = odeToVectorField(eqn,cond)
V =
$\left(\begin{array}{c}{Y}_{2}\\ x\end{array}\right)$
Input Arguments
collapse all
Higher-order differential equations, specified as a symbolic differential equation or an array of symbolic differential equations. Use the == operator to create an equation. Use the diff function to indicate differentiation. For example, represent d2y(t)/dt2 = t y(t) by entering the following command.
syms y(t)
eqn = diff(y,2) == t*y;
Output Arguments
collapse all
First-order differential equations, returned as a symbolic expression or a vector of symbolic expressions. Each element of this vector is the right side of the first-order differential equation Y[i]′ = V[i].
Substitutions in first-order equations, returned as a vector of symbolic expressions. The elements of the vector represent the substitutions, such that S(1) = Y[1], S(2) = Y[2],….
Tips
• To solve the resulting system of first-order differential equations, generate a MATLAB® function handle using matlabFunction with V as an input. Then, use the generated MATLAB function handle as an input for the MATLAB numerical solver ode23 or ode45.
• odeToVectorField can convert only quasi-linear differential equations. That is, the highest-order derivatives must appear linearly. For example, odeToVectorField can convert y*y″(t) = –t2 because it can be rewritten as y″(t) = –t2/y. However, it cannot convert y″(t)2 = –t2 or sin(y″(t)) = –t2.
Algorithms
To convert an nth-order differential equation
${a}_{n}\left(t\right){y}^{\left(n\right)}+{a}_{n-1}\left(t\right){y}^{\left(n-1\right)}+\dots +{a}_{1}\left(t\right){y}^{\prime }+{a}_{0}\left(t\right)y+r\left(t\right)=0$
into a system of first-order differential equations, odetovectorfield makes these substitutions.
$\begin{array}{l}{Y}_{1}=y\\ {Y}_{2}={y}^{\prime }\\ {Y}_{3}={y}^{″}\\ \dots \\ {Y}_{n-1}={y}^{\left(n-2\right)}\\ {Y}_{n}={y}^{\left(n-1\right)}\end{array}$
Using the new variables, it rewrites the equation as a system of n first-order differential equations:
$\begin{array}{l}{Y}_{1}{}^{\prime }={y}^{\prime }={Y}_{2}\\ {Y}_{2}{}^{\prime }={y}^{″}={Y}_{3}\\ \dots \\ {Y}_{n-1}{}^{\prime }={y}^{\left(n-1\right)}={Y}_{n}\\ {Y}_{n}{}^{\prime }=-\frac{{a}_{n-1}\left(t\right)}{{a}_{n}\left(t\right)}{Y}_{n}-\frac{{a}_{n-2}\left(t\right)}{{a}_{n}\left(t\right)}{Y}_{n-1}-...-\frac{{a}_{1}\left(t\right)}{{a}_{n}\left(t\right)}{Y}_{2}-\frac{{a}_{0}\left(t\right)}{{a}_{n}\left(t\right)}{Y}_{1}+\frac{r\left(t\right)}{{a}_{n}\left(t\right)}\end{array}$
odeToVectorField returns the right sides of these equations as the elements of vector V and the substitutions made as the second output S.
Compatibility Considerations
expand all
Warns starting in R2019b | 2020-05-30T08:33:34 | {
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https://mathematica.stackexchange.com/questions/23955/how-to-find-the-index-of-a-square-matrix-in-mathematica-quickly | # How to find the index of a square matrix in Mathematica quickly?
Let $A$ be an $n\times n$ complex matrix. The smallest nonnegative integer $k$ such that $\mathrm{rank}(A^{k+1})=\mathrm{rank}(A^{k})$, is the index fo $A$ and denoted by $\mathrm{Ind}(A)$. I would like to compute $\mathrm{Ind}(A)$ quickly in Mathematica (I am using V8).
Let us as a very simple example consider
$$A=\left(\begin{array}{rrrrrr} 1 & -1 & 0 & 0 & 0 & 0 \\ -1 & 1 & 0 & 0 & 0 & 0 \\ -1 & -1 & 1 & -1 & 0 & 0 \\ -1 & -1 & -1 & 1 & 0 & 0 \\ -1 & -1 & -1 & 0 & 2 & -1 \\ -1 & -1 & 0 & -1 & -1 & 2 \end{array}\right)$$
then it is clear that $\mathrm{Ind}(A)=2$. For computing this in Mathematica, I used
Solve[MatrixRank[MatrixPower[A, k + 1]] == MatrixRank[MatrixPower[A, k]] && k > 0,
k, Integers]
But unfortunately, I cannot get the result. I will be grateful if anyone could design a way to compute $\mathrm{Ind}(A)$ for random matrices of the size $n\times n=200\times 200$ in Mathematica.
• Hmmm, can't you just compute MatrixRank[MatrixPower[A, k + 1] for incresing k and just compare the previous two entries. Sorry don't have time for a full answer now... – Ajasja Apr 24 '13 at 12:32
• In theory, one could look carefully at the Jordan block that goes with the eigenvalue of zero. In practice this will not scale very well, and the numeric methods involving MatrixRank in responses below should be better. – Daniel Lichtblau Apr 24 '13 at 14:00
• @whuber (part 2) JordanDecomposition is not a good thing to try on non-exact input. I was never strongly in favor of extending it to approximate number input. To this day we find and fix bugs from that functionality. I'm not sure if we gained any advantage from having it. – Daniel Lichtblau Apr 24 '13 at 15:33
• @whuber Okay, I think I see the issue. I'll post what might be a viable workaround. – Daniel Lichtblau Apr 24 '13 at 16:14
• Dear J.M., I must compute the index for both types of matrices. Specially for inexact matrices. It would also be nice if an algorithm could produce $k$, $A^k$ and $A^{k+1}$ at the end of its run for all types of matrices. – Fazlollah Apr 25 '13 at 18:38
The idea behind the Jordan normal form does the trick, even though JordanDecomposition does not. (Incidentally, this suggests there may be a more reliable, stable algorithm to obtain Jordan decompositions than is implemented in Mathematica...) The resulting solution is very short, efficient, and numerically stable when applied to floating-point matrices. The following begins with a brief explanation of how it works, followed by several examples to show it actually does work.
Let's consider how any matrix can have an index exceeding $1$. Because the rank of $A^2$ is less than that of $A$, the nullspace of $A$ must be non-trivial. If the rank of $A^3$ further decreases, it must be because there is a sequence of vectors mapped by $A$, $v_2 \to v_1 \to 0$, with $v_1 \ne 0$. The index is the maximal length of such a sequence.
This provides a numerically stable, relatively quick way to compute the index. Beginning with computation of the nullspace $N_0$, find all vectors mapped by $A$ into the nullspace. In other words, find a basis for the solutions to $A x \in N_0.$ As far as I can tell, Mathematica does not have a direct method to do this, so let's reduce it to one it does have. NullSpace will provide a basis $e_1, \ldots, e_k$ of $N_0$. At the next step we seek a basis for the set of solutions
$$A x - \lambda_1 e_1 - \ldots - \lambda_k e_k = 0$$
This means the augmented vector $(x_1, x_2, \ldots, x_n; -\lambda_1, \ldots, -\lambda_k)$ lies in the nullspace of the matrix formed by augmenting $A$ with the columns $e_1, \ldots, e_k$. Among the solutions of this equation will be the original nullspace.
Iterating this procedure creates a flag of vector spaces $N_0 \subset N_1 \subset \cdots \subset N_i = N_{i+1} = \cdots$; $i+1$ is then the index. Provided we use a numerically stable procedure like LinearSolve or NullSpace, we can expect this algorithm to work even for large floating-point matrices. First I will provide an implementation and then illustrate it with several examples.
To help us study what's going on, this version returns a "generalized index" consisting of a sequence of bases of the flag: it gives far more information about the structure of $A$ than just its index, but obviously its index is easily derivable from the output: it is just its length. You should be able to recognize the iteration in NestWhileList, the augmentation of $A$ via Join, and the computation of nullspaces with NullSpace (including the initial nullspace, which is why this command appears twice).
index[a_] :=
With[{k = NullSpace[a]},
If[k == {}, {},
NestWhileList[
NullSpace[Join[a, Transpose[#[[All, 1 ;; Length@a]]], 2]] &, k,
Length[#1] != Length[#2] &, 2, Length@a, -1]]];
The test for termination is when the dimensions of the solutions stabilize (as computed by applying Length to their bases). Because termination occurs when $\dim(N_i) = \dim(N_{i+1})$, the final -1 in the NestWhileList command throws away the superfluous basis for $N_{i+1}$.
(Edit A special test has to be made for nonsingular matrices, for then Transpose fails when the nullspace is empty.)
To test this solution on large-ish matrices, let's generate some with known indexes. A good way to do this is to start with a bunch of Jordan blocks of zero eigenvalue: the index is one more than the longest contiguous string of superdiagonal ones (as is easily checked). Conjugating this by some random matrix (which is almost surely invertible) creates a non-sparse matrix for testing.
n = 30;
p = SparseArray[{Band[{1, 1}] -> 0, Band[{1, 2}] -> 1}, {n, n}];
j = Floor[n/3]; p[[j, j + 1]] = 0;
q = Rationalize[RandomReal[{0, 1}, {n, n}], 10^-2];
a = Inverse[q] .p . q;
ArrayPlot[p]
This is a plot of the Jordan normal form of $a$.
A direct calculation of its index will determine the ranks of the matrix powers. Because this matrix is designed to have an index of $20$, we can hard-code this into the check:
(ranks = MatrixRank[MatrixPower[a, #]] & /@ Range[20] ) // AbsoluteTiming
$\{7.1404084,\{28,26,24,22,20,18,16,14,12,10,9,8,7,6,5,4,3,2,1,0\}\}$
After seven seconds, we find that indeed the index is exactly $20$: $A^{19} \ne 0$ but $A^{20} = A^{21} = \cdots = 0$.
Of course, the corresponding numerical (floating point) calculation is far faster--but it gets the wrong answer:
(nRanks = MatrixRank[MatrixPower[N@a, #]] & /@ Range[Length@a] ) // AbsoluteTiming
$\{0.0100006,\{28,26,24,22,20,18,16,14,12,11,9,9,8,7,7,8,8,8,7,30,28,30,28,30,28,30,28,30,28,30\}\}$
Problems crop up around the $12^\text{th}$ power. This is evident in a plot of the two tests:
ListPlot[{ranks, nRanks}, PlotStyle -> PointSize[0.015], AxesLabel -> {"Power", "Rank"}]
Clearly the numerical calculations are producing garbage well before the correct index is reached.
I don't dare apply index to a itself: when using exact arithmetic, it's too slow! But let's see how it performs on the floating point version of a:
n = index[N@a]; // AbsoluteTiming
$\{0.0100006,\text{Null}\}$
It's as fast as the numerical brute-force rank-of-matrix-powers solution was. What about accuracy?
Length@n
$20$
It gets the right answer! But perhaps this was only luck? Let's check by restricting $A$ to the flag returned by index:
n0 = Reverse@(Last@n)[[All, 1 ;; Length[a]]];
u = Transpose[PseudoInverse[n0]] . a. Transpose[n0];
$u$ is just $A$ restricted to $N_{20}$, expressed in a different basis. Here are portraits of its powers through the $20^\text{th}$:
GraphicsGrid[{(MatrixPower[u, #]//Chop//ArrayPlot) & /@ Range@Length@n}, ImageSize -> 1000]
Up until the very end, the powers are nonzero, then finally the $20^\text{th}$ power is the zero matrix: this demonstrates the index of $A$ was at least $20$.
As another example, set $n=120$ in the previous one. My trials produce the generalized index (that is, the entire flag of $80$ subvectorspaces) in one-half to one second and consistently calculate the correct index of $80$. (I haven't tested with $n$ any larger than $120$ because it starts taking a long time just to create $a$.)
Finally, let's apply this solution to the matrix of the question:
a = {{1, -1, 0, 0, 0, 0}, {-1, 1, 0, 0, 0, 0}, {-1, -1, 1, -1, 0, 0},
{-1, -1, -1, 1, 0, 0}, {-1, -1, -1, 0, 2, -1}, {-1, -1, 0, -1, -1, 2}};
MatrixForm /@ index[a]
$$\left\{\left( \begin{array}{cccccc} 0 & 0 & 1 & 1 & 1 & 1 \end{array} \right),\left( \begin{array}{ccccccc} 1 & 1 & 0 & 0 & 0 & 0 & 2 \\ 0 & 0 & 1 & 1 & 1 & 1 & 0 \end{array} \right)\right\}$$
The first element of the list is a basis of the nullspace of $A$, which is one-dimensional. The second element is a basis of the nullspace of an augmented version of $A$. We are interested only in the first six entries in each row. They form a basis for $N_1$. They include (at the bottom) the previous basis for $N_0$. As one last check, let's verify that $A$ sends the second basis into the span of the first:
a . Transpose@(Last@index[a])[[All, 1 ;; 6]] // Transpose // MatrixForm
$$\left( \begin{array}{cccccc} 0 & 0 & -2 & -2 & -2 & -2 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{array} \right)$$
Sure enough, the second basis vector is killed (it was in the nullspace) and the first is sent to a multiple ($-2$) of the second. This detailed additional information about precisely how $A$ achieves its index is potentially useful in applications.
• Overkill for the problem at hand but as the first to upvote this I will say it has an aesthetic appeal as well as independent interest. Could perhaps be adapted to a Jordan decomposition algorithm, for example, since finding those nontrivial block eigenspaces is the hard part. – Daniel Lichtblau Apr 25 '13 at 13:36
• Excellent idea. Can you check that why your technique cannot produce the result for a simple matrix in floating points arithmetic as follows: $n=100; a = RandomReal[{}, {n, n}]$. I faced with some errors such as "Transpose::nmtx: "The first two levels of the one-dimensional list {} cannot be transposed."". – Fazlollah Apr 25 '13 at 16:28
• Thanks: I tacitly assumed $A$ is singular; if not, the index is $0$. I will add a test for this. However, the expression RandomReal[,n,n] is not even syntactically correct. The message you got indicates you created a vector rather than a matrix. An expression that will give you an actual $n$ by $n$ matrix would be RandomReal[{0,1},{n,n}]. For almost all such random matrices the index is $0$, so they don't make a very good test. Besides, you don't know in advance what the index is! Test instead with the code I provide, which produces exact matrices of known index. – whuber Apr 25 '13 at 17:39
Try this:
Length@NestWhileList[A.# &, A, MatrixRank[#] != MatrixRank[#2] &, 2] - 1
This'll keep multiplying $A$ together and checking the matrix rank at each step.
• f@x is a shorthand for f[x] and # and & are used in writing pure functions. The rest should be easy to find in the documentation. – Szabolcs Apr 24 '13 at 12:37
• Since MatrixRank is relatively costly (compared to Dot) this could be improved by caching the rank of of the matrix between steps. – Mr.Wizard Apr 24 '13 at 12:42
• @Mr.Wizard Yes, there's often a tradeoff between simplicity and performance. Why don't you post the faster solution? – Szabolcs Apr 24 '13 at 12:43
• @Szabolcs try testing your code with A={{1, 1, -1, 2}, {-1, 1, 0, 0}, {0, 1, -1, 2}, {1, 2, -1, 0}}. In my machine the kernel keeps on running.... – PlatoManiac Apr 24 '13 at 12:57
• I think the code is not working well, Mr. Wizard is right down here. For a random matrix, it sould give 0 while produces 5 as the $Ind(A)$. – Fazlollah Apr 24 '13 at 13:25
Here is the bad type of example mentioned in comments by @whuber.
SeedRandom[1111];
n = 3; p =
SparseArray[{Band[{1, 1}] -> 0, Band[{1, 2}] -> 1}, {n, n}]; q =
Rationalize[RandomReal[{0, 1}, {n, n}], 10^-3]; a = Inverse[q].p.q;
na = N[a];
MatrixRank[MatrixPower[na, #]] & /@ Range[Length@na]
(* Out[318]= {2, 1, 3} *)
To see the source of trouble we'll have a look at singular values.
svlist =
Map[SingularValueList[#, Tolerance -> 10^(-6)] &,
FoldList[Dot, na, Table[na, {n - 1}]]]
9* Out[319]= {{13.1845622926,
1.63446803072}, {21.5497455662}, {2.03322204001*10^-14,
2.76424896137*10^-15, 1.86215833311*10^-16}} *)
Notice that the tolerance setting does nothing for that last list (in this example it does nothing for the others either, but that need not be the case in general). The reason is that all of them are small, and none are a factor of 10^6 or more smaller than the largest.
What we want in this situation is to have also an absolute threshold for removing them, as Tolerance only uses a relative threshold. Chop can do this. I will illustrate with a larger example.
SeedRandom[1111];
n = 100;
p = SparseArray[{Band[{1, 1}] -> 0, Band[{1, 2}] -> 1}, {n, n}]; q =
Rationalize[RandomReal[{0, 1}, {n, n}], 10^-3]; a = Inverse[q].p.q;
na = N[a];
Timing[
svlist = Map[SingularValueList[#, Tolerance -> 10^(-6)] &,
FoldList[Dot, na, Table[na, {n - 1}]]];]
(* Out[340]= {0.180000, Null} *)
Map[Length, Chop[svlist, 10^(-6)] /. 0 :> Sequence[]]
(* Out[341]= {99, 98, 97, 96, 95, 94, 93, 92, 91, 90, 89, 88, 87, 86, \
85, 84, 83, 82, 81, 80, 79, 78, 77, 76, 75, 74, 73, 72, 71, 70, 69, \
68, 67, 66, 65, 64, 63, 62, 61, 60, 59, 58, 57, 56, 55, 54, 53, 52, \
51, 50, 49, 48, 47, 46, 45, 44, 43, 42, 41, 40, 39, 38, 37, 36, 35, \
34, 33, 32, 31, 30, 29, 28, 27, 26, 25, 24, 23, 22, 21, 20, 19, 18, \
17, 16, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0} *)
The upshot is you might want to experiment with setting Tolerance, and postprocessing using Chop, on SingularValueList[] of these powers.
I believe this is correct but I'm running out of time to check it or develop it further. I believe Szabolcs intented to write != where he wrote ==, but there is still the matter from starting with power zero.
Update #3:
f1 = {#, MatrixRank@#} &;
f2[m_][{_, {a2_, r2_}, n_}] := {{a2, r2}, f1[m.a2], n + 1}
f3[m_] := NestWhile[f2[m], {{-1, -1}, f1 @ MatrixPower[m, 0], -1}, #[[1, 2]] != #[[2, 2]] &]
Use:
A = {{1, -1, 0, 0, 0, 0}, {-1, 1, 0, 0, 0, 0}, {-1, -1, 1, -1, 0, 0},
{-1, -1, -1, 1, 0, 0}, {-1, -1, -1, 0, 2, -1}, {-1, -1, 0, -1, -1, 2}};
f3[A]
Output is in the form: {{matrix1, rank1}, {matrix2, rank2}, power}
You can use Part to extract whatever you need.
• I think Mr. Wizard answer needs some revision. I mean the answers are not the same. For instance, check the following sample examples: Clear["Global*"] SeedRandom[1234]; n = 500; A = RandomReal[{}, {n, n}]; Length@NestWhileList[A.# &, A, MatrixRank[#] == MatrixRank[#2] &, 2] // AbsoluteTiming Length@NestWhileList[{A.#[[1]], MatrixRank@#[[1]]} &, {A, -1}, #[[2]] == #2[[2]] &, 2] // AbsoluteTiming – Fazlollah Apr 24 '13 at 12:56
• @FazlollahSoleymani Sorry, let me see if I can fix that. – Mr.Wizard Apr 24 '13 at 13:00
• @FazlollahSoleymani I don't know that I understand the question. Can you confirm that the output of Szabolcs's code is in fact the result that you want? – Mr.Wizard Apr 24 '13 at 13:07
• @FazlollahSoleymani please take for example SeedRandom[12345]; A = RandomReal[{}, {500, 500}]; then Table[MatrixRank[MatrixPower[A, i]], {i, 0, 5}] is {500, 500, 500, 500, 500, 494} -- wouldn't the index be either zero or one? Szabolcs's code yields five. – Mr.Wizard Apr 24 '13 at 13:11
• @FazlollahSoleymani I think my updated answer now agrees with the first one, but I still don't think I understand the question. If the Rank of both A^1 and A^2 is 500, why isn't the answer one? – Mr.Wizard Apr 24 '13 at 13:17
We can exploit the fact that one can compute the index of a matrix from its minimal polynomial, so a good method is contingent on finding an efficient method for generating a matrix's minimal polynomial. Here, I present a not too fast method, but this can be sped up with a better algorithm for the minimal polynomial (e.g. by computing it directly from JordanDecomposition[]):
MatrixMinimalPolynomial[a_?MatrixQ, x_] := Module[{qu, a0, mnm},
a0 = IdentityMatrix[Length[a]]; mnm = {Flatten[a0]};
While[Length[qu] == 0,
a0 = a0.a;
AppendTo[mnm, Flatten[a0]];
qu = NullSpace[Transpose[mnm]]];
Fold[(#1 x + #2) &, 0, Reverse[First[qu]]]]
MatrixIndex[a_?MatrixQ] /; TrueQ[Det[a] != 0] := 0;
MatrixIndex[a_?MatrixQ] := First[Cases[FactorSquareFreeList[
MatrixMinimalPolynomial[a, \[FormalX]]], {\[FormalX], k_Integer} :> k]]
Using OP's example:
MatrixIndex[{{1, -1, 0, 0, 0, 0}, {-1, 1, 0, 0, 0, 0}, {-1, -1, 1, -1, 0, 0},
{-1, -1, -1, 1, 0, 0}, {-1, -1, -1, 0, 2, -1}, {-1, -1, 0, -1, -1, 2}}]
2
` | 2019-12-08T11:28:43 | {
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https://math.stackexchange.com/questions/53708/precise-definition-of-weaker-and-stronger/53727 | # Precise definition of “weaker” and “stronger”?
If I say that $A$ is stronger than $B$, do I mean that $A \Rightarrow B$, or that $B \Rightarrow A$? (Or something else?)
I feel like I have seen both usages in literature, which is confusing.
Thoughts based on intuition:
$A \Rightarrow B$ means $A$ is a special case of $B$ -- $B$ is more general. This would seem to imply that $B$ is "stronger". (Example: $n$ is an integer implies $n$ is a real number.)
$A \Rightarrow B$ also means that whenever $A$ holds, $B$ must hold. This would seem to imply that $A$ is "stronger".
• A stronger theorem is one which has a weaker hypothesis and/or a stronger conclusion. – GEdgar Jul 25 '11 at 17:54
• In your example "n is an integer implies n is a real number", you are thinking of the predicates involved. The integers are a proper subset of the real numbers. That might be the source of your confusion. When logicians talk about A being stronger than B, they're talking about A and B as distinct statements without regard to the predicates they contain, as Asaf explains. – MikeC Jul 25 '11 at 18:25
• Thanks for the explanations. So when I say "A is stronger than B", precisely what can A and B be? Only theorems? Logical sentences? – usul Jul 25 '11 at 21:41
• Let me also share the following Google Buzz that Terence Tao wrote which addresses some of these issues. – Willie Wong Jul 25 '11 at 23:11
• @bo1024: It means $A$ implies more sentences, then $B$. This is equivalent to $A \Rightarrow B$ because $\Rightarrow$ is a preorder. – beroal Jul 26 '11 at 20:36
If $A\Rightarrow B$, then for every $C$, if $B\Rightarrow C$ we have that $A\Rightarrow C$. Therefore $A$ implies at least the same propositions that $B$ implies.
We have two options from here:
1. $B\Rightarrow A$, in which case $A$ is equivalent to $B$, and they imply the same things.
2. $B\nRightarrow A)$, that is $B$ does not imply $A$. We have if so that $A$ is stronger than $B$ since $A\implies A$, but $B$ does not.
In essence "$A$ is stronger than $B$" is when $\{C\mid B\Rightarrow C\}\subsetneq\{C\mid A\Rightarrow C\}$, and equivalent is when the sets are equal.
• I am being somewhat inaccurate for didactic reasons, and I am not mentioning any underlying theory which may be required for some of the implications. – Asaf Karagila Jul 25 '11 at 17:47
• Could you add an example or two? – Samy Bencherif Dec 28 '17 at 4:44
• Take the language of equality. The empty theory is weaker than the theory stating that there are at least two different elements. – Asaf Karagila Dec 28 '17 at 6:26
Let's make the simplifying assumption that $\lnot(B \implies A)$.
Then $A \implies B$ can be informally expressed as "$A$ is (strictly) stronger than $B$."
It is certainly possible that in this situation, at some time, someone has instead written "$B$ is stronger than $A$." Stuff happens. We all have written $x<y$ when we meant $y<x$. And interchange of "necessary condition" and "sufficient condition" happens so (relatively) often that it may be best to avoid these terms.
But "$A$ is stronger than $B$" has only one correct interpretation in terms of the direction of the implication (with disagreement, possibly, in the case of equivalence.)
However, suppose that we have proved theorem $X$, $\:$(a) under the assumption $A$, and $\:$(b) under the assumption $B$. Then the result (b) is considered to be a stronger result than the result (a). That is perfectly consistent with the ordinary meaning of "stronger," since $$(B \implies X) \implies (A\implies X).$$ | 2019-04-24T07:53:00 | {
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https://www.physicsforums.com/threads/mathematical-induction-question.562103/ | # Homework Help: Mathematical induction question
1. Dec 22, 2011
### pc2-brazil
1. The problem statement, all variables and given/known data
Suppose a and b are real numbers with 0 < b < a. Show that, if n is a positive integer, then
$$a^n - b^n \leq na^{n-1}(a-b)$$
2. Relevant equations
3. The attempt at a solution
I'm trying to show this by induction.
Let P(n) be the proposition that $a^n - b^n \leq na^{n-1}(a-b)$
I've already verified that P(1) is true, which completes the basis step.
Inductive step:
I must show that, if P(k), then P(k+1).
So, I first assume that this is true for an arbitrary k: $a^k - b^k \leq ka^{k-1}(a-b)$
Then, I must show that, if P(k) is true, it follows that $a^{k+1} - b^{k+1} \leq (k+1)a^k(a-b)$.
This is where I'm having trouble.
I'm trying to find that $a^{k+1} - b^{k+1}$ is less than or equal to an expression involving $a^k - b^k$, so that I can use the expression for P(k) to derive an inequality for $a^{k+1} - b^{k+1}$.
I've tried several ways, like trying to rewrite $a^{k+1} - b^{k+1}$ as $aa^k - bb^{k}$ and then writing that $aa^k - bb^k \geq aa^k - ab^k = a(a^k - b^k)$ (since a > b), but this doesn't help, because I'm looking for something that $a^{k+1} - b^{k+1}$ is less than or equal to, not greater than or equal to.
2. Dec 22, 2011
### Dansuer
EDIT: nevermind. i though i had it but i did not
Last edited: Dec 22, 2011
3. Dec 22, 2011
### Deveno
if b < a, then a = b + c for some positive number c.
$$(b+c)^n - b^n \leq n(b+c)^{n-1}c$$
now expand the powers of b+c with the binomial formula.
4. Dec 22, 2011
### SammyS
Staff Emeritus
It can be instructive to see how the inequality works for the first few values of n. This isn't good enough for a proof, but it may give you some ideas for the proof.
For n=2:
$\displaystyle a^2-b^2\ \ ?\,\leq\,? \ \ 2a^1(a-b)$
Factor the left side.
$\displaystyle (a-b)(a+b)\ \ ?\,\leq\,? \ \ 2a(a-b)$
Of course, (a+b) < 2a because a < b .
So, this case doesn't appear to need induction.
For n=3:
We have a somewhat similar result with the difference of cubes giving a3-b3 = (a-b)(a2+ab+b2). Then the inequality holds if a2+ab+b2 ≤ 3a2, and this is true for a < b. Again, this doesn't require induction.
For general n:
Did you know that $\displaystyle a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+a^{n-3}b^{2}+\dots+a^{((n-1)-k)}\,b^{k}+\dots +a^{2}b^{n+3}+a\,b^{n-2}+b^{n-1})\,?$
Each of the n terms in the last factor is less than or equal to an-1.
For instance: $\displaystyle a^{12}-b^{12}=(a-b)(a^{11}+a^{10}b+a^{9}b^{2}+\dots+a^{5}\,b^{6}+ \dots +a^{2}b^{9}+a\,b^{10}+b^{11})$
5. Dec 23, 2011
### Curious3141
Building on Sammy's post, you might find it easier to put x = b/a, where 0<x<1
Use polynomial long division to see that for positive integral n,
$$(1-x^n) = (1-x)(1+x+x^2+...x^{n-1})$$
Alternatively you can just multiply the terms on the RHS, cancel out lots of adjacent terms and end up with the LHS. This is one of the polynomial identities which it's really good to know.
Now,
$$1+x+x^2+...x^{n-1}$$ has n terms, each of which is less than or equal to 1. Hence the whole sum is less than or equal to n.
So,
$$a^n - b^n = a^n(1 - x^n) = a^n(1-x)(1+x+x^2+...+x^{n-1}) = a(1-x).a^{n-1}.(1+x+x^2+...+x^{n-1}) = (a-b).a^{n-1}.(1+x+x^2+...+x^{n-1}) \leq (a-b).n.a^{n-1}$$
and you're done.
Last edited: Dec 23, 2011
6. Dec 28, 2011
### pc2-brazil
Very interesting idea.
I knew that result for an - bn, but I didn't think of using it.
This leads directly to the desired result, not requiring induction. | 2018-05-22T06:50:47 | {
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https://math.stackexchange.com/questions/954733/notation-for-summation-while-skipping-elements | # Notation for summation while skipping elements
Suppose I have a summation like so:
$\sum_{i =0}^n l^i$
Except I don't want to compute for all $0 \leq i \leq n$. I just want to compute it for the arithmetic sequence: $1, 3, 5, 7, 9...$ How do I write this in the summation notation?
Use $2i+1$ instead of $i$ in the expression to be summed.$$\sum_{i=0}^k l^{2i+1}$$ Where $2k+1=n$
• It depends on the context. $$\sum_{\text{i odd}}l^i$$ is used in some places. $$\sum_{i \in A}l^i \quad (A = \{1, 3, 5, \ldots \})$$ is also used. The important thing is to be clear. Often this is most easily achieved using words rather than flooding your mathematics with esoteric notation. Apr 13, 2015 at 15:30
Like this: $$\sum^n_{i=0}_\text{i is odd} \text{or just} \sum^n_{i=0}_\text{odd}$$ And in general you'd just replace "i is odd" with whatever criteria you have.
Or you could mention it outside the sum. As in, immediately after the sum write "where $i$ is odd" or "where $i \in \{x \in \Bbb N\ |\ x = 2k \wedge k \in \Bbb N\}$". | 2022-08-09T04:28:31 | {
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https://www.physicsforums.com/threads/matrix-representation-of-function-composition.871089/ | # Matrix representation of function composition
Am I on the right path here?
1. Homework Statement
i. Prove that ##T_{a}## and ##T_{b}## are linear transformations.
ii. Compose the two linear transformations and show the matrix that represents that composition.
2. The attempt at a solution
i. Prove that ##T_{a}## and ##T_{b}## are linear transformations.
i. ##T_{a} \begin{bmatrix}x \\ y\end{bmatrix} = \begin{bmatrix}-x \\ x+y\end{bmatrix}##
##x =\begin{bmatrix}-1\\1\end{bmatrix}+y\begin{bmatrix}0\\1 \end{bmatrix}##
##\begin{bmatrix}-1 & 0 \\ 1 & 1 \end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}##
##T_{a}## = Linear transformation.
##T_{b} \begin{bmatrix}x \\ y \end{bmatrix} = \begin{bmatrix}x+y \\ x -y \end{bmatrix}##
##x \begin{bmatrix}1\\1 \end{bmatrix} + y \begin{bmatrix}1\\-1 \end{bmatrix} = \begin{bmatrix}1 & 1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix}x\\y \end{bmatrix}##
##T_{b}## = Linear transformation.
ii. Compose the two linear transformations and show the matrix that represents that composition.
##T_{a} {\circ} T_{b} = \left[T_{a}\right]\left[T_{e}\right] = \begin{bmatrix}-1 & 0 \\ 1 & 1 \end{bmatrix} \begin{bmatrix}1 & 1 \\ 1 & -1 \end{bmatrix}##
##= \begin{bmatrix}-1 & -1 \\ 2 & 0 \end{bmatrix}##
andrewkirk
Homework Helper
Gold Member
That looks correct. However, from the way the question is written, they expect you to not just produce the matrix but also state the the transformation in the same form as that in which the original two were given, ie this form
$$T_{a} \begin{bmatrix}x \\ y\end{bmatrix} = \begin{bmatrix}-x \\ x+y\end{bmatrix}$$
Sociomath
That looks correct. However, from the way the question is written, they expect you to not just produce the matrix but also state the the transformation in the same form as that in which the original two were given, ie this form
$$T_{a} \begin{bmatrix}x \\ y\end{bmatrix} = \begin{bmatrix}-x \\ x+y\end{bmatrix}$$
##\left[T_{a}\right]\left[T_{b}\right] = \begin{bmatrix}-1 & -1 \\ 2 & 0 \end{bmatrix} \begin{bmatrix} x\\y \end{bmatrix} = \begin{bmatrix}-x -y\\ 2x \end{bmatrix}##
andrewkirk
$$(T_a\circ T_b)\begin{bmatrix} x\\y \end{bmatrix} = \begin{bmatrix}-x -y\\ 2x \end{bmatrix}$$ | 2022-07-06T03:20:39 | {
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https://math.stackexchange.com/questions/408095/finding-the-upper-and-lower-limit-of-the-following-sequence | # Finding the upper and lower limit of the following sequence.
$\{s_n\}$ is defined by $$s_1 = 0; s_{2m}=\frac{s_{2m-1}}{2}; s_{2m+1}= {1\over 2} + s_{2m}$$
The following is what I tried to do.
The sequence is $$\{0,0,\frac{1}{2},\frac{1}{4},\frac{3}{4},\frac{3}{8},\frac{7}{8},\frac{7}{16},\cdots \}$$
So the even terms $\{E_i\} = 1 - 2^{-i}$ and the odd terms $\{O_k\} = \frac{1}{2} - 2^{-k}$ and each of them has a limit of $1$ and $\frac{1}{2}$, respectively.
So, the upper limit is $1$ and the lower limit is $1\over 2$, am I right ?
Does this also mean that $\{s_n\}$ has no limits ?
Is my denotation $$\lim_{n \to \infty} \sup(s_n)=1 ,\lim_{n \to \infty} \inf(s_n)={1 \over 2}$$ correct ?
• Looks right to me! – Peter Košinár May 31 '13 at 23:26
• Answers to your questions in order: yes, yes and yes. (but it is "notation", not "denotation") – DonAntonio May 31 '13 at 23:39
## 1 Answer
Shouldn't it be $E_i = \frac{1}{2} - 2^{-i}$ and $O_i = 1 - 2^{1-i}$? That way $E_i = 0, \frac{1}{4}, \frac{3}{8}...$ and $E_i = 0, \frac{1}{2}, \frac{3}{4}...$, which seems to be what you want.
Your conclusion looks fine, but you might want to derive the even and odd terms more rigorously. For example, the even terms $E_i$ are defined recursively by $E_{i+1} = s_{2i+2} = \frac{s_{2i+1}}{2} = \frac{E_1 + \frac{1}{2}}{2}$, and $\frac{1}{2} - 2^{-i}$ also satisfies this recursion relation. $E_1 = 0$, and $\frac{1}{2} - 2^{-1} = 0$, hence they have the same first term. By induction the two sequences are the same.
If we partition a sequence into a finite number of subsequences then the upper and lower limit of the sequence are equal to the maximum upper limit and minimum lower limit of the subsequences; in this case you're partitioning into even and odd terms.
$\{s_n\}$ has a limit iff the upper and lower limits are the same (this is proved in most analysis books), so in this case $\{s_n\}$ has no limits.
• Thank you. Unfortunately my book (Rudin) says "it's obvious that $\{s_n\}$ has a limit iff the upper and lower limits are the same" so it never stuck to me. – hyg17 Jun 1 '13 at 0:50 | 2019-09-23T11:04:41 | {
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https://mathematica.stackexchange.com/questions/101070/how-to-typeset-and-evaluate-u-big-ab | # How to typeset and evaluate $u \big|_a^b$
Sometimes I need to evaluate an expression at the end points. e.g. the right hand side of $\int_a^bf(x)\textrm{d}x=F(x)\big|_a^b$. $F(x)$ could be complicated so I can't just substitute the values by hand. I currently do it this way:
(m[a]-m[b] /. m[x_] -> u)
But this introduces a new variable m and doesn't look very elegant. I'm looking for a built-in notation and/or a function for this purpose.
• How about defining a function r[u_, a_, b_] := u[b] - u[a] and applying it to replace rule u[x] /. u[x] -> r[u, a, b]. It will yield: -u[a] + u[b] or for different function u[x] /. u[x] -> r[F, a, b] it will yield -F[a] + F[b] – Wojciech Artichowicz Dec 3 '15 at 11:18
• @WojciechArtichowicz: Actually I don't have FUNCTION F[x], it's just an EXPRESSION u. I would want some function like diff[u,{x,a,b}] more. – Shou Ya Dec 3 '15 at 11:47
• Could you provide an actual simple example of such expression? – Wojciech Artichowicz Dec 3 '15 at 11:52
• First@Differences[expr /. {{x->a}, {x->b}}]? Or (expr /. x->b) - (expr /. x->a}? – Michael E2 Dec 3 '15 at 12:08
diff = Function[{expr},
Subtract @@ (expr /. {{#1 -> #3}, {#1 -> #2}})] &
longComplexExpr[x] // diff[x, 1, 3]
(* - longComplexExpr[1] + longComplexExpr[3] *)
So for integrals:
Integrate[x^2, x] // diff[x, 1, 3]
(* 26/3 *)
Integrate[x^2, {x, 1, 3}]
(* 26/3 *)
I've had a play to try and get a slightly more stylistic solution by creating a new template for the form you want and then assigning it an InputAlias, following closely the work here.
The code below will allow you to access the template by simply typing escbarEvalesc.
It will then evaluate the function at the limits and take the difference, as defined in the BarEvaluate function:
Some limitations are:
1. I've been a little hacky in trying to get a longer vertical bar, maybe someone can come up with a better way of producing it? (On my system I can get a proper LaTex one through the MaTeX package).
2. You might need to adapt it for cases when your function F takes more than just the single parameter.
Code:
SetAttributes[BarEvaluate, HoldAll]
BarEvaluate[f_, limits_] := f[limits[[1]]] - f[limits[[2]]]
BarEvaluate /: MakeBoxes[BarEvaluate[f_, {a_, b_}], TraditionalForm] :=
TemplateBox[{ToBoxes[f], ToBoxes[a], ToBoxes[b]},
"conditionalProduct",
DisplayFunction :> (RowBox[{#,
SubsuperscriptBox[
StyleBox["\[VerticalSeparator]", "Subsubtitle"],
InterpretationFunction :> (RowBox[{"BarEvaluate", "[",
RowBox[{#, ",", "{", #2, ",", #3, "}"}], "]"}] &)]
aliases = Options[EvaluationNotebook[], InputAliases];
newAliases =
Join[InputAliases /.
aliases, {"barEval" ->
TemplateBox[{"\[SelectionPlaceholder]", "\[Placeholder]",
"\[Placeholder]"}, "barEvaluate",
DisplayFunction :> (RowBox[{#,
SubsuperscriptBox[
StyleBox["\[VerticalSeparator]", "Subsubtitle"],
InterpretationFunction :> (RowBox[{"BarEvaluate", "[",
RowBox[{#, ",", "{", #2, ",", #3, "}"}], "]"}] &)]}];
SetOptions[EvaluationNotebook[], InputAliases -> newAliases];
• Very useful answer for much broader cases than here. I wish I knew how to search for stuff like that without knowing the keyword InputAliases +1 ofc – LLlAMnYP Dec 3 '15 at 14:44
Here is another version that should typeset and evaluate the way you want. The first step is to use a symbol that supports vertical spanning, and looks like a bar. There are several such symbols, |, \[VerticalSeparator], \[VerticalLine], \[VerticalLine], \[RightBracketingBar] and perhaps some others. I think \[VerticalLine] works best, so I will use that symbol. Next, one needs to restrict the spanning, so that only the selected portion influences the spanning. So, the basic plan is to use the boxes:
RawBoxes @ SubsuperscriptBox[
RowBox[{"\[SelectionPlaceholder]", "\[VerticalLine]"}],
"\[Placeholder]",
"\[Placeholder]"
]
An example:
RawBoxes @ SubsuperscriptBox[
RowBox[{FractionBox[SuperscriptBox["x","2"],"3"],"\[VerticalLine]"}],
"1",
"2"
]
By including the bar inside of a SubsuperscriptBox, the spanning will only consider whatever is placed inside \[SelectionPlaceholder]. For example:
RawBoxes @ RowBox[{
FractionBox[SuperscriptBox["x","3"],"4"],
"+",
SubsuperscriptBox[RowBox[{"x","\[VerticalLine]"}],"1","2"]
}]
You will notice that the min size of the bar and the spacing is a bit off, so the following version should look better (I also turned off the SpanSymmetric option):
CellPrint @ Cell[
BoxData @ SubsuperscriptBox[
RowBox[{
FractionBox[SuperscriptBox["x","2"],"3"],
StyleBox["\[VerticalLine]",
SpanSymmetric->False,
SpanMinSize->1.5
]
}],
RowBox[{"\[MediumSpace]", "1"}],
RowBox[{"\[MediumSpace]", "2"}]
],
"Input"
]
This produces a box structure that should look the way you want. Now, to make it evaluatable, let's define an EvaluatedAt function:
EvaluatedAt[expr_, Automatic, min_, max_] := EvaluatedAt[
expr,
Replace[ReduceFreeVariables[expr], {{v_,___}->v, _->None}],
min,
max
]
EvaluatedAt[expr_, x_, min_, max_] := (expr /. x->max) - (expr /. x->min)
A couple examples:
EvaluatedAt[1/x^2, x, 1, 2]
EvaluatedAt[1/x, Automatic, 1, 2]
-3/4
-1/2
In the second example with Automatic, EvaluatedAt looks for the first free variable in expr and then does the desired arithmetic.
Now, we are ready to create an input alias that both typesets as desired, and evaluates as desired:
CurrentValue[EvaluationNotebook[], {InputAliases,"at"}] = TemplateBox[
{"\[SelectionPlaceholder]", "Automatic", "\[Placeholder]", "\[Placeholder]"},
"EvaluatedAt",
DisplayFunction->(
SubsuperscriptBox[
RowBox[{
#1,
StyleBox[
"\[VerticalLine]",
SpanMinSize->1.5,
SpanSymmetric->False
]
}],
RowBox[{"\[MediumSpace]", #3}],
RowBox[{"\[MediumSpace]", #4}]
]&
)
];
And, here's another version where you specify the variable to be replaced:
CurrentValue[EvaluationNotebook[], {InputAliases,"at2"}] = TemplateBox[
{"\[SelectionPlaceholder]", "\[Placeholder]", "\[Placeholder]", "\[Placeholder]"},
"EvaluatedAt",
DisplayFunction->(
SubsuperscriptBox[
RowBox[{
#1,
StyleBox[
"\[VerticalLine]",
SpanMinSize->1.5,
SpanSymmetric->False
]
}],
RowBox[{"\[MediumSpace]", #2, "=", #3}],
RowBox[{"\[MediumSpace]", #4}]
]&
)
];
And here is an animation showing both in action:
Finally, all the necessary code in one code block:
EvaluatedAt[expr_, Automatic, min_, max_] := EvaluatedAt[
expr,
Replace[ReduceFreeVariables[expr], {{v_,___}->v, _->None}],
min,
max
]
EvaluatedAt[expr_, x_, min_, max_] := (expr /. x->max) - (expr /. x->min)
CurrentValue[EvaluationNotebook[], {InputAliases,"at"}] = TemplateBox[
{"\[SelectionPlaceholder]", "Automatic", "\[Placeholder]", "\[Placeholder]"},
"EvaluatedAt",
DisplayFunction->(
SubsuperscriptBox[
RowBox[{
#1,
StyleBox[
"\[VerticalLine]",
SpanMinSize->1.5,
SpanSymmetric->False
]
}],
RowBox[{"\[MediumSpace]", #3}],
RowBox[{"\[MediumSpace]", #4}]
]&
)
];
CurrentValue[EvaluationNotebook[], {InputAliases,"at2"}] = TemplateBox[
{"\[SelectionPlaceholder]", "\[Placeholder]", "\[Placeholder]", "\[Placeholder]"},
"EvaluatedAt",
DisplayFunction->(
SubsuperscriptBox[
RowBox[{
#1,
StyleBox[
"\[VerticalLine]",
SpanMinSize->1.5,
SpanSymmetric->False
]
}],
RowBox[{"\[MediumSpace]", #2, "=", #3}],
RowBox[{"\[MediumSpace]", #4}]
]&
)
];
diff = {x, a, b} \[Function] expr \[Function] #2 - # & @@ (Function @@ {x, expr}) /@ {a, b}
The usage of diff is the same as that in LLlAMnYP's answer. | 2020-06-01T05:22:23 | {
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https://math.stackexchange.com/questions/3580858/is-there-a-largest-open-interval-for-an-open-set-not-necessarily-bounded | Is there a largest open interval for an open set (not necessarily bounded)?
If $$G$$ is an open subset of $$R$$, and if $$x\in G$$, show that there exists a largest open interval $$I_x$$ containing $$x$$ s.t $$I_x$$ is the subset of $$G$$.
My idea:
Let $$x\in (a_x,b_x)$$ where
$$a_x=\inf\{a and
$$b_x=\sup\{b>x|(x,b)\subset G \}$$.
Let $$I_x=(a_x,b_x)$$.
I want to show $$a_x$$, $$b_x$$ can not belong to G, hence $$I_x$$ is the largest interval.
Assume $$a_x\in G$$, this contradicts the fact that $$a_x$$ was $$\inf$$. so $$a_x$$ is not in $$G$$. Likewise for $$b_x$$.
I think if it was said that $$G$$ is bounded, I could confidently use the proof idea above. But it is NOT. So what if G is unbounded? Then I may not have finite $$a_x$$ and $$b_x$$. Or do I need to be worried about this at all?
You could change your approach to defining $$a_x$$ and $$b_x$$ ever so slightly, to make your proof correct.
• If the set $$\{a is bounded below, then let $$a_x$$ be the infimum of that set. Otherwise let $$a_x = -\infty$$.
• If the set $$\{b>x \mid (x,b)\subset G \}$$ is bounded above, then let $$b_x$$ be the supremum of that set. Otherwise let $$b_x = +\infty$$.
Now you should go through the remainder of your proof and check carefully for any changes that are required by having changed the definitions of $$a_x$$ and $$b_x$$.
• Thank you very much. – BesMath Mar 14 at 16:39
• Not that I disagree with this solution, but just in order to not mislead the OP, in this case $a_x$ and $b_x$, when they are equal to $\infty$ or $-\infty$, are not real values anymore. Indeed they become just symbols helping to define the interval. – almaus Mar 14 at 16:39
Every open set in $$\mathbf{R}$$ can be written as a countable union of pairwise disjoint open intervals. So we get $$G=\bigcup_{i=1}^{\infty}I_i$$ Let $$x \in G \implies \exists! n\in\mathbf{N}$$ so that $$x\in I_n$$
Now you can check that this interval will be your maximal interval containing $$x$$ that's contained in $$G$$.
• You are using a bazooka theorem here... harder to show that than the original problem. – almaus Mar 14 at 16:18
• @almaus I agree but it's good to see different ways to prove things – guy3141 Mar 14 at 16:19
You can just say that as $$G$$ is open, it means that there exists an open interval included in $$G$$ around each of its points, thus there exists at least $$a_0$$ and $$b_0$$ such that $$x \in (a_0, b_0) \subset G$$.
And then you can consider the set of all the intervals in $$G$$ including $$x$$, and take the biggest.
• That is exactly where I have a problem with. since G is open for each element x in G, there are point less and greater that x, so no sup or inf has a place of meaning here. so how to make the largest interval? I am really confused. – BesMath Mar 14 at 16:27
• You need not be afraid of infinites here, as $(b, \infty)$ or $(-\infty, a)$ are perfectly valid open intervals of $\mathbb{R}$. – almaus Mar 14 at 16:29
• So just make two cases (for each boundary): when there exists an upper (resp. lower) bound, and when there is not. In the case there is not, that just means that your upper (resp. lower) boundary is $\infty$ (resp $-\infty$). – almaus Mar 14 at 16:32
Your proof is fine when situated in the appropriate framework.
Although it might not be accepted in the context of a class, I think the best way to handle this problem is to use the extended real line $$\bar{\mathbb R} = \mathbb R \cup \{-\infty,\infty\}$$ where we can order this set in the obvious way and define infimum and supremum from the order - and likewise, can define open intervals as usual, with the observation that $$(-\infty,x)$$ are honest intervals in this view that coincide with the usual definitions. The importance of this change is that every set has a supremum and infimum in the extended reals - so you do not need to worry about boundedness at all.
Basically, with this change of context, you just say that you have some subset of $$\mathbb R$$ and let $$a_x$$ and $$b_x$$ be the infimum and supremum of that set in $$\bar{\mathbb R}$$, and then just finish your argument exactly as you did - except that you might, for completeness, observe that if $$a_x$$ and $$b_x$$ are real, they are not in the set for the reasons you observe, and if they are not, they are not in the set because the set is a subset of $$\mathbb R$$.
One often finds that analysis questions such as this one are much clearer if you work with $$\pm \infty$$ within the domain of mathematics, rather than, as is common, saying that every expression involving $$\infty$$ is specially defined and requires casework - because often the extended reals unify the theory with no need for extra work. | 2020-08-11T07:14:53 | {
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https://math.stackexchange.com/questions/965550/highest-product-of-the-numbers-that-sum-to-100 | # highest product of the numbers that sum to $100$
what is the highest product of the numbers that sum to 100
for example $100 = 1+1+1+1+1+1+1+\ldots+1$ the product of these is just $1^{100} = 1$
$100 = 99 + 1$ the product of these is $99\times 1$
the numbers have to be positive integers
do the numbers all have to be the same - for example I think it is $2^{50}$
• $3\cdot3$ by itself is "better" than $2\cdot2\cdot2$ because it yields $9$ instead of $8$, while both options "consume" $6$ out of the sum. So there's got to be something better than $2^{50}$. Oct 9, 2014 at 18:20
• $3^{33}\times 1>2^{50}$ Oct 9, 2014 at 18:21
• Best one I found is $3^{32}\cdot4$. Oct 9, 2014 at 18:24
• is there a method for this rather than just trial and error Oct 9, 2014 at 18:30
• @Hamou: to get multidigit exponents, put them in braces, so 2^{50} gives $2^{50}$ instead of $2^50$ This works lots of places in $\LaTeX$ Oct 9, 2014 at 19:17
The arithmetic-geometric-mean equation says that $\sum_{i=1}^{n}{\frac{a_i}{n}}\geq\sqrt[n]{\prod_{i=1}^n{a_i}}$.
If you enter your condition on the left side and look at different values for $n$, you might find a solution.
Edit: This method works well to guess a solution, which is that most of your numbers will be $3$, and by trying that we can find $3^{32}\cdot 2^2$. Now we can go about proving that this indeed the maximum:
Assume that any $a_k$ of your numbers is greater than $4$. We could then substitute this number by $\frac{a_k}{2}+\frac{a_k}{2}$, which have a product larger than $a_k$, since $\frac{a_k}{2}^2\geq a_k \Leftrightarrow a_k\geq 4$.
(If $a_k$ is odd, we can do the same thing with two numbers with difference $1$.)
From this we can conclude that there can be no numbers greater than $3$ in your sum. Also, there can be no $1$s, for obvious reasons.
Hence, we have a selection of $a$ $3$s, and $b$ $2$s, where $3a+2b\leq 100$. Formally, we have to look at the cases $3a+2b=100, 3a+2b=99$, since if the sum was smaller than $98$ we could simply add a $2$.
Thus, we can write our product as $3^a2^{\frac{100-3a}{2}}$. If we look at this function and its maximum/monotony properties the only possible values for a maximum are $a=32, a=33$. Comparison leads to the maximum being at $a=32, b=2$.
Another possibility with less calculus would be to look at $3\cdot 3\geq 6$, immediately telling us that there can be no more than $3$ $2$s, else we could, again, substitute them with $2$ $3$s. In hindsight, this way might actually be a lot quicker and requires no differentiation...
In any case, with both variants we are done, and you have your maximum being what quite some people have pointed out so far.
• +1 Very interesting proof. It's always a little surprising to see discrete theorems with proofs involving differentiation. But you're right that the latter method of proof would be more elegant. Oct 10, 2014 at 5:06
If we write a continuous version $f(x)=x^{\frac{100}{x}}$ it maximum is when $x=e$, the closest natural number is 3 so the maximum is $3^{32}*4$.
I will add that $f(x)=x^{\frac{b}{x}};\ \forall b\geq 1$ have a maximum on $x=e$.
We need to see now that any composition will be lesser number that just an unique exponential. I write $$f(x's)=x_0^{\frac{b-\sum_{i}c_i}{x_0}}\cdot x_1^{\frac{c_1}{x_1}}\cdot x_2^{\frac{c_2}{x_2}}\cdots x_n^{\frac{c_n}{x_n}};\ c_i\neq c_j,\ \sum c_i<b\\ x_i,b,c_i>1$$
From above we can see that the maximum for any multiplier is when it base is $x_i=e$ so you can see that for any $c_i$ we choose it maximum value will be related to a base $x_i=e$ so doesnt exist any composition with $x_i\neq e$ that make the function to have a greater maximum choosing any $c_i$ decomposition that you want.
This translated to natural numbers put the maximum with $x=3$, i.e. no closest natural number to $e$ that $3$ than lead to a number f(n) closest to f(e):
$$|f(e)-f(3)|<|f(e)-f(n)|\ \forall n\in \Bbb N-\{3\}$$
And we can add that if $0\not\equiv b \mod 3$ then the next number close to $f(e)$ is $f(2)$ so we must compose the number with powers of base 3 and 2. I.e.: $$r\equiv b\mod 3\ \rightarrow f(b)=3^{\frac{b-c}{3}}2^{\frac{c}{2}}\begin{cases}r\equiv 0, f(b)=3^{\frac{b}{3}}\\r\equiv 1, f(b)=3^{\frac{b-4}{3}}*4\\r\equiv 2, f(b)=3^{\frac{b-2}{3}}*2\end{cases}$$
• Sorry for going into a bit more detail here, since your argument is definitely valid and leads to the correct solution. But I do not see how the conclusion "the closest natural number is 3, so the maximum is" would possibly hold, being examined more closely? The assumption is correct, and you can prove it fairly easily, but I fail to see how this is a formal proof. Oct 9, 2014 at 18:39
• Im not familiarized with the requirements of proofs but I think its work as a formal proof but we need to add the proof that the product of different and complementary product is a lesser number of an exponential, i.e., $(x-a)a < x^2$ and extend this notion for any division of x. Oct 9, 2014 at 18:45
• doesn't your function assume that all the numbers have to be the same - why can't they all be different or some be different Oct 9, 2014 at 18:55
• I do not quite see where to go from there, but that might just be me. Have added my variation of a formal proof from there - I think the problem with your first comment lies that, while we do get an upper bound for the problem, we do not know what the function does on any values other than the maximum. We might not have all the same numbers in the result. @zebra, we get an upper bound for our product, not an exact solution. The upper bound you get by assuming all numbers are equal but not necessarily in $\mathbb{N}$, since that's where in AM-GM equality holds. Oct 9, 2014 at 18:57
• @SomeMathStudent I still do not understand why making all the numbers equal gives the upper bound and how this relates to the AM GM inequality - isn't this to do with the nth root of the product Oct 9, 2014 at 19:00
Let's look at a factorisation containing the term $n$. We have $2(n-2)=2n-4$ so if $n\ge 4$ we have $2(n-2)= 2n-4 \ge n$, and replace $n$ with $2, n-2$, which does not reduce the product and may increase it.
If $1$ appears in the sum we can add it to another summand, which clearly increases the product without changing the sum.
So we get a sum consisting of terms which are either $2$ or $3$. Then we note Barak's observation that $3\cdot 3 \gt 2\cdot 2 \cdot 2$
Let's say a set $S=\{x_i\}:i\in \{1\to n\}$ for some $n\in\mathbb{N}$ is the optimal set. Since $\forall x \geqslant 4 : x * 2 \geqslant x + 2$, the set doesn't have any element $x \geqslant 4$ because we can take it out of the set and add a two and a $(x-2)$ instead. now for elements strictly smaller than 4: adding a $1$ want help since $x*1 = x$. However, comparing 2 to 3, we can find that $3*2 = 2*3$ which means adding 3 two times is equal to add 2 three times but $3*3>2*2*2$ which means we should switch as many sets of (2*2*2) to (3*3) as possible. Here we have three different cases which are the values for sum modulo 3. The case of 100 is 1 but let's speak generally.
• for 0 the number can be divided by 3 and we are done.
• for 2 we can add a 2 at the end with affecting threes so the answer is a 2 and $x/3$ threes.
• The case of one is a bit tricky that when divided by 3 we have a rest of one so we even ignore it since $x*1 = x$ or we can take a three out if $x > 3 (x\neq1)$ and add two twos so we're comparing 2*2 to 1*3 and 2*2 is bigger so in this case we have 2 twos and a rest of threes.
• In conclusion the answer to your question should be 2 twos and $(96\div3=32)$ threes | 2022-07-04T13:27:47 | {
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https://math.stackexchange.com/questions/3005764/why-does-int-1-sqrt2-frac1x-ln-left-frac2-2x2x42x-2x2x3-right | # Why does $\int_1^\sqrt2 \frac{1}{x}\ln\left(\frac{2-2x^2+x^4}{2x-2x^2+x^3}\right)dx$ equal to $0$?
In this question, the OP poses the following definite integral, which just happens to vanish: $$\int_1^\sqrt2 \frac{1}{x}\ln\bigg(\frac{2-2x^2+x^4}{2x-2x^2+x^3}\bigg)dx=0$$ As noticed by one commenter to the question, the only zero of the integrand is at $$x=\sqrt[3]{2}$$, meaning that the integral of the integrand from $$x=1$$ to $$x=\sqrt[3]{2}$$ is the additive inverse of the integral of the integrand from $$x=\sqrt[3]{2}$$ to $$x=\sqrt{2}$$.
This suggests some sort of symmetry obtainable by a substitution, but I cannot find an appropriate substitution or cancellation. It seems like the answer should be much simpler than those posted to the linked question.
Any ideas?
EDIT: I believe that this more general integral also vanishes: $$\int_1^{\sqrt{t}}\frac{1}{x}\ln\bigg(\frac{t-sx^2+x^4}{tx-sx^2+x^3}\bigg)dx=0$$
• @Rócherz Of course it isn't symmetric about $\sqrt[3]{2}$; I simply mean that perhaps an appropriate substitution might lead to a cancellation. – Frpzzd Nov 20 '18 at 0:47
We can indeed prove the result by symmetry. Put in other words we desire to show that: $$I=\int_1^\sqrt2 \frac{1}{x}\ln\bigg(\frac{x^4-2x^2+2}{x^2-2x+2}\bigg)dx=\color{green}{\int_1^\sqrt 2\frac{1}{x}\ln x dx=\frac18\ln^2 2}$$ First let us take the LHS integral and split it in two parts: $$I=\color{red}{\int_1^\sqrt2 \frac{\ln(x^4-2x^2+2)}{x}dx}-\color{blue}{\int_1^\sqrt2 \frac{\ln(x^2-2x+2)}{x}dx}$$ For the second one $$(I_2)$$ we will substitute $$\displaystyle{x=\frac{2}{t}\rightarrow dx=-\frac{2}{t^2}dt}$$: $$I_2=\int_{\sqrt 2}^2 \frac{\ln\left(\frac{2(t^2-2t+2)}{t^2}\right)}{\frac{2}{t}}\frac{2}{t^2}dt\overset{t=x}=\int_{\sqrt 2}^2 \frac{\ln(x^2-2x+2)}{x}dx+\int_{\sqrt 2}^2\frac{\ln 2 -2\ln x}{x}dx$$ Adding with the original $$I_2$$ leads to: $${2I_2=\int_1^2\frac{\ln(x^2-2x+2)}{x}dx+\int_{\sqrt 2}^2\frac{\ln 2 -2\ln x}{x}dx}$$ $${\Rightarrow I_2=\frac12\int_1^2 \frac{\ln(x^2-2x+2)}{x}dx-\frac{1}{8}\ln^22}\overset{x=t^2}=\color{blue}{\int_1^\sqrt{2}\frac{\ln(t^4-2t^2+2)}{t}dt-\frac{1}{8}\ln^2 2}$$ $$I=\color{red}{\int_1^\sqrt2 \frac{\ln(x^4-2x^2+2)}{x}dx}-\color{blue}{\int_1^\sqrt{2}\frac{\ln(x^4-2x^2+2)}{x}dx+\frac{1}{8}\ln^2 2}=\color{green}{\frac18\ln^2 2}$$
Your conjecture is indeed also correct since by the exact same method we can show that: $$\int_1^{\sqrt{t}}\frac{1}{x}\ln\bigg(\frac{t-sx^2+x^4}{t-sx+x^2}\bigg)dx=\int_1^\sqrt{t} \frac{1}{x}\ln xdx=\frac{1}{8}\ln^2t$$ And this time after we split the LHS integral into two parts, we will substitute in the second integral $$\displaystyle{x=\frac{t}{y}}$$ ($$t$$ is a constant here), followed by an addition with the original $$I_2$$ from there and the result follows. Of course it is valid in the following form too: $$\int_1^{\sqrt{t}}\frac{1}{x}\ln\bigg(\frac{t+sx^2+x^4}{t+sx+x^2}\bigg)dx=\int_1^\sqrt{t} \frac{1}{x}\ln xdx$$ | 2019-05-22T05:47:07 | {
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https://math.stackexchange.com/questions/414776/what-is-the-use-of-the-dot-product-of-two-vectors/415174 | What is the use of the Dot Product of two vectors?
Suppose you have two vectors a and b that you want to take the dot product of, now this is done quite simply by taking each corresponding coordinate of each vector, multiplying them and then adding the result together. At the end of performing our operation we are left with a constant number.
My question therefore is what can we do with this number,why do we calculate it so to speak? I mean it seems almost useless to me compared with the cross product of two vectors (where you end up with an actual vector).
• You can use it to find the angle between any two vectors. $\mathbf{a}\cdot\mathbf{b}=|\mathbf{a}||\mathbf{b}|\cos\theta$ where $\theta$ is the angle between the two vectors. This is a better approach than using the cross product as the cross product can only be defined in a few dimensions (normally only 3 dimensions). Obviously this is just one simple use of the dot product which is a special case of a more general phenomenon known as an inner product en.wikipedia.org/wiki/Inner_product_space . Jun 8, 2013 at 15:35
• You should take a look at how it is derived as it can be very enlightening. Take a look at Lang's Linear Algebra (2E) pg.19-20.
– user70962
Jun 8, 2013 at 20:18
• I always found that the dot product is a good way to measure how "parallel" two vectors are Jun 9, 2013 at 7:15
Re: "[the dot product] seems almost useless to me compared with the cross product of two vectors ".
Please see the Wikipedia entry for Dot Product to learn more about the significance of the dot-product, and for graphic displays which help visualize what the dot product signifies (particularly the geometric interpretation). Also, you'll learn more there about how it's used. E.g., Scroll down to "Physics" (in the linked entry) to read some of its uses:
Mechanical work is the dot product of force and displacement vectors.
Magnetic flux is the dot product of the magnetic field and the area vectors.
You've shared the algebraic definition of the dot product: how to compute it as the sum of the product of corresponding entries in two vectors: essentially, computing $\;\mathbf A \cdot \mathbf B = {\mathbf A}{\mathbf B}^T.\;$
But the dot product also has an equivalent geometric definition:
In Euclidean space, a Euclidean vector is a geometrical object that possesses both a magnitude and a direction. A vector can be pictured as an arrow. Its magnitude is its length, and its direction is the direction the arrow points. The magnitude of a vector A is denoted by $\|\mathbf{A}\|.$ The dot product of two Euclidean vectors A and B is defined by
$$\mathbf A\cdot\mathbf B = \|\mathbf A\|\,\|\mathbf B\|\cos\theta,\quad\text{where \theta is the angle between A and B.} \tag{1}$$
With $(1)$, e.g., we see that we can compute (determine) the angle between two vectors, given their coordinates: $$\cos \theta = \frac{\mathbf A\cdot\mathbf B}{\|\mathbf A\|\,\|\mathbf B\|}$$
• I am not stating what the dot product signifies, in fact that is the essence of this question, I did not know that the dot product has an equivalent geometric definition, or that it could be used to calculate the angle between two vectors that is extremely useful. Jun 8, 2013 at 15:45
• I didn't intend to criticize. I'm sorry if that's how the post comes across! Indeed, I upvoted the question because it's a fine question, and you put thought into writing it. Jun 8, 2013 at 15:47
• The dot product is also the product $\bf A B^T$ of vector A with the transpose of vector B. Jun 8, 2013 at 15:49
• @amWhy: Thank you, I had an upvote mind - trigger finger! :-) Fixed! Jun 9, 2013 at 1:37
• @amWhy thank you for your suburb answer , and also thank you for adding that formula at the bottom for finding the cosine of theta this alone is extremely useful to me due to my pursuit of physics! Jun 9, 2013 at 21:24
The original motivation is a geometric one: The dot product can be used for computing the angle $\alpha$ between two vectors $a$ and $b$:
$a\cdot b=|a|\cdot|b|\cdot \cos(\alpha)$.
Note the sign of this expression depends only on the angle's cosine, therefore the dot product is
• $<0$ if the angle is obtuse,
• $>0$ if the angle is acute,
• $=0$ if the $a$ and $b$ are orthogonal.
Another important special case appears when $a=b$: The root of the scalar product of a vector with itself is the length of a vector:
$a\cdot a=|a|\cdot|a|\cdot1=|a|^2$.
There's another interesting application of the dot product, in combination with the cross product: If you have three vectors $a$, $b$ and $c$, they define a parallelepiped, and you can compute its (signed) volume $V$ as follows using the so-called scalar triple product:
$V=(a\times b)\cdot c$
(Note that this is a generalization of $|a\times b|$ being the area of the parallelogram defined by $a$ and $b$.)
• There's nothing special about 2d/3d here; this means of finding the angle between vectors applies to an arbitrary number of dimensions. Jun 8, 2013 at 15:50
• @Muphrid: Of course you can define something like an "angle" in arbitrary dimensions. I've changed "geometrical" to "visual", I think that makes more sense. Jun 9, 2013 at 6:12
• Perhaps. But two vectors define only a plane, so even in a 4d space or higher, the geometry basically isn't changing: you have two vectors in some common plane, and the dot product tells us how alike they are. I think you're making it out to be more complicated than it really is. Jun 9, 2013 at 6:17
• @Muprid: Good point. I've deleted the sentence now. Jun 9, 2013 at 6:19
Before addressing your question, i want to say that this is a very good question and you are right to expect that the dot product has use/significance.
First, it is important that you think about vectors separate from their coordinates. While it is true that we often represent vectors as a series of coordinates along well-defined axes, this is merely for computational reasons. A vector as an idea "exists" in a space without any predefined coordinate system. I say this because there are two definitions of the dot product, one is coordinate free (i.e. $\mathbf a\cdot\mathbf b = \|\mathbf a\|\,\|\mathbf b\|\cos\theta$) and the other is based on coordinates (i.e. $\mathbf a\cdot\mathbf b = \sum_i{a_i b_i}$). Of these two, it is best to think of the dot product in terms of the former as it does not depend on a coordinate system. (It is relatively easy to show that the latter may be derived from the former, but in that derivation is an implicit assumption that the coordinate system being used to represent the dot product is orthogonal.)
Second, given the coordinate-free definition, the fundamental idea of the dot product is that of projection. By this it gives a single number which indicates the component of a vector in the direction of another vector. Your observation of the dissimilarity between the dot and cross product is correct, however, the dot product is used to produce a vector as well, it just does it component-by-component. Let's suppose that we have a vector $\mathbf v$ represented by its components in a given coordinate system. Let's further suppose that we have an orthonormal basis defined in that same coordinate system as the set of column vectors $\{\mathbf u_1, \mathbf u_2, \ldots, \mathbf u_n\}$. Finally, suppose that we want to represent $\mathbf v$ in this basis as $\mathbf w$. The question is how do we do that? We use the dot product of course! So the first component of $\mathbf w$ would then be $w_1 = \mathbf u_1\cdot \mathbf v$, and the second component would be $w_2 = \mathbf u_2\cdot \mathbf v$ and so on. (Note that because $\|\mathbf u_i\| = 1$, we have $\mathbf u_1\cdot \mathbf v= \|\mathbf v\|\cos\theta_i$.) If we then think of the vector $\mathbf w$ defined as such we have
$$\mathbf w = \left[ \begin{array}{c} w_1 \\ w_2 \\ \vdots \\ w_n \end{array} \right] = \left[ \begin{array}{c} \mathbf u_1\cdot \mathbf v \\ \mathbf u_2\cdot \mathbf v \\ \vdots \\ \mathbf u_n\cdot \mathbf v \end{array} \right] = \left[ \begin{array}{c} \mathbf u_1^T \mathbf v \\ \mathbf u_2^T \mathbf v \\ \vdots \\ \mathbf u_n^T \mathbf v \end{array} \right] = \left[ \begin{array}{c} \mathbf u_1^T \\ \mathbf u_2^T \\ \vdots \\ \mathbf u_n^T \end{array} \right]\mathbf v = \left[ \begin{array}{cccc} \mathbf u_1 & \mathbf u_2 & \cdots &\mathbf u_n \end{array} \right]^T\mathbf v$$
Finally, we conclude that the dot product plays a key role in the transformation of a vector from one basis to another and that the dot product is hidden in the definition of matrix multiplication in that one view of a matrix-vector product is that each element in the product represents a dot product between a row of the left and a column of the right.
I hope this helps.
The dot product is an essential ingredient in matrix product. The product of the two matrices $A$ and $B$ (of compatible sizes, that is, the number of columns of $A$ equals the number of rows of $B$) is a matrix whose $(i, j)$ component is the dot product of the $i$-th row of $A$ and the $j$-th column of $B$.
Among the many applications, consider this simple one.
You have students $s_{1}, \dots, s_{n}$ taking courses $c_{1}, \dots, c_{m}$. Consider the matrix $A$ which has $0$ everywhere, except that the $(i,j)$ coefficient is $1$ if the student $s_{j}$ takes the course $c_{i}$. Now note that the dot product of the $i_{1}$-th row of $A$ with the $i_{2}$-th row gives the number of students that take both $c_{i_{1}}$ and $c_{i_{2}}$. In other words, the matrix $A A^{t}$ has in its $(i, j)$ position the number of students taking both $c_{i}$ and $c_{j}$.
This matrix is of course useful in building a course timetable.
The geometric idea of the dot product has been touched upon, but there is a vast generalization of this product in geometric algebra, the algebra of not only oriented lines (vectors) but planes, volumes, and more (called blades).
In geometric algebra, we have a generalized dot product of a vector $a$ and another blade $B$ denoted $a \cdot B$. This product has a simple geometric interpretation as the part of $B$ orthogonal to the projection of $a$, with magnitude $|a||B|| \cos \theta|$, where $\theta$ is the angle $a$ forms with its projection in $B$.
(Note: as a point of fact, $a \cdot B$ is orthogonal to $a$ also, not just its projection into $B$, but thinking about this leads to some difficulties, while thinking about what's perpendicular to the projection does not.)
If $B$ is a 2-blade (also called a bivector), then you should be able to imagine this directly: if $a$ lies entirely in $B$, then $a \cdot B$ is just the vector perpendicular to $a$ in $B$. If $a$ does not lie entirely in $B$, then it can be decomposed into a tangential part and a normal part. We throw away the normal part, and the previous logic applies for the tangential part.
If $B$ is a 3-blade (a trivector), then in 3d space $a$ must lie in $B$ (for there is no 3d volume that a vector does not help span), and the product $a \cdot B$ is the "Hodge dual", or the plane perpendicular to $a$.
In this light, the dot product of vectors may actually be the most non-intuitive part of this reasoning. When you take the dot product, there's only a scalar left--there's no vector or other higher dimensional object left to be orthogonal to $a$. Again, this is why I emphasize that $a \cdot B$ is the part of $B$ orthogonal to the projection of $a$ onto $B$. When $B$ is a vector, it's clear there is no other vector or anything else that can be orthogonal to the projection of $a$, for $B$ and that projection are parallel, so the result is necessarily just a scalar.
When one calculates A.B, two measurements happen: measurement of how small the angle between them is, and how long A and B are. A.B basically means projection length of A on B, with this length then scaled by the absolute length of B.
One way to think about the interpretation of the dot product is to think how would one maximise or minimise the dot product between two vectors. Let's assume we are trying to maximise the dot product between two vectors that we can modify:
The dot product will be grow larger as the angle between two vector decreases. The dot product A.B will also grow larger as the absolute lengths of A and B increase. This is because as A gets larger, its projected length will be longer, and as B's length gets larger, the scaling of A's projection will grow larger, given that B's absolute length will act as a scaler of A's projection length.
Hence in problems where is desirable to maximise or minimise the size of vectors and minimise the deviation or angle between them, quantifying it using the dot product can be useful.
I don’t see enough plain English answers here, to be honest. One simple example: you can determine in a stealth game whether an object is within a 90 degree line of sight of something, or not. | 2023-02-03T23:07:39 | {
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https://math.stackexchange.com/questions/1834632/find-pairs-a-b-with-gcda-b-gcda-1-b-ldots-gcda-k-b-given | # Find pairs $(a,b)$ with $\gcd(a,b),\gcd(a + 1, b),\ldots, \gcd(a + k, b)$ given
Given a set of GCD's, how to find a set of numbers that satisfy all their criteria?
Suppose we are given a $k$ integers $\gcd(a,b),\gcd(a + 1, b),\ldots, \gcd(a + k, b)$ for some k.
How to get a and b from this? Given the GCD's, I need to find (a,b).
Example: if 4 GCD's are given 3,2,1,6, then the pair (a,b) which satisfy the above condition is (3,6).
Any help will be useful!
• Note that many other $(a,b)$ will also satisfy your $(3,2,1,6)$ report, and in particular $(3+12k, 6+12k)$ are solutions for $k\in \mathbb{Z}$ – Joffan Jun 21 '16 at 15:38
• But you don't necessarily have $a+k=b$ or indeed $a<b$ is that correct? – almagest Jun 21 '16 at 15:39
• no, i do not have a+k=b. But from the sample cases that i have, a<b is valid for all those cases. also, i just need to find a pair which satisfies the GCD's given. I need to select just one of the many possible solutions – Ashwath Narayan Jun 21 '16 at 15:41
Hints: Let $g_0,g_1,\ldots,g_k$ be the specified GCDs. Clearly one must have $\text{LCM}(g_0,\ldots,g_k) \mid b$. Show that if $(a,b)$ is any solution then we may replace $b$ by $b_0 := \text{LCM}(g_0,\ldots,g_k)$, so that $(a,b_0)$ is also a solution.
Therefore we are free to assume $b = b_0$. For each prime power factor $p^r$ of $b_0$, there must be at least one $g_i$ that is divisible by $p^r$. This uniquely determines $a$ mod $p^r$. Now Chinese Remaindering determines $a$ mod $b_0$.
There is at most one working choice of $a$ mod $b_0$. Prove that if any one value of $a$ satisfies the full set of constraints, then so does every $a$ in that congruence class. This provides all solutions of the specific form $(a,b_0)$; there are other solutions with larger values of $b$, and these are more complicated to describe — but you only are looking for one.
• @AshwathNarayan In your example, $b_0=6$ factors into prime powers $2\cdot 3$. $g_1$ is divisible by $2$, which forces $a$ to be odd. $g_0$ is divisible by $3$, which forces $a$ to be divisible by $3$. Therefore $a \equiv 3 \pmod 6$, which turns out to satisfy all the other constraints so it is a bona fide solution. – Erick Wong Jun 21 '16 at 16:05 | 2019-08-24T07:20:13 | {
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https://math.stackexchange.com/questions/3316241/parallel-system-functioning-problem | Parallel system functioning problem
I am currently solving the following problem about conditional probability:
"A parallel system functions whenever at least one of its components works. Consider a parallel system of n components, and suppose that each component works independently with probability $$\frac{1}{2}$$. Find the conditional probability that component 1 works given that the system is functioning."
I think that I do have the answer to this problem; however, since textbook does not contain answer to this one, I am sharing it with the community to poke holes in my logic, if any.
Let's call event "whole parallel system works" as $$W$$ and event "first component works" as $$W_1$$. Our task is to find $$P(W_1|W)$$. Let's use conditional probability definition to expand it: $$P(W_1|W) = \frac {P(W_1 \cap W)}{P(W)} = \frac {P(W | W_1)\cdot P(W_1)}{P(W)}$$ Now, $$P(W_1) = \frac{1}{2}$$. $$P(W) = 1 - P(\bar W) = 1 - (\frac{1}{2})^n = 1 - \frac{1}{2^n}$$. And $$P(W | W_1) = 1$$, since the whole system is active if first component is active. As the result, we have: $$P(W_1|W) = \frac{\frac{1}{2}}{1 - \frac {1}{2^n}} = \frac{\frac{1}{2}}{ \frac {2^n-1}{2^n}} = \frac{2^{n-1}}{2^n-1}$$
• I got the same answer, but the answer in textbook is given as $(1/2)/(1-1/2^{n-1})$. Don't know how they got $n-1$ instead of $n$. Aug 15 '21 at 11:17 | 2022-01-27T17:34:53 | {
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https://mathematica.stackexchange.com/questions/162597/making-calculation-of-correlation-dimension-faster | # Making calculation of correlation dimension faster
I am trying to make my calculation faster. I was hoping, when I compiled it, it would work much faster, but in the end its even slower. I don't know if I am using Compile correctly. I am open to any suggestion on how to make this computation faster.
I use the data:
DD = RandomReal[{0, 1}, {1000, 3}];
I have written the following function for calculating correlation dimension (D2):
CORRDIM[data_] := (
m =
ParallelTable[{r,
Total[
Table[
Length[Drop[Nearest[Drop[data, i - 1], data[[i]], {All, r}], 1]],
{i, 1, Length[data]}]]}, {r, 0.001, 0.011, 0.001}];
m);
CORRDIM works pretty fast, but for a large dataset it is still too slow.
CORRDIM[DD]; // AbsoluteTiming
{0.483119, Null}
I was hoping that Compile could help me speed up this calculation, especially that the options
CompilationTarget -> "C",
RuntimeAttributes -> {Listable},
Parallelization -> True
would somehow run the code in multiple threads, but the compile version works even slower than the interpreted version.
fcc =
Compile[{{x, _Real, 2}},
Table[
{r,
Total[
Table[
Length[Drop[Nearest[Drop[x, i - 1], x[[i]], {All, r}], 1]],
{i, 1, Length[x]}]]}, {r, 0.001, 0.011, 0.001}],
CompilationTarget -> "C",
RuntimeAttributes -> {Listable},
Parallelization -> True]
Map[fcc, {DD}]; // AbsoluteTiming
{0.787973, Null}
Is it there any way to make it faster? Am I making some mistakes in the use of Compile?
• Which definition of "correlation dimension" do you use? The one I found seems to be different than what you programmed. The correlation integral in that article, $C(N,r)$, can be computed with matrix arithmetic. (Much faster than using Nearest.) Dec 26, 2017 at 0:36
• Welcome to Mathematica.SE! I suggest the following: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Take the tour! 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign! Dec 26, 2017 at 4:06
• I believe your issue of having slower compiled code comes from the fact that Nearest isn’t compilable. Dec 26, 2017 at 15:03
• Related: (125543) Dec 26, 2017 at 20:39
Here is my take on this... I am following the definitions and computational steps given in the article
[1] James Theiler, "Efficient algorithm for estimating the correlation dimension from a set of discrete points", Phys. Rev. A 36, 4456 – Published 1 November 1987. DOI:https://doi.org/10.1103/PhysRevA.36.4456 .
Make data:
SeedRandom[33435]
n = 1000;
DD = Sort@RandomReal[{0, 1}, {n, 3}];
Dimensions[DD]
(* {1000, 3} *)
Compute the distance matrix and take the upper triangular values:
AbsoluteTiming[
dmat = DistanceMatrix[DD];
dvals = SparseArray[
SparseArray[UpperTriangularize[dmat, 1]]["NonzeroValues"]];
]
(* {0.051459, Null} *)
Define the correlation dimension integral function:
Clear[CDIntegral]
CDIntegral[dvals_SparseArray, n_Integer, r_?NumericQ] :=
Block[{res},
res = Total@UnitStep[r - dvals];
res*2./(n*(n - 1))
];
Note that the function CDIntegral is defined in such a way so dvals can be reused.
Compute the correlation integrals for a range of distance values.
rRange = Range[0.001, 0.011, 0.001];
AbsoluteTiming[
res =Table[CDIntegral[dvals, n, r], {r, rRange}];
]
(* {0.034141, Null} *)
This seems to be ~10 times faster, than using Nearest (on my recent MacBook Pro with \$Version 11.2.0.)
Using n=10000 the above computations are done for ~20 seconds.
Following [1] here is how the correlation dimension is estimated with the results obtained from the computations above:
lm = LinearModelFit[
Log@Select[Transpose[{rRange, res}], #[[2]] > 0 &], {1, x}, x]
lm["BestFitParameters"][[2]]
(* 2.68916 *)
Here is a plot of the found correlation integrals and the fit:
ListLogLogPlot[{Transpose[{rRange,res}], {#, Exp[lm["Function"][Log[#]]]} & /@ rRange},
Filling -> {1 -> Bottom}, PlotTheme -> "Detailed",
PlotLegends -> {"correlation integrals", "fit"}]
ClearAll[corrDIM]
corrDIM[data_] := Module[{nF = Nearest[data -> "Index"]},
Table[ {r, Tr[Table[Tr[UnitStep[nF[data[[i]], {All, r}]- i - 1]],
{i, 1, Length[data]}]]}, {r, 0.001, 0.011, 0.001}]]
Using Anton's example data:
SeedRandom[33435]
n = 1000;
DD = Sort@RandomReal[{0, 1}, {n, 3}];
First @ AbsoluteTiming[result = corrDIM[DD];]
0.044153
Comparing with OP's CORRDIM (on Wolfram Cloud where ParallelTable is not supported):
Quiet @ First @ AbsoluteTiming[result2 = CORRDIM[DD];]
0.192319
result == result2
True
Comparing with Anton's CDIntegral:
First @ AbsoluteTiming[dmat = DistanceMatrix[DD];
dvals = SparseArray[ SparseArray[UpperTriangularize[dmat, 1]]["NonzeroValues"]];
res = Table[CDIntegral[dvals, n, r], {r, rRange}];]
0.07153
2./(n*(n - 1)) result[[All, 2]] == res
True
• Nice, much faster than my implementation, especially with larger n, e.g. n=10000. Dec 27, 2017 at 15:41
• Thanks, can but I have a problem, I can not run your corrDIM function, ther is an error with "Index", what does this mean please ' Nearest[data -> "Index"] ' ? I couldnt find out. Dec 28, 2017 at 19:34
• Actually this helps me already, but originally I was trying to run computing of more of Datasets in threads, by using any of D2 estimating methods. I mean for example for all Datasets DD, DD1 DD2 ... DDn at once so it would take like the same time as for one DD. Is it possible ? I think using Compile it should work somehow, but I cant find out how. Dec 28, 2017 at 19:41
• @Mark, you can use Automatic in place of "Index" in versions before 10. Re parallelization / compile, Compile` might make Anton's method faster.
– kglr
Dec 28, 2017 at 20:07 | 2022-05-27T00:43:52 | {
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http://math.stackexchange.com/questions/424106/can-all-triangle-numbers-that-are-squares-be-expressed-as-sum-of-squares | # can all triangle numbers that are squares be expressed as sum of squares
I'm not sure if this is just a subset of Which integers can be expressed as a sum of squares of two coprime integers? which in turn points to http://en.wikipedia.org/wiki/Brahmagupta%E2%80%93Fibonacci_identity, but if so, I'm not seeing it.
Basically: looking at the numbers between 0 and 1000, if (but not iff) n is a square, then the nth triangle number (i.e. $$\frac{n(n+1)}{2}$$) can be expressed as the sum of two perfect squares . Does this hold for all squares, and can someone point me in the direction of why this is? (For what it's worth, I is an engineer but haven't really touched number theoryish stuff since college.)
(Also saw Prove that there are infinitely many natural numbers $n$, such that $n(n+1)$ can be expressed as sum of two positive squares in two distinct ways. which proves there are infinitely many for the similar $n(n+1)$ case but not that all squares will work unless I missed some aspect)
(Brute force approach to checking numbers, because brute force always works)
import math
maxsquare = 1001
squares = [i*i for i in xrange(maxsquare)]
for j in xrange(int(math.sqrt(maxsquare))):
i = j * j
n = i * (i+1) / 2
for s in squares:
f = n - s
if f in squares:
print "%d^2 + %d^2 = %d * %d+1 / 2" % (int(math.sqrt(f)),int(math.sqrt(s)),i,i)
break
else:
print "Could not find anything for i = %d" % i
-
I think the title of your question is phrased wrong. It says "all triangle numbers that are squares". But $\frac12 8(8+1)$ is a triangle number that is a square, but you are not interested in it. And obviously all triangle numbers that are squares are trivially sums of squares. Right? – MJD Jun 19 '13 at 2:01
correct... any suggestions on how to correct the title? – Foon Jun 19 '13 at 2:28
The answer is yes, it is true for all $n$.
Let me reformulate your question. You want to know if, whenever $n$ is a perfect square, it is the case that $\frac12 n(n+1)$ is a sum of two squares. If $n$ is a perfect square, then it has an integer square root, which we can call $m$, and write $n = m^2$. Then your question becomes whether $\frac12 m^2(m^2+1)$ is a sum of two squares for all integers $m$.
There is a very useful theorem about sums of two squares which says that a number $N$ is a sum of two squares if and only if every prime of the form $4k+3$ (such as $3, 7, 11, 19,$ etc.) appears in the prime factorization of $N$ an even number of times.
Now $m^2(m^2 + 1) = {\left(m^2\right)}^2 + m^2$ is obviously a sum of two squares. And dividing $m^2(m^2+1)$ by 2 can't possibly affect the number of times any prime of the form $4k+3$ appears in its factorization, so $\frac12 m^2(m^2+1)$ is also a sum of two squares.
(Addendum: Erick Wong points out below that we do not even need to invoke the Fermat $4k+3$ theorem.)
-
$n^2(n^2+1)=(n^2)^2+1$ is a sum of two squares without the machinery from the preceding paragraph. – Andres Caicedo Jun 19 '13 at 1:52
@AndresCaicedo I think you mean $(n^2)^2 + n^2$, but your point still stands. – anorton Jun 19 '13 at 1:54
@andres Is there a trick for showing that $a$ is a sum of squares if and only if $2a$ is, that doesn't depend on the $4k+1$ theorem? Because if not, I don't think I can abbreviate my answer very much. – MJD Jun 19 '13 at 2:00
@anorton Yes, of course. (How embarrassing!) – Andres Caicedo Jun 19 '13 at 2:05
@MJD Yes, there is a very simple trick: if $2a = x^2+y^2$, then $x$ and $y$ have the same parity and so $a = (\frac{x+y}{2})^2 + (\frac{x-y}{2})^2$. You may recognize this as unique factorization in $\mathbb Z[i]$ in disguise. – Erick Wong Jun 19 '13 at 2:16
If you run your program you should find $$0^2+1^2 = \frac{1\times 2}{2}\\ 1^2+3^2 = \frac{4\times 5}{2}\\ 3^2+6^2 = \frac{9\times 10}{2} \\ 6^2+10^2 = \frac{16\times 17}{2} \\ 10^2+15^2 = \frac{25\times 26}{2}$$ from which you might recognize the numbers in the sums as the triangular numbers again, in which case you can guess that $$\frac{j^2(j^2+1)}{2} = \left(\frac{j(j-1)}{2}\right)^2+\left(\frac{j(j+1)}{2}\right)^2$$ which is true. | 2014-03-12T20:13:23 | {
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# Factor table with sign-The useful tool to solve polynomial inequality
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Factor table with sign-The useful tool to solve polynomial inequality [#permalink]
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04 Dec 2016, 08:22
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Edit: How to solve quadratic equations - Factor quadratic equations? You may find this topic
First of all, let's try these questions.
Question 1.
Find the integer value of $$x$$ that $$(x-2)(x-4)<0$$
Question 2.
Solve for this inequation: $$x(x+1)(x-5)>0$$
Question 3.
Solve for this inequation: $$\frac{x(x-2)^3}{(x+1)(x+2)^2} \leq 0$$
To solve Question 1, we could quickly come to result $$x=3$$. However, to solve Question 2 and Question 3, how much time will we need? Almost much more time to solve them than to solve Question 1.
Basically, while solving polynomial inequalitie, we need to review each range value of $$x$$.
For example, to solve Question 1: $$(x-2)(x-4)<0$$, we need to review if $$x<2$$, if $$2 \leq x<4$$, and $$x \geq 4$$. Finally, we come to result $$2 <x<4$$.
This method could be used with basic polynomial inequalitie, but it seems unuseful to solve complex polynomial inequalitie. Now, I'm going to introduce you an useful tool to solve these kinds of complex polynomial inequalitie, called: Factor table with sign.
These are the steps to solve polynomial inequalities like $$(x-a_1)(x-a_2)...(x-a_n) < 0$$. (note that sign "$$<$$" could be "$$>$$", "$$\leq$$", "$$\geq$$")
Step 1. Change the polynomial/expression into factorization form.
Step 2. Make factor table with sign to solve the inequatility
Step 2.1. Find all values of $$x$$ those make each factor equals to 0. List them in the first row of table in ascending order.
Step 2.2. In next rows of table, review the signs of each factor of the polynomial/expression based on each range value of $$x$$
Step 2.3. In the fianl row of table, review the sign of the polynomial/expression based on each sign of each factor in each range value of $$x$$.
Step 3. Finally, come to solution.
Now, let's solve the Question 1.
Step 1. We already have $$(x-2)(x-4)<0$$
Step 2. Make factor table with sign like this.
Attachment:
Capture question 1.PNG [ 2.18 KiB | Viewed 1988 times ]
Step 2.1. Find all values of $$x$$ those make each factor equals to 0.
$$x-2=0 \iff x=2$$
$$x-4=0 \iff x=4$$
In the first row, list all these values in ascending order: $$-\infty , 2, 4, +\infty$$.
Step 2.2. In next 2 rows, review the signs of each factor.
This step based simply on this rule: If $$x<a \implies x-a<0$$. If $$x>a \implies x-a>0$$.
For example, $$x-2=0 \implies x=2$$. Hence, any value of $$x<2$$, then $$x-2$$ will be negative (-); any value of $$x>2$$, then $$x-2$$ will be positive (+).
We dont need to care about what is the value of $$x-2$$ if $$x=4$$ or what is the value of $$x-4$$ if $$x=2$$. These values are expressed as sign "|".
Step 2.3. In final row, review the sign of the expression.
The sign of the expression simply based on this rule:
$$(+) \times (+) = (+)$$
$$(-) \times (-) = (+)$$
$$(+) \times (-) = (-)$$
$$(-) \times (+) = (-)$$
Step 3. Come to solution.
We need to find sign (-) of the expression. Based on the table, we simply have $$2<x<4$$.
That's it.
Now let's solve Question 2.
$$x=0$$
$$x+1=0 \implies x=-1$$
$$x-5=0 \implies x=5$$
The factor table with sign:
Attachment:
Capture question 2.PNG [ 2.97 KiB | Viewed 1982 times ]
Hence, we simply have $$x \in (-1,0) \cup (5, +\infty)$$
Solution for Question 3.
$$x=0$$
$$x-2=0 \implies x=2$$
$$x+1=0 \implies x=-1$$
$$x+2=0 \implies x=-2$$
The factor table with sign
Attachment:
Capture question 3.PNG [ 9.54 KiB | Viewed 1977 times ]
In the table above, I've divided into 4 parts. Part 2 contains every factor without power for simplified purpose. Then I come to Part 3 that contains every factor with its power. In part 4 or the final row, I come to the sign of the expression.
Also note that the expression is undefined when $$x=-2$$ or $$x=-1$$. The solution is $$x \in (-\infty, -2) \cup (-2,-1) \cup [0,2]$$
Now come to another example.
Question 4. Find the product of the integer values of x that satisfy the inequality $$\frac{(x^2-4)^3}{(x-5)^5(x^2-9)^4}$$
This question belongs to e-GMAT http://gmatclub.com/forum/wavy-line-met ... l#p1727256
You could find the solution in http://gmatclub.com/forum/wavy-line-met ... l#p1771328
Question 5. Find the value of $$x$$ that $$x^4+5x^3-7x^2-41x-30 \leq 0$$.
Solution.
$$x^4+5x^3-7x^2-41x-30 =(x-3)(x+1)(x+2)(x+5)$$
The factor table with sign
Attachment:
Capture question 4.PNG [ 4.26 KiB | Viewed 1986 times ]
The solution is $$x \in [-5,-2] \cup [-1,3]$$
This tool is really useful when we need to find the value of x satisfied the polynomial inequalities. I hope this tool could assist you in solve DS/PS GMAT questions in the future.
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Re: Factor table with sign-The useful tool to solve polynomial inequality [#permalink]
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Re: Factor table with sign-The useful tool to solve polynomial inequality [#permalink] 08 Dec 2017, 06:16
Display posts from previous: Sort by | 2018-04-26T07:54:48 | {
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https://old.learntla.com/tla/sets/ | # Sets
You’ve seen sets before. {1, 2} is a set. 1..N is the set of all numbers between 1 and N. x \in S checks whether S contains x, unless you use it in a PlusCal variables statement or a with (where it creates new behaviors). Like everything else, you can nest them: {{{}}} is a set containing a set containing an empty set.
Often, we’re interested in some transformation of the set, such as “the set of male participants under 60” or “The owners of the pets.” In this section we’ll cover three ways to transform a set and how they are combined.
### Filtering
We can filter a set with {x \in S: P(x)}, which is the set of all x \in S where P(x) is true. For example, {x \in 1..8 : x % 2 = 1} is the set {1, 3, 5, 7}. P must return a boolean, so { x \in 1..4 : x * x } raises an error.
If S is a set of tuples, you can filter on some relationship between the elements of the tuple by instead using <<...>> \in S. If you want the set of ordered pairs, you could do {<<x, y>> \in S \X S : x >= y}.
As always, you can nest filters, and {x \in {y \in S : P(y)} : Q(x)} will filter on the filtered list. Generally, though, {x \in S: P(x) /\ Q(x)} is easier to understand.
### Mapping
{P(x): x \in S} applies P to every element in the set. { x * x : x \in 1..4 } is the set {1, 4, 9, 16}. { x % 2 = 1:x \in 1..8 } is the set {TRUE, FALSE}.
You can also write {P(x, y, ...) : x \in S, y \in T, ...}. { x + y : x \in 0..9, y \in { y * 10 : y \in 0..9} } is the first hundred numbers, in case you wanted to obfuscate 0..99.
Given DOMAIN Tuple is the set of numbers Tuple is defined over, write an operator that gives you the values of the Tuple, ie the range.
Range(T) == { T[x] : x \in DOMAIN T }
This is a useful enough operator that I’ll assume it’s available for all other examples and exercises.
### CHOOSE
CHOOSE x \in S : P(x) is some x where P(x) is true. CHOOSE x \in 1..8 : x % 2 = 1 will be one of 1, 3, 5, 7. TLC does not branch here; while the number it chooses is arbitrary, it will always return that number. This is similar to how CASE statements work: CHOOSE x \in S : TRUE is some element of S, but TLC won’t check all of them.
TLC assumes that you always intend for there to be at least one element to choose. If there aren’t any (trivial example: CHOOSE x \in S : FALSE), it will consider this a problem in your spec and raise an error. TLC will also raise if S is infinite because TLC can’t evaluate P on an infinite number of elements. There still may be a problem with your spec, though, and it’s a good idea to try it on a finite subset.
## Set Operators
Finally, there are extra operations for working with sets:
logic operator TRUE FALSE
in set \in 1 \in {1, 2} 1 \in {{1}, 2}
not in set \notin 1 \notin {} {1} \notin {{1}}
is subset \subseteq {1, 2} \subseteq {1, 2, 3} {1, 2} \subseteq {1, 3}
Write an operator that takes two sets S1 and S2 and determines if the double of every element in S1 is an element of S2.
IsDoubleSubset(S1, S2) == {x * 2 : x \in S1} \subseteq S2. If you wanted to check both ways (doubles of S2 are in S1), you could write two expressions with \/.
operator operation example
\union Set Union {1, 2} \union {2, 3} = {1, 2, 3}
\intersect Set Intersection {1, 2} \intersect {2, 3} = {2}
S1 \ S2 The elements in S1 not in S2 {1, 2} \ {2, 3} = {1}, {2, 3} \ {1, 2} = {3}
SUBSET S The set of all subsets of S SUBSET {1, 2} = {{}, {1}, {2}, {1, 2}}
UNION S Flatten set of sets UNION {{1}, {1, 2}, {5}} = {1, 2, 5}
Given a sequence of sets, write an operator that determines if a given element is found in any of the sequence’s sets. IE Op("a", <<{"b", "c"}, {"a", "c"}>>) = TRUE.
InSeqSets(elem, Seq) == elem \in UNION Range(Seq)
If you add EXTENDS FiniteSets, you also get the following operators:
operator operation
IsFiniteSet(S) TRUE iff S is finite
Cardinality(S) Number of elements of S, if S is finite
Given a set, write an operator that returns all subsets of length two. IE Op(1..3) = {{1, 2}, {1, 3}, {2, 3}}.
Op(S) == { subset \in SUBSET S : Cardinality(subset) = 2 } | 2022-08-09T16:44:25 | {
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https://math.stackexchange.com/questions/2241040/what-is-the-symbolic-form-of-there-does-not-exist-a-largest-natural-number/2241055 | # What is the symbolic form of "there does not exist a largest natural number "
Other students in office hour said this is the correct form $(\forall x)(\exists y)(y>x)$ { for all x natural number, there exists y such that y is greater than x }
But "there does not exist a largest natural number " $\neg(\exists x)(x\text{ largest natural number})$
Am I even close ?
• You have said "it is not true that for all $x$ there exists $y$ such that $y>x$". Think about whether that is an appropriate translation.
– Ian
Apr 19, 2017 at 0:17
• $\forall x \exists y (y > x)$ for $\forall x \exists y (y > x)$. Formatting tips here.
– Em.
Apr 19, 2017 at 3:30
• This was edited to say something completely different from what it originally said. Don't do that, because it makes the answers based on the original question seem nonsensical. Apr 20, 2017 at 6:31
The formula $\forall x \exists y (y > x)$ reads "for any $x$ there is a $y$ which is larger than $x$" which says that any $x$ is not largest and therefore no $x$ is largest. You can also use the formula $\forall \equiv \neg\exists\neg$ to get
\begin{align*} \forall x \exists y (y > x) &\equiv \neg\exists x \neg \exists y (y > x) \\ &\equiv \neg \exists x \forall y \neg(y > x) \\ &\equiv \neg \exists x \forall y (y \le x) \end{align*}
which says that "there is no $x$ such that every $y$ is less than or equal to $x$". This better fits with the "there does not exist a largest natural number" phrasing.
The statement as you have written it is false. If you read it out loud, $x$ has to be chosen before $y$ and for all $x$ you can find a greater $y$. You want to say that if you choose $y$. You want to choose $y$ first, then say that it is not greater than all $x$.
• so, (Ay)(Ex)(y>x)??? Apr 19, 2017 at 0:26
• No, given $y$ there has to be a larger $x$, so it should be $x \gt y$ Apr 19, 2017 at 1:07
• @QuencyCaroline That statement says every natural number has one smaler than it, which is not true. Apr 19, 2017 at 1:37
• $(\forall x)(\exists y)(y>x)$ is false? Assuming the universe of discourse is the natural numbers, can you give a counterexample? Apr 19, 2017 at 17:59
• @jwodder: when the answer was written there was a not sign before the sentence. As you have written it and the question now has it, it is true. Apr 19, 2017 at 18:41
# Yes, you are close.
So far, you have:
$\neg \exists x \,\,(x\text{ largest natural number})$
Since you're looking for the "symbolic form", your next step is to convert "largest natural number" to symbols.
• Note that "largest natural number" is the same thing as "number that is greater than all other natural numbers";
• To make things easier, note that the above is the same thing as "number that is greater than or equal to all natural numbers" (therefore, including itself without a problem!);
• Further rephrase it as "number such that all numbers are less than or equal to it", so we have:
$x$ is a number such that all numbers are less than or equal to $x$
• Observe that this last sentence can be easily translated to symbols as
$$\forall n \,\,\, n \le x$$
Now, just plug that into your original sentence, obtaining:
$\neg \exists x \,\,(\forall n \,\,\, n \le x)$
Note 1: This is not the only correct way to do this. It is possible to express the same fact differently.
Note 2: in the context of this question, it is clear that we are talking about natural numbers. But generally, it would be better to specify this, by writing:
$\neg \exists x \in \mathbb{N} \,\,(\forall n \in \mathbb{N} \,\,\, n \le x)$
Bonus: The students you mentioned are correct too. They chose to rephrase the sentence in a different sentence (but still equivalent). Instead of saying "there is no largest natural number", they are saying "all natural numbers have the property of being smaller than some number", which is the same thing, in the end.
It depends on whether with largest natural number you mean a greatest or a maximal element, though the difference would be important only for a non-total order.
If $(X,<)$ is a partially ordered set, we say that $a\in X$ is
• a greatest element if $\forall x\in X\colon (x< a\lor x=a)$
• a maximal element if $\forall x\in X\neg(a<x)$
(In a totally ordered set, exactly one of $x<a$, $x=a$, $x>a$ must be true, hence therre the notions of greatest and maximal element coincide).
Hence "There is no $a$ such that $a$ is a greatest/maximal element" translates "literally" to either $$\neg\exists a\in \Bbb N\colon \forall x\in\Bbb N\colon (x<a\lor x=a)$$ or $$\neg\exists a\in \Bbb N\colon \forall x\in\Bbb N\colon \neg(a<x)$$
If you make use of "$\neg\exists=\forall\neg$" and "$\neg\forall=\exists\neg$", you might equivalently write $$\forall a\in \Bbb N\colon \exists x\in\Bbb N\colon (x\not <a\land x\ne a)$$ and $$\forall a\in \Bbb N\colon \exists x\in\Bbb N\colon (a<x),$$ respectively.
In English: "there is no natural n, for which all natural k would be smaller than n".
$\lnot \exists n \forall k (k, n) \in \mathbb{N}^2 \land k < n$
I was going to write this in a comment, but I think it's better if I gave an answer.
I'm going to begin by pointing out things about the two statements in the question.
The first statement does have the intended meaning. I think it can still be made more precise.
The second statement is circular, or ambiguous at best. There is no way of saying what the largest natural number is until it has been decided definitely whether it exists or not.
Finally, although the domain has been described in the worded statement, for the logical statement to make sense, it must include the correct domain.
Since T. Gunn's answer very nicely covers the question, I will instead give an example of a correct statement which is useful in the study of modern algebra, how number (and other) sets are constructed, the relations between them, and the rigorous framework used to describe these ideas:
$$\forall n \in \mathbb N,n + 1 > n \land n + 1 \in \mathbb N$$
Though, to be precise this is a logically equivalent statement, not an identical one. And I have assumed, the addition operator and ordering relation ">" have already been defined.
This statement makes the intended meaning, and its demonstration, more explicit. It describes a function which guarantees the required property will be satisfied.
• If the addition operator has already been defined (as a function $\mathbb N\times\mathbb N\to\mathbb N$), then the proposition $n+1\in\mathbb N$ is redundant.
– user856
Apr 19, 2017 at 22:27
• @Rahul Fair point. I just thought writing it out would make the underlying framework more apparent. Apr 19, 2017 at 23:28
Since you are interested in formalisms, I would suggest to rephrase the statement to be more formal. In Mathematical logic, one usually uses quantors (similarly, the negation operator) and parentheses in the following way: $$\forall x (\; \text{logical statement} \;)$$ So, nesting this, your statement becomes $$\forall x \in \mathbb{N}\left(\,\exists y \in \mathbb{N}\left(\,y>x\right) \right)$$ and the other statement, the one you were asking for, becomes $$\neg (\,\exists M \in \mathbb{N} \,(\,\forall l \in \mathbb{N} \,(\,M \geq l))$$ where I replaced your variable-symbol $x$ by $M$ and used $l$ as the variable-symbol for your natural number. | 2022-05-22T04:59:53 | {
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https://www.jiskha.com/questions/56882/Write-the-following-as-a-fraction-and-show-your-work-0-83-the-three-is-repeating | # pre-Algebra
Write the following as a fraction and show your work. 0.83 (the three is repeating) I don't know how to do it with only one of the numbers repeating and it is not in my book.
1. 👍 0
2. 👎 0
3. 👁 105
1. Here is a neat algorithm to change any repeating decimal into an exact fraction
Assume you have only the decimal portion of your number, ie, for 4.5676767... consider only the 5676767..
For the numerator, write down all the digits to the end of your repeat, subtract from that the part that does not repeat.
For the denominator, write down a 9 for each digit in the repeating loop, followed by the number of digits that don't repeat.
e.g. 0.45676767..
=(4567-45)/9900
=4522/9900, now reduce to
2261/9900
.833333. = (83-8)/90
= 75/90
= 5/6
1. 👍 0
2. 👎 0
posted by Reiny
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More Similar Questions | 2019-05-23T16:07:20 | {
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https://www.physicsforums.com/threads/integral-of-gaussian-function-for-squared-x.763598/ | # Integral of Gaussian function, for squared x
1. Jul 27, 2014
### Avinto
1. The problem statement, all variables and given/known data
I am trying to compute an integral, as part of the expected value formula (using a Gaussian PDF)
$$\int_{-∞}^{∞} (x)^2 p(x) dx$$
Where p(x) is the Gaussian probability density function:
$$\frac{1}{\sigma \sqrt(2 \pi)} \exp(\frac{-x^2}{2 \sigma^2})$$
My aim after this is to be able to compute for all even x^n in the above formula. For all odd x^n, the positive and negative componets cancel out, with an computation of zero for all odd functions.
2. Relevant equations
Wikipedia lists two equations that relate to this:
[1]https://en.wikipedia.org/wiki/List_of_integrals_of_Gaussian_functions
[1]
$$\int_{-∞}^{∞} (x)^2 \phi(x)^n dx = \frac{1}{\sqrt(n^3 (2 \pi)^{n-1})}$$
and
[2]https://en.wikipedia.org/wiki/List_of_integrals_of_exponential_functions
[2]
The range here is zero to infinity, however as this function is even, the result for minus infinity to infinity should be twice what the below computes.
$$\int_{0}^{∞} (x)^n \exp(-\alpha x^2) dx= \frac{(2 k -1)!!}{2^{k+1} \alpha^k} \sqrt(\frac{\pi}{\alpha})$$
Where
$$n=2 k, k integer, \alpha > 0$$
3. The attempt at a solution
Using [1], I can compute a answer equalling 1, which is apparently the right answer. However when using [2] (where k=1), I compute a different anwser:
Compute:
Let $$\alpha = \frac{1}{2 \sigma^2}$$
and
$$k=1$$
then:
$$\int_{0}^{∞} (x)^2 \exp(-\alpha x^2) dx = \frac{(2 k -1)!!}{2^{k+1} \alpha^k} \sqrt(\frac{\pi}{\alpha})$$
$$=\frac{1!!}{4 \alpha}\sqrt(\frac{\pi}{\alpha})$$
$$=\frac{\sigma^2}{2}\sqrt(2 \sigma^2 \pi)$$
$$=\frac{\sigma^2}{2} \sigma \sqrt(2 \pi)$$
$$=\frac{\sigma^3 \sqrt(2 \pi)}{2}$$
Now, this is where I think I messed up. I previously removed
$$\frac{1}{\sigma \sqrt(2 \pi)}$$
from inside the integral, and attempted to multiply it by the obtained result (and then doubled everything, as it is an even function)
$$2 \frac{1}{\sigma \sqrt(2 \pi)} \frac{\sigma^3 \sqrt(2 \pi)}{2}$$
Then by cancelling:
$$=\sigma$$
I'm thinking this is because I took part of the PDF out of the integral, and then changed the limits. However, I'm not sure how to go about working this out while leaving it in there.
I will keep browsing around for a solution to this problem (and the more general x^n), and would really appreciate any hints on this.
Thanks.
2. Jul 27, 2014
### Ray Vickson
First get
$$I_n = \int_0^{\infty} x^n \phi(x) \, dx, \:\: \phi(x) = \frac{1}{\sqrt{2 \pi}} e^{-x^2/2},$$
then obtain the general result by re-scaling the variable. Note that $x \phi(x) \, dx = d(-\phi(x))$ so we can use integration by parts:
$$\int x^n \phi(x) \, dx = \int u dv,\\ u = x^{n-1}, \: dv = x \phi(x) \, dx = d(-\phi(x))$$
3. Jul 27, 2014
### Orodruin
Staff Emeritus
Alternatively, change variables to $t = x^2$ and use the definition of the Gamma function.
4. Jul 28, 2014
### Avinto
Ray, I will attempt to work it out using your way later today, and see how it goes.
$$t = x^2$$
$$\Gamma(z) = \int_0^{∞} t^{z-1} exp(-t) dt$$
$$\Gamma(3) = \int_0^{∞} t^{3-1} \exp(-t) dt$$
$$\Gamma(3) = (3-1)! = 2$$
Then for the range ${-∞,0}$, this answer should be the same, so I get an answer of 4 overall.I'm working under the assumption that the answer should be 1
Last edited: Jul 28, 2014
5. Jul 28, 2014
### Orodruin
Staff Emeritus
Well, the answer cannot be 1. It should be the variance of the distribution (since the expectation value is zero). In fact, you had it (almost) right in your first post... Also note that in the substitution, z is not 3. If you do it correctly, z should have a half-integer value (which is why you end up with a semi-factorial and 2^(k+1)).
Edit: Well, of course the answer would be one if the variance is ... :)
6. Jul 28, 2014
### Avinto
Ok, I see what you mean about the variance, the definition of $\phi$ on the wiki page assumes $\sigma=1$. Using $\sigma=1$, my calculation did evaluate to one, which is where I must have gotten that assumption from.
I redid my calculations with including $\sigma$, and realised I made a silly error in my cancellation.
$$2 \frac{1}{\sigma \sqrt(2 \pi)} \frac{\sigma^3 \sqrt(2 \pi)}{2}$$
Should evaluate to $\sigma^2$, not $\sigma$.
I will look more into the Gamma function as well, and try to understand it from that direction. Thanks | 2018-03-18T18:23:09 | {
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https://math.stackexchange.com/questions/593943/how-to-find-the-minimum-of-the-function | # How to find the minimum of the function?
How to find the minimum of the following function
$${\rm f}\left(w\right) = {1 \over 2}\sum_{i = 1}^{n}\left({1 \over 1 + {\rm e}^{-x_{i}\,w}} -y_{i}\right)^{2}$$
where $x_{i}, y_{i} \in \left(0, 1\right)$ are constants, $w\in \mathbb{R}$?
Could you find a analytic or computational way to get the minimum of the function ?.
• Please try to make use of MathJax here :) Dec 5 '13 at 12:13
• Ok, I will try it next time.@Shaun Dec 5 '13 at 12:15
• Check the following link ---> dlmf.nist.gov/software Dec 6 '13 at 4:20
• Thanks for your software index.@Felix Marin Dec 6 '13 at 5:03
I suppose that this is a nonlinear least square fit problem in which you have data points [x(i) , y(i)] and you want to adjust the parameter w. If you establish the derivative of f(w) with respect to w, you have one (not too complex) equation to solve but it can easily be done using Newton method. The problem is to start with a reasonable value; you can have one rewriting x(i) as a function of y(i). Going to logarithms, you will see that x(i) is along a straight line of Log[y(i) / (1-y(i)] and the slope of this line is w. So, you have everything to start.
• Thank you for your answer. With your method, I can find several points that the derivative of f(w) are 0. But can I find every point that the derivative of f(w) are 0? So that I can find the minimum of the whole function. Dec 5 '13 at 13:10
• No, this is not the way. Since your data y(i) are in error, you search for the value of "w" which minimizes your function f(w) that is to say which is the solution of f'(w)=0. The solution is unique. As I told you (with a typo), a look of the plot of Log[y(i) / (1 - y(i)] versus x(i) will give you an estimate of "w" which is the root you look for. If you want, post a few points for me and I shall enter into details in such a way you can continue by yourself. If you like my answer, you can accept it. Dec 5 '13 at 15:12
• To correct some typos (I am almost blind), plot Log[1 - 1 / y(i)] against x(i). The graph will look like a straight line the slope of which being "- w". Make a visual estimation of that (or use Excel plot with regression in the transformed space). Now, solve f'(w)=0 using Newton. You can also visualize ploting f(w) versus "w". Dec 5 '13 at 15:24
• I don't understand why you think there is only one solution of f'(w)=0. Could you proof it prove it? And I will appreciate your answer. Change y(i) into Log[1 - 1 / y(i)] can make it into a linear regression problem. But it's meaningless for me because I need to estimate the error with the form y(i).@Claude Leibovici Dec 6 '13 at 3:24
To try to clarify things, I generated 10 values [x(i) = i / 10] and the corresponding values y[i] are [0.58, 0.65, 0.72, 0.78, 0.83, 0.87, 0.90, 0.92, 0.94, 0.96].
Based on these, I wrote function f(w) as given in your post (sum of 10 terms). When plotted as a function of w, f(w) exhibits (as totally normal) a paraboloid shape with a marked minimum around 3.15 (to give you an idea, for f(2.0)=0.0337713, f(2.5)=0.00852217, f(3.0)=0.000357495, f(3.5)=0.00181296, f(4.0)=0.00843496. The absolute minimum corresponds to w=3.13938 for which f(w)=0.000031942 and this is the solution.
Using the second approach, I wrote function f'(w). Ploted against w, this function has a very nice shape and becomes exactly zero at w=3.13938 and this is the solution.
You must remember than solving an equation is much simpler than minimizing a function (almost if not constrained). For illustration purposes, to solve f'(w)=0, I used Newton method starting at w=2.0 (value which is far away from the one I suggested you to generate using some changes). The successive iterates for w are 2.45293, 2.75188, 2.93280, 3.03609, 3.09582, 3.13997.
• Thanks for your example. My colleagues also tried many examples and haven't find a case that has more than one relative extremum. I wonder could it be proved analytically that the function f(w) have exactly one relative extremum, or have a sufficiently large probability that has only one relative extremum, which you proved experimentally. Dec 6 '13 at 6:50
• @YangzheLau.In the case of your model, which is extremely simple, only one extremum can exist. Forgetting mathematics, you search for a value of "w" which gives you the best compromise between the y[i] and the model. This minimum is then unique. If your model was more complex such y = a / (b + Exp[-w x]), the story would be very different and in this case, except if you can generate good estimates, several minima could exist and global optimization could be required. But, this is NOT the case for your problem. Dec 6 '13 at 7:06
• @ClaudeLeibovici As I showed in my comment below on my answer, the functional need not have a global minimum (I.e. It is at $\pm\infty$). There are also times where it is I feasible to calculate the derivative due to its complexity, let alone root find with it! There is a reason that a great deal of research has been performed on minimisation techniques. Dec 6 '13 at 7:22
• @Daryl x[i] and y[i] belong to (0,1). We didn't find a counterexample yet. Dec 6 '13 at 7:34
• @ClaudeLeibovici You are right at least in any application case. Ok, I just consider it as a theorem and continue my work. Dec 6 '13 at 7:45
From a numerical point of view, your problem looks quite iteresting and I played with it, my concern being to get a quick estimate of the parameter "w" which has to be adjusted to your data.
What I did was to start a Newton procedure at w=0. As a result, what I obtained is that a rough estimate of "w" can be obtained using the following formula for the estimate
w = 2 (2 S3 - S1) / S2
in which S1 is the sum of the x(i), S2 is the sum of x(i)^2 and S3 is the sum of x(i) * y(i).
For an exact value of w = 1, 2 or 3, the corresponding estimates are is 0.95, 1.66 and 2.11. The corresponding "experimental" data were x(i) = i / 10 and the y(i) were calculated using the exact formula.
In the case I previously illustrated (with noise in the y data), this procedure would lead to an estimate equal to 2.16 for an excat value of 3.14.
• Your estimate can be used as an good start value of w for me, and I 'm surprised at the succinct form of the derivative and second derivative of the function when w=0. By the way, I'm considering a two dimensional case. $${\rm f}\left(w_{1},w_{2}\right) = {1 \over 2}\sum_{i = 1}^{n}\left({1 \over 1 + {\rm e}^{-x_{i}w_{1}-z_{i}w_{2}}} -y_{i}\right)^{2}$$ Now there are more than one local extremum. Dec 8 '13 at 12:51
• @YangzheLau. Now, we are in the serious case with two dimensions. However, you can first perform a linear regression (no intercept) for the model log(1/y - 1) = - w1 x - w2 z. This will give you very reasonable estimates and, starting from these, I am almost ready to bet that you will reach THE optimum. Dec 8 '13 at 13:30
• Thanks for your suggestion, I will do experiments on it. Dec 8 '13 at 14:43
If you are finding the minimum with respect to $w$, the domain of $w$ will effect what your minimum will be. So is $w\in \mathbb{R}$ or is it that $w$ can only take a particular set of values?
• Sorry, I forgot to notice that $w\in \mathbb{R}$. Dec 5 '13 at 13:02
• In that case you can it is clear that the minimum value should be zero. Since $w=\frac{1}{x_i}\ln\left(\frac{y_i}{1-y_i}\right)$ will make all the terms inside the summation zero.
– user112535
Dec 5 '13 at 13:06
• But there is only one w with several $x_i$ and $y_i$.@LinearAlgebra Dec 5 '13 at 13:17
• Oh I am sorry. I mixed up your data points as fixed constant values. :)
– user112535
Dec 5 '13 at 13:47
Given your function $f(w)$, its derivative wrt $w$ is given by \begin{align} \frac{df}{dw}&=\sum\limits_{i=1}^n\left(\frac{1}{1+\exp(-x_iw)}-y_i\right)\cdot\frac{x_i\exp(-x_iw)}{\left(1+\exp(-x_iw)\right)^2}\\ &=\sum\limits_{i=1}^n\frac{\left(1-y_i\left(1+\exp(-x_iw)\right)\right)\cdot x_i\exp(-x_iw)}{\left(1+\exp(-x_iw)\right)^3}, \end{align} which is a quite complicated expression when attempting to find the root analytically.
A number of computational methods exist for finding the minimum of $f(w)$. The class of solvers which is applicable to this problem solve unconstrained non-linear optimisation problems, of which a simple google search reveals many possibilities.
• Thank you for your suggestion, I'm searching on Google. It' will be appreciated if you or other mathematicians show me more clues on solving this problem as unconstrained non-linear optimization. Dec 6 '13 at 4:17
• @YangzheLau. You may adress the problem is two different manners : the first one would be an unconstrained minimization problem (minimize f(w) for w); the second is much simpler : back to definition, solve f'(w) = 0 for w. The second approach is much simpler. Dec 6 '13 at 4:39
• @ClaudeLeibovici There are multiple w where f'(w) = 0. If you can find all these w, the second approach is much simpler. Dec 6 '13 at 5:13
• @YangzheLau. I think we misunderstand : there is only one value of w which makes f'(w)=0. Please see my next answer with details on a built example. Dec 6 '13 at 5:17
• @Daryl Thank you for your link. Dec 6 '13 at 6:54 | 2021-12-01T06:53:21 | {
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https://mathoverflow.net/questions/208484/packing-obtuse-vectors-in-mathbbrd/208569 | # Packing obtuse vectors in $\mathbb{R}^d$
I came across this attractive theorem:
• Theorem. In $\mathbb{R}^d$, there can be at most $d+1$ vectors that form an obtuse angle with one another.
This was proved1 as a corollary of a lemma about irreducible matrices. I am wondering if anyone knows of an alternative, more geometric proof that somehow more directly captures the sense that one cannot "pack" more than $d+1$ obtuse vectors in $\mathbb{R}^d$.
1Lipeng Ning, Tryphon T. Georgiou, Allen Tannenbaum, Stephen P. Boyd. "Linear models based on noisy data and the Frisch scheme." SIAM Review. 57(2) 2015. arXiv preprint.
• Thanks to all for the clever and educational alternative proofs. I'd like to select them all, but I must choose one, and have done so. – Joseph O'Rourke Jun 5 '15 at 19:05
You can prove it by induction.
The case $d=1$ is obvious. Assume it's true for dimension $d \geq 1$, and consider any set of $d+3$ vectors in $\mathbb{R}^{d+1}$. Fix one of these, say $\vec{v}_0$, and consider the hyperplane $H$ perpendicular to it (which we can identify with $\mathbb{R}^d$). Any one of the remaining $d+2$ vectors must lie on the far side of H (i.e., not the same side as $\vec{v}_0$). No vector other than $\vec{v}_0$ can be perpendicular to $H$, since then no third vector could form an obtuse angle with both. Thus all of the remaining $d+2$ vectors lie on the same side of $H$ and each has a nonzero projection onto $H$. By hypothesis, two of them, say $\vec{v}_1$ and $\vec{v}_2$, have projections forming a non-obtuse angle. Since they lie on the same side of $H$ and their projections onto $H$ form a non-obtuse angle, the angle between them is non-obtuse. (To see that this last statement is true, consider the dot product $\vec{v}_1 \cdot \vec{v}_2$. If you compute this using a basis consisting of $\vec{v}_0$ and vectors in $H$, then the $\vec{v}_0$ term is positive (they're on the same side of $H$), and the rest of the dot product is non-negative (their projections form a non-obtuse angle).)
I don't know if this is exactly what you're looking for, but to me this proof is pretty intuitive: you can picture your vectors all pointing more or less away from $\vec{v}_0$, and then if there are too many two of them get crammed close together. The exact value of "too many" is then easily computed by induction.
I always give this as an exercise in my linear algebra class in the form:
If $k+1$ vectors in $\mathbb{R}^n$ form obtuse angles with one another, then after removing any one vector from that collection, the remaining vectors are linearly independent.
Proof. Assume that $v_1,\ldots,v_k$ are the remaining vectors, and that they are linearly dependent. Then $c_1v_1+\cdots+c_kv_k=0$ for some $c_1,\ldots,c_k$ not all of which are equal to zero. Without the loss of generality, $c_1,\ldots,c_m$ are positive, and $c_{m+1},\ldots,c_k$ are negative (we can remove the vectors with zero coefficients in front of them). Computing the scalar product with the vector $v$ we removed, we note that we cannot have $m=k$, so $m<k$. Now, $$c_1v_1+\cdots+c_mv_m=-c_{m+1}v_{m+1}-\cdots-c_kv_k.$$ Therefore, $$(c_1v_1+\cdots+c_mv_m,c_1v_1+\cdots+c_mv_m)= (c_1v_1+\cdots+c_mv_m,-c_{m+1}v_{m+1}-\cdots-c_kv_k)<0,$$ a contradiction.
In the context of spherical codes, this is an instance of Rankin's bound. The largest minimum distance between $m$ points on the $(n-1)$-sphere is achieved by:
(1) The vertices of a $(m-1)$-simplex inscribed in a great $(m-2)$-sphere when $m\le n+1$.
(2) Any $m$ of the $2n$ vertices of a cross-polytope inscribed in the sphere, if $n+1<m\le 2n$.
Therefore, any $n+2$ points will have to have at least one pair at a right angle or smaller.
Here is the reference to the original paper of Rankin: R. A. Rankin (1955). The Closest Packing of Spherical Caps in n Dimensions. Proceedings of the Glasgow Mathematical Association, 2, pp 139-144.
The proof I know of this result is essentially the same as Vladimir Dotsenko's, but in terms of Radon's lemma.:
Assume such a set exists with at least $d+2$ points. By Radon's lemma, given any $d+2$ points in $\mathbb{R}^d$ there is a partition of them into two sets $A$, $B$ such that $conv(A) \cap conv(B) \neq \emptyset$. Let $p$ be a point of this intersection. If we write $p$ as a convex combination of $A$ and one of $B$, and using the fact that the dot product of any two points in the original set is negative, we immediately obtain $p \cdot p < 0$, a contradiction.
Without loss of generality, the vectors are in general position. Take $d$ vectors with pairwise obtuse angles and rotate them so that they form the columns of a $d\times d$ upper-triangular matrix with nonnegative diagonal (à la $QR$ factorization). Then every entry on the diagonal is strictly positive, since otherwise the vectors would be linearly dependent, thereby violating general position. Next, in order for the columns to be pairwise obtuse, every entry above the diagonal must be strictly negative. At this point, observe that every vector which is obtuse with these $d$ columns necessarily has all strictly negative entries, but two vectors of this sort necessarily have a positive inner product with each other. | 2020-08-07T19:09:35 | {
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https://math.stackexchange.com/questions/1859269/how-to-calculate-variance-of-w-find-the-probability-distribution-of-w | # How to calculate variance of W? Find the probability distribution of W?
$W=Y-X$
I have figured out that $E(W)=0.3$ by using this formula $E(X+Y)=E(X)+E(Y)$.
I tried using the same formula with $E(X^2)$ and $E(Y^2)$ to find $E(W^2)$.
I also tried using $V(X+Y)=V(X)+V(Y)+2Cov(X,Y)$, but changing all the positive signs to negative, to find the Variance of W.
Here is the joint distribution of $X$ and $Y$:
$$\begin{array}{c||c|c|c} & Y=0 & Y=1 & Y=2 \\\hline X=0 & 0.1 & 0.1 & 0.2 \\ X=1& 0.3 & 0.2 & 0.1 \end{array}$$
What I think I got right.
• $E(X) = 0.6$
• $E(Y) = 0.9$
• $E(X^2) = 0.6$
• $E(Y^2) = 1.5$
• $X$ and $Y$ are not independent
I am also definitely mixing when I can use what formula and when I can't.
• $E(W^2) = 1.3$
• $Var(W)=1.21$
• Need more information about the relationship between $X$ and $Y$. Also what is the first itemized list? $\bullet XY - 0.0 - 1.0$ etc.? – jdods Jul 14 '16 at 14:09
• It is the probability distribution table with 2 variables, I think it's called. – David Lund Jul 14 '16 at 14:11
• ah. the edit messed up your text. I'll go in and fix it with proper tabular format. – jdods Jul 14 '16 at 14:12
• Does the edit now seem to reflect your wishes? Feel free to edit it more to your liking. Hope this helps. – jdods Jul 14 '16 at 14:24
• Yes, thank you :) Perfect! I am going to have to learn do edit like that myself soon. – David Lund Jul 14 '16 at 14:25
The easiest thing to do is to first compute the probability distribution of $W$ from the joint distribution of $X$ and $Y$; then use this to compute $\operatorname{Var}[W]$ directly.
to this end, simply create a table for $W$ using the table for $X$ and $Y$:
$$\begin{array}{|c|c|c|c|} \hline x & y & w & \Pr[(X,Y) = (x,y)] \\ \hline 1 & 0 & -1 & 0.3 \\ \hline 1 & 1 & 0 & 0.2 \\ \hline 1 & 2 & 1 & 0.1 \\ \hline 0 & 0 & 0 & 0.1 \\ \hline 0 & 1 & 1 & 0.1 \\ \hline 0 & 2 & 2 & 0.2 \\ \hline \end{array}$$
Then collapse this table for distinct values of $w$, that is to say, add the rightmost column values for each row with the same value of $w$:
$$\begin{array}{|c|c|} \hline w & \Pr[W = w] \\ \hline -1 & 0.3 \\ \hline 0 & 0.2 + 0.1 = 0.3 \\ \hline 1 & 0.1 + 0.1 = 0.2 \\ \hline 2 & 0.2 \\ \hline \end{array}$$
This gives the desired probability distribution of $W$
Now the expectation and variance are trivially computed from this table: $$\operatorname{E}[W] = -1(0.3) + 0(0.3) + 1(0.2) + 2(0.2) = 0.3 \\ \operatorname{E}[W^2] = (-1)^2 (0.3) + 0^2 (0.3) + 1^2 (0.2) + 2^2 (0.2) = 1.3 \\ \operatorname{Var}[W] = \operatorname{E}[W^2] - \operatorname{E}[W]^2 = 1.21.$$
If you calculated the variance from the joint distribution of $X$ and $Y$ directly, then you'd need to go back to get the probability distribution of $W$ for the second part of your question, anyway.
• Hi, you don't think you know about any videos that can maybe explain how to make that table. I don't quite get it. My book does not explain it at all, I think. I see that you take Y-X, and you get W, literally 0-1=-1. But I don't get much more than that. – David Lund Jul 14 '16 at 17:19
• @DavidLund All I did in the first table was rearrange the table in your question. The first, second, and fourth columns in my first table are taken directly from your table. The third column is just the calculation of $w = y - x$ from the first two columns. Then, in the second table I created, I just dropped the $x$ and $y$ columns, and for the $w$ column, I added together the two rows from the first table for which $w = 0$; and I also added together the two rows from the first table for which $w = 1$. – heropup Jul 14 '16 at 17:28
• Okey, I did not realize that, but now I am so happy. Thanks. – David Lund Jul 14 '16 at 17:56
Now you calculate $\mathbb E(W^2)$
$\mathbb E(W^2)=\sum_{i=1}^2 \sum_{j=1}^3 (w_{ij})^2 \cdot p(w_{ij})$
$\mathbb E(W^2)=\sum_{i=1}^2 \sum_{j=1}^3 (x_i-y_j)^2 \cdot p(x_i,y_j)$
$=0^2\cdot 0.1+(0-1)^2\cdot 0.1+(0-2)^2\cdot 0.2+(1-0)^2\cdot 0.3+(1-1)^2\cdot 0.2+(1-2)^2\cdot 0.1=1.3$
And finally
$Var(X-Y)=Var(W)=\mathbb E(W^2)-[\mathbb E(W)]^2=1.3-(0.6-0.9)^2=1.21$
with $\mathbb E(W)=\mathbb E(X)-\mathbb E(Y)$
Remark
You can use the formula $V(X+Y)=V(X)+V(Y)+2Cov(X,Y)$ as well.
$Cov(aX,bY)=abCov(X,Y)$
In your case $a=1$ and $b=-1$. Thus $Cov(X,-Y)=-Cov(X,Y)$
And $Cov(X,Y)=\sum_{y=0}^2 \sum_{x=0}^1 p(x,y)\cdot (x-E(x))\cdot (y-E(y))$
$=0.1\cdot (-0.6)\cdot (-0.9)+(-0.1\cdot 0.6\cdot 0.1)+0.2\cdot (-0.6)\cdot 1.1+0.3\cdot 0.4\cdot (-0.9)+0.2\cdot 0.4\cdot 0.1+0.1\cdot 0.4\cdot 1.1=-0.14$
Thus $Cov(X,-Y)=0.14$
And $Var(X)=\mathbb E(X^2)-[\mathbb E(X)]^2=0.6-0.6^2=0.24$
$Var(Y)=\mathbb E(Y^2)-[\mathbb E(Y)]^2=1.5-0.9^2=0.69$
Therefore $Var(X-Y)=0.24+0.69+2\cdot 0.14=1.21$
• So I also got -0.14. So if one of them, either X or Y is negative, the Cov will become negative as well? Cov(X,−Y)=−Cov(X,Y) – David Lund Jul 14 '16 at 17:11
• Not in general. Here you have an example. X is postive and -Y is negative, but the covariance is positive. Maybe you have meant the following:".So if one of them, either $X$ or $Y$ is negative, the Cov will have the opposite sign of $Cov(X,Y)$". You statement $Cov(X,-Y)=-Cov(X,Y)$ is right. – callculus Jul 14 '16 at 17:17
• Yea, that is what I meant. I realized that I needed 0.14, but I got -0.14. So is it like that always? What if both are negative? I haven't been studying this for a long time. – David Lund Jul 14 '16 at 17:22
• @DavidLund Just use the formula $Cov(aX,bY)=abCov(X,Y)$. What do you get for $Cov(-X,-Y)$ ? – callculus Jul 14 '16 at 17:23
• That's a handy formula. + :D I am going to write that down, definitely. Thanks – David Lund Jul 14 '16 at 17:26 | 2019-07-18T19:17:07 | {
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https://grocid.net/page/4/ | # Problem of dice
The initial purpose of this post was give a proper proof of a problem posted on Twitter by @jamestanton (it is hard within the 140 char limit), but the post was later extended to cover some other problems.
Show that a four-sided and a nine-sided dice cannot be used to simulate the probability distribution of the product of outcomes when using two six-sided dice. In the original wording:
Is there a 4-sided die & a 9-sided die that together roll each of the products 1,2,3,…,30,36 w the same prob as two ordinary 6-sided dice?
We make an argument by contradiction, considering only the possible outcomes without taking the actual probabilities into account.
Obviously, to reach the same outcomes as for two normal dice $\{1,2,3,4,5,6\} \times \{1,2,3,4,5,6\}$, we need both dice to have the identity $\{1\}$ (otherwise, we will not be able to reach $1 \cdot 1 = 1$). So, $\{1,*,*,*\} \times \{1,*,*,*,*,*,*,*,*\}$.
Now, consider the prime $5$. It must be on both dice, or we would have $\mathbb{P}(5^2\cdot b)>0, b>1$. So, $\{1,5,*,*\} \times \{1,5,*,*,*,*,*,*,*\}$. Also, since $5$ appears on both dice, no dice can contain some product of the primes $\{2,3,5\}$ and their powers (e.g $2^2 \cdot 3$) that does not exist on the original dice, because then impossible products could be reached.
Hence, $6$ must be on both dice, giving $\{1,5,6,*\} \times \{1,5,6,*,*,*,*,*,*\}$. There are $6$ sides left on the larger die but we have more even products, so $2$ must also be on each die. $\{1,5,6,2\} \times \{1,5,6,2,*,*,*,*,*\}$. Now, there is no space left for $3$ on the smaller die. This means that $3^2$ must be one the larger die, but then $\mathbb{P}(3^2\cdot 5)>0$, which is a contradiction.
(@FlashDiaz gave a shorter proof)
Project Euler 205
Peter has nine four-sided (pyramidal) dice, each with faces numbered $1, 2, 3, 4$.
Colin has six six-sided (cubic) dice, each with faces numbered $1, 2, 3, 4, 5, 6$.
Peter and Colin roll their dice and compare totals: the highest total wins. The result is a draw if the totals are equal.
What is the probability that Pyramidal Pete beats Cubic Colin? Give your answer rounded to seven decimal places in the form $0.abcdefg$.
The probability functions of the nine four-sided dice and the six six-sided dice are given by the generating functions $\frac{1}{4^9} \cdot (x^1+x^2+x^3+x^4)^9$ and $\frac{1}{6^6} \cdot (y^1+y^2+y^3+y^4+y^5+y^6)^6$, respectively.
Let $X_1,...,X_9$ be i.i.d random variables taking values in the range $[1,4]$ and let $Y_1,...,Y_6$ taking values in the range $[1,6]$. We want to determine the probability that $\rho = \mathbb{P}(X_1+...+X_9 > Y_1+...+Y_6)$.
The distributions can be computed as
def rec_compute_dist(sides, nbr, side_sum):
global dist
if nbr == 1:
for i in range(1, sides+1):
dist[side_sum+i] += 1
else:
for i in range(1, sides+1):
rec_compute_dist(sides, nbr-1, side_sum+i)
dist = [0]*37
rec_compute_dist(4,9,0)
dist_49 = dist
dist = [0]*37
rec_compute_dist(6,6,0)
dist_66 = dist
To determine $\rho$, we may express it as
$\begin{array}{rl} \rho = & \sum_{t=6}^{36} \mathbb{P}(X_1+...+X_9 > t| Y_1+...+Y_6 = t)\cdot \mathbb{P}(Y_1+...+Y_6 = t) \\\\ = & \sum_{t=6}^{36} \mathbb{P}(X_1+...+X_9 > t)\cdot \mathbb{P}(Y_1+...+Y_6 = t) \end{array}$.
Computing the sum using the following code,
probability = 0
for i in range(6,36+1):
for j in range(i+1,36+1):
probability += dist_66[i]*dist_49[j]
print 1.0 * probability/(6**6 * 4**9)
we obtain the answer. Great 🙂
Project Euler 240
There are $1111$ ways in which five six-sided dice (sides numbered $1$ to $6$) can be rolled so that the top three sum to $15$. Some examples are:
$\begin{array}{rcl} D_1,D_2,D_3,D_4,D_5 &=& 4,3,6,3,5\\ D_1,D_2,D_3,D_4,D_5 &=& 4,3,3,5,6\\ D_1,D_2,D_3,D_4,D_5 &=& 3,3,3,6,6\\ D_1,D_2,D_3,D_4,D_5 &=& 6,6,3,3,3 \end{array}$
In how many ways can twenty twelve-sided dice (sides numbered $1$ to $12$) be rolled so that the top ten sum to $70$?
Let us first consider the simpler problem $\left\{ d_1+..+d_{10}=70 \right\}$. If we restrict the remaining ten dice to be less than or equal to the minimum value of the ten dice, we then can compute the cardinality. Let $n_i$ denote the number of $i$‘s we got. Then, $n_1 \cdot 1 + n_2 \cdot 2 + ... + n_{12} \cdot 12 = 70$ where $n_1 + n_2 + ... + n_{12} = 10, n_i \geq 0$.
All histograms of top-ten dice can be computed with
from copy import copy
d = [0] * 12
possible = []
def rec_compute(i, j, sum):
global d
if j == 0:
if sum == 70:
possible.append(copy(d))
return
while i > 0:
if sum + i <= 70:
d[i - 1] += 1
rec_compute(i, j - 1, sum + i)
d[i - 1] -= 1
i -= 1
rec_compute(12, 10, 0)
The code exhausts all solutions in 200ms. Call any solution $H$. For instance H = [0, 0, 0, 0, 0, 0, 10, 0, 0, 0, 0, 0]. The remaining dice can take any values in the range $[1, j]$, where $j$ is the left-most non-zero index (starting from $1$). The number of configurations for this particular solution is then given by
$20! \cdot \left((10+H_7)!H_6!H_5!H_4!H_3!H_2!H_1!\right)^{-1}$, where $\sum^7_1 H_i = 10$.
Unfortunately, there is no good analytical way of computing this. So, the easiest way is to enumerate all possible $H_i$. Disregarding $H_7$, we compute all permutations of a given histogram in the same way (hence, we can make the search space a lot smaller) and the using the multiplicity to determine the exact number. All and all, the following code gives our answer:
def configurations(i, j, x, s, l):
if sum(x) == s:
# we can stop here as the sum cannot get smaller
multiplicity = fact(l) / fact(l-len(x)) / \
reduce(lambda m, n: m * n, \
[fact(y) for y in \
Counter(x).values()])
return fact(DICE) * multiplicity / \
reduce(lambda m, n: m * n, \
[fact(y) for y in x])
if j == 0 or i == 0: return 0
return configurations(i-1, j, x, s, l) + \
configurations(i, j-1, x + [i], s, l)
S = 0
for H in possible_top_dice:
min_index = next((i for i, \
x in enumerate(H) if x), None)
for j in range(0, REMAINING+1):
u = reduce(lambda m, n: m * n, \
[fact(y) for y in H])
if j < REMAINING:
q = configurations(REMAINING-j, min_index, \
[], REMAINING-j, min_index) / u
else:
q = fact(DICE) / u
H[min_index] += 1
S += q
print S
# Breaking affine ciphers – a matrix approach
An affine cipher is a one the remnants of classical cryptography, easily broken with todays computational power. The cipher defines a symbol mapping from $f :\{A,B,\ldots,\} \mapsto \mathbb{Z}_n$. Each cipher symbol is then computed as $a \cdot x + b \rightarrow y$, where $a \in \mathbb{Z}^*_n$ and $b \in \mathbb{Z}_n$. Decryption is then done by computing $x= (y - b) \cdot a^{-1}$.
In this blog post, I will show how to break this cipher in time faster than trying all keys. Let us first sketch the general idea. Consider an expected distribution $\hat{P}$ of the symbols and a given distribution $P$, the integral $\int (\hat{P}(x) - P(x))^2 dx$ defines a statistical distance between the distributions (this would correspond to the Euclidian distance), which we would like to minimize. Now, clearly
$(\hat{P}(x) - P(x))^2 = \hat{P}(x)^2 - \hat{P}(x)P(x) + P(x)^2$.
Trivially, $\hat{P}(x)^2$ and $P(x)^2$ remains constant over any keypair $(a,b)$, so instead of minimizing the above, we can maximize $\hat{P}(x)P(x)$. Therefore, the minimization problem can be turned into a maximization problem $\max_{a,b} \int \hat{P}(x)P_{a,b}(x) dx$. Cool.
In terms of our cipher, which is discrete, the minimization problem is a sum $\max_{a,b} \sum \hat{P}(x)P_{a,b}(x)$. The observant reader may notice that this looks like a term in a matrix multiplication. There is just one caveat; the indices corresponding appear only in one term. There is an easy way to get around this. Instead of applying transformations on only $P$, we may split them among the two. So by instead computing
$\max_{a,b} \sum \hat{P}_a(x) P_{b}(x)$,
we have achieved what we desired. This means that we shuffle $\hat{P}$ with $a$ and ${P}$ with $b$. Let us interpret this as Python. The expected distribution of an alphabet ABCDEFGHIJKLMNOPQRSTUVWXYZ ,. may be as follows (depending on the observation):
P_hat = {' ': 0.05985783763561542, ',': 0.0037411148522259637, '.': 0.0028058361391694723, 'A': 0.0764122708567153, 'C': 0.02600074822297044, 'B': 0.012065095398428732, 'E': 0.11878039655817432, 'D': 0.03974934530490086, 'G': 0.018892630003741116, 'F': 0.020856715301159744, 'I': 0.0651889263000374, 'H': 0.05695847362514029, 'K': 0.00720164609053498, 'J': 0.0014029180695847362, 'M': 0.02254021698466143, 'L': 0.03769173213617658, 'O': 0.07023943135054246, 'N': 0.06313131313131314, 'Q': 0.0009352787130564909, 'P': 0.01805087916199027, 'S': 0.05920314253647587, 'R': 0.0560231949120838, 'U': 0.025813692480359144, 'T': 0.08473625140291807, 'W': 0.022072577628133184, 'V': 0.00916573138795361, 'Y': 0.01842499064721287, 'X': 0.0014029180695847362, 'Z': 0.0006546950991395436}
The transformations are done by computing the matrices
# compute first matrix for transformed P_hat
for i in range(1, N):
for element in priori_dist:
X[i, (look_up.index(element) * i) % N] = priori_dist[element]
# compute second matrix for transformed P
for j in range(N):
for element in dist:
Y[(look_up.index(element) - j) % N, j] = dist[element]
Here, the $i$th row in $X$ corresponds to $\hat{P}$ transformed by $a = i$. Moreover, the $j$th row in $Y$ corresponds ${P}$ transformed by $b = j$.
For some distribution, they may look like
As we can see, $X$ is only shifted (by the addition), while in $Y$ the indices are reordered by multiplication with row index $i$. Taking advantage of the matrix multiplication property, we may now compute $Z=XY$.
Any entry in $Z$ is $Z_{a,b} = \sum_x X_{a,x} Y_{x,b}$ so finding a maximum element in $Z$ is equivalent to saying
$\max_{a,b} \sum_x X_{a,x} Y_{x,b}$.
Looks familiar? It should. This is our maximization problem, which we stated earlier.
Therefore, we may solve the problem using
Z = numpy.dot(X, Y)
a, b = numpy.unravel_index(Z.argmax(), Z.shape)
This breaks the affine cipher.
Some notes on complexity
So, what is the complexity of the matrix approach? Computing the matrices takes $O(N^2)$ modular operations. The matrix multiplication takes naively $O(N^3)$ operations, but for large $N$ this can be achieved faster. For instance Strassen takes $O(N^{2.807})$ but faster algorithms exist. Also, taking advantage of symmetry and structure could probably decrease the complexity further. This is the total complexity of this approach.
Compare this with brute-force guessing of the key (taking $O(N^2)$ guesses) and for each guess, compute the distance taking $O(N)$ operations, which in total yields $O(N^3)$. It should be noted that complexity of this approach may be reduced by picking $a,b$ in an order which minimizes the number of tries.
Example implementation for the id0-rsa.pub: github
# Custom terminal for Vagrant/SSH
Short story: I wanted to distinguish my terminal windows between local sessions, ssh sessions and vagrant sessions.
SSH_THEME="SSH"
VAGRANT_THEME="Vagrant"
set_th () {
osascript -e "tell app \"Terminal\" to set current settings of first window to settings set \"$1\"" } set_id () { osascript -e "tell app \"Terminal\" to set current settings of first window to$1 $2$3 $4" #$@ does not work!
}
get_id () {
cur_id=$(osascript -e "tell app \"Terminal\" to get current settings of first window") } ssh(){ #!/bin/sh get_id set_th$SSH_THEME
/usr/bin/ssh "$@" set_id$cur_id
}
vagrant(){
#!/bin/sh
if [ $1 = "ssh" ]; then get_id set_th$VAGRANT_THEME
/opt/vagrant/bin/vagrant "$@" set_id$cur_id
else
/opt/vagrant/bin/vagrant "$@" fi } The code creates a temporary variable of the current theme before switching. So, when ending the session, the original theme changes back instead of a fixed one. Putting the above code in your .bash_profile: gives the following nice behavior: Color coding your sessions is a great way to visualize things and make sure you do not take any unwanted action by mistake 🙂 Of course, the code can be used to wrap any application. For instance, one could use it to make the interactive interpreter of Python/Sage or terminal sessions using torsocks appear in different colors or even fonts. # Re-mapping KBT Pure Pro in OS X For my everyday-use computer, I use a modded KBT Pure Pro; this is a small mechanical keyboard with aluminium base and background lightning, perfect for programming and typing. The size of the keyboard is 60 % of a normal one, making it suitable for spatially constrained workspaces. To my experience, it is also more ergonomic. Below is a comparison of the Pure Pro and a wireless Apple keyboard. For those being the in the process of buying a keyboard, I recommend this one 🙂 For quite a while, I have used Linux on this computer. But after installing OS X, the keyboard map went wack, so to speak. Many keys were mapped incorrectly. Using Ukulele, I created a customized layout with correct mapping (don’t mind the duplicate keys): The layout covers all keys and can be found here. NOTE: this is a layout for KBT Pure Pro with British ISO layout and not ANSI. # BackdoorCTF16 – Collision Course With 350 points and a description as follows: In today’s world, hash collisions are becoming more and more popular. That is why, one must rely on standardized hashing techniques, such as bcrypt. However, n00bster shall never learn, and he has implemented his own hash function which he proudly calls foobar. Attached is an implementation of the hash function and a file with which you are supposed to find a collision. He believes that you will not be able to find a collision for the file, especially since he hasn’t even given you the hashing algorithm, but has packaged it as a black box application. Prove to him that he is wrong. Note: Multiple collisions are possible, but only one of them is a valid flag. You will realize you’ve gotten it once you do. The hash is given as follows: So, we start off by looking at the binary. Using Hopper, we obtain the following pseudo code by decompilation: int hash(int input) { eax = _rotr(input ^ 0x24f50094, (input ^ 0x24f50094) & 0xf); eax = _rotl(eax + 0x2219ab34, eax + 0x2219ab34 & 0xf); eax = eax * 0x69a2c4fe; return eax; } int main() { esp = (esp & 0xfffffff0) - 0x20; puts(0x80486d0); gets(0x804a060); stack[2039] = "\nBar:"; puts(stack[2039]); while (stack[2039] < *(esp + 0x18)) { stack[2039] = *(stack[2039] + stack[2039] * 0x4); *(esp + 0x14) = *(esp + 0x14) ^ hash(stack[2039]); eax = _rotr(stack[2039], 0x7); printf("%08lx", stack[2039]); *(esp + 0x10) = *(esp + 0x10) + 0x1; } eax = putchar(0xa); return eax; } We sketch the above code as block scheme below: The first thing to note is that we can find an infinite number of collisions just by appending arbitrary data after 10 blocks. However, this is not interesting to us, but completely defeats the conditions for a safe cryptographic hash function. This Merkle-Damgård-like structure allows us to solve blocks iteratively, starting from the first. Here is how. Starting from the first block, we can find an input to the function $H$ such that when rotated 7 steps is equal to block 0 (here, denoted $B_0$). Hence, the problem we solve is to find an $x$ such that $H(x) \ll 7 = B_0$. This is a simple thing for Z3. Then, we take the next block and solve for $(H(x) \oplus B_0) \ll 7 = B_1$ and so forth. Implemented in Python/Z3, it may look like the following: from z3 import * import binascii, string, itertools bits = 32 mask = 2**bits - 1 allowed_chars = string.printable def convert_to_hex(s): return ''.join([hex(ord(x))[2:].zfill(2) for x in s[::-1]]) def convert_to_string(h): return ''.join([chr(int(x, 16)) for x in list(map(''.join, zip(*[iter(hex(h)[2:])]*2)))[::-1]]) def rot(val, steps): return (val << (bits-steps)) | LShR(val, steps) def hash_foobar(input): eax = rot(input ^ 0x24f50094, (input ^ 0x24f50094) & 0xf) eax = rot(eax + 0x2219ab34, bits - (eax + 0x2219ab34 & 0xf)) eax = eax * 0x69a2c4fe return eax & mask def break_iteratively(hashdata, i): if i == 0: prev_block = 0 else: prev_block = hashdata[i-1] s = Solver() j = BitVec('current_block', bits) eax = rot(prev_block ^ hash_foobar(j), 7) s.add(eax == hashdata[i]) block_preimages = [] while s.check() == sat: sol = s.model() s.add(j != sol[j].as_long()) block_string = convert_to_string(sol[j].as_long()) if all(c in allowed_chars for c in block_string): block_preimages.append(block_string) return block_preimages known = '9513aaa552e32e2cad6233c4f13a728a5c5b8fc879febfa9cb39d71cf48815e10ef77664050388a3' # this the hash of the file data = list(map(''.join, zip(*[iter(known)]*8))) hashdata = [int(x, 16) for x in data] print '[+] Hash:', ''.join(data) print '[+] Found potential hashes:\n' for x in itertools.product(*[break_iteratively(hashdata, i) for i in range(10)]): print ' * ' + ''.join(x) This code is surprisingly fast, thanks to Z3, and runs in 0.3 seconds. Taking all possible collisions into consideration… [+] Hash: 9513aaa552e32e2cad6233c4f13a728a5c5b8fc879febfa9cb39d71cf48815e10ef77664050388a3 [+] Found potential hashes: * CTFEC0nstra1nts_m4keth_fl4g} * CTFEC0nstra1nts_m4keth_nl4g} * CTFEC0nstra1nws_m4keth_fl4g} * CTFEC0nstra1nws_m4keth_nl4g} * CTFEC0nstra9nts_m4keth_fl4g} * CTFEC0nstra9nts_m4keth_nl4g} * CTFEC0nstra9nws_m4keth_fl4g} * CTFEC0nstra9nws_m4keth_nl4g} * CTF{C0nstra1nts_m4keth_fl4g} * CTF{C0nstra1nts_m4keth_nl4g} * CTF{C0nstra1nws_m4keth_fl4g} * CTF{C0nstra1nws_m4keth_nl4g} * CTF{C0nstra9nts_m4keth_fl4g} * CTF{C0nstra9nts_m4keth_nl4g} * CTF{C0nstra9nws_m4keth_fl4g} * CTF{C0nstra9nws_m4keth_nl4g} …we finally conclude that the flag is the SHA-256 of C0nstra1nts_m4keth_fl4g. # BackdoorCTF16 – Baby Worth 200 points, this challenge was presented with the following: z3r0c00l has a safe repository of files. The filename is signed using z3r0c00l’s private key (using the PKCS-1 standard). Anyone willing to read a file, has to ask for a signature from z3r0c00l. But z3r0c00l is currently unavailable. Can you still access a file named “flag” on z3rc00l’s repository? nc hack.bckdr.in 9001 Let us take a look at the public key… 3072 bits and public exponent $e = 3$. Hmm… having a small exponent is usually not a good practice. First, I tried computing the roots to $x^3 - s \bmod n$, where $s$ is the signature and $n$ is the modulus, but then I realized that this was not the way to go. What if we use non-modular squareroot, plain old Babylonian style? After looking around, I also realized that this is Bleicherbacher’s $e = 3$ attack, which I probably should have known about. There is a lot of information about this attack (therefore, I will not describe it here) and, of course, lots of people have already written code for this. Being lazy/efficient, I rewrote a functional code into the the following: from libnum import * from gmpy2 import mpz, iroot, powmod, mul, t_mod import hashlib, binascii, rsa, os def get_bit(n, b): """ Returns the b-th rightmost bit of n """ return ((1 << b) & n) >> b def set_bit(n, b, x): """ Returns n with the b-th rightmost bit set to x """ if x == 0: return ~(1 << b) & n if x == 1: return (1 << b) | n def cube_root(n): return int(iroot(mpz(n), 3)[0]) snelhest = hashlib.sha256('flag') ASN1_blob = rsa.pkcs1.HASH_ASN1['SHA-256'] suffix = b'\x00' + ASN1_blob + snelhest.digest() sig_suffix = 1 for b in range(len(suffix)*8): if get_bit(sig_suffix ** 3, b) != get_bit(s2n(suffix), b): sig_suffix = set_bit(sig_suffix, b, 1) while True: prefix = b'\x00\x01' + os.urandom(3072//8 - 2) sig_prefix = n2s(cube_root(s2n(prefix)))[:-len(suffix)] + b'\x00' * len(suffix) sig = sig_prefix[:-len(suffix)] + n2s(sig_suffix) if b'\x00' not in n2s(s2n(sig) ** 3)[:-len(suffix)]: break print hex(s2n(sig))[2:-1] Ok, so lets try it: Great! # Defcon CTF – b3s23 (partial?) The server runs a program (game of life) which has a $110 \times 110$ board with cells (bits). After a fixed number $n$ of iterations, the simulation stops and the program jumps to the first bit of the memory containing the board. We want to create an input which contains shellcode in this area after $n$ iterations. Obviously, we could choose any shellcode, and run game of life backwards. Cool, let us do that then! Uh-oh, inverting game of life is in fact a very hard problem… so it is not really feasible 😦 What to do, then? Game of life Game of life a cellular automata, found by Jon Conway, and is based on the following rules: 1. A cell is born if it has exactly 3 neighbours. Neighbors are defined as adjacent cells in vertical, horistontal and diagonal. 2. A cell dies if it has less than two or more than three neighbors. Stable code (still life) Still life consists of cell structures with repeating cycles having period 1. Here are the building blocks I used to craft the shellcode. Of course, the still life is invariant of rotation and mirroring. Shellcode So, I tried to find the shortest shellcode that would fit one line (110 bits). This one is 8 bytes. Great. 08048334 <main>: 8048334: 99 cltd 8048335: 6a 0b push$0xb
8048337: 58 pop %eax
8048338: 60 pusha
8048339: 59 pop %ecx
804833a: cd 80 int \$0x80
In binary, this translates to:
000001101001100101101010000010110101100001100000010110011100110110000000
Ok, so we note that
110101 ... 01110
cannot be constructed by our building blocks (there most certainly exist such blocks, but I didn’t consider them). So, I use a padding trick. By inserting an operation which does nothing specific
10110011 00000000 mov bl,0x0
we are able to use the blocks given in previous section. This Python code gives the binary (still-life-solvable) sequence:
from pwn import *
binary_data = ''.join([bin(ord(opcode))[2:].zfill(8) for opcode in shellcode])
context(arch = 'i386', os = 'linux')
print disasm(shellcode)
print binary_data[0:110]
which is
0000011010011001101100110000000001101010000010110101100001100000010110011100110110000000
The following cellular automata is stable, and the first line contains our exploit:
As can be seen in the animation below, we have found a still-life shellcode.
When feeding it to the program, we find that it remains in memory after any number of iterations:
Nice! Unfortunately, the code did not give me a shell, but at least the intended code was executed. I had a lot of fun progressing this far 🙂
# TU CTF – Secure Auth
This was a 150 point challenge with the description:
We have set up this fancy automatic signing server!
We also uses RSA authentication, so it’s super secure!
nc 104.196.116.248 54321
Connecting to the service, we get the following
Obviously, we cannot feed the message get_your_hands_off_my_RSA! to the oracle. So, we will only receive signatures, but no way to verify them; this means we don’t know either the public modulus, nor the public exponent. But, of course, we could guess the public exponent… there are a few standard ones: $3, 17, 65537...$
First, I obtained the signatures for $3$ and $4$ from the provided service. Denote these $s_3, s_4$, respectively. We note that given a correct public exponent $e$, we may compute $s_3^e = 3 + k \cdot N$ and $s_4^e = 4 + l \cdot N$. Inevitably, $\textnormal{gcd}(s_3^e-3,s_4^e-4) = \textnormal{gcd}(k,l)\cdot N$. Hoping for $\textnormal{gcd}(k,l)$ to be small, we can use serveral pairs until we find one that works.
Trying all the listed (guessed) public exponents, we find that $e = 65537$ (this was performed surprisingly fast in Sage with my Intel hexacore). Hence, we have now determined the modulus
$\begin{array}{rl} N = & 24690625680063774371747714092931245796723840632401231916590850908498671935961736 \\ &33219586206053668802164006738610883420227518982289859959446363584099676102569045 \\ &62633701460161141560106197695689059405723178428951147009495321340395974754631827 \\ &95837468991755433866386124620786221838783092089725622611582198259472856998222335 \\ &23640841676931602657793593386155635808207524548748082853989358074360679350816769 \\ &05321318936256004057148201070503597448648411260389296384266138763684110173009876\\ &82339192115588614533886473808385041303878518137898225847735216970008990188644891 \\ &634667174415391598670430735870182014445537116749235017327.\end{array}$
Now, note that
libnum.strings.s2n('get_your_hands_off_my_RSA!') % 3 == 0
OK, so we may split this message $m$ into a product of two message factors: $m_1 = 3$ and $m_2 = 166151459290300546021127823915547539196280244544484032717734177$ and sign them. Then, we compute the final signature $s = m^d = (m_1 \cdot m_2)^d = m_1^d \cdot m_2^d = s_1 \cdot s_2 \bmod N$. Mhm, so what now?
Phew 🙂
# TU CTF – Hash’n’bake
This challenge, worth 200 points, exhibits a trivial (and, obviously, non-secure) hash function with the objective to find a keyed hash. The description:
A random goat from Boston hashed our password!
Can you find the full output?
The hash function is defined as:
def to_bits(length, N):
return [int(i) for i in bin(N)[2:].zfill(length)]
def from_bits(N):
return int("".join(str(i) for i in N), 2)
CONST2 = to_bits(65, (2**64) + 0x1fe67c76d13735f9)
def hash_n_bake(mesg):
mesg += CONST
shift = 0
while shift < len(mesg) - 64:
if mesg[shift]:
for i in range(65):
mesg[shift + i] ^= CONST2[i]
shift += 1
return mesg[-64:]
def xor(x, y):
return [g ^ h for (g, h) in zip(x, y)]
The following computations will give the hash
PLAIN_1 = "goatscrt"
PLAIN_2 = "tu_ctf??"
def str_to_bits(s):
return [b for i in s for b in to_bits(8, ord(i))]
def bits_to_hex(b):
return hex(from_bits(b)).rstrip("L")
if __name__ == "__main__":
with open("key.txt") as f:
print PLAIN_1, "=>", bits_to_hex(hash_n_bake(xor(KEY, str_to_bits(PLAIN_1))))
print "TUCTF{" + bits_to_hex(hash_n_bake(xor(KEY, str_to_bits(PLAIN_2)))) + "}"
# Output
# goatscrt => 0xfaae6f053234c939
# TUCTF{****REDACTED****}
So, the problem is: we need to compute the hash without knowing the key (or brute forcing it). The first observation we make is that the hash function is a truncated affine function, i.e., $h(m) = f((m \cdot 2^{64} \oplus \texttt{CONST})\cdot \texttt{CONST}_2)$, with $f(a \oplus b) = f(a) \oplus f(b)$ . There is a quite simple relation emerging: $h(k \oplus m) = h(k) \oplus h(m) \oplus h(0)$ (note: $h(0)$ denotes empty input here). Using this relation, we can do the following. We know $h(k \oplus m_1)$ and $h(m_2)$ and want to determine $h(k \oplus m_2)$. Consider the following relation:
$\begin{array}{rl} h(k \oplus m_2) = & h(k) \oplus h(m_2) \oplus h(0) \\ = & h(k) \oplus h(m_1) \oplus h(0) \oplus h(m_1) \oplus h(m_2) \phantom{\bigg(} \\ = & h(k \oplus m_1) \oplus h(m_1) \oplus h(m_2). \end{array}$
All terms on the last line of the above equation are known. So, we can easily compute the hash, even without knowing the key. Ha-ha! Computing the above relation using Python can be done in the following manner:
xor(xor(to_bits(64, 0xfaae6f053234c939), hash_n_bake(str_to_bits(PLAIN_1))),hash_n_bake(str_to_bits(PLAIN_2)))
This gives the flag TUCTF{0xf38d506b748fc67}. Sweet 🙂
# TU CTF – Pet Padding Inc.
A web challenge worth 150 points, with description
We believe a rouge whale stole some data from us and hid it on this website. Can you tell us what it stole?
http://104.196.60.112/
Visiting the site, we see that there is a cookie youCantDecryptThis. Alright… lets try to fiddle with it.
We run the following command
curl -v --cookie "youCantDecryptThis=aaaa" http://104.196.60.112/
and we observe that there is an error which is not present compared to
when running it with the correct cookie is set, i.e.,
curl -v --cookie "youCantDecryptThis=0KL1bnXgmJR0tGZ/E++cSDMV1ChIlhHyVGm36/k8UV/3rmgcXq/rLA==" http://104.196.60.112/
Clearly, this is a padding error (actually, there is an explicit padding error warning but it is not shown by curl). OK, so decryption can be done by a simple padding oracle attack. This attack is rather simple to implement (basically, use the relation $P_i = D_K(C_i) \oplus C_{i-1}$ and the definition of PCKS padding, see the wikipedia page for a better explanation), but I decided to use PadBuster. The following (modified example) code finds the decryption:
class PadBuster(PaddingOracle):
def __init__(self, **kwargs):
self.session = requests.Session()
self.wait = kwargs.get('wait', 2.0)
def oracle(self, data, **kwargs):
while 1:
try:
response = self.session.get('http://104.196.60.112',
stream=False, timeout=5, verify=False)
break
except (socket.error, requests.exceptions.RequestException):
logging.exception('Retrying request in %.2f seconds...', self.wait)
time.sleep(self.wait)
continue
self.history.append(response)
return
The decrypted flag we get is TUCTF{p4dding_bec4use_5ize_m4tt3rs}! | 2021-07-24T23:58:55 | {
"domain": "grocid.net",
"url": "https://grocid.net/page/4/",
"openwebmath_score": 0.8094127774238586,
"openwebmath_perplexity": 2297.7335466269597,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9848109520836027,
"lm_q2_score": 0.8615382112085969,
"lm_q1q2_score": 0.8484522660367423
} |
https://math.stackexchange.com/questions/1829043/eigenvalues-of-periodic-lattice-laplacian/1829228 | # Eigenvalues of periodic lattice Laplacian?
Consider the graph given by taking a rectangular lattice with $m$ rows and $n$ columns and joining each vertex to its four nearest neighbors, where vertices on the boundary are connected periodically (for example, $(1,2)$ is connected to $(1,1),(1,3),(2,2)$ and $(m,2)$). Are the exact eigenvalues of its Laplacian matrix $L$ known?
I know how to handle the problem in the 1D case. Here except for the first and last rows, the matrix is tridiagonal, with its diagonal entries being $2$ and its superdiagonal and subdiagonal entries being $-1$. (Here I am using the positive semidefinite convention for the Laplacian, as usual in graph theory but reversed from the usual for PDE). Additionally there is a $-1$ in the top right and bottom left corners. This means that we have the circulant matrix corresponding to the vector $(2,-1,0,0,\dots,-1)$, where there are $n-3$ zeros. In this case as for any circulant matrix, the eigenvectors are the columns of the discrete Fourier transform and the eigenvalues can be read off by substitution.
This suggests that the columns of an appropriate 2D Fourier transform would be the eigenvectors of our matrix here...and in fact that is correct. To be specific, under the column major ordering of the vertices, the eigenvectors of the Laplacian matrix are the columns of $F_n \otimes F_m$ where $F_n$ and $F_m$ are the discrete Fourier matrices and $\otimes$ is the Kronecker product.
Is there an easy way to read off the actual eigenvalues from this representation of the eigenvectors? In other words, I can see that we have, for example
$$\lambda_i=\frac{\sum_{j=1}^{mn} L_{1j} (F_n \otimes F_m)_{ji}}{(F_n \otimes F_m)_{1i}}$$
but does this have some simple explicit formula?
• Put the 1D analog in your question to see how much you can repeat the same process. – AHusain Jun 16 '16 at 21:37
• @AHusain Done. Does it help? – Ian Jun 16 '16 at 22:18
• So you have essentially answered your own question now. You have a guess for the eigenvectors and now you just check their eigenvalues. – AHusain Jun 16 '16 at 23:01
• @AHusain You are right, I have done most of the problem. But is there any chance of a simple explicit formula for the eigenvalues? The above is a formula in terms of the entries, but I am not especially comfortable with the Kronecker product so I am not so sure if there is a simple formula for the five relevant entries of $(F_n \otimes F_m)$ (for a given $i$). – Ian Jun 16 '16 at 23:07
Yes, I believe the eigenvalues are of the form $$4 - 2 \cos \left( \frac{2 \pi j}{n} \right) - 2 \cos \left( \frac{2 \pi k}{m} \right)$$ for $1 \leq j \leq n$, and $1 \leq k \leq m$.
Let's denote your graph as $G$ and let $C_n$ be the cylic graph on $n$ vertices. Note that we can write your graph as $G = C_n \ \displaystyle \square \ C_m$ where $\displaystyle \square$ denotes the Cartesian product of graphs. In general, for graphs $G$ and $H$ with Laplacian matrices $L_G$ and $L_H$, we have that $L_{G \square H} = L_G \otimes I + I \otimes L_H$ where $\otimes$ is the Kronecker product. All eigenvectors of $L_{G \square H}$ are of the form $v_G \otimes v_H$ where $v_G$ and $v_H$ are eigenvectors of $L_G$ and $L_H$, respecitvely. Moreover, the eigenvalue associated with $v_G \otimes v_H$ is $\lambda_G + \lambda_H$ where $\lambda_G$ and $\lambda_H$ are eigenvalues of $v_G$ and $v_H$ , respectively. Thus, knowing all the eigenvalues of $L_G$ and $L_H$ will exactly give us all the eigenvalues of $L_{G \square H}$.
So to find the eigenvalues of $L_G$, we need only to find the eigenvalues of the Laplacian matrix of $C_n$. You can check that the Laplacian matrix of $C_n$ is a circulant matrix and that their eigenvalues are of a special form. In this case, using $\omega_j = \exp (\frac{2 \pi i j}{n})$, we have that the eigenvalues of $L_{C_n}$ are of the form, \begin{align} \lambda_j &= 2 - \omega_j - \omega_j^{n-1} \\ &= 2 - \exp \left(\frac{2 \pi i j}{n} \right) - \exp \left( \frac{2 \pi i j (n-1)}{n} \right) \\ &= 2 - \exp \left(\frac{2 \pi i j}{n} \right) - \exp \left( -\frac{2 \pi i j}{n} \right) \\ &= 2 - 2 \cos \left( \frac{2 \pi j}{n} \right) \end{align} Using this and the fact above, we have that the eigenvalues of $G$ are of the form $$4 - 2 \cos \left( \frac{2 \pi j}{n} \right) - 2 \cos \left( \frac{2 \pi k}{m} \right)$$ for $1 \leq j \leq n$, and $1 \leq k \leq m$, as desired. Please check my work as I could have easily made a mistake and please let me know if you'd like more explanation on any of the parts.
• Looks good. I think in your final simplification you made a simple mistake: you should have $\lambda_{j,k}(L_{G \square H})=\lambda_j(L_G)+\lambda_k(L_H)=4-\omega_n^j-\omega_n^{j(n-1)}-\omega_m^k-\omega_m^{k(m-1)}$, where $\omega_n=\exp(2 \pi i/n)$. Also these can be simplified to be expressed in terms of cosine: since $\omega_n^n=\omega_m^m=1$ you have $4-2\cos(2 \pi j/n)-2\cos(2 \pi k/m)$. – Ian Jun 17 '16 at 0:54
• Anyway, with this correction the result matches up with numerical tests including with a case of $m \neq n$. So once you make this modification I will accept. Thanks for your help. Also thanks for giving your help in a "generalizable" form (i.e. it is clear how to extend this to more than two dimensions). – Ian Jun 17 '16 at 0:59
• Sure thing - glad I could help! It looks like we both had the same answer for the eigenvalues but different definitions for $\omega_k$, which led to the confusion. I added a bit of working out the eigenvalue at the end, along with your nice simplification! – Chris Harshaw Jun 17 '16 at 1:16
• You still haven't adjusted the denominators to be $m$, which you do need to do in some places. (That is why I changed the notation, I needed the denominator and the exponent to both be variables.) – Ian Jun 17 '16 at 1:17 | 2021-03-01T02:02:54 | {
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https://www.physicsforums.com/threads/how-do-you-find-the-equation-of-a-cubic-function-given-5-points-no-zeros.565992/ | # Homework Help: How do you find the equation of a Cubic function given 5 points? (no zeros)?
1. Jan 8, 2012
### srizen
1. The problem statement, all variables and given/known data
what the questions asks is that i need to find the equation of a polynomial with these given points:
1,1
2,-3
3,5
4,37
5,105
i know that one way to solve is by creating 5 equations then solve for ax^3+bx^2+cx+d using the elimination/substitution method.
however is there another, much easier way of doing this question?
2. Relevant equations
ax^3+bx^2+cx+d
3. The attempt at a solution
1= a+b+c+d
-3=8a+4b+2c+d
5= 27a+9b+3c+d
37=64a+16b+4c+d
108=125a+25b+5c+d
fixed, yes, my mistake
Last edited: Jan 8, 2012
2. Jan 8, 2012
### eumyang
Three of the equations are wrong. They should be
1= a+b+c+d
-3=8a+4b+2c+d
5= 27a+9b+3c+d
37=64a+16b+4c+d
108=125a+25b+5c+d
Also, you don't need the last equation, because there are 4 unknowns.
As for other methods, there's the finite difference method, but I don't think it will help for this particular problem (because you already told us that this is a cubic).
Last edited: Jan 8, 2012
3. Jan 8, 2012
### srizen
fixed, i typed the equations too fast
4. Jan 8, 2012
### SammyS
Staff Emeritus
True, but I got the coefficients fairly quickly using a difference method.
Actually, after playing around a bit with this, I got the result with two different difference methods.
5. Jan 8, 2012
### srizen
what exactly is the difference method?
i ask because i've tried this question, and i kept getting it wrong.
6. Jan 8, 2012
7. Jan 8, 2012
### SammyS
Staff Emeritus
Make a table of differences
$$\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline \quad i \quad & \quad x_i \quad & \quad f(x_i) \quad &\quad (\Delta^1)_i \quad &\quad (\Delta^2)_i \quad & \quad (\Delta^3)_i \quad & \quad (\Delta^4)_i \quad \\ \hline & & & & & & \\ 1 & 1 & 1 & -4 & 12 & 12 & 0 \\ & & & & & & \\ 2 & 2 & -3 & 8 & 24 & & -- \\ & & & & & & \\ 3 & 3 & 5 & & & -- & -- \\ & & & & & & \\ 4 & 4 & 37 & & -- & -- & -- \\ & & & & & & \\ 5 &5 & 105 & -- & -- & -- & -- \\ & & & & & & \\ \hline \end{array}$$
Where: $(\Delta^1)_i=f(x_{i+1})-f(x_{i})\,,$
$(\Delta^2)_i=(\Delta^1)_{i+1}-(\Delta^1)_i$
etc.
See if you can fill in the rest.
If f(x) is truly a cubic function then the Δ3 column will all be the same.
Fill out a similar Table for g(x) = x3 . The Δ3 column will all be 6's.
What do you suppose that means about the x3 coefficient of f(x) ?
8. Jan 9, 2012
### srizen
OMG! i love you forever, i had no idea this method existed, i already solved through almost an hour of writing matrices, with this i solved it in 3 minutes. thank you!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook | 2018-06-24T22:09:23 | {
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https://mathematica.stackexchange.com/questions/262192/is-it-possible-to-find-a-non-zero-solution-of-an-ode | # Is it possible to find a non-zero solution of an ODE?
Is it possible to solve the 4th-order ode analytically?
op[r_] = (D[#, {r, 2}] - 1/r*D[#, r]) &;
DSolve[{op[r][op[r][f[r]]] == 0, f[b] == 0, f'[b] == 0, (f''[r] - 1/r*f'[r] - f[r]/(c - 1) /. r -> (1 + b)) == 0, (f'''[r] - 1/r*f''[r] + 1/r^2*f'[r] /. r -> (1 + b)) == 0}, f[r], r]
I am using v11 in which DSolve returns a trivial solution f[r] -> 0 only.
Any suggestion is welcome. Thank you!
For general values of $$b$$ and $$c$$, the trivial solution is indeed the only solution. However, there are special values of $$b$$ and $$c$$ which can yield non-trivial solutions.
To find these, we can start by telling Mathematica to find the general solution to the ODE:
soln = DSolve[{op[r][op[r][f[r]]] == 0}, f, r]
(* {f -> Function[{r}, (r^2 C[1])/2 - (r^2 C[2])/4 + (r^4 C[3])/4 + C[4] + 1/2 r^2 C[2] Log[r]]} *)
We can then look at the boundary conditions and see what they imply about the coefficients C[i]. We do this by storing the vanishing quantities in a list BCs; the actual equations are then BCs == 0.
BCs = { f[b],
f'[b],
(f''[r] - 1/r*f'[r] - f[r]/(c - 1) /. r -> (1 + b)),
(f'''[r] - 1/r*f''[r] + 1/r^2*f'[r] /. r -> (1 + b))};
quants = BCs /. First[soln]
The quantities in quants are four linear combinations of the coefficients C[1], C[2], C[3], and C[4], all of which must vanish. We can think of this set of simultaneous equations as the result of a matrix $$M$$ multiplying the vector $$\vec{v} = \{C_1, C_2, C_3, C_4\}$$. From basic results in linear algebra, we know that the only way for there to be a non-trivial solution for $$\vec{v}$$ is for the matrix $$M$$ to have a non-zero determinant. So we construct this matrix and take its determinant:
mat = Outer[Coefficient, quants, {C[1], C[2], C[3], C[4]}];
Simplify[Det[mat]]
(* -((b (1 + b) (-3 + 2 b + 4 c + 2 (1 + b)^2 Log[b] - 2 (1 + b)^2 Log[1 + b]))/(-1 + c)) *)
This implies that the trivial solution is the only solution to the ODE unless this last quantity (in terms of the constants $$b$$ and $$c$$) is zero, which occurs when $$b = 0$$, $$b = -1$$, or $$c = \frac{3 - 2b + 2 (1+b)^2 \ln( (1+b)/b )}{4}.$$
• thank you very much, but the question is actually not solved completely... I tried csoln = Solve[Det[mat] == 0, c] // Simplify and substituted the special c in the original system but the solution still includes one unknown constant. Please see DSolve[{op[r][op[r][f[r]]] == 0, f[b] == 0, f'[b] == 0, (f''[r] - 1/r*f'[r] - f[r]/(c - 1) /. {r -> (1 + b), csoln[[1, 1]]}) == 0, (f'''[r] - 1/r*f''[r] + 1/r^2*f'[r] /. r -> (1 + b)) == 0}, f[r], r], it also gives a warning _Unable to resolve some of the arbitrary constants _ Jan 20 at 9:20
• Yes, there will be unknown constants in the solution. They should correspond to the overall normalization of the solution; if $f(x)$ is a solution then so is $\alpha f(x)$ for any $\alpha$, because the equation is linear. Jan 20 at 12:13
• Prof. Seifert thanks a lot! Your comment is plausible. But with Det[mat]==0 we should have non-zero solution for C[i], however, when substituting the eigenvalue c back to mat, which represents vanishing quantities for the b.c.s, I got zero solution... Please try csoln = Solve[Det[mat] == 0, c] and LinearSolve[mat/.First[csoln], {0, 0, 0, 0}] Jan 21 at 2:41
A somewhat different way is as follows.
s = DSolve[{op[r][op[r][f[r]]] == 0, f[b] == 0, f'[b] == 0}, f, r]
{{f -> Function[{r}, (1/( 4 b^2))(-b^4 C[1] + 2 b^2 r^2 C[1] - r^4 C[1] + b^4 C[2] - b^2 r^2 C[2] - b^4 C[2] Log[b] - r^4 C[2] Log[b] + 2 b^2 r^2 C[2] Log[r])]}}
Resolve[Exists[{C[1], C[2]},Simplify[f'[1 + b]/(1 + b)^2 - f''[1 + b]/(1 + b) + f'''[1 + b] /.
s[[1]]]==0 && Simplify[-(f[1 + b]/(-1 + c)) - f'[1 + b]/(1 + b) + f''[1 + b] /.
s[[1]]]==0&& C[1]^2 + C[2]^2 != 0], Reals]
b > 0 && c == 1/4 (3 - 2 b - 2 Log[b] - 4 b Log[b] - 2 b^2 Log[b] + 2 Log[1 + b] + 4 b Log[1 + b] + 2 b^2 Log[1 + b])
The main difference from Michael Seifert's answer consists in the use of quantifiers instead of linear algebra. This way is more automatical.
• One could also start from DSolve[{op[r][op[r][f[r]]] == 0, f, r]. Jan 16 at 17:12 | 2022-05-23T14:35:46 | {
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http://www.rhondagilliam.com/bud747/page.php?9599e6=adding-and-subtracting-radicals-quizlet | Each positive number has two square roots. Adding and Subtracting Square Roots. Example 1: Add or subtract to simplify radical expression: $2 \sqrt{12} + \sqrt{27}$ Solution: Step 1: Simplify radicals Combining functions. Adding and Subtracting Like Radicals Simplify each expression. Simplify radicals. Improve your math knowledge with free questions in "Add and subtract radical expressions" and thousands of other math skills. Learn vocabulary, terms, and more with flashcards, games, and other study tools. Adding, Subtracting & Multiplying Radicals DRAFT. Break down the given radicals and simplify each term. Add or subtract the like radicals by adding or subtracting their coefficients. Simplify each radical completely before combining like terms. This calculator performs addition and subtraction of algebraic fractions. The steps in adding and subtracting Radical are: Step 1. Adding and Subtracting Radicals – Practice Problems Move your mouse over the "Answer" to reveal the answer or click on the "Complete Solution" link to reveal all of the steps required for adding and subtracting radicals. This gives mea total of five copies: That middle step, with the parentheses, shows the reasoning that justifies the final answer. Email. Learn vocabulary, terms, and more with flashcards, games, and other study tools. The number i is defined as the square root of -1, so that i squared equals -1. It includes four examples. Start studying Simplifying/Adding and Subtracting Radicals V2. Next What Are Radicals. Radical expressions may be combined by using addition or subtraction only if they are SIMILAR, that is, if they have the same radicand with the same index. 57% average accuracy. 900 seconds . If the indices or radicands are not the same, then you can not add or subtract the radicals. This video looks at adding and subtracting radical expressions (square roots). Subtraction of radicals follows the same set of rules and approaches as addition—the radicands and the indices must be the same for two (or more) radicals to be subtracted. Examples: 1. This game goes along with the game in the last section. You probably won't ever need to "show" this step, but it's what should be going through your mind. The amount of bacteria in an infection n hours after taking medication is (1/x 3) n. Write a simplified expression that represents the amount of bacteria in an infection 4 hours after taking medication. Which of the following is a like radical to 3x 5? Division of Radical Expressions with binomial divisor. SURVEY . Break down the given radicals and simplify each term. Identify the like radicals. 66 times. Q. Intro to combining functions. Any number that can be written in the form, Addition and subtraction of complex numbers. 3. In the three examples that follow, subtraction has been rewritten as addition of the opposite. Edit. If the index and radicand are exactly the same, then the radicals are similar and can be combined. How much longer is the side of a cube with a surface area of 180 square meters than a cube with the surface area of 120 square meters? Tags: Question 13 . This practice includes adding and subtracting radicals that have the same and different numbers under the radical. 2. This involves adding or subtracting only the coefficients; the radical part remains the same. Adding and subtracting rational expressions (not factored) Subtracting rational expressions. salexander10. • Sometimes it is necessary to simplify radicals first to find out if they can be added or subtracted. Multiplying functions. Start studying Adding and Subtracting Radicals. -3√75 - √27. Cube roots, fourth roots, and so on are denoted by using an index and a radical. If the divisor ( the denominator and write the result in standard form not have like radicals adding! 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http://math.stackexchange.com/questions/173099/find-all-solutions-other-than-2-for-12x3-23x2-3x2-0 | # Find all solutions, other than $2$ for $12x^3-23x^2-3x+2=0$
Find all solutions, other than $2$ for $12x^3-23x^2-3x+2=0$
I started off by taking out an $x$ and got $$x(12x^2-23x-3)+2=0$$
I do not know if this is the correct first step, if it is, then am I able to use the quadratic formula or complete the square to get the answers. Can anyone give me general hints. Please do not solve this for me in anyway. Just give me hints.
-
That first step doesn't work very well. It's good if you can find something to factor out of the entire expression so that you end up with something like $AB=0$, because then you can break up the problem into "$A=0$ or $B=0$". But changing it to $AB+C=0$ does not usually simplify the problem. – Arturo Magidin Jul 20 '12 at 3:39
The step is technically not incorrect. But it is not at all useful. – André Nicolas Jul 20 '12 at 5:59
Since $12x^2-23x-3$ at $x=2$ is $-1$, then $x=2$ is a solution of the polynomial. Then divide the original polynomial by $x-2$ to get a quadratic which can be easily solved for the other roots 1/4 and -1/3. – i. m. soloveichik Jul 22 '12 at 14:33
HINT
The great bit here is to note that $2$ is a solution. By the factor theorem, we know that the linear factor $(x - 2)$ divides our polynomial.
So perhaps you should divide out $(x-2)$. You'll be left with a quadratic, which we know how to solve very quickly.
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So, I synthetically divide and then use the quadratic formula for the rest? – Austin Broussard Jul 20 '12 at 3:39
That sounds like a great strategy to me. Good luck! – mixedmath Jul 20 '12 at 3:40
I'll recomment with my answer and hopefully it'll be right! Thanks for the help! – Austin Broussard Jul 20 '12 at 3:40
@Austin: You don't need to ask. You could just plug in those values and check whether they satisfy your equation. – Javier Jul 20 '12 at 3:49
@Austin: Hey that's great! Although, unless my math is off, I happen to get $1/4$ and $-1/3$, as if one of us dropped a negative somewhere. – mixedmath Jul 20 '12 at 3:55
The preceding answers have provided you with sufficient information. However the following might help you with future endeavors:
How would you factor the polynomial if 2 wasn't given as a root?
There exists a useful theorem: the Rational Root Theorem, that helps factor polynomials (indirectly). It gurantees that any root of the polynomial will have a numerator that has $c$, the constant, as a factor and a denominator that has $a_n$, the leading coefficient, as a factor.
Therefore by investigating all possible conbinations of the factors of the constant and the factors of the leading coefficient you can eventually arrive at a valid root. The polynomial can then be further factored into a quadratic through synthetic division which can be factored (as you know) through the quadratic formula.
The interesting thing about this method is its large scope of applicability: it not only works on cubic polynomials but on a polynomial with any degree.
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The reciprocals of the roots $\rm\,r,s,1/2\:$ are roots of the reversed polynomial $\rm\:2\, x^3\! -\! 3\, x^2\! -\! 23\, x\! +\! 12.\:$ Thus by Vieta's Formulas $\rm\:r\!+\!s\!+\!1/2 = 3/2,\ rs/2 = -6,\:$ so $\rm\:r\!+\!s = 1,\ rs = -12,\:$ so $\rm\:r,s = \ldots$
Alternatively, by the Factor Theorem, $\rm\:f(2) = 0\:$ $\Rightarrow$ $\rm\:f(x)\:$ has $\rm\:x\!-\!2\:$ as a factor. Comparing coef's
$$\rm (x-2)(a\, x^2 + b\, x + c)\ =\ 12\, x^3 - 23\,x^2 -3\,x + 2$$
$\rm\qquad x^3\:$ coef $\rm\:\Rightarrow\: a = 12$
$\rm\qquad x^0\:$ coef $\rm\:\Rightarrow\: -2\,c = 2\:\Rightarrow\: c = -1$
$\rm\qquad x^1\:$ coef $\rm\:\Rightarrow\: -3 = c-2b = -1-2b\:\Rightarrow\: b = 1$
So the quadratic factor is $\rm\: 12\,x^2 + x - 1,\:$ which can be solved by either the Quadratic Formula or Rational Root Test, or $\rm\:(c\,x\!-\!1)\,(d\,x\!+\!1) = 12\,x^2\!+\!x\!-\!1\:$ $\Rightarrow$ $\rm\:cd=12,\ c\!-\!d = 1,\:$ so $\rm\:c,d = \ldots$
-
$c=4$ and $d=3$? – Austin Broussard Jul 20 '12 at 4:07
@Austin Indeed, they satisfy the equations. – Bill Dubuque Jul 20 '12 at 4:12 | 2015-07-29T14:21:17 | {
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https://math.stackexchange.com/questions/2617592/will-this-series-with-radical-converge | # Will this series with radical converge?
I'm trying to test the following series for convergence: $$\sum_{n=2}^{\infty}\frac{n}{\sqrt{n^5-n^3}}$$
I've progressed through several tests but am having trouble developing an intuition of how to approach a problem like this. I've checked the following cases so far:
• Divergence Test ($\lim a_n = 0$, so not helpful)
• Geometric Series Test (I can't find a straightforward way to find a common ratio)
• p-Series Test (it does not appear to be a p-Series)
• Limit Comparison and Comparison Tests (I can't find another series with which to prove convergence or divergence)
• Integral Test (I'm unable to integrate the expression)
Obviously I'm missing something here, but I'm just not sure which it is.
• Terms are roughly on the order of $\frac{n}{n^{5/2}} = \frac{1}{n^{3/2}}$ which should converge. How you get there is combination of comparison and $p$-series tests. Jan 23, 2018 at 14:28
• OK. It appears that I can just use the comparison test with something like $\frac{1}{x^\left(1.1\right)}$. Thank you! Jan 23, 2018 at 14:31
• Oftentimes, limit comparison is easier to use than comparison because you don't have to care if the "right" series dominates. Jan 23, 2018 at 14:33
• You should try to work it all out and answer your own question here. Would be good for you. Jan 23, 2018 at 14:33
• @Randall The trouble I have with limit comparison is that I have a hard time ending up with a constant. Seems I'm always getting infinity or infinity over infinity. Adding in the radical in the denominator makes it even easier to get lost down dead-ends. Jan 23, 2018 at 14:46
First of all, you seem to have misused some of the tests. For example, the divergence test only says that the series $\sum a_n$ diverges if $\lim a_n\neq 0$, which is not true here since $$\lim_{n\to\infty}\frac{n}{\sqrt{n^5-n^3}} = 0.$$
Second of all, you can make a comparison test:
$$\frac{n}{\sqrt{n^5-n^3}} = \frac{n}{\sqrt{n^5}\sqrt{1 - \frac{1}{n^2}}}=\frac{1}{\sqrt{n^3}}\cdot \frac{1}{\sqrt{1-\frac{1}{n^2}}}\leq 2\cdot \frac{1}{\sqrt{n^3}}$$
and of course, $$\sum_{n=1}^\infty \frac{2}{\sqrt{n^3}}$$ converges as it is two times a $p$-series.
• Thanks for the explanation and for pointing out my error with the divergence test. I'd just been checking if the sequence increases, but now I I see what I was doing wrong there. As for implementing the comparison test, I think I just need more practice so that I can "see" what you're seeing without having to try a dozen other strategies first. I appreciate your help. Jan 23, 2018 at 14:52
• @AlexJohnson In general, you just need to find the biggest factor of the numerator and denominator. In this case, the numerator is $n$, and the denominator, for large $n$, is dominated by $\sqrt{n^5}$. Since $\frac{5}{2} - 1> 1$, you have convergence.
– 5xum
Jan 23, 2018 at 14:54
You forgot the equivalence test: $$\sqrt{n^5-n^3\strut}\sim_\infty n^{5/2},\enspace\text{hence }\;\frac n{\sqrt{n^5-n^3\strut}}\sim_\infty \frac n{n^{5/2}}=\frac 1{n^{3/2}},$$ which is a convergent Riemann series.
Let $a_n$ be the $n$th term of your series and let $b_n= \frac{1}{n^{3/2}}$. Why is this my choice of $b_n$? See my first comment under your question.
Now, limit compare by $\lim_{n \to \infty} \frac{b_n}{a_n}$. Do the algebra and eventually get $$\frac{b_n}{a_n}= \frac{\sqrt{n^5-n^3}}{n^{5/2}} = \frac{\sqrt{n^5-n^3}}{\sqrt{n^5}} = \sqrt{\frac{n^5-n^3}{n^5}} = \sqrt{1-\frac{1}{n^2}}.$$ Now take the limit, and notice that it's easily $1$. Limit comparison now tells you that your series converges because $\sum_n \frac{1}{n^{3/2}}$ is a convergent $p$-series.
• I fudged a step or two but was able to get to the same place as you. This was all a tremendous help. Thank you again. Jan 23, 2018 at 15:00
• Excellent. You're well on your way. Jan 23, 2018 at 15:00
Simply note that
$$\frac{n}{\sqrt{n^5-n^3}}\sim \frac{1}{n^\frac32}$$
thus $\sum_{n=2}^{\infty}\frac{n}{\sqrt{n^5-n^3}}$ converges by comparison with
$$\sum_{n=2}^{\infty}\frac{1}{n^\frac32}$$ | 2022-05-28T19:42:25 | {
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https://stacks.math.columbia.edu/tag/07NA | Lemma 10.115.7. Let $R \to S$ be an injective finite type ring map. Assume $R$ is a domain. Then there exists an integer $d$ and a factorization
$R \to R[y_1, \ldots , y_ d] \to S' \to S$
by injective maps such that $S'$ is finite over $R[y_1, \ldots , y_ d]$ and such that $S'_ f \cong S_ f$ for some nonzero $f \in R$.
Proof. Pick $x_1, \ldots , x_ n \in S$ which generate $S$ over $R$. Let $K$ be the fraction field of $R$ and $S_ K = S \otimes _ R K$. By Lemma 10.115.4 we can find $y_1, \ldots , y_ d \in S$ such that $K[y_1, \ldots , y_ d] \to S_ K$ is a finite injective map. Note that $y_ i \in S$ because we may pick the $y_ j$ in the $\mathbf{Z}$-algebra generated by $x_1, \ldots , x_ n$. As a finite ring map is integral (see Lemma 10.36.3) we can find monic $P_ i \in K[y_1, \ldots , y_ d][T]$ such that $P_ i(x_ i) = 0$ in $S_ K$. Let $f \in R$ be a nonzero element such that $fP_ i \in R[y_1, \ldots , y_ d][T]$ for all $i$. Then $fP_ i(x_ i)$ maps to zero in $S_ K$. Hence after replacing $f$ by another nonzero element of $R$ we may also assume $fP_ i(x_ i)$ is zero in $S$. Set $x_ i' = fx_ i$ and let $S' \subset S$ be the $R$-subalgebra generated by $y_1, \ldots , y_ d$ and $x'_1, \ldots , x'_ n$. Note that $x'_ i$ is integral over $R[y_1, \ldots , y_ d]$ as we have $Q_ i(x_ i') = 0$ where $Q_ i = f^{\deg _ T(P_ i)}P_ i(T/f)$ which is a monic polynomial in $T$ with coefficients in $R[y_1, \ldots , y_ d]$ by our choice of $f$. Hence $R[y_1, \ldots , y_ d] \subset S'$ is finite by Lemma 10.36.5. Since $S' \subset S$ we have $S'_ f \subset S_ f$ (localization is exact). On the other hand, the elements $x_ i = x'_ i/f$ in $S'_ f$ generate $S_ f$ over $R_ f$ and hence $S'_ f \to S_ f$ is surjective. Whence $S'_ f \cong S_ f$ and we win. $\square$
Comment #4307 by Rankeya on
Does the proof use that S is a domain anywhere?
Comment #4468 by on
It looks like it doesn't. If you need this for something, then I'll change it.
Comment #4481 by Rankeya on
I had occassion to use this lemma when $S$ is not a domain, so I would appreciate if you can change this. It allows me to refer to this lemma without having to say "the proof does not need $S$ to be a domain." Thanks!
Comment #4641 by Andy on
$R[y_1,\ldots,y_n]$ on the $4$-th line from the bottom should be a $R[y_1,\ldots,y_d]$
There are also:
• 2 comment(s) on Section 10.115: Noether normalization
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar). | 2022-01-20T14:57:58 | {
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https://mathoverflow.net/questions/7881/sum-of-n-vectors-in-mathbb-z-nk/7890 | # Sum of $n$ vectors in $(\mathbb Z/n)^k$
Let $n,k$ be positive integers. What is the smallest value of $N$ such that for any $N$ vectors (may be repeated) in $(\mathbb Z/(n))^k$, one can pick $n$ vectors whose sum is $0$?
My guess is $N=2^k(n-1)+1$. It is certainly sharp: one can pick our set to be $n-1$ copies of the set $(a_1,...,a_k)$, with each $a_i=0$ or $1$. The case $k=1$ is some math competition question (I think, but can't remember the exact reference). Does anyone know of some references? Thanks.
Thank you all! I wish I could accept all the answers, they are very helpful!
• In the title of the question, don't you mean the sum of $N$ (not $n$) vectors? Dec 5, 2009 at 18:18
• @José: I think the lowercase n is correct. It's slightly long, but a good title would be, "Minimum N st any size N set in (Z/n)^k has a size n subset summing to 0." Dec 5, 2009 at 18:29
• Indeed, the same example has no size n-1 subsets summing to 0 vector, which explains why the problem is about size n subsets. Dec 5, 2009 at 23:54
• I got a bit confused by the notations $\mathbb Z/n$ and $\mathbb Z/(n)$. Perhaps better: Given two strictly positive integers $n,k$, find the smallest integer $N\geq n$ such that every set of $N$ elements in $\mathbb Z^k$ contains a subset of $n$ elements with sum in $n\mathbb Z^k$. Apr 25 at 16:27
Your guess is correct for k=1 and 2, but when k is bigger, things get more complicated. For instance, when k=n=3, N=19. For a summary of some known results, see:
Elsholtz, C. Lower Bounds For Multidimensional Zero Sums. Combinatorica 24, 351–358 (2004).
The case k = 1 is the Erdős-Ginzburg-Ziv Theorem. Take a look at this Wikipedia article which has links to some surveys of the large literature of similar results. (The particular generalizations I'm aware of ask for a set of vectors summing to 0 whose size is the cardinality of the group, not its exponent.)
The case k = 2 is the Kemnitz conjecture, as pointed out by Kristal, and as Ricky mentioned (and also as mentioned on the Wikipedia page I linked to) it was proved by Reiher in 2003.
There was some discussion of the case n=3 in sci.math around 1994. There is a card game called Set with an 81 card deck so that each card is naturally a point in $$(\mathbb Z/3)^4$$. Several cards are dealt out, and your task is to identify triples of cards called Sets which form a line, or equivalently, which add up to the 0 vector. A natural question is how many points you can deal out without the existence of a line. It's not too hard to construct 9 distinct points in affine 3-space, or 20 distinct points in affine 4-space over $$\mathbb Z/3$$ so that there is no line contained in the points, and these are the maximums. These correspond to $$N=19$$ for $$(n,k) = (3,3)$$ and $$N=41$$ for $$(n,k) = (3,4)$$, as in the reference Ricky Liu linked, by repeating each point twice.
The maximal configurations are highly symmetric. The 9 points in dimension 3 correspond to a nondegenerate conic, which is unique up to symmetry. The 20 points in dimension 4 actually correspond to a nondegenerate conic containing 10 points in projective 3-space viewed as lines passing through the origin in affine 4-space.
For example, there are 9 points in dimension 3 satisfying $$z=x^2 + y^2:$$ $$\{(0,0,0),(\pm1,0,1),(0,\pm1,1),(\pm1,\pm1,-1)\}$$ and this set contains no lines.
While I dont know the answer to your specific question it seems related to a well known problem where you only insist that the number of simmands is non zero. Much is known about it and there are interesting open problems. When n is a prime you need (n-1)k+1 vectors (and this is sharp). This is "Olson's theorem" and it can be proved by Chevaley's theorems on non zero solutions for polynomial equations when the number of variables exceeds the degree. Maybe similar techniques (at least for the n=prime case) will work for your problem.)
The reference for Olson is: J. E. Olson, A combinatorial problem on nite abelian groups I, J. Number Theory 1 (1969), 8-10.
For the non prime power look at the paper by R. Meshulam: An uncertainty inequality and zero subsums. Discrete Math. 84 (1990), no. 2, 197--200.
For a general relevant method: N. Alon, Combinatorial Nullstellensatz. Combin. Probab. Comput. 8 (1999), no. 1-2, 7--29.
The case $$k=2$$ is handled (extremal sequences identified) in this paper of Gao and Goldinger, which was published in the journal Integers.
For $$k$$ greater than one the smallest value will be equal or less than $$n^{k-1}(2n-2)+1$$. To see this use the pigeonhole principle for the first $$k-1$$ coordinates there are only $$n^{k-1}$$ sets of possible values so one set of these coordinates must have $$2n-1$$ elements we can choose $$n$$ of these that have coordinate $$k$$ sum to zero by the Erdős–Ginzburg–Ziv theorem then since the first $$k-1$$ coordinates are fixed they will sum to zero as well and we have the desired set of vectors summing to zero.
For $$k=2$$ there is the Kemnitz conjecture that it is $$4n-3$$.
I now see this conjecture is proved. See:
Reiher, C. On Kemnitz’ conjecture concerning lattice-points in the plane. Ramanujan J 13, 333–337 (2007).
We can then apply the argument of the first paragraph and get for $$k=2$$ or more the smallest value must be less than or equal to $$n^{k-2}(4n-4)+1$$.
For the pattern to continue the next case would be for $$k=3$$ it would be $$8m-7$$ there is a counterexample in fact for all odd $$k$$ and $$n$$ greater than $$3$$ it is not true. The following paper was mentioned in another answer; the last sentence of the abstract has the general result.
Elsholtz, C. Lower Bounds For Multidimensional Zero Sums. Combinatorica 24, 351–358 (2004).
There is a factor of $$1.125$$ to the $$d/3$$ power so there is an exponent greater than two as a lower bound. | 2022-08-19T14:29:51 | {
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https://math10blog.wordpress.com/tag/wiki/ | You are currently browsing the tag archive for the ‘Wiki’ tag.
Wikipedia calls this a classic paradox of elementary probability theory.
Taken from Wikipedia.org:
## Card version
Suppose you have three cards:
• black card that is black on both sides,
• white card that is white on both sides, and
• mixed card that is black on one side and white on the other.
You put all of the cards in a hat, pull one out at random, and place it on a table. The side facing up is black. What are the odds that the other side is also black?
The answer is that the other side is black with probability 2/3. However, common intuition suggests a probability of 1/2 because across all the cards, there are 3 white, 3 black. However, many people forget to eliminate the possibility of the “white card” in this situation (i.e. the card they flipped CANNOT be the “white card” because a black side was turned over).
In a survey of 53 Psychology freshmen taking an introductory probability course, 35 incorrectly responded 1/2; only 3 students correctly responded 2/3.[1]
### Preliminaries
To solve the problem, either formally or informally, we must assign probabilities to the events of drawing each of the six faces of the three cards. These probabilities could conceivably be very different; perhaps the white card is larger than the black card, or the black side of the mixed card is heavier than the white side. The statement of the question does not explicitly address these concerns. The only constraints implied by the Kolmogorov axioms are that the probabilities are all non-negative, and they sum to 1.
The custom in problems when one literally pulls objects from a hat is to assume that all the drawing probabilities are equal. This forces the probability of drawing each side to be 1/6, and so the probability of drawing a given card is 1/3. In particular, the probability of drawing the double-white card is 1/3, and the probability of drawing a different card is 2/3.
In our question, however, you have already selected a card from the hat and it shows a black face. At first glance it appears that there is a 50/50 chance (ie. probability 1/2) that the other side of the card is black, since there are two cards it might be: the black and the mixed. However, this reasoning fails to exploit all of your information; you know not only that the card on the table has a black face, but also that one of its black faces is facing you.
### Solutions:
#### Intuition
Intuition tells you that you are choosing a card at random. However, you are actually choosing a face at random. There are 6 faces, of which 3 faces are white and 3 faces are black. Two of the 3 black faces belong to the same card. The chance of choosing one of those 2 faces is 2/3. Therefore, the chance of flipping the card over and finding another black face is also 2/3. Another way of thinking about it is that the problem is not about the chance that the other side is black, it’s about the chance that you drew the all black card. If you drew a black face, then it’s twice as likely that that face belongs to the black card than the mixed card.
#### Labels
One solution method is to label the card faces, for example numbers 1 through 6.[2] Label the faces of the black card 1 and 2; label the faces of the mixed card 3 (black) and 4 (white); and label the faces of the white card 5 and 6. The observed black face could be 1, 2, or 3, all equally likely; if it is 1 or 2, the other side is black, and if it is 3, the other side is white. The probability that the other side is black is 2/3.
#### Bayes’ theorem
Given that the shown face is black, the other face is black if and only if the card is the black card. If the black card is drawn, a black face is shown with probability 1. The total probability of seeing a black face is 1/2; the total probability of drawing the black card is 1/3. By Bayes’ theorem,[3] the conditional probability of having drawn the black card, given that a black face is showing, is
$\frac{1\cdot1/3}{1/2}=2/3.$
#### Eliminating the white card
Although the incorrect solution reasons that the white card is eliminated, one can also use that information in a correct solution. Modifying the previous method, given that the white card is not drawn, the probability of seeing a black face is 3/4, and the probability of drawing the black card is 1/2. The conditional probability of having drawn the black card, given that a black face is showing, is
$\frac{1/2}{3/4}=2/3.$
#### Symmetry
The probability (without considering the individual colors) that the hidden color is the same as the displayed color is clearly 2/3, as this holds if and only if the chosen card is black or white, which chooses 2 of the 3 cards. Symmetry suggests that the probability is independent of the color chosen. (This can be formalized, but requires more advanced mathematics than yet discussed.)
#### Experiment
Using specially constructed cards, the choice can be tested a number of times. By constructing a fraction with the denominator being the number of times “B” is on top, and the numerator being the number of times both sides are “B”, the experimenter will probably find the ratio to be near 2/3.
Note the logical fact that the B/B card contributes significantly more (in fact twice) to the number of times “B” is on top. With the card B/W there is always a 50% chance W being on top, thus in 50% of the cases card B/W is drawn, card B/W virtually does not count. Conclusively, the cards B/B and B/W are not of equal chances, because in the 50% of the cases B/W is drawn, this card is simply “disqualified”.
Here's Monty!
First, visit — Monty Hall Problem. Don’t read past ‘Solution.’
Attempt to solve. Write down method of solution.
Carry out ‘Simulation.’
Set Operations (A Wiki Rip)
There are ways to construct new sets from existing ones. Two sets can be “added” together. The union of A and B, denoted by A ∪ B, is the set of all things which are members of eitherA or B.
The union of A and B, or A ∪ B
Examples:
• {1, 2} ∪ {red, white} = {1, 2, red, white}.
• {1, 2, green} ∪ {red, white, green} = {1, 2, red, white, green}.
• {1, 2} ∪ {1, 2} = {1, 2}.
Some basic properties of unions are:
• A ∪ B = B ∪ A.
• A ∪ (B ∪ C) = (A ∪ B) ∪ C.
• A ⊆ (A ∪ B).
• A ∪ A = A.
• A ∪ ∅ = A.
• A ⊆ B if and only if A ∪ B = B.
### Intersections
A new set can also be constructed by determining which members two sets have “in common”. The intersection of A and B, denoted by A ∩ B, is the set of all things which are members of both A and B. If A ∩ B = ∅, then A and B are said to be disjoint.
The intersection of A and B, orA ∩ B.
Examples:
• {1, 2} ∩ {red, white} = ∅.
• {1, 2, green} ∩ {red, white, green} = {green}.
• {1, 2} ∩ {1, 2} = {1, 2}.
Some basic properties of intersections:
• A ∩ B = B ∩ A.
• A ∩ (B ∩ C) = (A ∩ B) ∩ C.
• A ∩ B ⊆ A.
• A ∩ A = A.
• A ∩ ∅ = ∅.
• A ⊆ B if and only if A ∩ B = A.
### Complements
Two sets can also be “subtracted”. The relative complement of A in B (also called the set theoretic difference of B and A), denoted by B \A, (or B −A) is the set of all elements which are members of B, but not members of A. Note that it is valid to “subtract” members of a set that are not in the set, such as removing the element green from the set {1, 2, 3}; doing so has no effect.
In certain settings all sets under discussion are considered to be subsets of a given universal set U. In such cases, U \ A is called the absolute complement or simply complement of A, and is denoted by A′.
The relative complement
of A in B.
The complement of A in U.
Examples:
• {1, 2} \ {red, white} = {1, 2}.
• {1, 2, green} \ {red, white, green} = {1, 2}.
• {1, 2} \ {1, 2} = ∅.
• {1, 2, 3, 4} \ {1, 3} = {2, 4}.
• If U is the set of integers, E is the set of even integers, and O is the set of odd integers, then the complement of E in U is O, or equivalently, E′ = O.
Some basic properties of complements:
• A ∪ A′ = U.
• A ∩ A′ = ∅.
• (A′)′ = A.
• A \ A = ∅.
• U′ = ∅ and ∅′ = U.
• A \ B = A ∩ B.
### Cartesian product
A new set can be constructed by associating every element of one set with every element of another set. The Cartesian product of two sets A and B, denoted by A × B is the set of all ordered pairs (ab) such that a is a member of A and b is a member of B.
Examples:
• {1, 2} × {red, white} = {(1, red), (1, white), (2, red), (2, white)}.
• {1, 2, green} × {red, white, green} = {(1, red), (1, white), (1, green), (2, red), (2, white), (2, green), (green, red), (green, white), (green, green)}.
• {1, 2} × {1, 2} = {(1, 1), (1, 2), (2, 1), (2, 2)}.
Some basic properties of cartesian products:
• A × ∅ = ∅.
• A × (B ∪ C) = (A × B) ∪ (A × C).
• (A ∪ B) × C = (A × C) ∪ (B × C).
Let A and B be finite sets. Then
• | A × B | = | B × A | = | A | × | B |.
A visual of a Cantor Set
Step One: Gentle intro
“Are rational numbers countable?”
Step Two: The Hotel Infinity
“Related to Cantor’s argument that rational numbers are countable?”
Step Three: Self Similarity
“Can you write out details of constructing the Cantor Set?” | 2018-01-16T09:11:05 | {
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https://math.stackexchange.com/questions/1931807/why-use-taylor-series-instead-of-maclaurin?noredirect=1 | # Why use Taylor series instead of Maclaurin? [duplicate]
I know that Maclaurin series are a special case of Taylor series where we set $a=0$, and it is useful for exponential functions (for example). But my question is: Are Taylor series (evaluate at some other point different to zero) really useful in practice?
• There are many reasons, ranging from solving limits, creating linear approximations and approximating functions in general, studying asymptotic growth (generally through Laurent series), and so forth. This question is very broad as it stands... What do you have in mind? Try to be a bit more specific Sep 18 '16 at 18:08
• my question is principal for approximations. Why do we care to approximate a function at some point different to zero. I mean why evaluate for example f(8) = cos x with a = 5 (or other number different to zero) with taylor formula wikimedia.org/api/rest_v1/media/math/render/svg/… Sep 18 '16 at 18:21
• The simplest answer to that is because you might want to approximate a function elsewhere than around zero. A McLaurin series is usually fine if you want to estimate the value of a function at $x=0.5$, but is often terrible at $x=500.5$. If you want to estimate the function there a Taylor series about $x=500$ might be very useful! Sep 18 '16 at 18:45
• You might then ask why we don't just get the series around $x=500.5$ instead of $x=500$. The key to this is ease of calculation. Imagine you are messing with a square root function and trying to approximate $\sqrt{36.458}$. You'll get a pretty nice estimation if you say it's about $\sqrt{36}=6$. Notice that we are estimating a function by a nearby point not equal to zero.... A Taylor series just makes this more rigorous, algorithmic, and accurate! Sep 18 '16 at 18:49
• thank you @BrevanEllefsen that's a good explication Sep 18 '16 at 19:20
An important reason is that the function might not be defined at $$0$$. Consider the $$log$$ function.
As mentioned in the comments, you might need to approximate the function in a region not near $$0$$. The Maclaurin series might converge very slowly or not at all. You could look for nearer points st which it was easy to calculate the Taylor coefficients.
Taylor series can be used to approximate function at some point (not always $0$). For example, this is a powerful tool for calculating limits. | 2022-01-25T21:03:30 | {
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https://math.stackexchange.com/questions/1137040/integrating-int-frac1-sqrtx24-dx-using-trigonometric-substitution | # Integrating $\int \frac{1}{\sqrt{x^2+4}}\,dx$ using Trigonometric Substitution
I'm reviewing integration by trigonometric substitution in anticipation of covering it in class next week. I seem to be a bit rusty and keep catching myself making various mistakes. On this particular problem I keep getting the same answer which is very close to being correct. However, I somehow end up dividing by two where I should not. I'm hoping another set of eyes can quickly set me right so I can stop frustrating myself reworking the problem to the same apparently wrong answer repeatedly! Thanks in advance!
$$\int \frac{1}{\sqrt{x^2+4}}\,dx$$
$$ln\lvert x + \sqrt{x^2+4} \rvert + C$$
Somehow I keep getting:
$$ln\Bigg|\frac{\sqrt{x^2+4}}{2} + \frac{x}{2}\Bigg| + C$$
Here's my work:
$$\int \frac{1}{\sqrt{x^2+4}}\,dx = \int \frac{1}{\sqrt{4(\frac{1}{4}x^2+1)}}\,dx = \frac{1}{2}\int\frac{1}{\sqrt{\frac{1}{4}x^2+1}}\,dx = \frac{1}{2}\int\frac{1}{\sqrt{(\frac{1}{2}x)^2+1}}\,dx$$
At this point I substitute as follows:
$$\frac{1}{2}x = \tan\theta$$ $$x = 2\tan\theta$$ $$dx = 2\sec^2\theta$$
So I continue on with:
$$\frac{1}{2}\int\frac{2\sec^2\theta}{\sqrt{\tan^2\theta+1}}\,d\theta = \int\frac{\sec^2\theta}{\sqrt{\sec^2\theta}}\,d\theta = \int\frac{\sec^2\theta}{\sec\theta}\,d\theta = \int\sec\theta\,d\theta = ln\lvert\sec\theta + \tan\theta\rvert + C$$
Finally, to get the answer in terms of x I essentially draw a right triangle and use the fact that $\tan\theta = \frac{x}{2}$. The side opposite $\theta$ I take to be x, the side adjacent $\theta$ is 2, and the hypotenuse is $\sqrt{x^2+4}$. So $\sec\theta = \frac{\sqrt{x^2+4}}{2}$ and $\tan\theta = \frac{x}{2}$.
So, substituting these values back in, as mentioned, I end up with:
$$ln\Bigg|\frac{\sqrt{x^2+4}}{2} + \frac{x}{2}\Bigg| + C$$
Can anyone help me see where I'm going wrong or failing to understand something?
• Both answers are correct. Feb 6, 2015 at 21:58
• Thanks to everyone who answered! I greatly appreciate the help :) Feb 6, 2015 at 22:14
You are correct still. Notice that \begin{align} \ln\Bigg|\frac{\sqrt{x^2+4}}{2} + \frac{x}{2}\Bigg| + C &= \ln\Bigg|\frac{\sqrt{x^2+4}+x}{2}\Bigg| + C\\ &= \ln\Bigg|\sqrt{x^2+4}+x\Bigg|-\ln(2) + C\\ &= \ln\Bigg|\sqrt{x^2+4}+x\Bigg|+C' \end{align} where $C'$ is still an arbitrary constant.
$$\ln\left|{\sqrt{x^2+4}\over 2}+{x\over 2}\right|+C=\ln|\sqrt{x^2+4}+x|-\ln2+C.$$ | 2022-06-26T03:24:06 | {
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https://math.stackexchange.com/questions/872576/how-many-4-permutations-of-the-positive-integers-not-exceeding-100-contain-three | # How many 4-permutations of the positive integers not exceeding 100 contain three consecutive integers in the correct order
Question:
How many 4-permutations of the positive integers not exceeding $100$ contain three consecutive integers in the correct order
a.) where consecutive means in the usual order of the integers and where these consecutive integers can perhaps be separated by other integers in the permutation?
b.) where consecutive means both that the numbers be consecutive integers and that they be in consecutive integers and that they be in consecutive positions in the permutation?
My Attempt:
3 consecutive integers is, $k, k+1, k+2$, thus $k$ is limited to $98$,
a.) The number of ways we can place and the 2 others is $C(4, 3) = 4$. Since $3$ integers are already taken by $k, k+1, k+2$, the last integer can have 97 integers, thus
$$4*98*97 = 38,024$$
b.) Since $k, k+1, k+2$ have to be consecutive in placement, there's only two possible place for that, thus
$$2*98*97 = 19, 012$$
Problem:
Both of my answers are a bit off compared to the answer key. In the answer key, a.) $37,927$ b.) $18,915$. One might notice that if my answers are subtracted $97$, I get the answer key's answer. So how exactly do you arrive to these numbers? I'm out of ideas.
• Inclusion-exclusion principle? That is, make sure we don't double count the sequences of all 4 integers consecutively. – hardmath Jul 20 '14 at 13:27
• i feel you, bro!! i did the same stupid mistake – Godfather Oct 17 '16 at 10:13
• We interpret $3$ consecutive as meaning at least $3$ consecutive. You have counted $12,13,14,15$ twice in your way of counting, once as $12,13,14$ followed by $15$, and once as $13,14,15$ preceded by $12$. Every one of the $4$ consecutives has been counted twice. There are $97$ ways to have $4$ consecutives, which is why your answer for b) is greater by $97$ than the official (and correct) answer. To get the right answer, we do as you did, but being aware that one is double-counting, and subtracting $97$ at the end to compensate. (Cont) – André Nicolas Nov 6 '14 at 22:16 | 2019-08-24T12:33:43 | {
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http://secure.consumercreditagency.org/ez5a/master-theorem-latex.html | # Master theorem latex
### Master theorem latex
2. MIT One way is to model the algorithm in the form of a recurrence equation and then solve via a number of techniques. The ewcounter command uses a LaTeX internal command, and you can also use it: \@addtoreset{new-name}{master} (but remember that it needs to be between \makeatletter and \makeatother, or in a package of your own). Andrew McGregor Associate Professor If you would like to write your homework solutions in LaTeX, Mergesort, Master Theorem Section 1, 2. It will also automatically number theorems, lemmas, definitions and bibliographic entries. 1, 2. 3 of CLR). Master Theorem 373F19 - Karan Singh 34 •Here’s the master theorem, as it appears in CLRS Useful for analyzing divide-and-conquer running time If you haven’t already seen it, please spend some time understanding it It was typeset using the LaTeX language, with most diagrams done using Tikz. Monday, September 9, 2019. The Master Theorem Consider a function that, for all for all ,satisfies the recurrence relation: with ,integer , real , . 2 Letqbe the product of the ˆrstpnumbers. If , for some constant , and if for some constant and all sufficiently large , then. Seems like reading problem in the morning and thinking about it through out the day helped me a lot. ”, “Theorem (Bob). Master Theorem is used to determine running time of algorithms (divide and So , according to master theorem the runtime of the above algorithm can be 1. I want to cite a theorem using its label, but what should appear is both its theorem number and its name (the text in Stack Exchange Network Stack Exchange network consists of 175 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Abstract: We state and prove a quantum-generalization of MacMahon's celebrated Master Theorem, and relate it to a quantum-generalization of the boson-fermion correspondence of Physics. I have been working for the past week on improving the proofs of Master theorems, as well generalizing them. A table of contents will automatically be created, complete with clickable links to each section and subsection you create in your presentation. and typed all missing steps in LaTeX. Fix function: In the fix function, compare that node with its children. Cormen, C. Correctness 1. 1. 1: z-r+t=3 Lemma 1. Produce beautiful documents starting from our gallery of LaTeX templates for journals, conferences, theses, reports, CVs and much more. In my opinion, the closest to a "master theorem" is the criterion due to Terry Lyons, according to which a reversible Markov chain on a countable state space (in particular, the simple random walk on a locally finite graph) is transient if and only if there exists a flow of finite energy on the state space. Clicking on it will redirect you to their original webpages. 1/30/19 1 CS4102 Algorithms Spring 2019 Warm up Given any 5 points on the unit square, show there’s always a pair distance ≤" apart 1 1 1 1 1 1 2 1 2 2 If points # LaTeX is a document preparation system used to write scholarly articles, prepare presentations, and many other documents. So I'm curious whether there may be some way to define a new theorem environment that accepts the theorem title as an argument. The textbook Algorithms, 4th Edition by Robert Sedgewick and Kevin Wayne surveys the most important algorithms and data structures in use today. Peace is followed by disturbances; departure of evil men by their return. Proof. 2016. recurrence can not be solve by Master Theorem what is the difference between this two recurrences. Let P be a set of n points in d-dimensional space. Common techniques are master theorem, substitution, recurrence trees, The binary search algorithm can be seen as recurrences of dividing N in half with a comparison. Introduction Code Beamer Features More LATEX Disclaimer #1 I am NOT an expert in LATEX I am NOT an expert in Beamer Disclaimer #2 This talk is designed to introduce you to presentations in LATEX For that purpose, we firstly state and prove a \lambda-extension of Schwinger's Master Theorem (SMT), which turns out to be a useful mathematical tool for us, particularly as a generating function for the unitary-representation functions of the conformal group and for the derivation of the reproducing (Bergman) kernel of L^2_h(D_4,d u_\lambda). A score higher than 100% is Jon Wellner's lecture notes on the master theorem of parametric The Davis-Kahan theorem. 4. Master Theorem Example 3 25 56=3 6 2 +8 Case 1 Θ6BCDEF ≈Θ(6&. Lectures by Walter Lewin. ´) than is the case for an ordinary character. However, a much more powerful and flexible solution is achieved thanks to an auxiliary tool called BibTeX (which comes bundled as standard with LaTeX). Master Theorem: Practice Problems and Solutions. Fact (MacMahon Master theorem or Wronski relation). You can earn more than 100% on most problem sets. I am now quite content with the proofs. The last equation shows that f(n) is not (g +(n)) for any >0, which excludes case (3). See my brief slides on the "school method" for integer division. Extra Credit: This class will have extra credit assignments that can help you boost your exam scores (by up to 10%). If your work relies on important results in your field, including those theorems' proofs in your introduction and background sections makes sense. Master Theorem 373F19 - Nisarg Shah 32 •Here’s the master theorem, as it appears in CLRS Useful for analyzing divide-and-conquer running time If you haven’t already seen it, please spend some time understanding it The report document should be written in Latex and contain: problem description, benchmarking, figures, and discussion. No installation, real-time collaboration, version control, hundreds of LaTeX templates, and more. So T (n) = T (n/2) + 1. This is an extremely valuable theorem because it is a vast generalization of Dirichlet's Theorem on primes in an arithmetic progression. 3. Theorem There is no largest prime number. Fri, Sep 27: Prerequisites Review Heaps and Heapsort (continued) Chapter 6 (Heapsort), Section 6. 3rd edition. 1 Oct 2014 And I'm almost certain that everyone reading this blog has heard of the generalization of the master theorem due to Akra and Bazzi. Stein. If you can write basic LATEX, you can easily make a Beamer presentation. 3 (Bulding a Heap), Introduction to Algorithms (3rd Edition) by Cormen et al. Master Theorem 3. If a master-file has been specified, then it will scan that file instead of the current file. If, on the other hand, we know the characteristic function φ and want to find the corresponding distribution function, then one of the following inversion theorems can be used. Q&A for users of TeX, LaTeX, ConTeXt, and related typesetting systems Stack Exchange Network Stack Exchange network consists of 175 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. A group earns 30% bonus to the entire homework for each solution posted on the class website. Submitted my pdf at around 10 am. The Master Theorem can be applied to any recurrence of the form. It was typeset using the LaTeX language, with most diagrams done using Tikz. Master Theorem. 2: q=r How do I do that? Stack Exchange Network Stack Exchange network consists of 175 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. I managed to prove that master theorem is applicable, but I don't know how to go about using it. Aug 16, 2016 · To start with, a building can never stand without a base. As Raphael mentions in his comment, here you are encountering two common versions of the Master Theorem, one of which is stronger and can handle your case. In your algorithm, I believe there is one subproblem of size n/2, so a is 1 and b is 2. Algorithms AppendixII:SolvingRecurrences[Fa’13] Change is certain. It is nearly complete (and over 500 pages total!!), there were a few problems that proved some combination of more difficult and less interesting on the initial pass, so they are not yet completed. Determine a good fancy style for each sentence : box shape; colour; title ( yes/no , style) With "theorem" we can mean any kind of labelled enunciation that we want to look separated from the rest of the text and with sequential numbers next to it. Gauss’s law, also known as Gauss’s flux theorem, is a law relating the distribution of electric charge to the resulting electric field. Homework 1 handed out. Case : If is either of the same order as or exactly logarithmically larger than , then the two terms compound each other, I want to write a LaTeX code to show, Lemma 1 : x+y=z Lemma 1. Algorithm design. Define 1;p to be the first positive eigenvalue of the p-Laplace operator on M, then, for any p 2(1;1), the following sharp estimates holds: 1;p p 1 ˇ p d p; (1. Texts. Prepare your solutions in LaTeX and submit in Canvas, like you would do for a regular assignment, but remember not to put your name on your paper! Solve the following recurrence exactly (in other words, do not use asymptotic notation), where you may assume n is a power of 4. Posted 01-14-2020: In case you want to typeset your homework, Here is a nice Latex Introduction. Case : If is polynomially smaller than , then it is negligible, and the asymptotic behavior is the same as what you would have without it included. If you are using a program with a graphical interface to TeX, you probably need to click on a button. It describes the per-iteration overhead, which may depend on the problem size in each iteration. because like the master theorem it gives you a quick way to generate the desired answer (or at least a guess that you can plug in to the recurrence to check). Pearson Education. The aim of classes is to introduce undergraduate mathematics students to $\LaTeX$ in preparation for writing their second year essay or fourth year project. A lifetime of learning Get started with Brilliant’s course library as a beginner, or dive right into the intermediate and advanced courses for professionals and lifelong learners. In Section 5. Then 练习使用latex嘤嘤嘤 注意 以及 LaTeX will calculate the right numbering for the objects in the document; the marker you have used to label the object will not be shown anywhere in the document. This approach is commonly used for theorems in mathematics, but can be used for anything. In this article, you will learn with the help of examples the DFS algorithm, DFS pseudocode and the code of the depth first search algorithm with implementation in C++, C, Java and Python programs. Fermat's little theorem is the basis for the Fermat primality test and is one of the fundamental results of elementary number theory. Leiserson, R. Working on the O-notation was at first supposed to be a brief side-project to “make things right”, but then ended up taking quite a bit of time. 21 May 2019 Move both the theorem and the proof completely in appendix the-end/raw/ master/demo. J. And typeset documents with LaTeX with just the click of a button. Tardos. 6 Proof of the master theorem 97 5 Probabilistic Analysis and Randomized Algorithms 114 5. Master Theorem · Divide and Conquer Algorithm · Binary Search · Floyd-Warshall Algorithm · Strongly Connected Components · Rabin-Karp Algorithm. By the master theorem, this recurrence has values =, =, =, which is a "steady state" tree, and thus we use the steady state case that tells us that T ( n ) = Θ ( n k ⋅ log n ) = Θ ( log n ) . They will make you ♥ Physics. In order to untoggle this functionality resort to the command onfrenchspacing . Abstract. tex simple. : We develop the theory of N-homogeneous algebras in a super setting, with particular emphasis on the Koszul property. Here are the latex sources files of Homework 1. The theorem is named after Pierre de Fermat , who stated it in 1640. Question about Master Theorem was very intuitive. View courses, graduate and undergraduate programs, faculty and research interests, activities, events … Jan 23, 2010 · With " theorem " we can mean any kind of labelled enunciation that we want to look separated from the rest of the text and with sequential numbers next to it. The distance is calculated by taking the square root of the sum of the squared pair-wise distances of every dimension. Master theorems 14. when typesetting in LATEX a master or a doctoral thesis according to current re- quirements of TTU Graduate School mathematics, and sciences – who would like to use LATEX in typesetting their theses. The presentation should be made with about 10-12 slides generated with LaTeX (Beamer). It analyzes the theorem package can be used to customize theorem-like environments. It brings a minimalist and powerful approach to text editing. Master Theorem 2. Each of the exercises below involves a choice among the master theorem templates discussed in lecture. 31 Aug 2016 Master Theorem CSE235 Introduction Pitfalls Examples 4th Condition Master Theorem Slides by Christopher M. PS Contrary to The standard commands of LATEX also work in Beamer. A personal use latex package for highlight the mathematical logic in a text. master template. Once the external bibliography file is imported, the command \cite is used just as in the introductory example . So as you can see, the recurrence relation is gleaned from looking at the algorithm. The included files must have a file variable TeX-master or tex-main-file pointing to the master So we need to tell RefTeX that theorem and axiom are new labeled Latex-Suite is released under the Vim charityware license. Such recurrences should not constitute occasions for sadness but realities for awareness, so that one may be happy in the interim. If I'm wrong, what did I miss on? The answer is that you didn't miss anything. General results Master Theorem. The \glue" is f(n) = n5=2. BibTeX is a widely used bibliography management tool in LaTeX, with BibTeX the bibliography entries are kept in a separate file and then imported into the main document. the problem size must shrink by a constant factor, the subproblems must all have the same size, Then, once you have the recurrence you can analyze using the Master Theorem. If no case applies, simply state that fact; you are not required to attempt a solution when no master theorem case applies. The Quantum MacMahon Master Theorem By Stavros Garoufalidis, Thang TQ Le, Doron Zeilberger and X [maybe you!] Warning: The proof is incomplete, read below for a chance to win a prize and be co-author. $$T(n) = \sum_i a_i T(n/b_i) + g(n)$$. f(n)=lgn=O(n2−ϵ) for ϵ=1, since lgn=O(n). For the special case where the Hecke operator is the ordinary supersymmetry and typed all missing steps in LaTeX. There are four main functions associated with heapsort 1. 1: As part of your background section. If f(n) = O(nlogb a− ) for some constant > 0, then T(n) = Θ(nlogb a). Homework that is Chapter 7: The Prime Number Theorem for arithmetic progressions RefTeX wraps itself round four LaTeX macros: \label , \ref , \cite , and \index . Since there is always a power of 2 in the range $[\frac {n} {2}, n]$, recurrence master-theorem 追加された 27 9月 2013 〜で 02:06 著者 WSS , それ マスター定理を使用した次の再帰アルゴリズムの実行時間はどのくらいですか？ The DFS algorithm works as follows: Start by putting any one of the graph's vertices on top of a stack. There are 3 cases: 1. 2 it was indicated that kd-trees can also be used to store sets of points in higher-dimensional space. Results There Is No Largest Prime Number The proof uses . Welcome to the Dartmouth Mathematics Department located in Kemeny Hall. 4 Probabilistic analysis and further uses of indicator random variables 130 •Master Theorem 3. Note that the master theorem does not provide a solution for all f f f. Master Theorem: Practice Problems and Solutions Master Theorem The Master Theorem applies to recurrences of the following form: T(n) = aT(n/b)+f(n) where a ≥ 1 and b > 1 are constants and f(n) is an asymptotically positive function. The broad perspective taken makes it an appropriate introduction to the field. LATEX is preferred but not required. With "theorem" we can mean any kind of labelled enunciation that we want to look separated from the rest of the text and with sequential numbers next to it. 1 The hiring problem 114 5. Recommended for you Latex-Theorem. Show your derivation and then prove that your answer is correct. Dec 19, 2019 · Vim-LaTeX is an extension for Vim that lets you edit and compile LaTex documents. 11 Sep 2019 PDF | We use the method of brackets to evaluate quadratic and quartic type Ramanujan's Master Theorem and the Method of Brackets. T(n) = a · T(n/b) + f(n) that describes the complexity of some recursive functions. Dec 02, 2018 · Jupyter notebook extension which supports (some) LaTeX environments within markdown cells. T(n) = aT(n / b) + O(n d) where a, b, and d are constants. November 10: Midterm this week was quite easy. A score higher than 100% is bonus. . (There are some other formulations, but this above one handles the more common cases). Convergence conditions are also studied. In particular, if f f f is smaller or larger than n log b a n^{\log_b{a}} n lo g b a by less than a polynomial factor, then none of the three cases are satisfied. 5. It is recommended that students take both classes. – daleif Aug 10 '17 at 8:28 Also note that your example has no chance to work as \institute and \email is not defined in the article class – daleif Aug 10 '17 at 8:29 latex 定理环境，引理，定义，修改格式 Theorem, lemma, definition 时间 2017-03-11 标签 定理环境 latex 定理格式 定理冒号 定理换行 Jan 24, 2016 · For the Love of Physics - Walter Lewin - May 16, 2011 - Duration: 1:01:26. In CLRS terms In the analysis of algorithms, the master theorem provides a cookbook solution in asymptotic The master theorem concerns recurrence relations of the form:. The Master Theorem: Chapter 4 (Divide-and-Conquer), Section 4. If in a min-heap, swap the parent and the smallest child considering that the child is smaller than the parent itself. You may find this template useful. It lets you solve recurrence relations that can help you find the complexity of recursive algorithms. e. For example, there is a book titled "Synopsis of Elementary Results in Pure and Applied Mathematics: Containing Propositions, Formulae, And Methods Of Analysis, With Abridged Demonstrations" (Cambridge Library Collection - Mathematics) by George Shoobridge Carr, which I am told consists of theorem without proof, mostly. Kleinberg, E. Typically, running LaTeX means typing latex <filename> at the command prompt and running BibTeX means typing bibtex <filename>. Master Written (use Latex!) – Submit both zip and pdf ! 21 Dec 2005 The LaTeX template for CS thesis is based on version 2. Comment: AMS-LaTeX, 7 pages, revised and final version. characters directly instead of having to rely on Latex-Suite's method of displaying diacritics. pdf) to see what is possible, or generate it with. If you would like to write your homework solutions in LaTeX, here's a template 3 Sept, Preliminaries, Mergesort, Master Theorem, Slides, Section 1, 2. Week 1. LaTeX provides a command that will let you easily define any theorem-like enunciation. Create a list of that vertex's adjacent nodes. LaTeX. Recursive functions, Master Theorem LaTeX is beautiful and strongly recommended We will post our TeX source files, which you can use as templates. T(n1/2). tex the LATEX master file, to be used as a Note that for LIPIcs all numbered theorem-like environments should use one. Assume A is diagonalizible, then it is can be proved in the eigenbasis and is exactly the previous fact. This article explains how to use the standard verbatim environment as well as the package listings, which provide more advanced code-formatting features. For each, indicate which case applies and specify the asymptotic growth class of the function. 2 of the thesis template for \macs — Master of Applied Computer Science project report; \mcs — Master of diff simple. l) • Case 1: if Q6=7(6BCDU^_‘)for some constant a>0, then 56=Θ(6BCDU^) • Case 2: if Q6=Θ(6BCDU^), then 56=Θ(6BCDU^log6) • Case 3: if Q6=Ω(6BCDU^e‘)for some constant a>0, and if NQ O P ≤;Q(6)for some constant ;<1and all sufficiently large 6, then 56=Θ(Q6) 56=N5 6 S +Q(6) LaTeX Resources. As we will soon see, a polynomial of degree n in the complex number system will have n zeros. Mar 19, 2018 · 1 Answer 1. The current figures use images from external websites. We strongly recommend you use LaTeX to typeset your work so it is easy to read. Since the theorem-like structures use a beamercolorbox, using allowframebreaks option directly won't work in this case. Certainly Data Structures and Algorithms are the base to all programming languages and hence is a must learn… Theorem 5. If you use (b) We cannot apply the master theorem because of the square root, so we draw the recursion tree: T(n). You can easily create overlays and dynamic effects. T. Again, like I mentioned in class, please remember that the regular expression that is taught in class (CSC236) is different from the regular expression that is presented elsewhere. The extra credit assignments will involve doing machine-checked proofs in Coq , which you will learn about during discussion sections. bst only in that the names of all three authors of Arrow, Hurwicz, and Uzawa (1961) are listed in the first citation to that work, although not in the second, parenthetical, citation. Solutions to problems sets must be typeset using LaTeX . a= 4;b= 2 implies a reference function g(n) = nlog 2 4 = n2. When you run the LaTeX file through LaTeX and BibTeX (instructions below), you'll get output for the body of the document that differs from the output when you use te. Similarly, if x−k is a factor of f (x) , then the remainder of the Division Algorithm f (x) = (x−k)q (x)+r is 0. Theorem. Introduction Code Beamer Features More LATEX Disclaimer #1 I am NOT an expert in LATEX I am NOT an expert in Beamer Disclaimer #2 This talk is designed to introduce you to presentations in LATEX 1/30/19 1 CS4102 Algorithms Spring 2019 Warm up Given any 5 points on the unit square, show there’s always a pair distance ≤" apart 1 1 1 1 1 1 2 1 2 2 If points # Solutions to problems sets must be typeset using LaTeX . Jan 24, 2016 · For the Love of Physics - Walter Lewin - May 16, 2011 - Duration: 1:01:26. Miscelanea Defined terms on the margin; Date and time of compilation; Print labels on the margin (equation, section and theorem) Images in separate folders; Color all math; Poster Examples (external link) Bibliography Single (master) BibTeX file for all your projects; Definitions on the margin Writes the term being defined on the margin. - T-F-S/tcolorbox Mason–Stothers theorem (polynomials) Master theorem (analysis of algorithms) (recurrence relations, asymptotic analysis) Maschke's theorem (group representations) Matiyasevich's theorem (mathematical logic) Max flow min cut theorem (graph theory) Max Noether's theorem (algebraic geometry) Maximal ergodic theorem (ergodic theory) latex 定理环境，引理，定义，修改格式 Theorem, lemma, definition 时间 2017-03-11 标签 定理环境 latex 定理格式 定理冒号 定理换行 Jan 21, 2014 · A Sample Proof Using Mathematical Induction (playing with LaTeX) It’s been a long time since I used LaTeX regularly, and I discovered that I don’t have any leftover files from my days as a math student in Waterloo. Other guides can be found at Wikibooks and NYU. The Master theorem applies if f (n) are exactly of the same order of magnitude, or if their ratio grows at least like n , for some > 0, but there are possible values for a, b and f (n) such that neither case applies. Add the ones which aren't in the visited list to the top of stack. David Wu Guess and Check Method. Dec 08, 2015 · To do this, consider instead the recurrence $T (n) = T (n-1) + T (f (n)) + n$, where $f (n)$ is equal to the greatest power of 2 less than n. An introduction to the lab environment and tools used to write, compile, and run a program in either PHP (for those with prior Java experience) or Java (for those without prior Java experience). LaTeX Templates Theses An academic thesis, also known as a dissertation, is a substantial work produced by a graduate student to communicate their research and earn a degree. Application of Hall's Theorem: every doubly-stochastic matrix is a convex combination of permutation matrices matching markets: market clearing prices always exist (envy free allocation, maximizing total valuation); here are the pdf slides. GitHub is home to over 40 million developers working together to host and review code, manage projects, and build software together. This pair of implications is the Factor Theorem. The \newtheorem command may have at most one optional argument. [Val12b, main Theorem] Let M be a compact Riemannian manifold with nonnegative Ricci curvature and diameter d, and possibly with convex boundary. Homeworks •Hw1 due Wed, January 30 at 11pm –Start early! –Written (use Latex!) –Submit BOTH pdf and zip! Master Theorem 1. So: Aug 27, 2014 · [back to top] The Euclidean distance is a distance measure between two points or or vectors in a two- or multidimensional (Euclidean) space based on Pythagoras' theorem. T(n) = 2T(bn=4c) + n1 Together, these define the parameters f(n), a, and b in the Master-Theorem equation. 2 Indicator random variables 118 5. This is true even if the proofs are well known. Intro to algorithms. Master Theorem is good to know for programming. CLR calls this the “Master Theorem” (section 4. Together, these define the parameters f(n), a, and b in the Master-Theorem equation. 09. 2 2 : 7 Sept Euler Circle Spring Paper: Čebotarev Density Theorem In this paper, we do exactly what the title implies: prove the Čebotarev Density Theorem. Evaluate, simplify, solve, and plot functions without the need to master a complex syntax. LaTeX is widely used in science and programming has become an important aspect in several areas of science, hence the need for a tool that properly displays code. The law was formulated by Carl Friedrich Gauss (see ) in 1835, but was not published until 1867. Master Theorem 1. CS4102 Algorithms Spring 2019 Warm up Given any 5 points on the unit square, show there’s always a pair distance ≤" apart 1 1 1 All homeworks must be typed, preferably in LaTeX. Mar 04, 2016 · Please, check out some of my other uploads or sub! Open to suggestions for other lack (pvc, vinyl, latex) compilations. 3 Then q+1 is not divisible by any of them. 1/29/19 2 CLRS Readings •Chapter 4 4 Homeworks •Hw1 due Wed, January 30 at 11pm –Start early! –Written (use Latex!) –Submit BOTH pdf CS4102 Algorithms Spring 2019 Warm up Given any 5 points on the unit square, show there’s always a pair distance ≤" apart 1 1 1 Luckily, the lemma is nevertheles true for the weaker hypothesis, and surprise!, the theorem also holds with the strong hypthesis replaced by the weaker one. SMT is related to MacMahon's Master Theorem (MMT) and an extension of both in terms of Louck's SU(N) solid harmonics is also provided for completeness. Goodrich & Roberto Tamassia!Algorithms is a course required for all computer science majors, with a strong focus on theoretical topics. Thesis 14. We state and prove a quantum-generalization of MacMahon's celebrated Master Theorem, and relate it to a quantum-generalization of the boson-fermion correspondence of Physics. Comments: AMS-LaTeX, 7 pages, revised and final version. theorem. Course for 3rd year bachelor and master students - Fall 2016. In Mathematics most papers are written in a type setting markup language called LaTeX - which evolved from TeX. CLRS: p93 I've been looking at my algorithms and complexity course's suggested exercises for master theorem and I the only one I can't seem to solve is a case in which k is rational. Since f(n)=g(n) !1 For that purpose, we firstly state and prove a \lambda-extension of Schwinger's Master Theorem (SMT), which turns out to be a useful mathematical tool for us, particularly as a generating function for the unitary-representation functions of the conformal group and for the derivation of the reproducing (Bergman) kernel of L^2_h(D_4,d u_\lambda). Thursday, October 17, 2019 For that purpose, we firstly state and prove a \lambda-extension of Schwinger's Master Theorem (SMT), which turns out to be a useful mathematical tool for us, particularly as a generating function for the unitary-representation functions of the conformal group and for the derivation of the reproducing (Bergman) kernel of L^2_h(D_4,d u_\lambda). How would I figure out which case of Master's theorem this lie on? Would taking logarithms of both sides for comparison make sense since logarithm is a monotonic function? $$2\cdot \lg(\lg(n)) \quad\text{and} \quad (0. So all is well that ends well. Abstract This article provides useful tools to write a thesis with LATEX. Specifically, this means that. The chngcntr package encapsulates the \@addtoreset command into a command \counterwithin. 3 Randomized algorithms 122? 5. ”, and “Theorem (Will)”. Extracredit will be offered if you write the final project in a Jupyter notebook which includes the LaTeX and Python code. 3 Master theorem. There are two types of class: beginners and intermediate. A LaTeX package to create highly customizable colored boxes. LaTeX provides a convenient way to produce high-quality documents and it is the standard used for typesetting computer science papers. Introducing a NEW addition to our growing library of computer science titles, Algorithm Design and Applications, by Michael T. TODO: Identify all the sentences which require an " encapsulation box" and write a "phylosphical" pseudo-definition of them in the example file. 5) Brilliant guides you through problem solving strategies and challenges you to think outside the box. For that purpose, we firstly state and prove a \lambda-extension of Schwinger's Master Theorem (SMT), which turns out to be a useful mathematical tool for us, particularly as a generating function for the unitary-representation functions of the conformal group and for the derivation of the reproducing (Bergman) kernel of L^2_h(D_4,d u_\lambda). To any Hecke operator on a vector superspace, we associate certain superalgebras and generalizing the ordinary symmetric and Grassmann algebra, respectively. Master theorem states that, For any , If for some constant , then . The second equation shows the f(n) is not ( g(n)) and therefore precludes case (2). The solution will tell you the nature of the runtime, i. The various LaTeX processing programs can be found at CTAN. Fortunately, LaTeX has a variety of features that make dealing with references much simpler, including built-in support for citing references. tex 3a4,13 > %Theorem 20 Feb 2013 lipics-sample-article. \newtheorem*{mainA}{Theorem A }. Session 4 Bibliography Management, Theorems, Formatting your CV using LaTex Using Built-in Support If you are writing only one or two documents and aren't planning on writing more on the same subject for a long time, maybe you don't want to waste time creating a database of references you are never going to use. The time for dividing is O(1) and time for recombining is O(1) assuming the analysis is not in terms of bit operations. Lab 01 – Introduction, Gettings Started. 4 Thus q+1 is also prime and greater thanp. Presentations, otherwise known as seminars, talks or lectures, are given to an audience with the purpose of sharing information with a group of people. Also you might want to have a look in the amsthm manual to familiarize yourself with the ewtheorem syntax. The formula in the definition of characteristic function allows us to compute φ when we know the distribution function F (or density f). The master template is not applicable to this recurrence. When I searched on google I found tow questions related with this two recurrence First Question: The formulæ are written using $$\LaTeX$$ and the final render of the page is done by the Javascript library MathJax (currently the simplest way to use $$\LaTeX$$ in HTML). The necessary space above and below the statement of the theorem will automatically be generated by LaTeX. Guide: An introduction to LaTeX can be found here. The master theorem is a formula for solving recurrences of the form T(n) = aT(n/b)+f(n), where a ≥ 1 and b > 1 and f(n) is asymptotically Usually this will be the name of the master theorem. {\displaystyle T(n)=\Theta (n^{k}\cdot \log n)=\Theta (\log n). You write text and mathematics in your document without having to know a programming language or exit to an equation editor. An online LaTeX editor that's easy to use. Here's some intuition about the master theorem. It is much better at handling equations than Word type programs. The relation is then solved by either the master theorem or the substitution method or gleaned from drawing a graph of the recursions (recursion tree) and then proved rigorously with the substitution method. Felt so good. In the analysis of algorithms, the master theorem for divide-and-conquer recurrences provides an asymptotic analysis (using Big O notation) for recurrence relations of types that occur in the analysis of many divide and conquer algorithms. 631+\epsilon)\cdot \lg(n)$$ Then figure out what epsilon is Thank you in advance, kpark Sometimes an apparently weaker version of the theorem is equivalent to the full theorem with a short proof (for example, Hilbert's Nullstellensatz). Note that the f(n) function is also a complexity expression. All homeworks must be typed, preferably in LaTeX. If I have a lot of these, it becomes tedious having to always go back to the preamble to define a new theorem environment. Comment: LaTeX, 40 pages, three new Sections and six new references added. We strongly recommend typesetting solutions to the homework assignments using LaTeX. Master theorem solver (JavaScript) In the study of complexity theory in computer science, analyzing the asymptotic run time of a recursive algorithm typically requires you to solve a recurrence relation. T(n) = { LaTeX Original \documentclass{article} \begin{document} {\bf 6 Dec 2015 I think you used that method wrong! As mentioned in master theorem case 3. Induction. Homework 5 handed out; here are the tex sources. active oldest votes. See my slides on the homework assignment policy of this class. Rivest, C. We prove that these algebras are N-Koszul. They are free. ) ]: Let. Latex is mandatory for submitting papers to many journals and style files for many of the top journals are available. Also provides support for labels and crossreferences, document wide numbering, bibliography, and more Skip to main content Switch to mobile version Intuitively, an algorithm’s efficiency is a function of the amount of computational resources it requires, measured typically as execution time and the amount of space, or memory, that the algorithm uses. 6. Bourke Instructor: Berthe Y. 5 (The Master Method for Solving Recurrences), Introduction to Algorithms (3rd Edition) by Cormen et al. Submission in LaTeX will increase your chances. Submit both paper copy in class, and digital version by e-mail to the TAs. Stack Exchange network consists of 175 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Posted 02-06-2020: Here is the Master Theorem. If , then . LaTeX is capable of producing presentation slides using the Beamer class in a simple and easy-to-use way. Bibliography management with Bibtex. 2. Join GitHub today. In your case you should use case 3 which uses Ω-notation which describes 18 Apr 2017 For your first question. The Zero Theorem Latex Slick Shine TV incredible workes Master It's particularly useful when you have recurrences of the form. Theorem 1. The university of closed for Labor Day next Monday, September 4th, so Kyle will hold makeup office hours this Friday, September 1st at the usual time, 3:00pm to 4:00pm. A workaround would be to "simulate" your theorem-like heading with a beamercolorbox and then write the text outside the box with the allowframebreaks option: See how you defined theorem. $\begingroup$ From Wikipedia (in agreement with recall of readings many years ago): Historically, as a book-keeping device of covariant perturbation theory, the graphs were called Feynman–Dyson diagrams or Dyson graphs, because the path integral was unfamiliar when they were introduced, and Freeman Dyson's derivation from old-fashioned perturbation theory was easier to follow for physicists How does python multiply Big Integers? How to solve recurrences without remembering the Master Theorem. It is due Wednesday, September 6th, in class. } Depth first traversal or Depth first Search is a recursive algorithm for searching all the vertices of a graph or tree data structure. Use Vim-LaTeX to bring the best of both the worlds into your document editing processes. 1 Suppose pwere the largest prime number. 5 The master method for solving recurrences 93? 4. Take the top item of the stack and add it to the visited list. [e-studiegids To simplify grading it is preferable that you submit your homework in latex. rather than as “Theorem (Jim). CLRS Readings •Chapter 4 4. There are three reasons to include a proof in your master's thesis - two of them good, and one of them bad. This tells us that k is a zero. It's particularly useful when you have recurrences of the form. det (1 − tM) ( ∑iTrtiSi (M)) = 1 , where Si are symmetric tensor power of M. To reference the number assigned to that result in some other theorem, one gives the command \ref{name}; this will be replaced by the number automatically assigned by LaTeX to that result when the LaTeX program is executed. Instead, LaTeX will replace the string " \ref{ marker } " with the right number that was assigned to the object. Yihang: Apr, 11 R The scribe latex template is available Let A be a matrix. Lecture 4: Karatsuba, Induction, and Master Theorem. 1 •Master Theorem 3. The amount of computational resources can be a complex function of the size and structure of the input set. 6 [The Master Theorem (From page 268 of Goodrich and Tamassia. Instructions for checking out code from GitHub using Eclipse. The Master Theorem applies to recurrences of the following form: T(n) = aT(n/b) + f(n). MIT Instructs LaTex to abstain from inserting more space after a period (´. master theorem latex | 2020-02-22T15:38:00 | {
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https://tex.stackexchange.com/questions/575527/gather-multiple-equations-with-curly-bracket/575530 | # Gather multiple equations with curly bracket
I have two equations, both of which hold under the same condition (x<5):
\begin{align}
f(x) &= \sum_{n} a^i_j \Psi_n(x) &\text{for } x < 5
\\ g(x) &= \sum_{n} b^i_j \Phi_n(x) &\text{for } x < 5
\end{align}
That code results in this output:
Now, what I would like to do is to gather both equations with a curly bracket and this way write the condition only once. I tried this:
\begin{align}
\left.
\begin{array}{rl}
f(x) &= \sum_{n} a^i_j \Psi_n(x)
\\ g(x) &= \sum_{n} b^i_j \Phi_n(x)
\end{array}
\right\} &\text{ for } x < 5
\end{align}
Which results in the following output:
However, this doesn't look good to me: The equations are now too close to each other and the alignment of the left hand side with the equation marks is not convincing.
Do you have a hint how I can resolve that issue?
• You might be interested in rcases from mathtools. – leandriis Dec 18 '20 at 14:37
\documentclass{article}
\usepackage{mathtools}
\begin{document}
\begin{rcases} \begin{aligned} f(x) &= \sum_{n} a^i_j \Psi_n(x) \\ g(x) &= \sum_{n} b^i_j \Phi_n(x) \end{aligned} \end{rcases} x < 5
\end{document}
• Exactly what I needed, thanks. – Amos Egel Dec 18 '20 at 14:49
Another easy solution with empheq enables two equation numbers:
\documentclass{article}
\usepackage{empheq}
\begin{document} | 2021-05-06T11:54:22 | {
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http://math.stackexchange.com/questions/223238/what-is-the-average-of-rolling-two-dice-and-only-taking-the-value-of-the-higher | # What is the average of rolling two dice and only taking the value of the higher dice roll?
What is the average result of rolling two dice, and only taking the value of the higher dice roll?
To make sure the situation I am asking about is clear, here is an example: I roll two dice and one comes up as a four and the other a six, the result would just be six.
Would the average dice roll be the same or higher than just rolling one dice?
-
With regard to your final question: what does your intuition tell you? – Benjamin Dickman Oct 29 '12 at 3:51
The number of ways to roll a number $x$ under your definition would be $2(x-1) + 1$.
Therefore the expected value would be $$E[X] = \sum_{x=1}^6\frac{2(x-1)+1}{36}x = \frac{1}{36}\sum_{x=1}^6(2x^2 - x) = \frac{161}{36} \approx 4.47$$ So the average is considerably higher than the average of a single die, being $3.5$.
-
For $k=1,\dots,6$ there are $k^2$ ways to get two numbers less than or equal to $k$. To get two numbers whose maximum is $k$ I must get two numbers that are less than or equal to $k$, but not two numbers that are less than or equal to $k-1$, so there are $k^2-(k-1)^2=k^2-(k^2-2k+1)=2k-1$ ways to get two numbers whose maximum is $k$. Thus, the probability of getting a maximum of $k$ is
$$\frac{2k-1}{36}\;,$$
and the expected value of the maximum is
\begin{align*} \sum_{k=1}^6k\cdot\frac{2k-1}{36}&=\frac1{36}\sum_{k=1}^6\left(2k^2-k\right)\\ &=\frac1{18}\sum_{k=1}^6k^2-\frac1{36}\sum_{k=1}^6k\\ &=\frac{6\cdot7\cdot13}{18\cdot6}-\frac{6\cdot7}{36\cdot2}\\ &=\frac{91}{18}-\frac{21}{36}\\ &=\frac{161}{36}\\ &=4.47\overline{2}\;. \end{align*}
Of course this is larger than the expected value of $\frac72=3.5$ for a single roll of a die: picking the maximum of the two numbers can be expected to bias the result upwards.
-
@Matthew: Thanks for catching the typo. – Brian M. Scott Oct 29 '12 at 4:38
This is very much delayed, but consider the case with an $n$-sided die. As has already been observed, the expected value of the maximum of two $n$-sided die is
$${1 \over n^2} \sum_{k=1}^n (2k^2-k)$$
and we can write out this sum explicitly. In particular, we can expand to get
$${1 \over n^2} \left( \left( 2 \sum_{k=1}^n k^2 \right) - \sum_{k=1}^n k \right)$$ and recalling the formulas for those sums, this is
$${1 \over n^2} \left( {2n(n+1)(2n+1) \over 6} - {n(n+1) \over 2} \right)$$
or after some rearrangement
$${(n+1)(4n-1) \over 6n}.$$
In particular this is approximately $2n/3$. This could have been guessed if you know that the expectation of the maximum of two uniform random variables on $[0, 1]$ has the beta distribution $B(2,1)$, which has mean $2/3$.
- | 2015-07-01T05:19:55 | {
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https://rao.im/mathematics/2014/05/13/average-length-of-longest-arc/ | # Average Length of the Longest Arc in $S^1$
Suppose that we draw $n$ points $a_i \sim Uniform(S^1)$ for $i = 1 \ldots n$. These points determine a partition or set of disjoint arcs of $S^1$. What is the average length of the longest arc? To even measure the $n$ arc lengths, we need an ordering of the $\{ a_i \}$ , $a_{(1)}, \ldots, a_{(N)}$.
The arc lengths are:
$\begin{gather*} l_1 = |a_{(2)} - a_{(1)}| \\ \vdots \\ l_{n-1} = |a_{(n)} - a_{(n-1)}| \\ l_n = |a_{(1)} - a_{(n)}| \end{gather*}$
At this point we’d still have to order the arc lengths, to get the longest arc. Ultimately, we need the distribution of $l_{(n)}$, $P(l_{(n)} > x)$, to compute anything.
\begin{align*} P(l_{(n)} > x) &= P( l_1 > x, \text{ or } \ldots, \text{ or } l_n > x) \\ &= \sum^{n}_{k} \binom{n}{k} (-1)^{k-1} P( l_1 > x, \text{ and } \ldots, \text{ and } l_k > x), \end{align*} \begin{align*} P( l_1 > x, \text{ and } \ldots, \text{ and } l_k > x) & = \text{probability of } n-1 \text{ cuts in a } 2\pi - kx \text{ long arc}\\ & = \left( \frac{2 \pi - kx}{2\pi}\right)^{n-1} \end{align*}
But why!! The simple graphic below should make this clear. Essentially, we can transform the circle problem to one on a line, after the first cut: the unit circle problem with $n$ points is the same problem with $n-1$ points on a $2 \pi$ long line segment. The first cut is superfluous. From there, the argument boils down to recognizing that each cut is made in segments of total length $2\pi - kx$.
Lemma. A Suppose $X \sim p(x)$ is a non-negative random variable, then $E[X] = \int^{+\infty}_{0} P(X > x) dx.$
proof:
Define $F(x) = \int^{x}_{0} p(x) dx = P(X < x)$. Integrating by parts \begin{align*} \int^{b}_{0} x p(x) dx & = x \Big\lvert^{b}_{0} \int^{b}_{0}p(x)dx - \int^{b}_{0}F(x)dx \\ & = \int^{b}_{0} dx \int^{b}_{0}p(x)dx - \int^{b}_{0}F(x)dx \\ & = F(b)\int^{b}_{0} dx - \int^{b}_{0}F(x)dx \\ & = F(b)\int^{b}_{0} 1 - F(x)dx \\ & = F(b)\int^{b}_{0} P(X > x)dx \\ \end{align*}
Taking the limit as $b \longrightarrow + \infty$ finishes the argument.
q.e.d.
Using the Lemma,
\begin{align*} E[l_{(n)}] &= \int^{+\infty}_{0} P(l_{(n)} > x) dx \\ &= \sum^{n}_{k} \binom{n}{k} (-1)^{k-1} \int^{+\infty}_{0}P(l_k > x)dx \\ &= \sum^{n}_{k} \binom{n}{k} (-1)^{k-1} \int^{+\infty}_{0}\left( \frac{2 \pi - kx}{2\pi}\right)^{n-1}dx \\ \end{align*}
Focusing on the integral, set $u = \frac{2\pi - kx}{2\pi}$,
$\frac{2 \pi - kx}{2\pi} \geq 0 \quad \Longrightarrow \quad \frac{2\pi}{k} \geq x$ \begin{align*} \int^{+\infty}_{0}\left( \frac{2 \pi - kx}{2\pi}\right)^{n-1}dx &= \int^{\frac{2\pi}{k}}_{0}\left( \frac{2 \pi - kx}{2\pi}\right)^{n-1}dx\\ &= \frac{2\pi}{-k}\int^{u(\frac{2\pi}{k})}_{u(0)} u^{n-1}du\\ &= \frac{2\pi}{k}\int^{1}_{0} u^{n-1}du\\ &= \frac{2\pi}{kn} \end{align*}
Substituting back into the formula,
\begin{align*} E[l_{(n)}] &= \frac{2\pi}{n} \sum^{n}_{k} \binom{n}{k} \frac{(-1)^{k-1}}{k} \\ &= \frac{2\pi}{n} \sum^{n}_{k} \frac{1}{k}. \end{align*}
The last step applies a known binomial sum identity.
Awesome! We have an answer! But is it right? Lets check our work, by generating the empirical distribution of max-segment lengths. I wrote up some code to do this.
In [16]: data = array([ MaxArc(10) for x in range(10000)])
In [17]: data.mean()
Out[17]: 1.8404203486925401
In [18]: 2 * pi * harmonic(10)/10
Out[18]: 1.8403250298528779
Both sample mean and true mean are within a few thousands of each other, which I think validates the analysis above. I took $N = 10$, but other choices will yield similar results. | 2022-08-09T23:14:00 | {
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http://dlmf.nist.gov/4.7 | # §4.7 Derivatives and Differential Equations
## §4.7(i) Logarithms
4.7.1 $\frac{\mathrm{d}}{\mathrm{d}z}\ln z=\frac{1}{z},$ ⓘ Symbols: $\frac{\mathrm{d}\NVar{f}}{\mathrm{d}\NVar{x}}$: derivative of $f$ with respect to $x$, $\ln\NVar{z}$: principal branch of logarithm function and $z$: complex variable A&S Ref: 4.1.46 Permalink: http://dlmf.nist.gov/4.7.E1 Encodings: TeX, pMML, png See also: Annotations for 4.7(i), 4.7 and 4
4.7.2 $\frac{\mathrm{d}}{\mathrm{d}z}\operatorname{Ln}z=\frac{1}{z},$
4.7.3 $\frac{{\mathrm{d}}^{n}}{{\mathrm{d}z}^{n}}\ln z=(-1)^{n-1}(n-1)!z^{-n},$
4.7.4 $\frac{{\mathrm{d}}^{n}}{{\mathrm{d}z}^{n}}\operatorname{Ln}z=(-1)^{n-1}(n-1)!z% ^{-n}.$
For a nonvanishing analytic function $f(z)$, the general solution of the differential equation
4.7.5 $\frac{\mathrm{d}w}{\mathrm{d}z}=\frac{f^{\prime}(z)}{f(z)}$
is
4.7.6 $w(z)=\operatorname{Ln}\left(f(z)\right)+\hbox{ constant}.$ ⓘ Defines: $w$: solution (locally) Symbols: $\operatorname{Ln}\NVar{z}$: general logarithm function, $z$: complex variable and $f(z)$: non-vanishing analytic function Permalink: http://dlmf.nist.gov/4.7.E6 Encodings: TeX, pMML, png See also: Annotations for 4.7(i), 4.7 and 4
## §4.7(ii) Exponentials and Powers
4.7.7 $\frac{\mathrm{d}}{\mathrm{d}z}e^{z}=e^{z},$ ⓘ Symbols: $\frac{\mathrm{d}\NVar{f}}{\mathrm{d}\NVar{x}}$: derivative of $f$ with respect to $x$, $\mathrm{e}$: base of exponential function and $z$: complex variable A&S Ref: 4.2.49 Permalink: http://dlmf.nist.gov/4.7.E7 Encodings: TeX, pMML, png See also: Annotations for 4.7(ii), 4.7 and 4
4.7.8 $\frac{\mathrm{d}}{\mathrm{d}z}e^{az}=ae^{az},$ ⓘ Symbols: $\frac{\mathrm{d}\NVar{f}}{\mathrm{d}\NVar{x}}$: derivative of $f$ with respect to $x$, $\mathrm{e}$: base of exponential function, $a$: real or complex constant and $z$: complex variable A&S Ref: 4.2.50 (gives n-th derivative.) Permalink: http://dlmf.nist.gov/4.7.E8 Encodings: TeX, pMML, png See also: Annotations for 4.7(ii), 4.7 and 4
4.7.9 $\frac{\mathrm{d}}{\mathrm{d}z}a^{z}=a^{z}\ln a,$ $a\neq 0$.
When $a^{z}$ is a general power, $\ln a$ is replaced by the branch of $\operatorname{Ln}a$ used in constructing $a^{z}$.
4.7.10 $\frac{\mathrm{d}}{\mathrm{d}z}z^{a}=az^{a-1},$ ⓘ Symbols: $\frac{\mathrm{d}\NVar{f}}{\mathrm{d}\NVar{x}}$: derivative of $f$ with respect to $x$, $a$: real or complex constant and $z$: complex variable A&S Ref: 4.2.52 Permalink: http://dlmf.nist.gov/4.7.E10 Encodings: TeX, pMML, png See also: Annotations for 4.7(ii), 4.7 and 4
4.7.11 $\frac{{\mathrm{d}}^{n}}{{\mathrm{d}z}^{n}}z^{a}=a(a-1)(a-2)\cdots(a-n+1)z^{a-n}.$
The general solution of the differential equation
4.7.12 $\frac{\mathrm{d}w}{\mathrm{d}z}=f(z)w$
is
4.7.13 $w=\exp\left(\int f(z)\mathrm{d}z\right)+{\rm constant}.$
The general solution of the differential equation
4.7.14 $\frac{{\mathrm{d}}^{2}w}{{\mathrm{d}z}^{2}}=aw,$ $a\neq 0$,
is
4.7.15 $w=Ae^{\sqrt{a}z}+Be^{-\sqrt{a}z},$
where $A$ and $B$ are arbitrary constants.
For other differential equations see Kamke (1977, pp. 396–413). | 2017-09-26T10:55:56 | {
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http://www.eurisko.us/all_problems_iteration_2.html | # Problem 4-1¶
Location: assignment-problems/flatten.py
Grading: 1 point for passing test, and then (assuming it passes the test) 1 point for code quality
Write a function flatten which takes a nested dictionary and converts it into a flat dictionary based on the key names. You can assume that the nested dictionary only has one level of nesting, meaning that in the output, each key will have exactly one underscore.
Assert that your function passes the following test:
>>> colors = {
'animal': {
'bumblebee': ['yellow', 'black'],
'elephant': ['gray'],
'fox': ['orange', 'white']
},
'food': {
'apple': ['red', 'green', 'yellow'],
'cheese': ['white', 'orange']
}
}
>>> flatten(colors)
{
'animal_bumblebee': ['yellow', 'black'],
'animal_elephant': ['gray'],
'animal_fox': ['orange', 'white'],
'food_apple': ['red', 'green', 'yellow'],
'food_cheese': ['white', 'orange']
}
# Problem 4-2¶
Location: assignment-problems/convert_to_base_2.py
Grading: 1 point for passing test, and then (assuming it passes the test) 1 point for code quality
Write a function convert_to_base_2 that converts a number from base-10 to base-2. Assert that it passes the following test:
>>> convert_to_base_2(19)
10011
Hint: use $\log_2$ to figure out how many digits there will be in the binary number. Then, fill up the binary number, repeatedly subtracting off the next-largest power of 2 if possible.
# Problem 4-3¶
Location: assignment-problems/linear_encoding_cryptography.py
Grading: for each part, you get 1 point for passing test, and then (assuming it passes the test) 1 point for code quality
In Assignment 1, we encountered the trivial encoding function which maps
• ' ' $\rightarrow 0,$
• 'a' $\rightarrow 1,$
• 'b' $\rightarrow 2,$
and so on.
Using a linear encoding function $s(x) = 2x+3,$ the message 'a cat' can be encoded as follows:
1. Original message: 'a cat'
2. Trivial encoding: [1, 0, 3, 1, 20]
3. Linear encoding: [5, 3, 9, 5, 43]
a. Create a function encode(string,a,b) which encodes a string using the linear encoding function $s(x) = ax+b.$ Assert that your function passes the following test:
>>> get_encoding('a cat', 2, 3)
[5, 3, 9, 5, 43]
b. Create a function decode(numbers,a,b) which attempts to decode a given list of numbers using the linear encoding function $s(x) = ax+b.$
To do this, you should apply the inverse encoding $s^{-1}(x) = \dfrac{x-b}{a},$ to all the numbers in the list and then check if they are all integers in the range from $0$ to $26$ (inclusive). If they are, then return the corresponding letters; if they are not, then return False.
Assert that your function passes the following tests:
>>> decode([5, 3, 9, 5, 43], 2, 3)
'a cat'
for debugging purposes, here's the scratch work for you:
[(5-3)/2, (3-3)/2, (9-3)/2, (5-3)/2, (43-3)/2]
[1, 0, 3, 1, 20]
'a cat'
>>> decode([1, 3, 9, 5, 43], 2, 3)
False
for debugging purposes, here's the scratch work for you:
[(1-3)/2, (3-3)/2, (9-3)/2, (5-3)/2, (43-3)/2]
[-1, 0, 3, 1, 20]
False (because -1 does not correspond to a letter)
>>> decode([5, 3, 9, 5, 44], 2, 3)
False
for debugging purposes, here's the scratch work for you:
[(5-3)/2, (3-3)/2, (9-3)/2, (5-3)/2, (44-3)/2]
[1, 0, 3, 1, 20.5]
False (because 20.5 does not correspond to a letter)
c. Decode the message
[377,
717,
71,
513,
105,
921,
581,
547,
547,
105,
377,
717,
241,
71,
105,
547,
71,
377,
547,
717,
751,
683,
785,
513,
241,
547,
751],
given that it was encoded with a linear encoding function $s(x) = ax+b$ where $a,b \in \{ 0, 1, 2, \ldots, 100 \}.$
You should run through each combination of $a$ and $b,$ try to decode the list of numbers using that combination, and if you get a valid decoding, then print it out. Then, you can visually inspect all the decodings you printed out to find the one that makes sense.
# Problem 3-1¶
Location: assignment-problems/convert_to_base_10.py
Grading: 1 point for passing test, and then (assuming it passes the test) 1 point for code quality
Write a function convert_to_base_10 that converts a number from base-2 (binary) to base-10 (decimal). For example, the binary number $10011$ corresponds to the decimal number $$1 \cdot 2^{4} + 0 \cdot 2^3 + 0 \cdot 2^2 + 1 \cdot 2^1 + 1 \cdot 2^0 = 19.$$
Assert that your function passes the following test:
>>> convert_to_base_10(10011)
19
# Problem 3-2¶
Location: assignment-problems/make_nested.py
Grading: you get 1 point for passing the test, and then (assuming it passes the test) 1 point for code quality
Write a function make_nested which takes a "flat" dictionary and converts it into a nested dictionary based on underscores in the the key names. You can assume that all keys have exactly one underscore.
Assert that your function passes the following test:
>>> colors = {
'animal_bumblebee': ['yellow', 'black'],
'animal_elephant': ['gray'],
'animal_fox': ['orange', 'white'],
'food_apple': ['red', 'green', 'yellow'],
'food_cheese': ['white', 'orange']
}
>>> make_nested(colors)
{
'animal': {
'bumblebee': ['yellow', 'black'],
'elephant': ['gray'],
'fox': ['orange', 'white']
},
'food': {
'apple': ['red', 'green', 'yellow'],
'cheese': ['white', 'orange']
}
}
# Problem 3-3¶
Location: assignment-problems/stack.py
Grading: you get 0.5 points for passing each test, and then (assuming your code passes all the tests) 2 points for code quality.
Implement a stack. That is, create a class Stack which operates on an attribute data using the following methods:
• push: add a new item on top of the stack
• pop: remove the top (rightmost) item from the stack
• peek: return the top item without modifying the stack
Assert that your class passes the following sequence of 5 tests. (You should write 5 assert statements in total.)
>>> s = Stack()
>>> s.data
[]
>>> s.push('a')
>>> s.push('b')
>>> s.push('c')
>>> s.data
['a', 'b', 'c']
>>> s.pop()
>>> s.data
['a', 'b']
>>> s.peek()
'b'
>>> s.data
['a', 'b']
# Problem 2-1¶
Location: assignment-problems/union_intersection.py
a. (1 point for code quality; 1 point for passing test)
Write a function intersection that computes the intersection of two lists. Assert that it passes the following test:
>>> intersection([1,2,'a','b'], [2,3,'a'])
[2,'a']
b. (1 point for code quality; 1 point for passing test)
Write a function union that computes the union of two lists. Assert that it passes the following test:
>>> union([1,2,'a','b'], [2,3,'a'])
[1,2,3,'a','b']
# Problem 2-2¶
Location: assignment-problems/count_characters.py
(2 points for code quality; 2 points for passing test)
Write a function count_characters that counts the number of each character in a string and returns the counts in a dictionary. Lowercase and uppercase letters should not be treated differently.
Assert that your function passes the following test:
>>> countCharacters('A cat!!!')
{'a': 2, 'c': 1, 't': 1, ' ': 1, '!': 3}
# Problem 2-3¶
Location: assignment-problems/recursive_sequence.py
Consider the sequence defined recursively as
$$a_n = 3a_{n-1} -4, \quad a_1 = 5.$$
a. (1 point for code quality; 1 point for passing test)
Write a function first_n_terms that returns a list of the first $n$ terms of the sequence: $[a_1, a_2, a_3, \ldots, a_{n}]$
Assert that your function passes the following test:
>>> first_n_terms(10)
[5, 11, 29, 83, 245, 731, 2189, 6563, 19685, 59051]
b. (1 point for code quality; 1 point for passing test)
Write a function nth_term that computes the $n$th term of the sequence using recursion. Here's the video that you were asked to watch before class, in case you need to refer back to it: https://www.youtube.com/watch?v=zbfRgC3kukk
Assert that your function passes the following test:
>>> nth_term(10)
59051
# Problem 1-1¶
Getting started...
2. Create a bash repl named assignment-problems
3. Create a file assignment-problems/test_file.py
5. On repl.it, link assignment-problems to your github and push up your work to github. Name your commit "test commit".
6. After you complete this assignment, again push your work up to github. Name your commit "completed assignment 1".
# Problem 1-2¶
Location: assignment-problems/is_symmetric.py
Note: This problem is worth 1 point for passing both tests, plus another 1 point for code quality (if you pass the tests). So, the rubric is as follows:
• 0/2 points: does not pass both tests
• 1/2 points: passes both tests but code is poor quality
• 2/2 points: passes both tests and code is high quality
Write a function is_symmetric(input_string) that checks if a string reads the same forwards and backwards, and assert that your function passes the following tests:
>>> is_symmetric('racecar')
True
>>> is_symmetric('batman')
False
To be clear -- when you run is_symmetric.py, your code should print the following:
>>> python is_symmetric.py
testing is_symmetric on input 'racecar'...
PASSED
testing is_symmetric on input 'batman'...
PASSED
# Problem 1-3¶
Location: assignment-problems/letters_numbers_conversion.py
a. (1 point for passing test, 1 point for code quality)
Write a function convert_to_numbers(input_string) that converts a string to a list of numbers, where space = 0, a = 1, b = 2, and so on. Then, assert that your function passes the following test:
>>> letters2numbers('a cat')
[1,0,3,1,20]
b. (1 point for code quality, 1 point for passing test)
Write a function convert_to_letters(input_string) that converts a list of numbers to the corresponding string, and assert that your function passes the following test:
>>> convert_to_letters([1,0,3,1,20])
'a cat'
To be clear -- when you run letters_numbers_conversion.py, your code should print the following:
>>> python letters_numbers_conversion.py
testing convert_to_letters on input [1,0,3,1,20]...
PASSED
testing convert_to_numbers on input 'batman'...
PASSED
# Problem 1-4¶
(2 points for passing tests, 2 points for code quality)
Write a function is_prime(n) that checks if an integer input $n > 1$ is prime by checking whether $m | n$ for any integer $m \in \left\{ 2, 3, \ldots, \left\lfloor \dfrac{n}{2} \right\rfloor \right\}.$
• $m|n$ means "$m$ divides $n$"
• $\left\lfloor \dfrac{n}{2} \right\rfloor$ is called the "floor" of $\dfrac{n}{2},$ i.e. the greatest integer that is less than or equal to $\dfrac{n}{2}.$
(Hint: Check for divisibility within a for loop.)
Also, assert that your function passes the following tests:
>>> is_prime(59)
True
>>> is_prime(51)
False
To be clear -- when you run is_prime.py, your code should print the following:
>>> python is_prime.py
testing is_prime on input 59...
PASSED
testing is_prime on input 51...
PASSED
In [ ]: | 2022-05-17T17:08:04 | {
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https://math.stackexchange.com/questions/3407996/how-to-write-xmx-m-as-a-polynomial-in-xx-1 | # How to write $x^m+x^{-m}$ as a polynomial in $x+x^{-1}$.
I did find this question Proving that $x^m+x^{-m}$ is a polynomial in $x+x^{-1}$ of degree $m$. but it only shows by induction that this is possible, not the actual form of the solution. How, for example, would you write $$x^2+x^{-2}$$ in a polynomial of $$x+x^{-1}$$?
• Did you mean $x^2\color{red}+x^{-2}$? Because in that case it's $(x+x^{-1})^2-2$ Oct 25, 2019 at 4:18
• Yes I did, edited. How did you come up with that? Just from squaring $(x+x^-1)$ and subtracting off the extra $2$, or is there a more general way? Oct 25, 2019 at 4:21
• This answer math.stackexchange.com/a/1341510/42969 to the referenced question shows the explicit solution. Oct 25, 2019 at 4:34
It's just a simple system of equations (simple, if the degree is low...)
You want \begin{align} x^2+\frac1{x^2}&=a+b\left(x+\frac1x\right)+c\left(x+\frac1x\right)^2\\ &=a+2c+bx+\frac bx+cx^2+\frac c{x^2}. \end{align} And you see by inspection that $$a+2c=1$$, $$b=0$$, $$c=1$$. So $$a=-2$$. That is \begin{align} x^2+\frac1{x^2}&=-2+\left(x+\frac1x\right)^2\\ \end{align}
• Thanks, I will mark this as the accepted answer since it does generalize for $m>2$. I see it is very simple now, but thank you for the clear explanation. Oct 25, 2019 at 4:27
These are basically Chebyshev polynomials. One can derive an explicit formula for them via generating functions.
Let $$y=x+x^{-1}$$ and consider the formal power series $$F(x,t)=\sum_{n=0}^\infty(x^n+x^{-n})t^n.$$ Then \begin{align} F(x,t)&=\sum_{n=0}^\infty x^nt^n+\sum_{n=0}^\infty x^{-n}t^n =\frac1{1-xt}+\frac1{1-x^{-1}t}=\frac{(1-xt)+(1-x^{-1}t)}{(1-xt)(1-x^{-1}t)}\\ &=\frac{2-(x+x^{-1})t}{1-(x+x^{-1})t+t^2}=\frac{2-yt}{1-yt+t^2}=(2-yt) \frac1{1-(y-t)t}\\ &=(2-yt)\sum_{m=0}^\infty(y-t)^mt^m=(2-yt)\sum_{m=0}^n\sum_{k=0}^{m} (-1)^k\binom{m}{k}y^{m-k}t^{m+k}\\ &=(2-yt)\sum_{m=0}^\infty(y-t)^mt^m=(2-yt)\sum_{n=0}^\infty t^n\sum_{k:0\le k\le n/2}(-1)^k\binom{n-k}{k}y^{n-2k}. \end{align} Comparing coefficients of $$t^n$$ gives $$x^n+x^{-n}=2\sum_{k:0\le k\le n/2}(-1)^k\binom{n-k}{k}y^{n-2k} -\sum_{k:0\le k\le (n-1)/2}(-1)^k\binom{n-1-k}{k}y^{n-2k}.$$ One can simplify this a bit: $$2\binom{n-k}{k}-\binom{n-1-k}{k}=\binom{n-k}{k}+\binom{n-k-1}{k-1} =\binom{n-k}{k}+\frac{k}{n-k}\binom{n-k}{k}$$ to get $$x^n+x^{-n}=\sum_{k:0\le k\le n/2}(-1)^k\frac{n}{n-k} \binom{n-k}{k}(x+x^{-1})^{n-2k}.$$
• Very interesting. I have not studied formal power series in more than one variable before. Oct 25, 2019 at 6:26
In order to get $$x^2$$ and $$x^{-2}$$ terms, square $$(x+x^{-1}$$). The result is $$x^2+2+x^{-2}$$.
So subtract $$2$$ to get what you want: $$x^2+x^{-2}=(x+x^{-1})^2-2$$.
• Thanks, does this generalize to $x^m+x^{-m}$ for $m>2$? Oct 25, 2019 at 4:25
• $(x+x^{-1})^3=x^3+3x+3x^{-1}+x^{-3}$ so $x^3+x^{-3}=(x+x^{-1})^3-3(x+x^{-1})$ Oct 25, 2019 at 4:30
• $(x+x^{-1})^4=x^4+4x^2+6+4x^{-2}+x^{-4}$ so $x^4+x^{-4}=(x+x^{-1})^4-4(x^2+x^{-2})-6=(x+x^{-1})^4-4(x+x^{-1})^2+2$ Oct 25, 2019 at 4:35
• I get the point, you can stop now :) Oct 25, 2019 at 4:39
• :), okay, thanks, will do Oct 25, 2019 at 4:39
The inductive argument does tell you how to do this - and, it's worthwhile to see that induction very often gives you the answers you are after, especially since the fact that inductive proofs can be unrolled is often overlooked.
In the linked answer, it is noted that $$x^{k+1}+x^{-k-1} = (x^k +x^{-k})(x+x^{-1}) - (x^{k-1} + x^{-k+1}).$$ and an inductive argument is built from this, but remember that the inductive hypothesis is just that, for each $$k$$, there exists some polynomial $$P_k$$ such that $$P_k(x+x^{-1})= x^k+x^{-k}$$. If we fill out the proof more completely, the implication here is that $$x^{k+1}+x^{-k-1} = P_k(x+x^{-1})\cdot (x+x^{-1}) - P_{k-1}(x+x^{-1}).$$ And then we see that the left hand side is a polynomial in $$x+x^{-1}$$ as well - but we can be more explicit: Let $$P_{k+1}(z)=zP_k(z)-P_{k-1}(z)$$ where we start the sequence as $$P_0(z)=2$$ and $$P_1(z)=z$$. Then, we have $$P_k(x+x^{-1})=x^k+x^{-k}$$ due to the inductive argument. Note that this sequence is very easy to compute incrementally: $$P_2(z)=z\cdot P_1(z) - P_0(z) = z^2 - 2$$ $$P_3(z)=z\cdot P_2(z) - P_1(z) = z^3 - 3z$$ $$P_4(z)=z\cdot P_3(z) - P_2(z) = z^4 - 4z^2 + 2$$ and so on.
As pointed out in the comments, it is possible to write these terms in a general form, although it's a bit surprising that the form you get is actually a polynomial. I won't go into details, since I'm using the usual tools for solving linear homogenous recurrences:
$$P_k(z) = \frac{\left(z - \sqrt{z^2-4}\right)^k + \left(z + \sqrt{z^2-4}\right)^k}{2^k}$$
Note that the $$\sqrt{z^2-4}$$ terms cancel out due to the symmetry of the sum and this does always leave a polynomial.
(I also think this sequence has a name, but I don't know what it is)
• It might be worthwhile to expand this to actually obtain an explicit formula, which should be easy to do since we already have the recurrence (using e.g. generating functions). Oct 25, 2019 at 4:37
• That is very interesting. I have not thought to do this before. I will have to keep it in mind, thanks! Oct 25, 2019 at 4:41
• @YiFan Thanks, I added it, I'd thought that the explicit formula was a bit worse than it is when I was writing this. Oct 25, 2019 at 4:43 | 2022-07-04T06:57:59 | {
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https://math.stackexchange.com/questions/382174/ways-of-merging-two-incomparable-sorted-lists-of-elements-keeping-their-relative | # Ways of merging two incomparable sorted lists of elements keeping their relative ordering
Suppose that, for a real application, I have ended up with a sorted list A = {$a_1, a_2, ..., a_{|A|}$} of elements of a certain kind (say, Type-A), and another sorted list B = {$b_1, b_2, ..., b_{|B|}$} of elements of a different kind (Type-B), such that Type-A elements are only comparable with Type-A elements, and likewise for Type-B.
At this point I seek to count the following: in how many ways can I merge both lists together, in such a way that the relative ordering of Type-A and Type-B elements, respectively, is preserved? (i.e. that if $P_M(x)$ represents the position of an element of A or B in the merged list, then $P_M(a_i)<P_M(a_j)$ and $P_M(b_i)<P_M(b_j)$ for all $i<j$)
I've tried to figure this out constructively by starting with an empty merged list and inserting elements of A or B one at a time, counting in how many ways each insertion can be done, but since this depends on the placement of previous elements of the same type, I've had little luck so far. I also tried explicitly counting all possibilities for different (small) lengths of A and B, but I've been unable to extract any potential general principle in this way.
The merged list has length $|A|+|B|$. Once you know which $|A|$ positions in it are occupied by the elements of list A, you also know exactly how the whole thing has to be ordered, since the internal orders of the elements of A and the elements of B are already known. Thus, there are $$\binom{|A|+|B|}{|A|}=\binom{|A|+|B|}{|B|}$$ possible merged lists, one for each choice of $|A|$ positions for the elements of list A.
• Great answer! I've made an impl in C++ that generates all possible interleaved combinations of two lists, while maintaining relative order for anyone who is interested: github.com/domfarolino/algorithms/tree/master/src/… Oct 22, 2018 at 20:37
• Can this further generalize to 3 lists? To 4? To N? May 27, 2021 at 15:54
• Indeed, I believe this can further generalize to N lists, by viewing the overall operation as a series of (N-1) merges, so the total number of merged lists is computed as the product of (N-1) Binomial Coeffecients, where the i-th expression computes the selection of the positions of the i-th lists's elements from the available positions left after the first (i-1) lists have already been merged. May 27, 2021 at 21:38
• @BryanPendleton: Yes: it’s a multinomial coefficient, $$\binom{|A_1|+|A_2|+\ldots+|A_N|}{A_1,A_2,\ldots,A_N}\,,$$ which can be expanded as a product of binomial coefficients as shown at the link. May 28, 2021 at 3:57
• Great answer. Since we have to keep the relative ordering. This question is equal to "How many different ways are there to select n positions from m + n positions?“ Sep 17, 2021 at 5:47
$\frac{(m+n)!}{(m)!(n)!}$; where m,n is the length of the arrays. $(m+n)!$ is the total number of ways of arranging without any restriction. The condition given that you can not change the order of each array so you should not change the order of each array , so we need to divide all the permutations of each array!
For merging N sorted lists, here is a good way to see that the solution is $$\frac{(|A_1|+\dots|+|A_N|)!}{|A_1|!\dots |A_N|!}$$
All the $$(|A_1|+\dots|+|A_N|)$$ elements can be permuted in $$(|A_1|+\dots|+|A_N|)!$$ ways. Among these, any solution which has the ordering of the $$A_1$$ elements different from the given order have to be thrown out. If you keep the positions of everyone except $$A_1$$ fixed, you can generate $$|A_1|!$$ permutations by shuffling $$A_1$$. Only 1 among these $$|A_1|!$$ solutions is valid. Hence the number solutions containing the right order of $$A_1$$ is $$\frac{(|A_1|+\dots|+|A_N|)!}{|A_1|!}$$
Now repeat the argument for $$A_2$$, $$A_2, \dots A_N$$. | 2022-07-05T09:49:13 | {
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http://math.stackexchange.com/questions/503155/finding-coordinate-point-if-two-points-and-time-are-known | # Finding coordinate point if two points and time are known
$A$ is at coordinate $(x_1,y_1)$ at $6:00$ am and is at coordinate $(x_2,y_2)$ at $6:15$ am. How can I know where $A$ will be at $6:05$ am i.e $(x_3,y_3)$? The value for $x_1=392,y_1=455,x_2=512,y_2=452$ are known and is straight line.
I understand there is a formula to calculate this in coordinate mathematics. I will appreciate your answer and if you share the formula or term in mathematics.
-
In my answer, I assume $A$ moves at a constant speed. Is that what you need? – Rustyn Sep 24 '13 at 3:45
It takes $A$ $15$ minutes, to go $-3$ units downward and $120$ units rightward. So he's traveling, $$\sqrt{3^2 + 120^2} = 3\sqrt{1601}$$ units per $15$ minutes.
Thus, at $6:05$, he'll have gone $\sqrt{1601}$, units from is starting position, $\left(\tfrac{1}{3}\text{ the distance }\right)$. We know that the point $(x_3,y_3)$ is on the line: $$y-455=\frac{-3}{120}(x-392)$$ Thus, $y_3 = \frac{-3}{120}(x_3-392)+455$
AND: $$\sqrt{(x_3-x_1)^2+(y_3-y_1)^2} = \sqrt{(x_3-392)^2+(y_3-455)^2} = \sqrt{1601}$$ So, putting this all together we have: $$\sqrt{(x_3-392)^2+\left(\tfrac{-3}{120}(x_3-392)+455-455\right)^2} = \sqrt{1601} \Longrightarrow \dots$$ $$x_3=432, y_3 = 454$$ EDIT
As Ross Millikan states, we can avoid the square roots entirely. See his comment below for the simplest solution to this problem.
-
You needn't do the square roots. Just compute the $x$ velocity $v_x=\frac {-3}{15}$ and $y$ velocity $v_y=\frac {120}{15}=8$ The position at time $t$ (measured in minutes after $6:00$) is $(x_1+v_xt, y_1+v_yt)$ – Ross Millikan Sep 24 '13 at 3:48
@RossMillikan You are absolutely right. Except for I think you got them switched around? $v_x = 120/15$, $v_y = -3/15$ – Rustyn Sep 24 '13 at 3:49
Is A moving at a constant speed? Use the distance formula: http://en.wikipedia.org/wiki/Distance_formula#Geometry
At a constant speed, the distance traveled is proportional to the time spent traveling because of the distance = rate x time equation.
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http://math.stackexchange.com/questions/702616/in-a-commutative-ring-is-addition-necessarily-commutative | # In a Commutative Ring, is Addition Necessarily Commutative?
In A First Course in Abstract Algebra, Fraleigh writes on p. 172 that "a ring in which multiplication is commutative is a commutative ring". Of course, this raises the question: is addition necessarily commutative in a commutative ring? Is it commutative in any ring? Are there examples of rings in which addition is non-commutative?
As far as I could tell, commutativity of the underlying additive operation is not one of the defining properties of a ring. Or is it? I don't think the book has been entirely clear on these issues --- I would be thankful if someone could shed some light on this.
-
The definition of a ring $R$ is that it has two binary operations, $+$ which is an abelian group and $\cdot$ which is not necessarily commutative (and need not even be a group, merely associative). Of course, you have your extra distributive properties added as well. – mathematics2x2life Mar 7 '14 at 4:06
Possible duplicate of Why is ring addition commutative? – Najib Idrissi Nov 23 '15 at 18:59
The three basic properties of a ring are;
• The set under addition makes an abelian group,
• Multiplication is associative, and
• Left and right distributive laws hold.
Thus, by definition of "abelian group", the addition must be commutative. Hopes that help.
-
Oh. I totally glossed over the fact that the underlying group under addition is abelian. – Newb Mar 7 '14 at 4:11
@Newb A ring with not necessarily commutative addition is sometimes called a near-ring. I don't know a whole lot of study about them (but I'm generally ignorant of that sort of thing). – Mike Miller Mar 7 '14 at 4:25
In order to generalize rings to structures with noncommutative addiiton, one cannot simply delete the axiom that addition is commutative, since, in fact, other (standard) ring axioms force addition to be commutative (Hankel, 1867 [1]). The proof is simple: apply both the left and right distributive law in different order to the term $\rm\:(1\!+\!1)(x\!+\!y),\:$ viz.
$$\rm (1\!+\!1)(x\!+\!y) = \bigg\lbrace \begin{eqnarray} (1\!+\!1)x\!+\!(1\!+\!1)y\, =\, x \,+\, \color{#C00}{x\!+\!y} \,+\, y\\ \rm 1(x\!+\!y)\!+1(x\!+\!y)\, =\, x\, +\, \color{#0A0}{y\!+\!x}\, +\, y\end{eqnarray}\bigg\rbrace\:\Rightarrow\: \color{#C00}{x\!+\!y}\,=\,\color{#0A0}{y\!+\!x}\ \ by\ \ cancel\ \ x,y$$
Thus commutativity of addition, $\rm\:x+y = y+x,\:$ is implied by these axioms:
$(1)\ \ *\,$ distributes over $\rm\,+\!:\ \ x(y+z)\, =\, xy+xz,\ \ (y+z)x\, =\, yx+zx$
$(2)\ \, +\,$ is cancellative: $\rm\ \ x+y\, =\, x+z\:\Rightarrow\: y=z,\ \ y+x\, =\, z+x\:\Rightarrow\: y=z$
$(3)\ \, +\,$ is associative: $\rm\ \ (x+y)+z\, =\, x+(y+z)$
$(4)\ \ *\,$ has a neutral element $\rm\,1\!:\ \ 1x = x$
Said more structurally, recall that a SemiRing is that generalization of a Ring whose additive structure is relaxed from a commutative Group to merely a SemiGroup, i.e. here the only hypothesis on addition is that it be associative (so in SemiRings, unlike Rings, addition need not be commutative, nor need every element $\rm\,x\,$ have an additive inverse $\rm\,-x).\,$ Now the above result may be stated as follows: a semiring with $\,1\,$ and cancellative addition has commutative addition. Such semirings are simply subsemirings of rings (as is $\rm\:\Bbb N \subset \Bbb Z)\,$ because any commutative cancellative semigroup embeds canonically into a commutative group, its group of differences (in precisely the same way $\rm\,\Bbb Z\,$ is constructed from $\rm\,\Bbb N,\,$ i.e. the additive version of the fraction field construction).
Examples of SemiRings include: $\rm\,\Bbb N;\,$ initial segments of cardinals; distributive lattices (e.g. subsets of a powerset with operations $\cup$ and $\cap$; $\rm\,\Bbb R\,$ with + being min or max, and $*$ being addition; semigroup semirings (e.g. formal power series); formal languages with union, concat; etc. For a nice survey of SemiRings and SemiFields see [2]. See also Near-Rings.
[1] Gerhard Betsch. On the beginnings and development of near-ring theory. pp. 1-11 in:
Near-rings and near-fields. Proceedings of the conference held in Fredericton, New Brunswick, July 18-24, 1993. Edited by Yuen Fong, Howard E. Bell, Wen-Fong Ke, Gordon Mason and Gunter Pilz. Mathematics and its Applications, 336. Kluwer Academic Publishers Group, Dordrecht, 1995. x+278 pp. ISBN: 0-7923-3635-6 Zbl review
[2] Hebisch, Udo; Weinert, Hanns Joachim. Semirings and semifields. $\$ pp. 425-462 in: Handbook of algebra. Vol. 1. Edited by M. Hazewinkel. North-Holland Publishing Co., Amsterdam, 1996. xx+915 pp. ISBN: 0-444-82212-7 Zbl review, AMS review
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I would upvote this answer multiple times if I could. As usual, Bill, your answer is highly informative. – Newb Mar 7 '14 at 7:50
The definition of a ring is that it has two binary operations, $+$ and $\cdot$. The $+$ operation forms an abelian group and $\cdot$ need only be associative. The distributive laws need hold. However, notice that the distributive laws force $+$ to be abelian when $R$ has $1$!
$$(1+1)(x+y)=1(x+y)+1(x+y)=x+y+x+y$$
and
$$(1+1)(x+y)=(1+1)x+(1+1)y=x+x+y+y$$ so that $x+y+x+y=x+x+y+y$ then adding $-x$ and $-y$ ($+$ forms a group so has inverses $-x,-y$) on the left and right yields $y+x=x+y$.
-
A lot of people don't accept a ring without unity as a ring. – Tim Seguine Mar 7 '14 at 11:18
But they do indeed exist and can be useful, thought they do not appear often. So it is important to know they rings are not required to have such properties. Then one can happily continue working with rings with identity. – mathematics2x2life Mar 7 '14 at 21:16 | 2016-02-10T22:17:34 | {
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https://www.physicsforums.com/threads/apc-4-1-2-find-the-slope-at-x-4.1043355/ | # -apc.4.1.2 Find the slope at x=4
• MHB
Gold Member
MHB
$\tiny{apc.4.1.2}$
Find the slope of the tangent line to the graph of
$f(x)=-x^2+4\sqrt{x}$ at $x=4$
$a.\ 8\quad b.\ -10\quad c.\ -9\quad d.\ -5\quad e.\ -7$
$f'(x)=-2x+\frac{2}{\sqrt{x}}$
$m=f'(4)=-2(4)+\frac{2}{\sqrt{4}}=-7$ which is (e)
however not asked for here but I forgot how to find $b$ of $y=-7x+b$
Last edited:
skeeter
$y - f(4) = -7(x-4)$
clean it up ...
Gold Member
MHB
$y - f(4) = -7(x-4)$
clean it up ...
that didn't seem to be a tangent line to f{x}
$-8=-7(4)+b$
$-8+28=20=b$
$y=-7x+20$
Last edited:
skeeter
the equation in my previous post is in point-slope form, $y - y_1 = m(x - x_1)$ where $m = -7, \, x_1 = 4 \text{ and } y_1 = f(4)$
$f(4) = -8$
$y + 8 = -7(x - 4) \implies y = -7x + 20$
HOI
that didn't seem to be a tangent line to f{x}
$-8=-7(4)+b$
$-8+28=20=b$
$y=-7x+20$
Yes, it is. Since $f(x)= -x^2+ 4\sqrt{x}$ so $f(4)= -4^2+4\sqrt{4}= -16+ 8= -8$. That is, the expression skeeter gave, $y- f(4)= -7(x- 4)$ becomes $y- (-8)= y+ 8= -7x+ 28$ so, subtracting 8 from both sides, $y= -7x+ 20$ as you have.
TiffanyBK
Is the option wrong? I used cameramath and the answer is -8
HOI
Is the option wrong? I used cameramath and the answer is -8
View attachment 10874
The answer to what question? You want to find the equation of the tangent line to $f(x)= -x^2+ 4\sqrt{x}$ at x= 4. You have already determined that the derivative of f, so the slope of the tangent line, at x= 4 is -7. Thus you know that the equation of the tangent line is y= -7x+ b for some number, b. So what is the value of y at x= 4? $y= f(4)= -(4^2)+ 4\sqrt{4}= -16+ 4(2)= -8$ -8 is the value of f(4). But that is not necessarily b! Now we know that we must have y= -7(4)+ b= -28+ b= -8. Adding 28 to both sides, b= 20.
The equation of the tangent line at x= 4 is y= -7x+ 20.
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MHB
well i think the confusion was f(4) vs f'(4) | 2023-03-22T02:52:57 | {
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https://math.stackexchange.com/questions/2713020/why-is-the-monotone-convergence-theorem-more-famous-than-its-stronger-cousin | # Why is the Monotone Convergence Theorem more famous than it's stronger cousin?
I am reading Stein & Shakarchi. On page 62 we have the Monotone Convergence Theorem:
Suppose $\{f_n\}$ is a sequence of non-negative measurable functions with $f_n\nearrow f.$ Then $\displaystyle \lim_{n \to \infty} \int f_n = \int f$.
However, just before this theorem we have this much more powerful corrolary of Fatou's lemma:
Suppose $f$ is a non-negative measurable function, and $\{f_n\}$ a sequence of non-negative measurable functions with $f_n(x) \le f(x)$ and $f_n(x) \to f(x)$ for a.e. $x$. Then $\displaystyle \lim_{n \to \infty} \int f_n = \int f$.
To me, this second corollary seems strictly better than the Monotone Convergence Theorem, yet it is the M.C.T. that has a name and is used often. Am I misunderstanding the theorems, or is there a reason why the M.C.T. is more popular?
Does this corollary have a name?
• Are you sure that corollary is true? I think you would still require integrability of $f$. – Calculon Mar 29 '18 at 9:55
• @Calculon I copied it word for word; I will recheck the proof – Ovi Mar 29 '18 at 9:57
• @Calculon $f$ is required to be non-negative and measurable, so the worst that could happen is that $\int f=+\infty$, in which case the inequality holds trivially. – Arnaud D. Mar 29 '18 at 9:58
• This question seems related : math.stackexchange.com/questions/2468412/… – Arnaud D. Mar 29 '18 at 10:08
• @ArnaudD. Nice catch! – Ovi Mar 30 '18 at 14:57
Some people do call this corollary monotone convergence theorem. Moreover, some instructora introduce this corollary as M.C.T., rather than the origianl version of M.C.T., for instance, Professor Mark J. Schervish and Professor Alessandro Rinaldo at CMU (Lecture notes). It's fine because this corollary is stronger and more practical.
However, from the perspective of mathematics, I think it's better to be traditional and use the original version of M.C.T. and just regard that corollary as a useful corollary.
Let's take a look at the Lebesgue's dominant convergence theorem:
Suppose $\{f_n\}$ and $f$ are measurable functions. If there exists a non-negative function $g$ such that $\int g <\infty$ and $|f_n|\leq g$ a.e., for all $n$. Then $f_n\overset{a.e.}{\rightarrow} f$ or $f_n\overset{\mu}{\rightarrow} f$ implies that $$\lim_{n \to \infty} \int f_n = \int f.$$
Now, compare this with the result you give:
Suppose $f$ is a non-negative measurable function, and $\{f_n\}$ a sequence of non-negative measurable functions with $f_n(x) \le f(x)$ and $f_n(x) \to f(x)$ for a.e. $x$. Then $\displaystyle \lim_{n \to \infty} \int f_n = \int f$.
We can find that when $\int f <\infty$, the result is implied by the Lebesgue's D.C.T. When $\int f =\infty$, by Fatou's lemma we have
$$\infty = \int f = \int \liminf_{n\rightarrow \infty} f_n \leq \liminf_{n\rightarrow \infty} \int f_n\Rightarrow \int f_n\rightarrow \infty.$$
In summary, this result seems more similar to D.C.T than M.C.T. This is one reason why it cannot take the place of M.C.T.
The other reason is more fundamental. If we take a look at the proof of the Fatou's lemma, we may find that it relies on the monotonicity of $g_{k} = \inf_{n\geq k}f_n$ and $g_k\nearrow \liminf_{n\rightarrow \infty} f_n$. That is to say, the Fatou's lemma is actually implied by the M.C.T.
In other words, M.C.T. is the most fundamental theorem, then the Fatou's lemma and finally the D.C.T. Therefore from the perspective of mathematician, it's definitely better to use the "naive version" (actually the only correct version) of M.C.T., but not the seemingly stronger result you give, which is implied by D.C.T. and Fatou's lemma.
Nevertheless, it is true that the result you give here is much more convenient to use than either M.C.T. or D.C.T. That's why some instructors prefer to introduce this result as a substitution for M.C.T.
• Your statement of Lebesgue's dominant convergence theorem doesn't make sense. (Typos?). What does $g$ do? – DanielWainfleet Mar 30 '18 at 0:42
• Thanks a lot for the reminder! Yes it's a typo haha. @DanielWainfleet Now already fixed. – Wanshan Mar 30 '18 at 0:46
• So just to make sure I understand: the result I posted is in fact implied by D.C.T and Fatou's Lemma? Because the top voted answer in the math overflow question gives an example where D.C.T. doesn't work but this theorem works; so I am thinking that that example does not work specifically for Lebesgue measure? mathoverflow.net/questions/296312/… – Ovi Apr 25 '18 at 16:06
• Thanks for the reference and sorry for the late reply. I think for his example we can still use DCT and Fatou to prove that claim. If $\int_X f d\mu<\infty$, we use DCT, otherwise we can use Fatou following the line in my answer to prove that the right hand side of his equality tends to infinity. He said that equality can be used for the definition of Lebesgue integral, but I think it does not make sense because any of the integral theorem depends on the definition of integral, and we cannot use a higher level result to define a lower level concept. @Ovi – Wanshan May 13 '18 at 21:54 | 2019-07-19T14:50:13 | {
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https://byjus.com/question-answer/assertion-if-the-two-pair-of-lines-displaystyle-x-2-2m-xy-y-2-0/ | Question
# Assertion :If the two pair of lines $$\displaystyle x^{2}-2m\:xy-y^{2}=0$$ and $$\displaystyle x^{2}-2n\:xy-y^{2}=0$$ are such that one of them represents the bisector of angle between the other, then $$\displaystyle 1+mn=0$$. Reason: Equation of bisector of angle between the pair of lines $$\displaystyle ax^{2}+2hxy+by^{2}=0$$ is given by $$\displaystyle \frac{x^{2}-y^{2}}{a-b}=\frac{xy}{h}.$$
A
Both Assertion & Reason are individually true & Reason is correct explanation of Assertion.
B
Both Assertion & Reason are individually true but Reason is not the correct (proper) explanation of Assertion.
C
Assertion is true but Reason is false.
D
Assertion is false but Reason is true.
Solution
## The correct option is A Both Assertion & Reason are individually true & Reason is correct explanation of Assertion.Given $$x^2-2mxy-y^2=0$$The pair if bisector of given pair is $$\dfrac{x^2-y^2}{1-(-1)}=\dfrac{xy}{-m}$$$$\dfrac{x^2-y^2}{2}=\dfrac{xy}{-m}$$ or $$x^2+\dfrac{2xy}{m}-y^2=0$$ which is identical to $$x^2-2nxy-y^2=0$$$$\therefore \dfrac{1}{m}=-n \Rightarrow mn+1=0$$Maths
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View More | 2022-01-20T14:17:13 | {
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https://math.stackexchange.com/questions/2746802/value-of-the-game-from-payoff-matrix | # Value of the game from payoff matrix
I am absolutely new to decision theory . I came across this following payoff matrix in the book.(Math. Stats : John E Freund).
Player A
I II
Player B 1 7 -4
2 8 10
The value of the game is given as 8 units.However i . have a question . I agree the optimal strategy for Play B is "2" . But i want to argue that the optimal choice for player A is 2 . This is because in decision theory the assumption is made that "each player must choose a strategy without knowing what the opponent is going to do and that once a player has made a choice it can not be changed" ..Going by this logic, the optimal strategy for player A should be II , because i see in that case , the loss would be less overall for the moves of player B . This , logic if true gives the value of the game to be 10 , which is not correct. I want to know , where i am wrong.
• Is this a zero-sum game? Which player is the minimiser and which is the maximiser? – Theoretical Economist Apr 21 '18 at 2:48
The statement that "each player must choose a strategy without knowing what the opponent is going to do and that once a player has made a choice it can not be changed" just means the choice in each time the game is played are made without information of what the opponent played this time. The players are allowed to examine the whole payoff matrix and think about what the opponent might do. In this game Player B finds that move $2$ dominates move $1$. No matter what A does, B is better off playing $2$ than playing $1$, so he should play $2$ all the time. For A, neither strategy dominates the other, but A "knows" B will play $2$, so by playing $1$ A can lose $8$ instead of $10$. If B were so foolish to play $1$ it is true that A could do better by playing $2$, but we don't count on foolishness.
• Thanks a lot.. Your answer is a perfect clarification. – warrior_monk Apr 21 '18 at 2:55
The basic assumptions in a game theory are:
1) Each player is rational and tries to maximize his/her payoff in the game.
2) Each player knows that his/her opponent is rational and tries to maximize his/her payoff in the game.
You can solve this problem by Reducing by Dominance.
Since the entries in Row 2 are greater (or equal to) the corresponding ones in Row 1 (it implies Row 2 dominates Row 1, i.e. Player 2 is better off with strategy 2), we can eliminate Row 1 to get: $$\begin{array}{cc|cc} &&Player \ 1 \\ &&A&B \\ \hline Player \ 2 & 2 & 8 & 10 \end{array}$$ Since the entry in Column A is less than (or equal to) the corresponding one in Column B (it implies Column A dominates Column B, i.e. the Player 1 is better off with strategy A), we can eliminate Column B to get: $$\begin{array}{cc|c} &&Player \ 1 \\ &&A \\ \hline Player \ 2 & 2 & 8 \end{array}$$ So, the Player 2 has an advantage of $8$ units over Player 1.
• What if you begin with player 1(Player A in original ques)? In that case . "II" is the optimal strategy for him , which is Player 1 – warrior_monk Apr 22 '18 at 4:26
• Option 1: If P1 first chooses the strategy A, then P2 will choose the strategy 2 and P2 will win 8 units. Option 2: If P1 first chooses the strategy B, then P2 will choose the strategy 2 and P2 will win 10 units. Which option is rational (better) for P1? Certainly, Option 1 (because P1 loses less: 8 instead of 10). So, P1 knows that P2 is rational and P2 will maximize his/her payoff. For P2 the strategy 2 is better than strategy 1 (i.e. strategy 2 dominates strategy 1). Now, can you work out the two options when P2 starts and P1 ends and which option is rational for P2? – farruhota Apr 22 '18 at 5:22 | 2020-07-14T20:58:55 | {
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https://math.stackexchange.com/questions/1203251/how-many-different-permutations-are-there-of-the-sequence-of-letters-in-mississ/1203346 | # How many different permutations are there of the sequence of letters in “MISSISSIPPI”?
There are 11 letters in the word.
M - 1 I - 4 S - 4 P - 2
so the number of different permutations is $\dfrac{11!}{1! 4 !4!2!}$
Is this correct solution?
• Yes, this is exactly it. – Jimmy R. Mar 23 '15 at 20:46
Yes! Here's why it's correct:
If you numbered each of the $I$s, $S$s, and $P$s, there would be 11! permutations of $$(M, I_1, I_2, I_3, I_4, S_1, S_2, S_3, S_4, P_1, P_2).$$
But, because you don't care about the numbers of the $P$s, each permutations of $$(M, I_1, I_2, I_3, I_4, S_1, S_2, S_3, S_4, P, P)$$ is actually double-counted, once with $P_1$ before $P_2$ and once with $P_2$ before $P_1$. For example, in the original 11! permutations $$M,I_1,S_1,S_2,I_2,S_3,S_4,I_3,P_1,P_2,I_1$$ looks the same as $$M,I_1,S_1,S_2,I_2,S_3,S_4,I_3,P_2,P_1,I_1$$ if we ignore the numbers of the $P$s.
So, $\frac{11!}{2!}$ is the number of permutations of $$(M, I_1, I_2, I_3, I_4, S_1, S_2, S_3, S_4, P, P).$$
For the same reason, we find 24-times the number of permutations of $$(M, I_1, I_2, I_3, I_4, S_1, S_2, S_3, S_4, P, P)$$ as $$(M, I, I, I, I, S_1, S_2, S_3, S_4, P, P),$$ one for each of the $4! = 24$ orderings of $(I_1, I_2, I_3, I_4).$ Thus there are $\frac{11!}{2!4!}$
And again there are 24-times the number of permutations of $$(M, I, I, I, I, S_1, S_2, S_3, S_4, P, P)$$ as $$(M, I, I, I, I, S, S, S, S, P, P),$$ one for each of the $4! = 24$ orderings of $(S_1, S_2, S_3, S_4).$ Thus there are $\frac{11!}{2!4!4!}$ total.
An alternative solution:
• Choose $1$ out of $11$ places for the M
• Choose $4$ out of $10$ remaining places for the I's
• Choose $4$ out of $6$ remaining places for the S's
• Choose $2$ out of $2$ remaining places for the P's
$$\binom{11}{1}\cdot\binom{10}{4}\cdot\binom{6}{4}\cdot\binom{2}{2}$$
Please note that you can apply this process in any order you'd like.
• I think it might be worth explicitly pointing that the result you get is the same as OP's and the ones on other answers. – Wojowu Mar 23 '15 at 21:32
• @Wojowu: Isn't that obvious from the opening statement an alternative solution? – barak manos Mar 23 '15 at 21:34 | 2021-05-08T21:40:29 | {
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http://mathhelpforum.com/trigonometry/102139-prove-identity.html | # Thread: prove this identity
1. ## prove this identity
2sin(x)/[(cos(x)+sin(x)] = tan(2x) + 1 - sec(2x)
First, I turn tan(2x) into [sin(2x) / cos(2x)] then I turn sec(2x) into 1/cos(2x)
So, I end up with... [sin(2x) / cos(2x)] + 1 - (1/(cos2x))
I multiple and get the GCD of cos(2x)..
[sin(2x) + cos(2x) - 1] / cos(2x)
Now things get kind of fuzzy.. I probably didn't do this the best way thus far but I can't really advance much further.. I need help, please.
2. Originally Posted by zodiacbrave
2sin(x)/[(cos(x)+sin(x)] = tan(2x) + 1 - sec(2x)
First, I turn tan(2x) into [sin(2x) / cos(2x)] then I turn sec(2x) into 1/cos(2x)
So, I end up with... [sin(2x) / cos(2x)] + 1 - (1/(cos2x))
I multiple and get the GCD of cos(2x)..
[sin(2x) + cos(2x) - 1] / cos(2x)
Now things get kind of fuzzy.. I probably didn't do this the best way thus far but I can't really advance much further.. I need help, please.
Start with right side with the identy
$tan(2x)=\frac{2tanx}{1-tan^2x}$ and $sec(2x)=\frac{1}{cos(2x)}=\frac{1+tan^2x}{1-tan^2x}$
Then the right side reduces to
$\frac{2tanx+2}{1-tan^2x}$
Chage tan to terms of sin and cos and you should be able to get there.
3. Hello, zodiacbrave!
Another approach . . .
Prove: . $\frac {2\sin x}{\cos x+\sin x} \:=\: \tan(2x) + 1 - \sec(2x)$
Multiply the left side by: . $\frac{\cos x - \sin x}{\cos x - \sin x}$
$\frac{2\sin x}{\cos x + \sin x}\cdot\frac{\cos x - \sin x}{\cos x - \sin x}$
. . $=\;\frac{\overbrace{2\sin x\cos x}^{\sin(2x)} - \overbrace{2\sin^2\!x}^{1-\cos(2x)}}{\underbrace{\cos^2\!x - \sin^2\!x}_{\cos(2x)}}$
. . $= \;\frac{\sin(2x) + \cos(2x) - 1}{\cos(2x)}$
. . $=\;\frac{\sin(2x)}{\cos(2x)} + \frac{\cos x}{\cos x} - \frac{1}{\cos(2x)}$
. . $=\;\tan(2x) + 1 - \sec(2x)$
4. Soroban's LATEXing technique is superb and very instructive, thanks MHF expert | 2016-10-21T11:54:45 | {
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http://math.stackexchange.com/questions/669348/questions-about-continuous-functions | # Questions about continuous functions.
Recently when working with my thesis, I've got 2 questions.
1. Let $S_n$ be the set $\{x=(x_1,x_2,\cdots,x_n)\in\mathbb{R}^n\mid x_1+x_2+\cdots+x_n=1~\mbox{and}~0\leq x_i~\mbox{for}~ i=1,2,\cdots,n\}$. Is $S_n$ compact or convex, or closed? Or does it possess any topological property?
2. Let $s$ be a point of the product $S_{n_1}\times S_{n_2}\times\cdots S_{n_k}$, where $S_{n_i}$ is the subset of Euclidean space $\mathbb{R}^{n_i}$ satisfying the description above.
$~~~~~~~~$Let $f$ and $g$ be two continuous functions(linear) from $S_{n_1}\times S_{n_2}\times\cdots S_{n_k}$ to $\mathbb{R}$, and define $\phi:S_{n_1}\times S_{n_2}\times\cdots S_{n_k}\rightarrow\mathbb{R}$ by $\phi(s)=\max\{0,f(s)-g(s)\}$. Then is $\phi$ continuous? Why?
Can anyone help me with these questions?
(Sorry to bring these questions with long long descriptions. :'( )
Added: For the last part, I think my real question is why the function, max, is continuous. Any suggestion? And thanks for other answers and advices. :)
-
The objects $S_n$ are the standard n-Simplices. See en.wikipedia.org/wiki/Simplex#The_standard_simplex for example. The properties you mention can be proven elementary. – canaaerus Feb 9 at 9:05
Yes, $S_n$ is compact, convex and closed : all three. Your map is continuous also, since $\sf max$ is continuous.
To see why $S_n$ is closed, write it as a finite intersection (in fact, $n+1$) of closed sets.
$S_n$ is compact because it is both closed and bounded.
$S_n$ is convex because it is defined by convex conditions.
-
Each of your $S_n$ sets is the intersection of the convex and closed positive orthant $\{x \geq 0 \}$ with a level set $\{f=1\}$ of the convex function $f(x) = |x_1+ \dots +x_n|$. Its convexity follows from an application of the triangle inequality: $$\left|\sum \lambda x_i + (1-\lambda)y_i\right| = \left|\lambda \sum x_i + (1-\lambda)\sum y_i\right|\leq \lambda \left|\sum x_i\right| + (1-\lambda) \left|\sum y_i\right|.$$ As level sets of convex functions are also convex and closed, $S_n$ is convex and closed.
They are also compact. To see it you only need to notice that the value of each coordinate of any point in $S_n$ is bounded below by $0$ and above by $1$. Since a closed and bounded set in $\mathbb R^n$ is compact, $S_n$ is compact.
Finally, considering that $f-g$ and $\max\{0,\cdot\}$ are continuous functions, the function $\phi$ defined in your question is a composition of continuous functions, and therefore continuous.
- | 2014-10-22T12:59:12 | {
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http://math.stackexchange.com/questions/838314/what-is-this-notation-similar-to-the-binomial-coefficient | What is this notation, similar to the binomial coefficient?
I've come accross this notation:
$$\left\{\begin{eqnarray}n\\m\end{eqnarray}\right\}$$
The only other info I have about this notation is that $\left\{\begin{eqnarray}4\\2\end{eqnarray}\right\}=7$
What's the name of this notation and what is it used for?
Thanks
-
I find it slightly disappointing that with the vast knowledge we have here, all three answers do little more than link to wikipedia. – user1729 Jun 18 '14 at 10:33
There is a Stirling numbers, which I added to your post. If you are interested in them then you might want to browse through the questions there. – user1729 Jun 18 '14 at 10:44
Maybe you're using Stirling numbers of the second kind, where $\displaystyle{n\brace k}$ denotes the number of ways to partition a set of $n$ objects into $k$ non-empty subsets.
You can use {n\brace k} to produce ${n\brace k}$ and \displaystyle{n\brace k} to produce $\displaystyle{n\brace k}$.
An example of a 'real world' application of these numbers can be found in this MSE question which asks how many rooks can be placed on a triangular chessboard so that none of them are attacking each other.
-
Any chance of an application? It would make your answer complete...(for example, they can be used to find certain generating sets of certain semigroups). – user1729 Jun 18 '14 at 10:25
Can you elaborate your answer by providing a real life example of its usage? Thanks... – Tunk-Fey Jun 18 '14 at 10:29
Stirling numbers of second kind, the number of partitions of a set of $n$ objects into $k$ non-empty subsets.
- | 2015-01-30T21:38:30 | {
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Over the past 7 weeks, the Smith family had weekly grocery bills of $74,$69, $64,$79, $64,$84, and $77. What was the Smiths' average (arithmetic mean) weekly grocery bill over the 7-week period? A.$64
B. $70 C.$73
D. $74 E.$85
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Over the past 7 weeks, the Smith family had weekly grocery bills of $7 [#permalink] ### Show Tags 27 Jun 2018, 18:28 7 9 Bunuel wrote: Over the past 7 weeks, the Smith family had weekly grocery bills of$74, $69,$64 $79,$64, $84, and$77. What was the Smiths' average (arithmetic mean) weekly grocery bill over the 7-week period?
A. $64 B.$70
C. $73 D.$74
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29 Jun 2018, 06:58
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First, choose a nice, round number, such as 70, within the range of values.
Then calculate the average with the following formula:
Average = Nice number + Average of differences from the nice number
Average = 70 + (4 - 1 - 6 + 9 - 6 + 14 + 7)/7 = 70 + 21/7 =73
Answer: C
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Re: Over the past 7 weeks, the Smith family had weekly grocery bills of $7 [#permalink] ### Show Tags 25 Jun 2018, 05:10 Solution Given: • Over the past 7 weeks, the Smith family had weekly grocery bills of$74, $69,$64, $79,$64, $84, and$77
To find:
• Smith’s average weekly grocery bill over the 7-week period
Approach and Working:
• Smith’s total bill amount over the 7-week period = (74 + 69 + 64 + 79 + 64 + 84 + 77) = 511
• Therefore, Smith’s average bill amount over the same period = $$\frac{511}{7}$$ = 73
Hence, the correct answer is option C.
Answer: C
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### Show Tags
28 Jun 2018, 18:07
1
Bunuel wrote:
Over the past 7 weeks, the Smith family had weekly grocery bills of $74,$69, $64,$79, $64,$84, and $77. What was the Smiths' average (arithmetic mean) weekly grocery bill over the 7-week period? A.$64
B. $70 C.$73
D. $74 E.$85
We can determine the average using the formula: average = sum / number:
average = (74 + 69 + 64 + 79 + 64 + 84 + 77)/7 = 511/7 = 73
Answer: C
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Re: Over the past 7 weeks, the Smith family had weekly grocery bills of $7 [#permalink] ### Show Tags 28 Aug 2018, 03:42 Bunuel wrote: Over the past 7 weeks, the Smith family had weekly grocery bills of$74, $69,$64, $79,$64, $84, and$77. What was the Smiths' average (arithmetic mean) weekly grocery bill over the 7-week period?
A. $64 B.$70
C. $73 D.$74
E. $85 NEW question from GMAT® Official Guide 2019 (PS07369) Easy way out is consider avg 70 - so diff for all terms is 4, -1 , -6, 9 , -6,14, 7 = sum is 21 = divided by 7 = equal to 3 so 70 +3 = 73 avg Board of Directors Status: QA & VA Forum Moderator Joined: 11 Jun 2011 Posts: 4868 Location: India GPA: 3.5 WE: Business Development (Commercial Banking) Re: Over the past 7 weeks, the Smith family had weekly grocery bills of$7 [#permalink]
### Show Tags
10 Sep 2018, 07:34
Bunuel wrote:
Over the past 7 weeks, the Smith family had weekly grocery bills of $74,$69, $64,$79, $64,$84, and $77. What was the Smiths' average (arithmetic mean) weekly grocery bill over the 7-week period? A.$64
B. $70 C.$73
D. $74 E.$85
NEW question from GMAT® Official Guide 2019
(PS07369)
$$\frac{( 70 + 4 ) + ( 70 - 1 ) + ( 60 + 4 ) + ( 80 - 1 ) + ( 60 + 4 ) + ( 80 + 4 ) + ( 70 + 7 )}{7}$$
= $$\frac{( 70 + 70 + 60 + 80 + 60 + 80 + 70 ) + ( 4 - 1 + 4 - 1 + 4 + 4 + 7 )}{7}$$
= $$\frac{490 + 21}{7}$$
= $$\frac{490}{7} + \frac{21}{7}$$
= $$70 + 3$$
= $$73$$, Answer must be (C)
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02 Oct 2018, 00:09
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This solution might be too blunt for some people. Sum the unit digits and you will have 41.
Now there are 7 numbers, which mean if x multiply by 7 must will result a number with unit digit of 1, so the x is 3.
The only answer with 3 as its unit digit is C.
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### Show Tags
27 Jan 2020, 12:45
Notice that all options' values have different unit digit;
To Solve: Add the unit digits of each values(4+9+4+9+4+4+7) = 41 and check the unit digit of output (unit digit 1).
for division by 7, 3 should be unit digit of the quotient.
Only Option C satisfies, it is $73! _________________ Unable to Give Up! Re: Over the past 7 weeks, the Smith family had weekly grocery bills of$7 [#permalink] 27 Jan 2020, 12:45
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https://forum.azimuthproject.org/discussion/2344/new-free-booklet-on-applied-category-theory | #### Howdy, Stranger!
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# New Free Booklet on Applied Category Theory
edited September 2018
Tai-Danae Bradley has a new free "booklet" on applied category theory. It grew out of the workshop Applied Category Theory 2018, and I think it makes a great complement to Fong and Spivak's books Seven Sketches:
Abstract. This is a collection of introductory, expository notes on applied category theory, inspired by the 2018 Applied Category Theory Workshop, and in these notes we take a leisurely stroll through two themes (functorial semantics and compositionality), two constructions (monoidal categories and decorated cospans) and two examples (chemical reaction networks and natural language processing) within the field.
Check it out!
• Options
1.
Comment Source:Woohoo more fun reading! Thanks for the link!
• Options
2.
Looking forward to reading it! Still need to close many blind spots in this class, but this reading will further help :) Happy to see so much material approachable to layman persons! :)
Comment Source:Looking forward to reading it! Still need to close many blind spots in this class, but this reading will further help :) Happy to see so much material approachable to layman persons! :)
• Options
3.
In page 24 there is helpful insight on the hints of the end paragraph of previous lecture. Efforts like this aimed at a wider categorical outreach out of closed circles shine. Thanks to the author!
Comment Source:In page 24 there is helpful insight on the hints of the end paragraph of previous [lecture](https://forum.azimuthproject.org/discussion/2342/lecture-75-chapter-4-the-grand-synthesis#latest). Efforts like this aimed at a wider categorical outreach out of closed circles shine. Thanks to the author!
• Options
4.
edited September 2018
The links provided in the further reading section are also great especially this blog which was mentioned during John's lectures as well. Now with more understanding of the subject the blog seems to be extremely enlightening.
Also, I came across a paper written by John Baez and James Dolan called From Finite Sets to Feynman Diagrams. I found it to be very illuminating as well. I never thought too deeply about $$\mathbf{FinSet}$$ but it is a great place to start for newbies and I regret not doing so earlier.
Comment Source:The links provided in the further reading section are also great especially [this blog](http://www.math3ma.com) which was mentioned during John's lectures as well. Now with more understanding of the subject the blog seems to be extremely enlightening. Also, I came across a paper written by John Baez and James Dolan called [From Finite Sets to Feynman Diagrams](https://arxiv.org/pdf/math/0004133.pdf). I found it to be very illuminating as well. I never thought too deeply about \$$\mathbf{FinSet}\$$ but it is a great place to start for newbies and I regret not doing so earlier.
• Options
5.
Great!! This was very helpful in explaining -- in a mathematically entertaining way -- the pith of some nice topics in applied category theory.
Comment Source:Great!! This was very helpful in explaining -- in a mathematically entertaining way -- the pith of some nice topics in applied category theory.
• Options
6.
=D> !
Comment Source:=D> !
• Options
7.
To help me digest these ideas, I thought I'd write a summary of Tai-Danae Bradley's booklet. I've added my own reactions and I've tried link the ideas to what we've discussed in this course.
## Two Themes
Tai-Danae starts with two basic themes in applied category. (1) Functorial semantics and (2) compositionality.
### Functorial Semantics
Modeling natural human language is a good example (and Tai-Danae discusses this example at length). Roughly speaking, language consists of words ("semantics") and ways to combine to words to create sentences/new words ("grammar"). This distinction can be captured by a functor between two categories!
$$F : C \to D$$ The category $$C$$ specifies the grammar (how words act, how they can be combined; see compositionality) while the functor assigns meaning to diagrams in $$C$$ (where the category $$D$$ represents meaning). So assigning individual words their meaning.
Another good example in John's Lectures and Seven Sketches in Compositionality is Chapter 3: Database Schemas. A database schema was a category presented by a graph. The schema specified objects (data) and some morphisms (relations; functions).
The database schema tells us the grammar of our database. It says what data is required, what actions can be performed, how different types of data are related, etc. But it doesn't say what the data actually is (or more abstractly, how we should interpret the objects and morphisms). We get an actual database by constructing a functor which sends our database schema to the category $$\mathbf{Set}$$ (assigning sets to objects; and functions to morphisms).
I think the most important idea here (especially for modeling/ theoretical science) is that we can separate how complex systems combine (this is the grammar) and how those complex systems function (this is the semantics). And we can interpret the same grammar with different semantics (which allows us to be somewhat agnostic about the fundamentals of a complex system).
Electric circuits are a good example of this last point. We can model a electric circuit using a labeled graph, then construct a category where these labeled graphs are morphisms so we can compose them (this is done with cospans). Each edge in the graph is thought of as some basic component (like a resistor). These constructions don't tell us how voltages or currents distribute over the circuit (i.e., they don't tell us what a resistor really means), just how circuits combine. Choosing how components relate currents and voltages (Ohm's law for linear resistors) is then a functor which sends the labeled graphs to some other category that says how the circuit behaves.
But the grammar of electric circuits constrains how the circuits can behave, because whatever semantics we choose, the semantics has to preserve the functor axioms (intuitively, it has to distribute over our grammar). So if a resistor behaves like a linear relation between two vector spaces, then sticking resistors together still has to be a linear relation on vector spaces.
### Compositionality
This idea is relatively straight-forward. Essentially, complex things can be constructed out of simpler parts. But as pointed out in Chapter 1 of Seven Sketches, how you combine parts to form a whole matters! The whole is not necessary the sum of its parts (but it might be up to isomorphism!). A system's components might be behave a bit differently when studied separately than when glued together.
Biology has many, many examples of this phenomenon! Chemical reactions studied in vitro (in a beaker) aren't the same as in vivo (in a cell). The environment/context matters! The cell has many other reactions that influence the reaction under consideration. Making measurements of system components might not agree with measurements of the system as a whole.
But this problem affects other fields! Quantum mechanics has this very issue with quantum interference (probabilities depend on the whole system; and can't be added up from measuring decoupled components; this is very roughly speaking as I am not really much of a physicist! ). Even the more hands-on case of electric circuits has this same problem. Adding components like resistors changes what you measure elsewhere in the network.
A goal in applied category theory is to identify the right algebraic structure for systematically building a system out of some component pieces. Using functorial semantics, algebraic structure in our "syntax" category is preserved. So how systems can be combined partially determines how they behave.
### The Yoneda Lemma (bonus theme)
Tai-Danae mentions the Yoneda Lemma is her discussion of natural language modeling. The Yoneda Lemma gets the heart of why category theory is useful for modeling! I won't give the formal statement, but I will also quote John Firth:
You shall know a word by the company it keeps.
Essentially this is what the Yoneda lemma says: "You shall know an object by the company it keeps." Or "You shall know an object by how it interacts with other objects." In other words, you can define/study an object by how it interacts with other objects. And in category theory, "how it interacts with other objects" means "the morphisms between this object and others".
To quote from the movie Forrest Gump:
Stupid is as stupid does
Similarly, "$$X$$ is as $$X$$ does" is a wonderful encapsulation of the Yoneda Lemma.
What does this mean for modeling? Sometimes you don't need to know what a thing is, just what it does!
## Two constructions
Tai-Danae discusses two constructions: monoidal categories and decorated cospans. I won't really summarize these constructions, but discuss how they could be used in modeling.
### Monoidal Categories
A monoidal category gives us the syntax/grammar to talk about systems where we can talk about composition in series (via morphisms) and in parallel (via tensoring). We've have many, many examples in this course:
• Monoidal preorders: e.g. integers with order $$\le$$ (as morphisms) under addition (as tensoring)
• Resource monoidal preorder: stuff (like pie ingredients) and mixtures of stuff (tensoring) with processes/reactions turning stuff/mixtures into other stuff/mixtures (morphisms)
• Database schema: data types with combined data types (tensoring) and relationships/actions between types (morphisms)
• Co-design diagrams
along more traditional math examples (Tai-Danae gives some good examples).
Another way to think about the monoid part of a monoidal category is to notice that in several examples the monoidal product lets us combine or mix together basic stuff. For example, chemical reactions turn mixtures of certain substances to other substances. We can describe reactions as morphisms (morphisms are good for processes) while mixing stuff together is tensoring.
### Decorated Cospans
Let's imagine a pipe (or a wire). Pipes have no natural distinction between "input" and "output". Water can go in either direction. But morphisms are always directional! If we want to turn "direction agnostic" processes into a morphisms, we have to find a way to make distinction between input and output not really matter. Decorated cospans are the answer (well an answer).
A cospan is the following diagram: $$X \rightarrow N \leftarrow Y$$ where the object $$N$$ represents our pipe (potentially bidirectional process). What these objects are depends on the category (finite sets is a common choice since we will have to be able to compute colimits for any objects to get a category; and finite sets always have colimits).
The cospan can be decorated with some extra structure (which might say things about how our pipe can be combined with other pipes). Decorating our cospan is the same idea as the functorial semantics.
Comment Source:To help me digest these ideas, I thought I'd write a summary of Tai-Danae Bradley's booklet. I've added my own reactions and I've tried link the ideas to what we've discussed in this course. ## Two Themes Tai-Danae starts with two basic themes in applied category. (1) Functorial semantics and (2) compositionality. ### Functorial Semantics Modeling natural human language is a good example (and Tai-Danae discusses this example at length). Roughly speaking, language consists of words ("semantics") and ways to combine to words to create sentences/new words ("grammar"). This distinction can be captured by a functor between two categories! $F : C \to D$ The category \$$C\$$ specifies the grammar (how words act, how they can be combined; see compositionality) while the functor assigns meaning to diagrams in \$$C\$$ (where the category \$$D\$$ represents meaning). So assigning individual words their meaning. Another good example in John's Lectures and *Seven Sketches in Compositionality* is Chapter 3: Database Schemas. A database schema was a category presented by a graph. The schema specified objects (data) and some morphisms (relations; functions). The database schema tells us the grammar of our database. It says what data is required, what actions can be performed, how different types of data are related, etc. But it doesn't say what the data actually is (or more abstractly, how we should interpret the objects and morphisms). We get an actual database by constructing a functor which sends our database schema to the category \$$\mathbf{Set}\$$ (assigning sets to objects; and functions to morphisms). I think the most important idea here (especially for modeling/ theoretical science) is that we can separate how complex systems combine (this is the grammar) and how those complex systems function (this is the semantics). And we can interpret the same grammar with different semantics (which allows us to be somewhat agnostic about the fundamentals of a complex system). Electric circuits are a good example of this last point. We can model a electric circuit using a labeled graph, then construct a category where these labeled graphs are morphisms so we can compose them (this is done with cospans). Each edge in the graph is thought of as some basic component (like a resistor). These constructions don't tell us how voltages or currents distribute over the circuit (i.e., they don't tell us what a resistor really means), just how circuits combine. Choosing how components relate currents and voltages (Ohm's law for linear resistors) is then a functor which sends the labeled graphs to some other category that says how the circuit behaves. But the grammar of electric circuits constrains how the circuits can behave, because whatever semantics we choose, the semantics has to preserve the functor axioms (intuitively, it has to distribute over our grammar). So if a resistor behaves like a linear relation between two vector spaces, then sticking resistors together still has to be a linear relation on vector spaces. ### Compositionality This idea is relatively straight-forward. Essentially, complex things can be constructed out of simpler parts. But as pointed out in Chapter 1 of *Seven Sketches*, how you combine parts to form a whole matters! The whole is not necessary the sum of its parts (but it might be up to isomorphism!). A system's components might be behave a bit differently when studied separately than when glued together. Biology has many, many examples of this phenomenon! Chemical reactions studied *in vitro* (in a beaker) aren't the same as *in vivo* (in a cell). The environment/context matters! The cell has many other reactions that influence the reaction under consideration. Making measurements of system components might not agree with measurements of the system as a whole. But this problem affects other fields! Quantum mechanics has this very issue with quantum interference (probabilities depend on the whole system; and can't be added up from measuring decoupled components; this is very roughly speaking as I am not really much of a physicist! ). Even the more hands-on case of electric circuits has this same problem. Adding components like resistors changes what you measure elsewhere in the network. A goal in applied category theory is to identify the right algebraic structure for systematically building a system out of some component pieces. Using functorial semantics, algebraic structure in our "syntax" category is preserved. So how systems can be combined partially determines how they behave. ### The Yoneda Lemma (bonus theme) Tai-Danae mentions the Yoneda Lemma is her discussion of natural language modeling. The Yoneda Lemma gets the heart of why category theory is useful for modeling! I won't give the formal statement, but I will also quote John Firth: > You shall know a word by the company it keeps. Essentially this is what the Yoneda lemma says: "You shall know an object by the company it keeps." Or "You shall know an object by how it interacts with other objects." In other words, you can define/study an object by how it interacts with other objects. And in category theory, "how it interacts with other objects" means "the morphisms between this object and others". To quote from the movie Forrest Gump: > Stupid is as stupid does Similarly, "\$$X\$$ is as \$$X\$$ does" is a wonderful encapsulation of the Yoneda Lemma. What does this mean for modeling? Sometimes you don't need to know what a thing is, just what it does! ## Two constructions Tai-Danae discusses two constructions: monoidal categories and decorated cospans. I won't really summarize these constructions, but discuss how they could be used in modeling. ### Monoidal Categories A monoidal category gives us the syntax/grammar to talk about systems where we can talk about composition in series (via morphisms) and in parallel (via tensoring). We've have many, many examples in this course: + Monoidal preorders: e.g. integers with order \$$\le\$$ (as morphisms) under addition (as tensoring) + Resource monoidal preorder: stuff (like pie ingredients) and mixtures of stuff (tensoring) with processes/reactions turning stuff/mixtures into other stuff/mixtures (morphisms) + Database schema: data types with combined data types (tensoring) and relationships/actions between types (morphisms) + Co-design diagrams along more traditional math examples (Tai-Danae gives some good examples). Another way to think about the monoid part of a monoidal category is to notice that in several examples the monoidal product lets us combine or mix together basic stuff. For example, chemical reactions turn mixtures of certain substances to other substances. We can describe reactions as morphisms (morphisms are good for processes) while mixing stuff together is tensoring. ### Decorated Cospans Let's imagine a pipe (or a wire). Pipes have no natural distinction between "input" and "output". Water can go in either direction. But morphisms are always directional! If we want to turn "direction agnostic" processes into a morphisms, we have to find a way to make distinction between input and output not really matter. Decorated cospans are the answer (well an answer). A cospan is the following diagram: $X \rightarrow N \leftarrow Y$ where the object \$$N\$$ represents our pipe (potentially bidirectional process). What these objects are depends on the category (finite sets is a common choice since we will have to be able to compute colimits for any objects to get a category; and finite sets always have colimits). The cospan can be decorated with some extra structure (which might say things about how our pipe can be combined with other pipes). Decorating our cospan is the same idea as the functorial semantics. | 2019-03-21T01:30:19 | {
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https://math.stackexchange.com/questions/3043161/prove-that-r-subseteq-x-times-x-is-an-equivalence-relation-and-construct-its-e | # Prove that $R\subseteq X\times X$ is an equivalence relation and construct its equivalence class
Prove that relation $$R\subseteq X\times X$$, where $$X= \mathbb{R}\times\mathbb{R}$$, is an equivalence relation and construct its equivalence class. $$R$$ is defined as: $$\langle x_1, y_1\rangle R \langle x_2, y_2\rangle \Longleftrightarrow x_1^2+y_1^2 = x_2^2+y_2^2$$
The way I did prove that it is an equivalence relation was based on equality relationship between two or more quantities. To make it easier to read, we can also substitute $$a = x_1^2+y_1^2$$, $$b = x_2^2+y_2^2$$, $$c = x_3^2+y_3^2$$, at least I think we can.
The first property is reflexivity:
$$\langle x_1, y_1\rangle R \langle x_1, y_1\rangle \Longleftrightarrow x_1^2+y_1^2 = x_1^2+y_1^2$$, which is always true, as $$a=a$$.
Then we check for symmetry:
$$\langle x_1, y_1\rangle R \langle x_2, y_2\rangle \Longrightarrow \langle x_2, y_2\rangle R \langle x_1, y_1\rangle$$, so if $$x_1^2+y_1^2 = x_2^2+y_2^2$$, then $$x_2^2+y_2^2 = x_1^2+y_1^2$$. Using the substitution we can say that $$(a=b) \Rightarrow (b=a)$$, which is also true.
The last one is transitivity:
$$\langle x_1, y_1\rangle R \langle x_2, y_2\rangle \wedge \langle x_2, y_2\rangle R \langle x_3, y_3\rangle \Longrightarrow \langle x_1, y_1\rangle R \langle x_3, y_3\rangle$$, which means $$x_1^2+y_1^2 = x_2^2+y_2^2 \wedge x_2^2+y_2^2 = x_3^2+y_3^2 \Longrightarrow x_1^2+y_1^2 = x_3^2+y_3^2$$, using substitutions it is $$(a=b \wedge b=c) \Longrightarrow (a=c)$$.
Up to this moment I think that my line of thinking is quite correct, but I am not sure whether that proof is good enough and it is what I would like to know.
When it comes to equivalence class of $$R$$, I would say that the relation tells us about points on the common circle, but I have to write it up with the definition, which is:
Equivalence class of element $$a\in A$$ in regard to equivalence relation $$R\subseteq A\times A$$ is a set $$[a]_R = \{b\in A\: |\: a R b\}$$.
$$[\langle a,b \rangle]_R = \{\langle a, b\rangle \in \mathbb{R} \times \mathbb{R}\:|\: a^2+b^2=x^2+y^2\}$$ I would like you to tell me whether my proof (also usage of substitutions) and construction of equivalence class is right, also how could I get grasp of it, as lectures I attend are not good enough.
• You first part(the relation part) looks good, the second part it should be $[\langle a,b \rangle]_R = \{\langle x, y\rangle \in \mathbb{R} \times \mathbb{R}\:|\: a^2+b^2=x^2+y^2\}$: this is the set of all elements of $A\times A$ that are in relation with $\langle a,b\rangle$(here $A$ is indeed $\Bbb R\times \Bbb R$) – Holo Dec 16 '18 at 21:14
I like your proof, but I'd write the equivalence classes as follows:
$$[(a,b)]_R=\{(x,y)\in \mathbb{R}^2|a^2+b^2=x^2+y^2\}$$
which translates to: "The equivalence class of $$(a,b)$$ is the set of all points $$(x,y)$$ having the same distance from 0 as $$(a,b)$$.
Your idea is quite good and can be better formalized.
Consider a set $$X$$ and a map $$f\colon X\to Z$$, $$Z$$ any set. Then you can define a relation $$\sim_f$$ on $$X$$ by decreeing that $$x\sim_f y\quad\text{if and only if}\quad f(x)=f(y)$$ Then $$\sim_f$$ is clearly an equivalence relation: the proof is easily based on the properties of equality, which is essentially what you did just with more complicated symbols.
For $$a\in X$$, its equivalence class is $$[a]_{\sim_f}=\{x\in X:f(x)=f(a)\}$$.
In your case, $$X=\mathbb{R}\times\mathbb{R}$$, $$Z=\mathbb{R}$$ and $$f(x,y)=x^2+y^2$$ (square of the distance of $$(x,y)$$ from the origin).
Thus the equivalence class of $$(a,b)\in X$$ is the set of all points in $$X=\mathbb{R}\times\mathbb{R}$$ that share the same distance from the origin as $$(a,b)$$.
• Is it more correct to say "and a map $f\colon X\to Z$, where $Z$ is any set"? – manooooh May 23 at 19:14
• @manooooh As correct. – egreg May 23 at 19:32 | 2019-07-19T08:05:10 | {
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http://gateoverflow.in/93184/linear-search-expected-number-of-searches | 185 views
Given an array of 1,000,000 distinct integers you are going to perform 100,000 searches, one at a time, using the linear search algorithm. Each element you are searching for is in the array. What is the expected, total number of comparisons performed by the 100,000 linear searches?
Given Ans: 5x10^10 + 5x10^4
Expected number of searches for a single linear search of $n$ elements
$$= 1.\frac{1}{n} + 2.\frac{1}{n} + \dots n.\frac{1}{n}$$
as elements are distinct and the probability of any element being equal to the searched element is $\frac{1}{n}$ as element is guaranteed to be present
$= \frac{n+1}{2}$.
Here, $n = 1,000,000$. So, expected number of comparisons $= 1,000,001/2$
Expected no. of comparisons for 100,000 such searches $= 100,000 \times 1,000,001/2 \\=5 \times 10^{10} + 5 \times 10^4$
selected
good question and good explaination ,thanks :)
@Arjun Sir, after seeing the question,my approach was like:
For 1st search, the number of comparisons would be O(n).
For 2nd search, the number of comparisons will be O(n-1) since now, 1 element has been decreased from the array.
Similarly, the searches would continue in the form of O(n)+O(n-1)+O(n-2)+O(n-3)+........+1=O(n^2).
So, for 100,000 searches, Total comparisons= (n^2)*100,000
Given, n=1000,000, so, Total Comparisons= 100,000 *100,000 *100000=10^15.
Kindly guide me where am I going wrong since this is not the answer. :(
@Bongbirdie, the question does not mention that once an element is found in the array, it is removed from it. So even after x number of searches are performed, the array still has 1,000,000 entries. Even if you keep track of the indices at which your search succeeded, to skip those indices in subsequent searches, you have to make a comparison anyway. For example, suppose the first search succeeded at index 200 and you remember this fact for performing the second search. Now while performing the second search, if you decide that I will skip the index 200, then each index has to be compared to 200 to decide whether or not to skip it. So instead of reducing the number of comparisons, you end up increasing it.
For the above two reasons, the number of possible comparison counts does not decrease in consecutive iterations of the linear search.
Hope this helps :) | 2017-09-19T13:27:31 | {
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https://casmusings.wordpress.com/tag/calculus/ | # Tag Archives: calculus
## Many Roads Give Same Derivative
A recent post in the AP Calculus Community expressed some confusion about different ways to compute $\displaystyle \frac{dy}{dx}$ at (0,4) for the function $x=2ln(y-3)$. I share below the two approaches suggested in the original post, proffer two more, and a slightly more in-depth activity I’ve used in my calculus classes for years. I conclude with an alternative to derivatives of inverses.
### Two Approaches Initially Proposed
1 – Accept the function as posed and differentiate implicitly.
$\displaystyle \frac{d}{dx} \left( x = 2 ln(y-3) \right)$
$\displaystyle 1 = 2*\frac{1}{y-3} * \frac{dy}{dx}$
$\displaystyle \frac{dy}{dx} = \frac{y-3}{2}$
Which gives $\displaystyle \frac{dy}{dx} = \frac{1}{2}$ at (x,y)=(0,4).
2 – Solve for y and differentiate explicitly.
$\displaystyle x = 2ln(y-3) \longrightarrow y = 3 + e^{x/2}$
$\displaystyle \frac{dy}{dx} = e^{x/2} * \frac{1}{2}$
Evaluating this at (x,y)=(0,4) gives $\displaystyle \frac{dy}{dx} = \frac{1}{2}$ .
### Two Alternative Approaches
3 – Substitute early.
The question never asked for an algebraic expression of $\frac{dy}{dx}$, only the numerical value of this slope. Because students tend to make more silly mistakes manipulating algebraic expressions than numeric ones, the additional algebra steps are unnecessary, and potentially error-prone. Admittedly, the manipulations are pretty straightforward here, in more algebraically complicated cases, early substitutions could significantly simplify work. Using approach #1 and substituting directly into the second line gives
$\displaystyle 1 = 2 * \frac{1}{y-3} * \frac{dy}{dx}$.
At (x,y)=(0,4), this is
$\displaystyle 1 = 2 * \frac{1}{4-3}*\frac{dy}{dx}$
$\displaystyle \frac{dy}{dx} = \frac{1}{2}$
The numeric manipulations on the right side are obviously easier than the earlier algebra.
4 – Solve for $\frac{dx}{dy}$ and reciprocate.
There’s nothing sacred about solving for $\frac{dy}{dx}$ directly. Why not compute the derivative of the inverse and reciprocate at the end? Differentiating first with respect to y eventually leads to the same solution.
$\displaystyle \frac{d}{dy} \left( x = 2 ln(y-3) \right)$
$\displaystyle \frac{dx}{dy} = 2 * \frac{1}{y-3}$
At (x,y)=(0,4), this is
$\displaystyle \frac{dx}{dy} = \frac{2}{4-3} = 2$, so
$\displaystyle \frac{dy}{dx} = \frac{1}{2}$.
### Equivalence = A fundamental mathematical concept
I sometimes wonder if teachers should place much more emphasis on equivalence. We spend so much time manipulating expressions in mathematics classes at all levels, changing mathematical objects (shapes, expressions, equations, etc.) into a different, but equivalent objects. Many times, these manipulations are completed under the guise of “simplification.” (Here is a brilliant Dan Teague post cautioning against taking this idea too far.)
But it is critical for students to recognize that proper application of manipulations creates equivalent expressions, even if when the resulting expressions don’t look the same. The reason we manipulate mathematical objects is to discover features about the object in one form that may not be immediately obvious in another.
For the function $x = 2 ln(y-3)$, the slope at (0,4) must be the same, no matter how that slope is calculated. If you get a different looking answer while using correct manipulations, the final answers must be equivalent.
### Another Example
A similar question appeared on the AP Calculus email list-server almost a decade ago right at the moment I was introducing implicit differentiation. A teacher had tried to find $\displaystyle \frac{dy}{dx}$ for
$\displaystyle x^2 = \frac{x+y}{x-y}$
using implicit differentiation on the quotient, manipulating to a product before using implicit differentiation, and finally solving for y in terms of x to use an explicit derivative.
1 – Implicit on a quotient
Take the derivative as given:\$
$\displaystyle \frac{d}{dx} \left( x^2 = \frac{x+y}{x-y} \right)$
$\displaystyle 2x = \frac{(x-y) \left( 1 + \frac{dy}{dx} \right) - (x+y) \left( 1 - \frac{dy}{dx} \right) }{(x-y)^2}$
$\displaystyle 2x * (x-y)^2 = (x-y) + (x-y)*\frac{dy}{dx} - (x+y) + (x+y)*\frac{dy}{dx}$
$\displaystyle 2x * (x-y)^2 = -2y + 2x * \frac{dy}{dx}$
$\displaystyle \frac{dy}{dx} = \frac{-2x * (x-y)^2 + 2y}{2x}$
2 – Implicit on a product
Multiplying the original equation by its denominator gives
$x^2 * (x - y) = x + y$ .
Differentiating with respect to x gives
$\displaystyle 2x * (x - y) + x^2 * \left( 1 - \frac{dy}{dx} \right) = 1 + \frac{dy}{dx}$
$\displaystyle 2x * (x-y) + x^2 - 1 = x^2 * \frac{dy}{dx} + \frac{dy}{dx}$
$\displaystyle \frac{dy}{dx} = \frac{2x * (x-y) + x^2 - 1}{x^2 + 1}$
3 – Explicit
Solving the equation at the start of method 2 for y gives
$\displaystyle y = \frac{x^3 - x}{x^2 + 1}$.
Differentiating with respect to x gives
$\displaystyle \frac{dy}{dx} = \frac {\left( x^2+1 \right) \left( 3x^2 - 1\right) - \left( x^3 - x \right) (2x+0)}{\left( x^2 + 1 \right) ^2}$
Equivalence
Those 3 forms of the derivative look VERY DIFFERENT. Assuming no errors in the algebra, they MUST be equivalent because they are nothing more than the same derivative of different forms of the same function, and a function’s rate of change doesn’t vary just because you alter the look of its algebraic representation.
Substituting the y-as-a-function-of-x equation from method 3 into the first two derivative forms converts all three into functions of x. Lots of by-hand algebra or a quick check on a CAS establishes the suspected equivalence. Here’s my TI-Nspire CAS check.
Here’s the form of this investigation I gave my students.
### Final Example
I’m not a big fan of memorizing anything without a VERY GOOD reason. My teachers telling me to do so never held much weight for me. I memorized as little as possible and used that information as long as I could until a scenario arose to convince me to memorize more. One thing I managed to avoid almost completely were the annoying derivative formulas for inverse trig functions.
For example, find the derivative of $y = arcsin(x)$ at $x = \frac{1}{2}$.
Since arc-trig functions annoy me, I always rewrite them. Taking sine of both sides and then differentiating with respect to x gives.
$sin(y) = x$
$\displaystyle cos(y) * \frac{dy}{dx} = 1$
I could rewrite this equation to give $\frac{dy}{dx} = \frac{1}{cos(y)}$, a perfectly reasonable form of the derivative, albeit as a less-common expression in terms of y. But I don’t even do that unnecessary algebra. From the original function, $x=\frac{1}{2} \longrightarrow y=\frac{\pi}{6}$, and I substitute that immediately after the differentiation step to give a much cleaner numeric route to my answer.
$\displaystyle cos \left( \frac{\pi}{6} \right) * \frac{dy}{dx} = 1$
$\displaystyle \frac{\sqrt{3}}{2} * \frac{dy}{dx} = 1$
$\displaystyle \frac{dy}{dx} = \frac{2}{\sqrt{3}}$
And this is the same result as plugging $x = \frac{1}{2}$ into the memorized version form of the derivative of arcsine. If you like memorizing, go ahead, but my mind remains more nimble and less cluttered.
One final equivalent approach would have been differentiating $sin(y) = x$ with respect to y and reciprocating at the end.
### CONCLUSION
There are MANY ways to compute derivatives. For any problem or scenario, use the one that makes sense or is computationally easiest for YOU. If your resulting algebra is correct, you know you have a correct answer, even if it looks different. Be strong!
## CAS and Normal Probability Distributions
My presentation this past Saturday at the 2015 T^3 International Conference in Dallas, TX was on the underappreciated applicability of CAS to statistics. This post shares some of what I shared there from my first year teaching AP Statistics.
MOVING PAST OUTDATED PEDAGOGY
It’s been decades since we’ve required students to use tables of values to compute by hand trigonometric and radical values. It seems odd to me that we continue to do exactly that today for so many statistics classes, including the AP. While the College Board permits statistics-capable calculators, it still provides probability tables with every exam. That messaging is clear: it is still “acceptable” to teach statistics using outdated probability tables.
In this, my first year teaching AP Statistics, I decided it was time for my students and I to completely break from this lingering past. My statistics classes this year have been 100% software-enabled. Not one of my students has been required to use or even see any tables of probability values.
My classes also have been fortunate to have complete CAS availability on their laptops. My school’s math department deliberately adopted the TI-Nspire platform in part because that software looks and operates exactly the same on tablet, computer, and handheld platforms. We primarily use the computer-based version for learning because of the speed and visualization of the large “real estate” there. We are shifting to school-owned handhelds in our last month before the AP Exam to gain practice on the platform required on the AP.
The remainder of this post shares ways my students and I have learned to apply the TI-Nspire CAS to some statistical questions around normal distributions.
FINDING NORMAL AREAS AND PROBABILITIES
Assume a manufacturer makes golf balls whose distances traveled under identical testing conditions are approximately normally distributed with a mean 295 yards with a standard deviation of 3 yards. What is the probability that one such randomly selected ball travels more than 300 yards?
Traditional statistics courses teach students to transform the 300 yards into a z-score to look up in a probability table. That approach obviously works, but with appropriate technology, I believe there will be far less need to use or even compute z-scores in much the same way that always converting logarithms to base-10 or base-to use logarithmic tables is anachronistic when using many modern scientific calculators.
TI calculators and other technologies allow computations of non-standard normal curves. Notice the Nspire CAS calculation below the curve uses both bounds of the area of interest along with the mean and standard deviation of the distribution to accomplish the computation in a single step.
So the probability of a randomly selected ball from the population described above going more than 300 yards is 4.779%.
GOING BACKWARDS
Now assume the manufacturing process can control the mean distance traveled. What mean should it use so that no more than 1% of the golf balls travel more than 300 yards?
Depending on the available normal probability tables, the traditional approach to this problem is again to work with z-scores. A modified CAS version of this is shown below.
Therefore, the manufacturer should produce a ball that travels a mean 293.021 yards under the given conditions.
The approach is legitimate, and I shared it with my students. Several of them ultimately chose a more efficient single line command:
But remember that the invNorm() and normCdf() commands on the Nspire are themselves functions, and so their internal parameters are available to solve commands. A pure CAS, “forward solution” still incorporating only the normCdf() command to which my students were first introduced makes use of this to determine the missing center.
DIFFERENTIATING INSTRUCTION
While calculus techniques definitely are NOT part of the AP Statistics curriculum, I do have several students jointly enrolled in various calculus classes. Some of these astutely noted the similarity between the area-based arguments above and the area under a curve techniques they were learning in their calculus classes. Never being one to pass on a teaching moment, I pulled a few of these to the side to show them that the previous solutions also could have been derived via integration.
I can’t recall any instances of my students actually employing integrals to solve statistics problems this year, but just having the connection verified completely solidified the mathematics they were learning in my class.
CONFIDENCE INTERVALS
The mean lead level of 35 crows in a random sample from a region was 4.90 ppm and the standard deviation was 1.12 ppm. Construct a 95 percent confidence interval for the mean lead level of crows in the region.
Many students–mine included–have difficulty comprehending confidence intervals and resort to “black box” confidence interval tools available in most (all?) statistics-capable calculators, including the TI-Nspire.
As n is greater than 30, I can compute the requested z-interval by filling in just four entries in a pop-up window and pressing Enter.
Convenient, for sure, but this approach doesn’t help the confused students understand that the confidence interval is nothing more than the bounds of the middle 95% of the normal pdf described in the problem, a fact crystallized by the application of the tools the students have been using for weeks by that point in the course.
Notice in the solve+normCdf() combination commands that the unknown this time was a bound and not the mean as was the case in the previous example.
EXTENDING THE RULE OF FOUR
I’ve used the “Rule of Four” in every math class I’ve taught for over two decades, explaining that every mathematical concept can be explained or expressed four different ways: Numerically, Algebraically, Graphically (including graphs and geometric figures), and Verbally. While not the contextual point of his quote, I often cite MIT’s Marvin Minsky here:
“You don’t understand anything until you learn it more than one way.”
Learning to translate between the four representations grants deeper understanding of concepts and often gives access to solutions in one form that may be difficult or impossible in other forms.
After my decades-long work with CAS, I now believe there is actually a 5th representation of mathematical ideas: Tools. Knowing how to translate a question into a form that your tool (in the case of CAS, the tool is computers) can manage or compute creates a different representation of the problem and requires deeper insights to manage the translation.
I knew some of my students this year had deeply embraced this “5th Way” when one showed me his alternative approach to the confidence interval question:
I found this solution particularly lovely for several reasons.
• The student knew about lists and statistical commands and on a whim tried combining them in a novel way to produce the desired solution.
• He found the confidence interval directly using a normal distribution command rather than the arguably more convenient black box confidence interval tool. He also showed explicitly his understanding of the distribution of sample means by adjusting the given standard deviation for the sample size.
• Finally, while using a CAS sometimes involves getting answers in forms you didn’t expect, in this case, I think the CAS command and list output actually provide a cleaner, interval-looking result than the black box confidence interval command much more intuitively connected to the actual meaning of a confidence interval.
• While I haven’t tried it out, it seems to me that this approach also should work on non-CAS statistical calculators that can handle lists.
(a very minor disappointment, quickly overcome)
Returning to my multiple approaches, I tried using my student’s newfound approach using a normCdf() command.
Alas, my Nspire returned the very command I had entered, indicating that it didn’t understand the question I had posed. While a bit disappointed that this approach didn’t work, I was actually excited to have discovered a boundary in the current programming of the Nspire. Perhaps someday this approach will also work, but my students and I have many other directions we can exploit to find what we need.
Leaving the probability tables behind in their appropriate historical dust while fully embracing the power of modern classroom technology to enhance my students’ statistical learning and understanding, I’m convinced I made the right decision to start this school year. They know more, understand the foundations of statistics better, and as a group feel much more confident and flexible. Whether their scores on next month’s AP exam will reflect their growth, I can’t say, but they’ve definitely learned more statistics this year than any previous statistics class I’ve ever taught.
COMPLETE FILES FROM MY 2015 T3 PRESENTATION
If you are interested, you can download here the PowerPoint file for my entire Nspired Statistics and CAS presentation from last week’s 2015 T3 International Conference in Dallas, TX. While not the point of this post, the presentation started with a non-calculus derivation/explanation of linear regressions. Using some great feedback from Jeff McCalla, here is an Nspire CAS document creating the linear regression computation updated from what I presented in Dallas. I hope you found this post and these files helpful, or at least thought-provoking.
## Calculus Humor
Completely frivolous post.
OK, my Halloween “costume” at school this year was pretty lame, but I actually did put a minute amount of thought into it.
In case you can’t read the sign, it says $\int 3(ice)^2d(ice)$. If you remember some calculus and treat ice as your variable, that works out to $ice^3$–An Ice Cube! Ha!
But it gets better. As there weren’t any bounds, adding the random constant of integration makes it $ice^3+C$ or “Ice Cube + C”, or maybe “Ice Cube + Sea”–I was really dressed up as an Iceberg. Ha! Ha! Having no idea how to dress like an iceberg, I wore a light blue shirt for the part of the iceberg above the water and dark blue pants for the part below the water. I tried to be clever even if the underlying joke was just “punny”.
Then a colleague posted another integral joke I’d seen sometime before. It has some lovely extensions, so I’ll share that, too.
What is $\int \frac{d(cabin)}{cabin}$?
At first glance, it’s a “log cabin.” Funny.
But notationally, the result is actually $ln(cabin)$, so the environmentalists out there will appreciate that the answer is really a “natural log cabin.” Even funnier.
The most correct solution is $ln(cabin)+C$. If you call the end “+ Sea”, then the most clever answer is that $\int \frac{d(cabin)}{cabin}$ is a “Houseboat”. Ha!
Hope you all had some fun.
## Traveling Dots, Parabolas, and Elegant Math
Toward the end of last week, I read a description a variation on a paper-folding strategy to create parabolas. Paraphrased, it said:
1. On a piece of wax paper, use a pen to draw a line near one edge. (I used a Sharpie on regular copy paper and got enough ink bleed that I’m convinced any standard copy or notebook paper will do. I don’t think the expense of wax paper is required!)
2. All along the line, place additional dots 0.5 to 1 inch apart.
3. Finally, draw a point F between 0.5 and 2 inches from the line roughly along the midline of the paper toward the center of the paper.
4. Fold the paper over so one of the dots on line is on tope of point F. Crease the paper along the fold and open the paper back up.
5. Repeat step 4 for every dot you drew in step 2.
6. All of the creases from steps 4 & 5 outline a curve. Trace that curve to see a parabola.
I’d seen and done this before, I had too passively trusted that the procedure must have been true just because the resulting curve “looked like a parabola.” I read the proof some time ago, but I consumed it too quickly and didn’t remember it when I was read the above procedure. I shamefully admitted to myself that I was doing exactly what we insist our students NEVER do–blindly accepting a “truth” based on its appearance. So I spent part of that afternoon thinking about how to understand completely what was going on here.
What follows is the chronological redevelopment of my chain of reasoning for this activity, hopefully showing others that the prettiest explanations rarely occur without effort, time, and refinement. At the end of this post, I offer what I think is an even smoother version of the activity, freed from some of what I consider overly structured instructions above.
CONIC DEFINITION AND WHAT WASN’T OBVIOUS TO ME
A parabola is the locus of points equidistant from a given point (focus) and line (directrix).
What makes the parabola interesting, in my opinion, is the interplay between the distance from a line (always perpendicular to some point C on the directrix) and the focus point (theoretically could point in any direction like a radius from a circle center).
What initially bothered me about the paper folding approach last week was that it focused entirely on perpendicular bisectors of the Focus-to-C segment (using the image above). It was not immediately obvious to me at all that perpendicular bisectors of the Focus-to-C segment were 100% logically equivalent to the parabola’s definition.
1. I knew without a doubt that all parabolas are similar (there is a one-to-one mapping between every single point on any parabola and every single point on any other parabola), so I didn’t need to prove lots of cases. Instead, I focused on the simplest version of a parabola (from my perspective), knowing that whatever I proved from that example was true for all parabolas.
2. I am quite comfortable with my algebra, geometry, and technology skills. Being able to wield a wide range of powerful exploration tools means I’m rarely intimidated by problems–even those I don’t initially understand. I have the patience to persevere through lots of data and explorations until I find patterns and eventually solutions.
I love to understand ideas from multiple perspectives, so I rarely quit with my initial solution. Perseverance helps me re-phrase ideas and exploring them from alternative perspectives until I find prettier ways of understanding.
In my opinion, it is precisely this willingness to play, persevere, and explore that formalized education is broadly failing to instill in students and teachers. “What if?” is the most brilliant question, and the one we sadly forget to ask often enough.
ALGEBRAIC PROOF
While I’m comfortable handling math in almost any representation, my mind most often jumps to algebraic perspectives first. My first inclination was a coordinate proof.
PROOF 1: As all parabolas are similar, it was enough to use a single, upward facing parabola with its vertex at the origin. I placed the focus at $(0,f)$, making the directrix the line $y=-f$. If any point on the parabola was $(x_0,y_0)$, then a point C on the directrix was at $(x_0,-f)$.
From the parabola’s definition, the distance from the focus to P was identical to the length of CP:
$\sqrt{(x_0-0)^2-(y_0-f)^2}=y_0+f$
Squaring and combining common terms gives
$x_0 ^2+y_0 ^2-2y_0f+f^2=y_0 ^2+2y_0f+f^2$
$x_0 ^2=4fy$
But the construction above made lines (creases) on the perpendicular bisector of the focus-to-C segment. This segment has midpoint $\displaystyle \left( \frac{x_0}{2},0 \right)$ and slope $\displaystyle -\frac{2f}{x_0}$, so an equation for its perpendicular bisector is $\displaystyle y=\frac{x_0}{2f} \left( x-\frac{x_0}{2} \right)$.
Finding the point of intersection of the perpendicular bisector with the parabola involves solving a system of equations.
$\displaystyle y=\frac{x_0}{2f} \left( x-\frac{x_0}{2} \right)=\frac{x^2}{4f}$
$\displaystyle \frac{1}{4f} \left( x^2-2x_0x+x_0 ^2 \right) =0$
$\displaystyle \frac{1}{4f} \left( x-x_0 \right) ^2 =0$
So the only point where the line and parabola meet is at $\displaystyle x=x_0$–the very same point named by the parabola’s definition. QED
Proof 2: All of this could have been brilliantly handled on a CAS to save time and avoid the manipulations.
Notice that the y-coordinate of the final solution line is the same $y_0$ from above.
MORE ELEGANT GEOMETRIC PROOFS
I had a proof, but the algebra seemed more than necessary. Surely there was a cleaner approach.
In the image above, F is the focus, and I is a point on the parabola. If D is the midpoint of $\overline{FC}$, can I conclude $\overline{ID} \perp \overline{FC}$, proving that the perpendicular bisector of $\overline{FC}$ always intersects the parabola?
PROOF 3: The definition of the parabola gives $\overline{FI} \cong \overline{IC}$, and the midpoint gives $\overline{FD} \cong \overline{DC}$. Because $\overline{ID}$ is self-congruent, $\Delta IDF \cong \Delta IDC$ by SSS, and corresponding parts make the supplementary $\angle IDF \cong \angle IDC$, so both must be right angles. QED
PROOF 4: Nice enough, but it still felt a little complicated. I put the problem away to have dinner with my daughters and when I came back, I was able to see the construction not as two congruent triangles, but as the single isosceles $\Delta FIC$ with base $\overline{FC}$. In isosceles triangles, altitudes and medians coincide, automatically making $\overline{ID}$ the perpendicular bisector of $\overline{FC}$. QED
Admittedly, Proof 4 ultimately relies on the results of Proof 3, but the higher-level isosceles connection felt much more elegant. I was satisfied.
TWO DYNAMIC GEOMETRY SOFTWARE VARIATIONS
Thinking how I could prompt students along this path, I first considered a trace on the perpendicular lines from the initial procedure above (actually tangent lines to the parabola) using to trace the parabolas. A video is below, and the Geogebra file is here.
http://vimeo.com/89759785
It is a lovely approach, and I particularly love the way the parabola appears as a digital form of “string art.” Still, I think it requires some additional thinking for users to believe the approach really does adhere to the parabola’s definition.
I created a second version allowing users to set the location of the focus on the positive y-axis and using a slider to determine the distances and constructs the parabola through the definition of the parabola. [In the GeoGebra worksheet (here), you can turn on the hidden circle and lines to see how I constructed it.] A video shows the symmetric points traced out as you drag the distance slider.
A SIMPLIFIED PAPER PROCEDURE
Throughout this process, I realized that the location and spacing of the initial points on the directrix was irrelevant. Creating the software versions of the problem helped me realize that if I could fold a point on the directrix to the focus, why not reverse the process and fold F to the directrix? In fact, I could fold the paper so that F touched anywhere on the directrix and it would work. So, here is the simplest version I could develop for the paper version.
1. Use a straightedge and a Sharpie or thin marker to draw a line near the edge of a piece of paper.
2. Place a point F roughly above the middle of the line toward the center of the paper.
3. Fold the paper over so point F is on the line from step 1 and crease the paper along the fold.
4. Open the paper back up and repeat step 3 several more times with F touching other parts of the step 1 line.
5. All of the creases from steps 3 & 4 outline a curve. Trace that curve to see a parabola.
This procedure works because you can fold the focus onto the directrix anywhere you like and the resulting crease will be tangent to the parabola defined by the directrix and focus. By allowing the focus to “Travel along the Directrix”, you create the parabola’s locus. Quite elegant, I thought.
As I was playing with the different ways to create the parabola and thinking about the interplay between the two distances in the parabola’s definition, I wondered about the potential positions of the distance segments.
1. What is the shortest length of segment CP and where could it be located at that length? What is the longest length of segment CP and where could it be located at that length?
2. Obviously, point C can be anywhere along the directrix. While the focus-to-P segment is theoretically free to rotate in any direction, the parabola definition makes that seem not practically possible. So, through what size angle is the focus-to-P segment practically able to rotate?
3. Assuming a horizontal directrix, what is the maximum slope the focus-to-P segment can achieve?
4. Can you develop a single solution to questions 2 and 3 that doesn’t require any computations or constructions?
CONCLUSIONS
I fully realize that none of this is new mathematics, but I enjoyed the walk through pure mathematics and the enjoyment of developing ever simpler and more elegant solutions to the problem. In the end, I now have a deeper and richer understanding of parabolas, and that was certainly worth the journey.
## The Value of Counter-Intuition
Numberphile caused quite a stir when it posted a video explaining why
$\displaystyle 1+2+3+4+...=- \frac{1}{12}$
Doug Kuhlman recently posted a great follow-up Numberphile video explaining a broader perspective behind this sum.
It’s a great reminder that there are often different ways of thinking about problems, and sometimes we have to abandon tradition to discover deeper, more elegant connections.
For those deeply bothered by this summation result, the second video contains a lovely analogy to the “reality” of $\sqrt{-1}$. From one perspective, it is absolutely not acceptable to do something like square roots of negative numbers. But by finding a way to conceptualize what such a thing would mean, we gain a far richer understanding of the very real numbers that forbade $\sqrt{-1}$ in the first place as well as opening the doors to stunning mathematics far beyond the limitations of real numbers.
On the face of it, $\displaystyle 1+2+3+...=-\frac{1}{12}$ is obviously wrong within the context of real numbers only. But the strange thing in physics and the Zeta function and other places is that $\displaystyle -\frac{1}{12}$ just happens to work … every time. Let’s not dismiss this out of hand. It gives our students the wrong idea about mathematics, discovery, and learning.
There’s very clearly SOMETHING going on here. It’s time to explore and learn something deeper. And until then, we can revel in the awe of manipulations that logically shouldn’t work, but somehow they do.
May all of our students feel the awe of mathematical and scientific discovery. And until the connections and understanding are firmly established, I hope we all can embrace the spirit, boldness, and fearless of Euler.
## Base-x Numbers and Infinite Series
In my previous post, I explored what happened when you converted a polynomial from its variable form into a base-x numerical form. That is, what are the computational implications when polynomial $3x^3-11x^2+2$ is represented by the base-x number $3(-11)02_x$, where the parentheses are used to hold the base-x digit, -11, for the second power of x?
So far, I’ve explored only the Natural number equivalents of base-x numbers. In this post, I explore what happens when you allow division to extend base-x numbers into their Rational number counterparts.
Level 5–Infinite Series:
Numbers can have decimals, so what’s the equivalence for base-x numbers? For starters, I considered trying to get a “decimal” form of $\displaystyle \frac{1}{x+2}$. It was “obvious” to me that $12_x$ won’t divide into $1_x$. There are too few “places”, so some form of decimals are required. Employing division as described in my previous post somewhat like you would to determine the rational number decimals of $\frac{1}{12}$ gives
Remember, the places are powers of x, so the decimal portion of $\displaystyle \frac{1}{x+2}$ is $0.1(-2)4(-8)..._x$, and it is equivalent to
$\displaystyle 1x^{-1}-2x^{-2}+4x^{-3}-8x^{-4}+...=\frac{1}{x}-\frac{2}{x^2}+\frac{4}{x^3}-\frac{8}{x^4}+...$.
This can be seen as a geometric series with first term $\displaystyle \frac{1}{x}$ and ratio $\displaystyle r=\frac{-2}{x}$. It’s infinite sum is therefore $\displaystyle \frac{\frac{1}{x}}{1-\frac{-2}{x}}$ which is equivalent to $\displaystyle \frac{1}{x+2}$, confirming the division computation. Of course, as a geometric series, this is true only so long as $\displaystyle |r|=\left | \frac{-2}{x} \right |<1$, or $2<|x|$.
I thought this was pretty cool, and it led to lots of other cool series. For example, if $x=8$,you get $\frac{1}{10}=\frac{1}{8}-\frac{2}{64}+\frac{4}{512}-...$.
Likewise, $x=3$ gives $\frac{1}{5}=\frac{1}{3}-\frac{2}{9}+\frac{4}{27}-\frac{8}{81}+...$.
I found it quite interesting to have a “polynomial” defined with a rational expression.
Boundary Convergence:
As shown above, $\displaystyle \frac{1}{x+2}=\frac{1}{x}-\frac{2}{x^2}+\frac{4}{x^3}-\frac{8}{x^4}+...$ only for $|x|>2$.
At $x=2$, the series is obviously divergent, $\displaystyle \frac{1}{4} \ne \frac{1}{2}-\frac{2}{4}+\frac{4}{8}-\frac{8}{16}+...$.
For $x=-2$, I got $\displaystyle \frac{1}{0} = \frac{1}{-2}-\frac{2}{4}+\frac{4}{-8}-\frac{8}{16}+...=-\frac{1}{2}-\frac{1}{2}-\frac{1}{2}-\frac{1}{2}-...$ which is properly equivalent to $-\infty$ as $x \rightarrow -2$ as defined by the convergence domain and the graphical behavior of $\displaystyle y=\frac{1}{x+2}$ just to the left of $x=-2$. Nice.
I did find it curious, though, that $\displaystyle \frac{1}{x}-\frac{2}{x^2}+\frac{4}{x^3}-\frac{8}{x^4}+...$ is a solid approximation for $\displaystyle \frac{1}{x+2}$ to the left of its vertical asymptote, but not for its rotationally symmetric right side. I also thought it philosophically strange (even though I understand mathematically why it must be) that this series could approximate function behavior near a vertical asymptote, but not near the graph’s stable and flat portion near $x=0$. What a curious, asymmetrical approximator.
Maclaurin Series:
Some quick calculus gives the Maclaurin series for $\displaystyle \frac{1}{x+2}$ : $\displaystyle \frac{1}{2}-\frac{x}{4}+\frac{x^2}{8}-\frac{x^3}{16}+...$, a geometric series with first term $\frac{1}{2}$ and ratio $\frac{-x}{2}$. Interestingly, the ratio emerging from the Maclaurin series is the reciprocal of the ratio from the “rational polynomial” resulting from the base-x division above.
As a geometric series, the interval of convergence is $\displaystyle |r|=\left | \frac{-x}{2} \right |<1$, or $|x|<2$. Excluding endpoint results, the Maclaurin interval is the complete Real number complement to the base-x series. For the endpoints, $x=-2$ produces the right-side vertical asymptote divergence to $+ \infty$ that $x=-2$ did for the left side of the vertical asymptote in the base-x series. Again, $x=2$ is divergent.
It’s lovely how these two series so completely complement each other to create clean approximations of $\displaystyle \frac{1}{x+2}$ for all $x \ne 2$.
Other base-x “rational numbers”
Because any polynomial divided by another is absolutely equivalent to a base-x rational number and thereby a base-x decimal number, it will always be possible to create a “rational polynomial” using powers of $\displaystyle \frac{1}{x}$ for non-zero denominators. But, the decimal patterns of rational base-x numbers don’t apply in the same way as for Natural number bases. Where $\displaystyle \frac{1}{12}$ is guaranteed to have a repeating decimal pattern, the decimal form of $\displaystyle \frac{1}{x+2}=\frac{1_x}{12_x}=0.1(-2)4(-8)..._x$ clearly will not repeat. I’ve not explored the full potential of this, but it seems like another interesting field.
CONCLUSIONS and QUESTIONS
Once number bases are understood, I’d argue that using base-x multiplication might be, and base-x division definitely is, a cleaner way to compute products and quotients, respectively, for polynomials.
The base-x division algorithm clearly is accessible to Algebra II students, and even opens the doors to studying series approximations to functions long before calculus.
Is there a convenient way to use base-x numbers to represent horizontal translations as cleanly as polynomials? How difficult would it be to work with a base-$(x-h)$ number for a polynomial translated h units horizontally?
As a calculus extension, what would happen if you tried employing division of non-polynomials by replacing them with their Taylor series equivalents? I’ve played a little with proving some trig identities using base-x polynomials from the Maclaurin series for sine and cosine.
What would happen if you tried to compute repeated fractions in base-x?
It’s an open question from my perspective when decimal patterns might terminate or repeat when evaluating base-x rational numbers.
I’d love to see someone out there give some of these questions a run!
## Number Bases and Polynomials
About a month ago, I was working with our 5th grade math teacher to develop some extension activities for some students in an unleveled class. The class was exploring place value, and I suggested that some might be ready to explore what happens when you allow the number base to be something other than 10. A few students had some fun learning to use their basic four algorithms in other number bases, but I made an even deeper connection.
When writing something like 512 in expanded form ($5\cdot 10^2+1\cdot 10^1+2\cdot 10^0$), I realized that if the 10 was an x, I’d have a polynomial. I’d recognized this before, but this time I wondered what would happen if I applied basic math algorithms to polynomials if I wrote them in a condensed numerical form, not their standard expanded form. That is, could I do basic algebra on $5x^2+x+2$ if I thought of it as $512_x$–a base-x “number”? (To avoid other confusion later, I read this as “five one two base-x“.)
Following are some examples I played with to convince myself how my new notation would work. I’m not convinced that this will ever lead to anything, but following my “what ifs” all the way to infinite series was a blast. Read on!
If I wanted to add $(3x+5)$$(2x^2+4x+1)$, I could think of it as $35_x+241_x$ and add the numbers “normally” to get $276_x$ or $2x^2+7x+6$. Notice that each power of x identifies a “place value” for its characteristic coefficient.
If I wanted to add $3x-7$ to itself, I had to adapt my notation a touch. The “units digit” is a negative number, but since the number base, x, is unknown (or variable), I ended up saying $3x-7=3(-7)_x$. The parentheses are used to contain multiple characters into a single place value. Then, $(3x-7)+(3x-7)$ becomes $3(-7)_x+3(-7)_x=6(-14)_x$ or $6x-14$. Notice the expanding parentheses containing the base-x units digit.
The last example also showed me that simple multiplication would work. Adding $3x-7$ to itself is equivalent to multiplying $2\cdot (3x-7)$. In base-x, that is $2\cdot 3(-7)_x$. That’s easy! Arguably, this might be even easier that doubling a number when the number base is known. Without interactions between the coefficients of different place values, just double each digit to get $6(-14)_x=6x-14$, as before.
What about $(x^2+7)+(8x-9)$? That’s equivalent to $107_x+8(-9)_x$. While simple, I’ll solve this one by stacking.
and this is $x^2+8x-2$. As with base-10 numbers, the use of 0 is needed to hold place values exactly as I needed a 0 to hold the $x^1$ place for $x^2+7$. Again, this could easily be accomplished without the number base conversion, but how much more can we push these boundaries?
Level 3–Multiplication & Powers:
Compute $(8x-3)^2$. Stacking again and using a modification of the multiply-and-carry algorithm I learned in grade school, I got
and this is equivalent to $64x^2-48x+9$.
All other forms of polynomial multiplication work just fine, too.
From one perspective, all of this shifting to a variable number base could be seen as completely unnecessary. We already have acceptably working algorithms for addition, subtraction, and multiplication. But then, I really like how this approach completes the connection between numerical and polynomial arithmetic. The rules of math don’t change just because you introduce variables. For some, I’m convinced this might make a big difference in understanding.
I also like how easily this extends polynomial by polynomial multiplication far beyond the bland monomial and binomial products that proliferate in virtually all modern textbooks. Also banished here is any need at all for banal FOIL techniques.
Level 4–Division:
What about $x^2+x-6$ divided by $x+3$? In base-x, that’s $11(-6)_x \div 13_x$. Remembering that there is no place value carrying possible, I had to be a little careful when setting up my computation. Focusing only on the lead digits, 1 “goes into” 1 one time. Multiplying the partial quotient by the divisor, writing the result below and subtracting gives
Then, 1 “goes into” -2 negative two times. Multiplying and subtracting gives a remainder of 0.
thereby confirming that $x+3$ is a factor of $x^2+x-6$, and the other factor is the quotient, $x-2$.
Perhaps this could be used as an alternative to other polynomial division algorithms. It is somewhat similar to the synthetic division technique, without its significant limitations: It is not limited to linear divisors with lead coefficients of one.
For $(4x^3-5x^2+7) \div (2x^2-1)$, think $4(-5)07_x \div 20(-1)_x$. Stacking and dividing gives
So $\displaystyle \frac{4x^3-5x^2+7}{2x^2-1}=2x-2.5+\frac{2x+4.5}{2x^2-1}$.
CONCLUSION
From all I’ve been able to tell, converting polynomials to their base-x number equivalents enables you to perform all of the same arithmetic computations. For division in particular, it seems this method might even be a bit easier.
In my next post, I push the exploration of these base-x numbers into infinite series. | 2017-06-23T05:02:18 | {
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http://chhh.makobene.de/two-headed-coin-probability.html | The hypotheses areH1 - the coin is two headed, and H2 the coin is fair. What is the probability mass function of X? What is the expected value of X?. probability of flipping nine heads on two headed coin = 1 probability of picking two headed coin and then flipping 9 straight heads =. a: a "biased" coin). You randomly take one of them out of your pocket without looking at it. 01, what is his posterior belief ? Note. Furthermore, if one wanted to determine whether the coin was fair or weighted, it would be difficult to do that without using inferential methods derived from measure theory. G is surprised to find that he loses the first ten times they play. (d) There are two coins in a box. What is the probability of getting two heads and a four? What is the probability of getting two heads and a four? Answer by stanbon(75874) ( Show Source ):. You reach one coin at random, toss it, and it lands up heads. 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The probability that the two-headed coin is selected out of the box is $P(E_1)=1/3$. A probability of zero means that an event is impossible. One is a two-headed coin, another is a fair coin, and the third is a biased coin that comes up heads 75 percent of the time. Consider an urn containing one fair coin and one two-headed coin. Let V denote the probability of heads of the selected coin, and Y the number of heads. This is, however, wrong, because given that heads came, it is more likely that the two-headed coin was chosen. You shouldn't believe or disbelieve anything. One is a two-headed coin (having a head on both faces), another is a biased coin that comes up heads 75% of the times, and third is also a biased coin that comes up tails 40% of the times. We use Wi to denote the event that the ith ball is white and Ri to denote the event that the ith ball is red. 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This is, however, wrong, because given that heads came up, it is more likely that the two-headed coin was chosen. 1 The while loop does Kerrich’s entire experiment in a split second! 2. The first coin is a fair coin, the second coin is a biased coin such that P(T) = 0:15, and the third coin is a two headed coin (a) What is the probability the coin lands on tails? (b) Given that the coin landed on heads, what is the probability it was the fair coin? 29. Instead of probability distributions, we use probability densities and integrate over ranges of possible. If the first 50 tosses of the coin are heads, what is the probability that it is the 2-headed coin. After all, real life is rarely fair. The correct reasoning is to calculate the conditional probability p= P(two-headed coin was chosen|heads came up) = P(two-headed coin was chosen and heads. There are three coins in a box. Given that you see 10 heads, what is the probability that the next toss of that coin is also a head?. What is the probability that it is the fair coin?. One is a two-headed coin, another is a fair coin, and the third is a biased coin that comes up heads 75 percent of the time. Case 2: One head. (a) What is the sample space of the experiment? (b) Given that both flips produce heads, what is the probability that Alice drew the two-headed coin from the urn?. What is the probability that it was the two-headed coin? 43. He is then either shouted at or not. If it is heads, he is willing. The first coin is two-headed. Problem 1 (20%) There are three coins in a box. A probability of one means that the event is certain. One is a two-headed coin; another is a fair coin; and the third is a biased coin that comes up heads 75% of the time. One is a two-headed coin (having head on both faces), another is a biased coin that comes up heads 75% of the times and the third is also a biased coin that comes up tails 40% of the time. 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It is equally likely to be a fair coin, to be a two-headed coin, to be a two-tailed coin, or any mixture of alloy that has one side heavier than the other. Byju's Coin Toss Probability Calculator is a tool which makes calculations very simple and interesting. A coin is chosen at random from the bag and tossed 2 times. Let H 1 first coin flip is heads H 2 second coin flip is heads The likelihood of a coin flip coming up heads is 0. , flipping a two-headed coin. b:if heads appear on the second toss,then it also appears on the first toss. the opposite face is either heads or tails, the desired probability is 1/2. (a) what is the probability that the lower face of the coin is a head?. I'm not a mathematician so please bear with me. To find the probability of two independent events occuring, we simply multiply together the probabilities associated with two individual events. A box contains three coins. Odds & Probability in Blackjack. 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(a) A gambler has a fair coin and a two-headed coin in his pocket. He selects one of the coins at random; when he flips it, it shows heads. Exercise 2: How many consecutive coin flips will it take for the subjective probability that it is a 2-headed coin to be creater than the subjective probability that it is a fair coin? Question: What happens to the posterior if you observe a single tail - even after a long string of heads?. EDIT #2: By "no retosses" I mean that your algorithm for obtaining the 1/3 probability can not have a "retoss until you get 1/3" rule which can theoretically cause you to toss infinitely many times. The probability of A and B is 1/100. One is a two headed coin (having head on both faces), another is a biased coin that comes up heads 75% of the time and third is an unbiased coin. Solution: a) A tree diagram of all possible outcomes. It is equally likely to be a fair coin, to be a two-headed coin, to be a two-tailed coin, or any mixture of alloy that has one side heavier than the other. Three of the outcomes show a head, of which two are from the double-headed coin. If X = 0, then the coin has to be the 1-headed one. b) Find the probability that of the first 4 marbles selected, exactly two are black. He selects one of the coins at random; when he flips it, it shows heads. two coins and one six sided number cube are tossed together. If heads appears both times, what is the probability that the coin is two-head. I think hugin is right: the probability of a 6th head is just the combination of the probability you have the double-headed coin (0. (a) What is the sample space of the experiment? (b) Given that both flips produce heads, what is the probability that Alice drew the two-headed coin from the urn?. The second one is a fair coin. You know that the coins you are given to test are either unfair coins with heads probability 1 4; unfair coins with heads probability 3 4; and fair coins. for the two-headed coin, the probability of getting two heads in two flips is 1. coin is chosen at random and flipped, and comes up heads. 1 Directed graphical models. A coin is chosen at random from the bag and tossed 2 times. If the experiment can be repeated potentially infinitely many times, then the probability of an event can be defined through relative frequencies. Published on June 14, 2016. We do not return it to the bin. You flip it and it comes up "heads". (d) There are two coins in a box. The first coin is two-headed. A fair coin has a 50-50 probability of coming up heads or tails; a double-headed coin always comes up heads. Given that heads show both times, what is the probability that the coin is the two-headed one?. There are three coins in a box. A two headed quarter is not something that was done at the mint, it is a novelty item, generally with high enough magnification you can see the seam that the two coins were joined together. Math Two-Headed Coin and Bayesian Probability Date: 04/21/2003 at 17:12:44 From: Maggie Subject: Probability In a box there are nine fair coins and one two-headed coin. What is the probability that there is a head on the OTHER side of this coin? Yes, it could be the fair coin or the two-headed coin, but they're not equally likely: because the fair coin COULD have come up tails, the two-headed coin is now twice as likely. Given that a tail appears on the third toss, then the probability that it is the two-headed coin is 0, so the probability that it is the fair coin is 1 in this case. I'm not a mathematician so please bear with me. I think hugin is right: the probability of a 6th head is just the combination of the probability you have the double-headed coin (0. Search Items > CARINA Grey Sleeveless Lace Dress & Cropped Jacket Formal/Wedding Outfit Size 12. There is a probability of 0. A gambler has in his pocket a fair coin and a two headed coin. It shows heads. It comes up heads every time. What is the probability that it is the fair coin?(b) Suppose that he flips the same coin a second time and, again, it shows heads. This already is a pretty good estimate of the real bias! But you might want an even better estimate. Randomness as a tool: graph theory; scheduling; internet routing. Answer on Question #26655 - Math - Statistics and Probability A bag contains three coins, one of which is coined with two heads, while the other two coins are normal and not biased. So the probability of choosing double head coin and get head is 1/3, while choosing single head coin and get head is 1/6. b:if heads appear on the second toss,then it also appears on the first toss. 2 Conditioning Independence, Conditional Probability Hypotheses, Total Probability Examples: Queen of Spades, Manufacturing Bayes, and Bridged Circuit 3. The outcome of the tosses comes up heads, heads, and heads. Answer: $$\frac{41}{72}$$. There are 3 coins in a box. The 2-headed coin is H, H, and the normal coin is h,t. R tosses a coin, and wins $1 if it lands on H or loses$1 on T. The same coin (from part (b)) is ipped again and it shows heads. In the very first Two-Face story, after Two-Face captured Batman and Robin and released them unharmed because the coin said so, he captured Batman again. One has two heads, one has two tails, and the other is a fair coin with one head and one tail. If you get it wrong, only one more coin toss is required to choose between the remaining two doors. Suppose your materials science roommate managed to make a two-headed coin. You have two coins, one of which is fair and comes up heads with a probability 1/2, and the other which is biased and comes up heads with prob You have two coins in a bag: one fair coin and one trick coin that has heads on both sides. Step 2 of 3: (b) It is given that the gambler flips the coin for the second time and again he gets a head. G is surprised to find that he loses the first ten times they play. What is the probability that two headed coin was selected? Denote with Ak the event that randomly selected coin lands heads up k times. 5 is the probability that the selected coin is a fair coin. a) What is the probability that it is the fair coin?. 6 LAB 1: some elementary (but creative) extensions. What is now the probability that it is the fair coin?. One is a two-headed coin, another is a fair coin, and the third is a biased coin that comes up heads 75 percent of the time. So the probability that you get three heads with a randomly chosen coin is:. You give the probability of a two-headed coin a 5% chance and probability of a two-tailed coin also a 5% chance. what is the probability that this is the two headed coin?. Compute the proportion of. What is the probability that it shows heads? b. If heads appears both times, what is the probability that the coin is two-head. Flipping a coin many times and getting heads each time, however, seems to go against the 50% chance of a normal coin landing on heads, that's when you might wonder if the coin is two-headed, but I still seem to reject that it actually affects the probability. What is the probability of throwing a head on 1 toss? Throwing a tail on one toss? Answer: B. Exercise 2: How many consecutive coin flips will it take for the subjective probability that it is a 2-headed coin to be creater than the subjective probability that it is a fair coin? Question: What happens to the posterior if you observe a single tail - even after a long string of heads?. (a) A gambler has in his pocket a fair coin and a two-headed coin. What is the probability that it is the fair coin? (b) Suppose that he flips the same coin a second time and again it shows heads. One of the three coins is chosen at random and tossed, it shows heads, what is the probability that it was the two heade. Answer on Question #26655 - Math - Statistics and Probability A bag contains three coins, one of which is coined with two heads, while the other two coins are normal and not biased. Given that you see 10 heads, what is the probability that the next toss of that coin is also a head?. A two-headed coin is worth very little, usually between $3 to$10, depending on how well the crafter made the coin and the face value of the coin. then another coin is selected from the two remaining coins and tossed. Now what is the probability that it is a fair coin?. You reach one coin at random, toss it, and it lands up heads. He selects one of the coins at random; when he flips it, it shows heads. What is the probability that it shows heads? b. Interview question for Quantitative Trader in Hong Kong. If you pick the two headed coin, you have a 100% probability of getting three heads. You give the probability of a two-headed coin a 5% chance and probability of a two-tailed coin also a 5% chance. If it's rainy and there is heavy traffic, I arrive late for work with probability 1 2. Consider an urn containing one fair coin and one two-headed coin. A gambler has in his pocket a fair coin and a two-headed coin. 2 Suppose that he flips the same coin a second time and, again, it shows heads. (a) What is the sample space of the experiment? (b) Given that both flips produce heads, what is the probability that Alice drew the two-headed coin from the urn?. There are three coins One is two headed coin, another is biased coin that comes up tails 25% of the times and the other is unbiased coin One of the three coins is chosen at random and tossed , it shows head What is the probability that - Math - Probability. Problem 1 (20%) There are three coins in a box. ” Now I flip a coin ten times, and ten times in a row it comes up heads. This discussion on There are three coins. What are the chances of 10 heads in a row? The probability is 1/1024. One of them is a 2-headed coin and the other one is a regular coin with Heads and a Tails. One coin is chosen at random and tossed twice. To find the probability of two independent events occuring, we simply multiply together the probabilities associated with two individual events. You have three coins in a bag. Prior and posterior beliefs are assessments of probability before and after seeing an outcome. Indeed, without throwing coins at all, there is a $1/5$ chance you get the double headed coin, and a $4/5$ chance you get a normal coin. One of these coins is selected at random and tossed three times. Problem 43: There are 3 coins in a box. 2 Conditioning Independence, Conditional Probability Hypotheses, Total Probability Examples: Queen of Spades, Manufacturing Bayes, and Bridged Circuit 3. given that it is a two-headed coin) Probability of heads coming up, given that it is a biased coin= 75%. Given that the coin is heads, find the conditional probability of each coin type. What is the probability that the transferred ball was white? 7. Suppose you pick one coin at random from the jar, flip it 10 times and get all heads. It doesn't matter whether any coin is tossed or or not. Probability problem: A generalization of an earlier two biased coins problem? starting with coin C1, P[2 heads in a row 999 coins and one two-headed coin. Then they do another, and another. Pair of Real Double Sided Quarters 1 Two Headed and 1 Two Tailed Coin - 1 x Double Headed Quarter + 1 x Double Tailed Quarter by QUICK PICK MAGIC. Thus, if an event can happen in m ways and fails to occur in n ways and m+n ways is equally likely to occur then the probability of happening of the event A is. (a) Find the probability that the blades of grass, when tied at random, form a ring. question_answer37) There are three coins. One coin has heads on both sides, one coin has tails on both sides, the third one has head on one side and tail on the other side. When the two-headed coin is picked, it always lands heads. Instead of probability distributions, we use probability densities and integrate over ranges of possible. What is the probability that it was the two-headed coin? c. • The first time this happens, it seems normal. What is the probability that this coin is the two-headed coin? Solution: This is a simple application of Bayes rule: there are 17 head faces, two of which belong to the two- headed coin. Ai having positive probability, then P(Aj|B) = P(B|Aj)P(Aj) Pn i=1 P(B| Ai)P(i) (b) One coin in a collection of 65 has two heads. if no more than five tosses each are allowed for a single game, find the probability that the person who tosses first will win the game. The Bayesian next takes into account the data observed and updates the prior beliefs to form a "posterior" distribution that reports probabilities in light of the data. Case 2: One head. In fact, the probability for most other values virtually disappeared — including the probability of the coin being fair (Bias = 0. One of the three coins is chosen at random and tossed. One coin is fair, the other has heads on both sides. You give the probability of a two-headed coin a 5% chance and probability of a two-tailed coin also a 5% chance. What is the probability that it is the fair coin? (b) Suppose that he °ips the same coin a second time and again it shows heads. I've read that if you flip a coin 10 times and it comes up heads every time then it's still a 50/50 chance to be a heads or tails on the 11th flip. given that it is a two-headed coin) Probability of heads coming up, given that it is a biased coin= 75%. One of the three coins is chosen at random and tossed, it shows heads, what is the probability that it was the two-headed coin?. the coin ip, they are now dependent: if you were to go on to discover that the coin has two heads, the hypothesis of psychic powers would return to its baseline probability { the evidence for psychic powers was \explained away" by the presence of the two-headed coin. Byju's Coin Toss Probability Calculator is a tool which makes calculations very simple and interesting. One is a two-headed coin, another is a fair coin, and the third is a biased coin that comes up heads 75 percent of the time. a) What is the probability that the coin chosen is the two-headed coin?. "Mathematical Expectation" is one of those few topics that is rarely discussed in details in any curriculum, but is nevertheless very important. Suppose that a bag contains 12 coins: 5 are fair, 4 are biased with probability of heads 1 3; and 3 are two-headed. chance of choosing one of the other coins, and getting two heads - 4/5*1/2*1/2 or 20% so there is a 40% chance of getting two heads in a row with any randomly chosen coin, so the probability that the 2 headed coin was chosen should be 20/40 or 50% so there was a 50% chance that it was the two-headed coin. and it first number is greater than or equal to (1-p) and second number is less than (1-p) then its t. Numerous people have tried to explain why they think the answer is 1/2, arguing that since both coins have a head then seeing a head doesn't rule out anything and thus it could be either coin with equal probability. I've found a reasonable negative filter is. What is the probability that it is the fair coin? Hint: It is given that. One is a two headed coin (having head on both faces), another is a biased coin that comes up heads 75% of the time and third is an unbiased coin. Suppose that one of these three coins is selected at random and flipped. MATH 264 PROBLEM HOMEWORK #2 Due to December 9, [email protected]:30 PROBLEMS 1. You reach one coin at random, toss it, and it lands up heads. Similarly, the probability of getting ailsT is 1 (1 2 p+ 1 2 q). What is the probability that it is the fair coin? (b) Suppose that he °ips the same coin a second time and again it shows heads. The hypotheses areH1 - the coin is two headed, and H2 the coin is fair. An urn contains 2 black balls and 3 white balls. A two-headed coin is worth very little, usually between $3 to$10, depending on how well the crafter made the coin and the face value of the coin. Depending on which coin you have, there is a 50% chance that the other side is tails (regular coin) and a 50% chance that the other side is heads (two-headed coin). A medical practice uses a \rapid in. what are the odds against A's losing if she goes first. You draw the normal coin and see tails. When one of the coins is selected at random and flipped, it shows heads. the opposite face is either heads or tails, the desired probability is 1/2. What are the chances of 10 heads in a row? The probability is 1/1024. In other words, it should happen 1 time in 4. There are 3 coins in a box. The Bayesian next takes into account the data observed and updates the prior beliefs to form a "posterior" distribution that reports probabilities in light of the data. There are three coins. H or T) of your coin to each other simultaneously. So P(X=1) = P(choose a 1 headed coin) x P(1 head, 1 tail obtained) = 4/6 x 1/2 x 1/2 = 1/6 If X = 2, there are 2 scenarios:. (a) He selects one of the coins at random, and when he °ips it, it shows heads. This is obviously an extreme example, but it illustrates that having a group of coins whose average probability of landing heads is 50% is not necessarily the. Numerous people have tried to explain why they think the answer is 1/2, arguing that since both coins have a head then seeing a head doesn't rule out anything and thus it could be either coin with equal probability. to bet you even money that it is the two headed coin. One coin has been specially made and has a head. He selects a coin at random and flips it twice. One coin is chosen at random and flipped, coming up heads. Perform the following experiment. on each side. If the same coin is tossed twice, find the probability that it is the two-headed coin. There are three coins One is two headed coin, another is biased coin that comes up tails 25% of the times and the other is unbiased coin One of the three coins is chosen at random and tossed , it shows head What is the probability that - Math - Probability. I've read that if you flip a coin 10 times and it comes up heads every time then it's still a 50/50 chance to be a heads or tails on the 11th flip. I'm not a mathematician so please bear with me. The hypotheses areH1-the coin is two headed, and H2 the coin is fair. Condtion that the face observed is already heads. Find the probability that heads appears twice. A coin is chosen at random and tossed 2 times. Remember that P(A given B) = P(A and B)/P(B) So let's say that A is the event that he chose the 2-headed coin, and B is an event denoted by H(N), which indicates that the coin was tossed N times, and came up heads each time, so the answer in our first case is P(A given H(1)), and the answer our last case is P(A given H(3)). What is the conditional prob-ability that both are boys given that at least one of them is a boy? 2. There are three coins. This is the currently selected item. (The fair coin lands on heads 50% of the time). All k times the coin landed up heads. One is a two-headed coin, another is a fair coin, and the third is a biased coin that comes up heads 75 percent of time. If you toss a coin, you cannot get both a head and a tail at the same time, so this has zero probability. What is the conditional probability that it is the fair coin. The rest are fair. onditional probability is a tool for updating conjectured view of the world using increasing amount of gradually incoming information. You have a jar containing 999 fair coins and one two-headed coin. Find the probability that three heads are obtained. Similar Questions. , in short (H, H) or (H, T) or (T, T) respectively; where H is denoted for head and T is denoted for tail. Find the pr Algebra -> Probability-and-statistics -> SOLUTION: A box contains 3 coins: one coin with two sides-head & tail, one coin two headed and one coin with probability of heads is 1/3. Notice that tails must be received in at least one of the first two flips. A box contains three coins: two regular coins and one fake, two{headed coin. coin is chosen at random and flipped, and comes up heads. If you combine the two sets the probability of getting a head is 7/9(3/7)+2/9(1/2)= 4/9. G is surprised to find that he loses the first ten times they play. Answer: $$\frac{41}{72}$$. You have a jar containing 999 fair coins and one two-headed coin. In the case of the coins, we understand that there's a $$\frac{1}{3}$$ chance we have a normal coin, and a $$\frac{2}{3}$$ chance it's a two-headed coin. Let's split problem into two parts: 1) What is the probability you picked the double-headed coin (now referred as D)? 2) What is the probability of getting a head on the next toss?. One of them is a fair coin, the second is a two-headed coin, and the third coin is weighted so that it comes up heads 75% of the time. Condtion that the face observed is already heads. find probability that a:heads appear on the second toss. There are three coins in a box. One is two headed coin (having head on both faces), another is a biased coin that comes up heads 75% of the time and third is an unbiased coin. One is a two-headed coin, another is a fair coin, and the third is a biased coin that comes up heads 75 percent of time. He selects one of the coins at random; when he flips it, it shows heads. 01, what is his posterior belief ? Note. Depending on which coin you have, there is a 50% chance that the other side is tails (regular coin) and a 50% chance that the other side is heads (two-headed coin). And if you roll a standard die, there’s a 1/6 probability that you’ll roll a six. This is, however, wrong, because given that heads came up, it is more likely that the two-headed coin was chosen. The probability that the coin is two-headed, given that it shows heads, is given by P (E1|A). If you pick the two headed coin, you have a 100% probability of getting three heads. A coin is randomly selected and ipped. (a) He selects one of the coins at random, and when he °ips it, it shows heads. In the case of the coins, we understand that there's a $$\frac{1}{3}$$ chance we have a normal coin, and a $$\frac{2}{3}$$ chance it's a two-headed coin. Two coins are available, one fair and the other two-headed. Given that a tail appears on the third toss, then the probability that it is the two-headed coin is 0, so the probability that it is the fair coin is 1 in this case. At the root (level 0) we choose randomly the first coin. a coin is selected at random and tossed. Figure:Probability of mother being carrier free, given n sons are disease free for n = 1(black), 2(orange),3(red),4(magenta), and5(blue), The vertical dashed line at p = 1=2is the case for the boxes, one with a fair coin and one with a two-headed coin. ip, they are now dependent: if you were to go on to discover that the coin has two heads, the hypothesis of psychic powers would return to its baseline probability { the evidence for psychic powers was \explained away" by the presence of the two-headed coin. Your intuition is not totally off-base here, just slow. The probability that the two-headed coin is selected out of the box is $P(E_1)=1/3$. A coin is selected at random and tossed. This is a basic introduction to a probability distribution table. b:if heads appear on the second toss,then it also appears on the first toss. What is the probability that the selected coin was the two-headed coin? Read solution. then another coin is selected from the two remaining coins and tossed. Given that heads comes up, what is the probability that you flipped the two headed coin?. One is a two-headed coin ( having head on both faces ), another is a biased coin that comes up heads 75% of the times and third is also a biased coin that comes up tails 40% of the times. | 2019-12-11T03:36:51 | {
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