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3,054,103 |
All numbers $1$ to $155$ are written on a blackboard, one time each. We randomly choose two numbers and delete them, by replacing one of them with their product plus their sum. We repeat the process until there is only one number left. What is the average value of this number? I don't know how to approach it: For two numbers, $1$ and $2$ , the only number is $1\cdot 2+1+2=5$ For three numbers, $1, 2$ and $3$ , we can opt to replace $1$ and $2$ with $5$ and then $3$ and $5$ with $23$ , or $1$ and $3$ with $7$ and then $2$ , $7$ with $23$ or $2$ , $3$ with $11$ and $1$ , $11$ with $23$ so we see that no matter which two numbers we choose, the average number remains the same. Does this lead us anywhere?
|
Claim: if $a_1,...,a_n$ are the $n$ numbers on the board then after n steps we shall be left with $(1+a_1)...(1+a_n)-1$ . Proof: induct on $n$ . Case $n=1$ is true, so assume the proposition holds for a fixed $n$ and any $a_1$ ,... $a_n$ . Consider now $n+1$ numbers $a_1$ ,..., $a_{n+1}$ . Suppose that at the first step we choose $a_1$ and $a_2$ . We will be left with $ n$ numbers $b_1=a_1+a_2+a_1a_2$ , $b_2=a_3$ ,..., $b_n=a_{n+1}$ , so by the induction hypothesis at the end we will be left with $(b_1+1)...(b_n+1)-1=(a_1+1)...(a_{n+1}+1)-1$ as needed, because $b_1+1=a_1+a_2+a_1a_2+1=(a_1+1)(a_2+1)$ Where did I get the idea of the proof from? I guess from the n=2 case: for $a_1,a_2$ you are left with $a_1+a_2+a_1a_2=(1+a_1)(1+a_2)-1$ and I also noted this formula generalises for $n=3$ So in your case we will be left with $156!-1=1\times 2\times...\times 156 -1$
|
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|
3,056,616 |
$P(x) = 0$ is a polynomial equation having at least one integer root, where $P(x)$ is a polynomial of degree five and having integer coefficients. If $P(2) = 3$ and $P(10)= 11$ , then prove that the equation $P(x) = 0$ has exactly one integer root. I tried by assuming a fifth degree polynomial but got stuck after that. The question was asked by my friend.
|
The assumption that $P$ has degree $5$ is irrelevant and unhelpful. If $r$ is a root of $P$ , we can write $P(x)=(x-r)Q(x)$ for some polynomial $Q$ . If $r$ is an integer, then $Q$ will also have integer coefficients (polynomial division never requires dividing coefficients if you are dividing by a monic polynomial). So, for any integer $a$ , $P(a)=(a-r)Q(a)$ must be divisible by $a-r$ . Taking $a=2$ and $a=10$ , we see easily that the only possible value of $r$ is $-1$ . Moreover, we can say that $P$ only has one integer root even counting multiplicity, because if $-1$ were a root of higher multiplicity, we could write $P(x)=(x+1)^2R(x)$ where $R(x)$ again has integer coefficients, so $P(2)$ would need to be divisible by $(2+1)^2=9$ .
|
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|
3,056,890 |
Question: Can we show that $$\phi=\frac{1}{2}+\frac{11}{2}\sum_{n=0}^\infty\frac{(2n)!}{5^{3n+1}(n!)^2} $$ ; where $\phi={1+\sqrt{5} \above 1.5pt 2}$ is the golden ratio ? Some background and motivation: Wikipedia only provides one series for the golden ratio - see also the link in the comment by @Zacky. I became curious if I could construct another series for the Golden Ratio based on a slight modification to a known series representation of the $\sqrt{2}.$ At first I considered $$\sqrt{2}=\sum_{n=0}^\infty(-1)^{n+1}\frac{(2n+1)!!}{(2n)!!}$$ ; which series can be accelerated via an Euler transform to yield $$\sqrt{2}=\sum_{n=0}^\infty(-1)^{n+1}\frac{(2n+1)!!}{2^{3n+1}(n!)^2}$$ This last series became the impetus to try and and get to the Golden ratio. Through trial and error I stumbled upon $$\frac{\sqrt{5}}{11}=\sum_{n=0}^\infty\frac{(2n)!}{5^{3n+1}(n!)^2}$$
|
First of all note that $$\frac1{\sqrt{1-4x}}=\sum_{n=0}^{\infty}\binom{2n}n x^n$$ Lets rewrite your sum as the following $$\sum_{n=0}^\infty\frac{(2n)!}{5^{3n+1}(n!)^2}=\frac15\sum_{n=0}^\infty\binom{2n}n\left(\frac1{5^3}\right)^n=\frac15\frac1{\sqrt{1-\left(\frac4{5^3}\right)}}=\frac{\sqrt 5}{11}$$ And therefore you can correctly conclude that $$\frac{1}{2}+\frac{11}{2}\sum_{n=0}^\infty\frac{(2n)!}{5^{3n+1}(n!)^2}=\frac12+\frac{11}2\frac{\sqrt 5}{11}=\frac{1+\sqrt 5}2$$ $$\therefore~\frac{1}{2}+\frac{11}{2}\sum_{n=0}^\infty\frac{(2n)!}{5^{3n+1}(n!)^2}=\phi$$
|
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|
3,057,924 |
I was reading Advanced Integration Techniques , and found that $$\int_{0}^{1}\sqrt{x}\sqrt{1-x}\,\mathrm dx =\frac{\pi}{8}$$ The book provides one method using residue theorem and Laurent expansion. However, I wonder if there are other techniques that I can use to evaluate this integral. The most direct method would be solving the integral and plug in the limits. $$\int_{0}^{1}\sqrt{x}\sqrt{1-x}\,\mathrm dx = \left[\frac{\arcsin(2x-1)+\sqrt{(1-x)x}(4x-2)}{8}\right]_{0}^{1}=\frac{\pi}{8}$$
|
Write $x=\sin^2 t$ so $dx=2\sin t\cos t dt$ . Your integral becomes $$\int_0^{\pi/2}2\sin^2 t\cos^2 t dt=\int_0^{\pi/2}\frac12 \sin^2 2t dt=\int_0^{\pi/2}\frac{1-\cos 4t}{4}dt=\left[\frac{t}{4}-\frac{1}{16}\sin 4t\right]_0^{\pi/2}=\frac{\pi}{8}.$$
|
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|
3,062,393 |
Given the property of the logarithm that $\log{xy} = \log{x} + \log{y}$ , how would one take the 'derivative' of this? To be more clear, $\log{xy} = \log{x} + \log{y}$ (property of $\log$ ) $D(\log{xy}) = D(\log{x} + \log{y})$ (i) (Take derivative on both sides) Now, $D(\log{x} + \log{y}) = D(\log{x}) + D(\log{y})$ (ii) (Derivative of sum is sum of derivatives) Combining (i) and (ii): $D(\log{xy}) = D(\log{x}) + D(\log{y})$ (iii) Implies: $\frac{1}{xy} = \frac{1}{x} + \frac{1}{y}$ (Evaluate derivative of logarithm using $D(\log{x}) = \frac{1}{x}$ $\frac{1}{xy} = \frac{1}{x} + \frac{1}{y}$ looks false to me; e.g. while $\log{6}$ does equal $\log{2} + \log{3}$ , $\frac{1}{6}$ does not equal $\frac{1}{2} + \frac{1}{3}$ . My first guess was that the issue was related to what variable I take the derivative with respect to, but I'd like to understand this a little more formally if someone could guide me. What's going wrong in this example?
|
Yes, as you noted, it matters which variable you take a derivative with respect to. Since we have two variables, if we assume they are independent, then we need to use partial derivatives . If we chose $x$ , then $$\frac{\partial \log(xy)}{\partial x}=\frac{y}{xy}=\frac1x$$ and $$\frac{\partial}{\partial x}(\log x+\log y)=\frac1x+0=\frac1x$$ So the answers agree. (the same thing happens if we chose $y$ instead). Alternatively, you could take a total derivative . $$\mathrm d(\log xy)=\frac{\partial \log xy}{\partial x}\mathrm dx+\frac{\partial \log xy}{\partial y}\mathrm dy=\frac1x\mathrm dx+\frac1y \mathrm dy$$ This agrees with $$d(\log x+\log y)=\frac1x\mathrm dx+\frac1y \mathrm dy$$ So there are no inconsistencies.
|
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|
3,063,610 |
I recently asked a question on this forum regarding why 3 points guaranteed the presence or absence of a unique equation representing a specific circle. (link here What do "3 different points" have to do with linear dependence in determining a unique circle? ) Shortly after this, I came across a question in my book that provided a picture of 4 red dots (image below) and asked, "How many ellipses do these 4 red points define". Having read the comments on my post with the circle, I thought that this was fairly straight forward. I chose " 1 ". This was wrong. The answer was infinite. This caught me as surprising as I didn't think of the equations for a circle and an ellipse as differing by much beyond a scaling factor for each quadratic term. I know that the general equation for an ellipse is as follows: $$\left(\frac{x-h}a\right)^2 + \left(\frac{y-k}b\right)^2 = 1$$ The only thing I can think of is that because of the added scaling factors, there are now technically two additional unknowns (for a total of 4 different unknowns... h, a, k, and b), and therefore I need 4 points to specify an unique ellipse. However, I thought to myself again, even if the ellipse is not centered at the origin, if all 4 given points happened to coincide with the intersection between the major axis and the ellipse and the minor axis and the ellipse, then certainly that would specify an unique ellipse. If this is true, then why does the arrangement of the points matter in determining whether or not an unique ellipse is specified? Visual explanations would be greatly appreciated!
|
The equation $\left(\frac{x-h}{a}\right)^2 + \left( \frac{y-k}{b}\right)^2 = 1$ is the equation for an ellipse with major and minor axes parallel to the coordinate axes. We expect such ellipses to be unchanged under horizontal reflection and under vertical reflection through their axes. In this equation, these reflections are effected by $x \mapsto 2h - x$ and $y \mapsto 2k -y$ . This means, if all you have is one point on the ellipse and the three reflected images of this point, you do not have $8$ independent coordinates; you have $2$ and uninformative reflections forced by the equation. We can see this by plotting two ellipses at the same center (same $h$ and $k$ ), intersecting at $4$ points, with, say, semiaxes of length $1$ and $2$ . These clearly have four points of intersection. But as soon as you know an ellipse is centered at the origin and contains any one of the four points of intersection, by the major and minor axis reflection symmetries, it contains all four. This is still true if you use generic ellipses, which can be rotated. Remember that the reflections are through the major and minor axes, wherever they are. Of course, there are other ways for two ellipses to intersect at four points. So just knowing those four points are on an ellipse cannot possibly tell you which one is intended. Returning to the first diagram, corresponding to the diagram you have where the four known points are the vertices of a square... Symmetries force the center of the ellipse to be the center of the square, but that's not a very strong constraint.
|
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|
3,063,611 |
I started studying Probability, and I'm not sure if I understand what is the meaning of "closed under union". A collection (say $F$ ) of subsets of a set (say $\Omega$ ) is said to be a $\sigma$ -algebra if: $\Omega \in F$ $F$ is closed under complement $F$ is closed under union Now, consider the following example: Given $\Omega = \{1, 2, 3, 4, 5\}$ , is $F = \{\emptyset, \{1, 2\}, \{3, 4, 5\}, \{5\}, \{1, 2, 3, 4\}, \{1, 2, 3, 4, 5\}\}$ closed under union? I'm asking because I know that $\bigcup_{i = 1}^{6} X_i \in F$ ; however it does not happen for any two elements, for example: $\{1,2\} \cup \{5\} \not\in F$ .
|
The equation $\left(\frac{x-h}{a}\right)^2 + \left( \frac{y-k}{b}\right)^2 = 1$ is the equation for an ellipse with major and minor axes parallel to the coordinate axes. We expect such ellipses to be unchanged under horizontal reflection and under vertical reflection through their axes. In this equation, these reflections are effected by $x \mapsto 2h - x$ and $y \mapsto 2k -y$ . This means, if all you have is one point on the ellipse and the three reflected images of this point, you do not have $8$ independent coordinates; you have $2$ and uninformative reflections forced by the equation. We can see this by plotting two ellipses at the same center (same $h$ and $k$ ), intersecting at $4$ points, with, say, semiaxes of length $1$ and $2$ . These clearly have four points of intersection. But as soon as you know an ellipse is centered at the origin and contains any one of the four points of intersection, by the major and minor axis reflection symmetries, it contains all four. This is still true if you use generic ellipses, which can be rotated. Remember that the reflections are through the major and minor axes, wherever they are. Of course, there are other ways for two ellipses to intersect at four points. So just knowing those four points are on an ellipse cannot possibly tell you which one is intended. Returning to the first diagram, corresponding to the diagram you have where the four known points are the vertices of a square... Symmetries force the center of the ellipse to be the center of the square, but that's not a very strong constraint.
|
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|
3,063,680 |
I have some statement, which is "proved" by numerical methods (particularly, using simulations), since it isn't possible to prove analytically. Is it legitimate to articulate this statement as a Theorem, if it isn't proved analytically. If no, what should I call it? P.S. More details regarding the statement. I have some data generating process and need to prove that the generated data as a function of some variable has some properties. For example, I have $x^1,...,x^n$ generated data (n=100000) and need to show that the function $x(\gamma)$ is positive defined over the domain. Therefore, on the generated data I apply interpolation technique in order to construct the function and I check the property.
|
In general, numerical methods don't constitute proofs. If all we have is an unknown blackbox function $f:\mathbb{R} \to \mathbb{R}$ that we know nothing about, and all we can do is compute its value at (finitely) many points, then we simply can't prove that $f$ is positive. However, in specific cases, we could have arguments based on numerical methods that are valid. Typically, we'd need to make a numerical approximation, and then prove, using non-numerical methods , that our numerical approximation is accurate enough for the theorem to follow. As such, how numerical methods can aid us in proving a statement is very statement-specific. Take, for example the following problem: Prove that $f(x) = x^2 + 1 > 0 \ \forall x \in [-1, 1]$ . Invalid proof: We computed $f(x)$ at $10^{10^{10000}}$ random points and used linear interpolation between them. Here's a plot. We can see that $f(x)$ is always positive. Valid proof 1: We computed $f(x)$ at points three points: $f(-1) = 2$ , $f(0) = 1$ , and $f(1)=2$ . Let $g(x)$ be the linear interpolation of the points $(-1, 2)$ , $(0, 1)$ , and $(1, 2)$ . $g$ attains its minimum at $g(0) = 1$ . Since $f^{\prime \prime} = 2$ , we can compute an error bound on our interpolation (see https://ccrma.stanford.edu/~jos/resample/Linear_Interpolation_Error_Bound.html ): $|f(x) - g(x)| \leq \frac{2}{8}$ . Therefore, we can conclude that $f(x) \geq \frac{3}{4} > 0$ . Note: Often, if we need to resort to numerical methods, if would be just as hard to compute derivatives. However, we don't need the actual derivatives, we just need an upper bound. The better the bound, the less points we would need to evaluate $f(x)$ at. Furthermore, bound to the first derivative is enough, but having second could also reduce the number of points needed. Valid proof 2: We know that $f(x)$ is convex. We use a numerical method to compute its minimum. find that $\min f(x) \approx 1.0000000075$ . We also have an (true, non-numerical ) error bound on our approximation: $|1.0000000075 - \min f(x)| < 0.001$ . Therefore, $f(x) > 1.0000000075 - 0.001 > 0$ . Finally, it doesn't really matter whether analytical proofs exist or not. The validity of any proof is only determined by that proof and no others. In fact, it has been proven that not all true statements can be proven. But that is no reason to reduce our standards of rigor.
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3,063,839 |
Solve the system of equation for real numbers \begin{split}
(a+b) &(c+d) &= 1 & \qquad (1)\\
(a^2+b^2)&(c^2+d^2) &= 9 & \qquad (2)\\
(a^3+b^3)&(c^3+d^3) &= 7 & \qquad (3)\\
(a^4+b^4)&(c^4+d^4) &=25 & \qquad (4)\\
\end{split} First I used the identity $$(a^2+b^2)(c^2+d^2)=(ac-bd)^2+(bc+ad)^2$$ Use this identity to (4) too
and simplify (3),
we obtain $$(a^2+b^2-ab)(c^2+d^2-cd)=7$$ And suppose $x=abcd$ use $ac=x /bd , bc=x/ad$ But got stuck...
|
Hint: $ac=x, bc=y, ad=u, bd=v$ , then the equations are $x+y+u+v=1$ $x^2+y^2+u^2+v^2=9$ $x^3+y^3+u^3+v^3=7$ $x^4+y^4+u^4+v^4=25$ Use Newton-Girard to compute the elementary polynomials.
Then you have the polynomial $P(z)= (z-x)(z-y)(z-u)(z-v)$ with variable $z$ .
Solve the quartic equation $P(z)=0$ , and there you have the values $x,y,u,v$ in some order.
Note that not in any order: $xv=yu$ must be true, see the definition of these variables. Of course, once you have $x,y,u,v$ , it is easy to compute $a,b,c,d$ . P.S. By this way we can get: $$\{x,y,u,v\}=\{-1,2,\sqrt2,-\sqrt2\},$$ which gives $abcd=-2.$ Up to symmetry, the solution is $(a,b,c,d)= (t, -\sqrt{2}t, -\frac{1}{t}, -\frac{\sqrt{2}}{t})$ for any $t\neq 0$ .
(By up to symmetry, I mean you can switch $a$ and $b$ , you can switch $c$ and $d$ , and you can switch the pair $(a,b)$ with $(c,d)$ , so there are $8$ symmetries.)
|
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3,063,845 |
If the symbol $A\sim B $ signifies that there is a bijection between A and B, and We take our sets to be dedekind infinite, then is the following correct? If not, what is the counter example?: $$A \sim B \Rightarrow A \sim B\cup\{x\}$$
|
Hint: $ac=x, bc=y, ad=u, bd=v$ , then the equations are $x+y+u+v=1$ $x^2+y^2+u^2+v^2=9$ $x^3+y^3+u^3+v^3=7$ $x^4+y^4+u^4+v^4=25$ Use Newton-Girard to compute the elementary polynomials.
Then you have the polynomial $P(z)= (z-x)(z-y)(z-u)(z-v)$ with variable $z$ .
Solve the quartic equation $P(z)=0$ , and there you have the values $x,y,u,v$ in some order.
Note that not in any order: $xv=yu$ must be true, see the definition of these variables. Of course, once you have $x,y,u,v$ , it is easy to compute $a,b,c,d$ . P.S. By this way we can get: $$\{x,y,u,v\}=\{-1,2,\sqrt2,-\sqrt2\},$$ which gives $abcd=-2.$ Up to symmetry, the solution is $(a,b,c,d)= (t, -\sqrt{2}t, -\frac{1}{t}, -\frac{\sqrt{2}}{t})$ for any $t\neq 0$ .
(By up to symmetry, I mean you can switch $a$ and $b$ , you can switch $c$ and $d$ , and you can switch the pair $(a,b)$ with $(c,d)$ , so there are $8$ symmetries.)
|
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|
3,064,917 |
Can someone give me a clue on how am I going to prove that this pattern is true or not as I deal with repeated division by 2? I tried dividing 360 by 2 repeatedly, eventually the digits of the result, when added repeatedly, always result to 9 for example: \begin{align*}
360 ÷ 2 &= 180 \text{, and } 1 + 8 + 0 = 9\\
180 ÷ 2 &= 90 \text{, and } 9 + 0 = 9\\
90 ÷ 2 &= 45 \text{, and } 4 + 5 = 9\\
45 ÷ 2 &= 22.5 \text{, and } 2 + 2 + 5 = 9\\
22.5 ÷ 2 &= 11.25 \text{, and } 1 + 1 + 2 + 5 = 9\\
11.25 ÷ 2 &= 5.625 \text{, and } 5 + 6 + 2 + 5 = 18 \text{, and } 1 + 8 = 9\\
5.625 ÷ 2 &= 2.8125 \text{, and } 2 + 8 + 1 + 2 + 5 = 18 \text{, and } 1 + 8 = 9
\end{align*} As I continue this pattern I found no error (but not so sure)... please any idea would be of great help!
|
Define the digit sum of the positive integer $n$ as the sum $d(n)$ of its digits. The reduced digit sum $d^*(n)$ is obtained by iterating the computation of the digit sum. For instance $$
n=17254,\quad
d(n)=1+7+2+5+4=19,
\quad
d(d(n))=1+9=10,
\quad
d(d(d(n)))=1+0=1=d^*(n)
$$ Note that $d(n)\le n$ , equality holding if and only if $1\le n\le 9$ . Also, $1\le d^*(n)\le 9$ , because the process stops only when the digit sum obtained is a one-digit number. The main point is that $n-d(n)$ is divisible by $9$ : indeed, if $$
n=a_0+a_1\cdot10+a_2\cdot10^2+\dots+a_n\cdot10^n,
$$ then $$
n-d(n)=a_0(1-1)+a_1(10-1)+a_2(10^2-1)+\dots+a_n(10^n-1)
$$ and each factor $10^k-1$ is divisible by $9$ . This extends to the reduced digit sum, because we can write (in the example above) $$
n-d^*(n)=\bigl(n-d(n)\bigr)+\bigl(d(n)-d(d(n))\bigr)+\bigl(d(d(n))-d(d(d(n)))\bigr)
$$ and each parenthesized term is divisible by $9$ . This works the same when a different number of steps is necessary. Since the only one-digit number divisible by $9$ is $9$ itself, we can conclude that $d^*(n)=9$ if and only if $n$ is divisible by $9$ . Since $360$ is divisible by $9$ , its reduced digit sum is $9$ ; the same happenso for $180$ and so on. When you divide an even integer $n$ by $2$ , the quotient is divisible by $9$ if and only if $n$ is divisible by $9$ . What if the integer $n$ is odd? Well, the digits sum of $10n$ is the same as the digit sum of $n$ . So what you are actually doing when arriving at $45$ is actually \begin{align}
&45 \xrightarrow{\cdot10} 450 \xrightarrow{/2} 225 && d^*(225)=9 \\
&225 \xrightarrow{\cdot10} 2250 \xrightarrow{/2} 1225 && d^*(1225)=9
\end{align} and so on. Note that the first step can also be stated as $$
45 \xrightarrow{\cdot5} 225
$$ Under these operations divisibility by $9$ is preserved, because you divide by $2$ or multiply by $5$ .
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3,064,922 |
I have the function $f=\{x,-y,0\}$ , and each of the derivatives together result in $\{1,-1,0\}$ . To calculate the divergence $\nabla\cdot$ of $f$ I'd have to do the dot (scalar) product of partial derivatives with $f$ , while for the curl $\nabla\times$ I have to do the cross product. I I know that the curl is $\{0,0,2\}$ but I do not really understand how this result comes up: the cross product of the partial derivatives with $f$ results in something like $\{0,0,x+y\}$ . Please bear in mind I'm very new to this topic and no one has taught me this, so it would help if someone could point out what has to be multiplied by what exactly . Thanks.
|
Define the digit sum of the positive integer $n$ as the sum $d(n)$ of its digits. The reduced digit sum $d^*(n)$ is obtained by iterating the computation of the digit sum. For instance $$
n=17254,\quad
d(n)=1+7+2+5+4=19,
\quad
d(d(n))=1+9=10,
\quad
d(d(d(n)))=1+0=1=d^*(n)
$$ Note that $d(n)\le n$ , equality holding if and only if $1\le n\le 9$ . Also, $1\le d^*(n)\le 9$ , because the process stops only when the digit sum obtained is a one-digit number. The main point is that $n-d(n)$ is divisible by $9$ : indeed, if $$
n=a_0+a_1\cdot10+a_2\cdot10^2+\dots+a_n\cdot10^n,
$$ then $$
n-d(n)=a_0(1-1)+a_1(10-1)+a_2(10^2-1)+\dots+a_n(10^n-1)
$$ and each factor $10^k-1$ is divisible by $9$ . This extends to the reduced digit sum, because we can write (in the example above) $$
n-d^*(n)=\bigl(n-d(n)\bigr)+\bigl(d(n)-d(d(n))\bigr)+\bigl(d(d(n))-d(d(d(n)))\bigr)
$$ and each parenthesized term is divisible by $9$ . This works the same when a different number of steps is necessary. Since the only one-digit number divisible by $9$ is $9$ itself, we can conclude that $d^*(n)=9$ if and only if $n$ is divisible by $9$ . Since $360$ is divisible by $9$ , its reduced digit sum is $9$ ; the same happenso for $180$ and so on. When you divide an even integer $n$ by $2$ , the quotient is divisible by $9$ if and only if $n$ is divisible by $9$ . What if the integer $n$ is odd? Well, the digits sum of $10n$ is the same as the digit sum of $n$ . So what you are actually doing when arriving at $45$ is actually \begin{align}
&45 \xrightarrow{\cdot10} 450 \xrightarrow{/2} 225 && d^*(225)=9 \\
&225 \xrightarrow{\cdot10} 2250 \xrightarrow{/2} 1225 && d^*(1225)=9
\end{align} and so on. Note that the first step can also be stated as $$
45 \xrightarrow{\cdot5} 225
$$ Under these operations divisibility by $9$ is preserved, because you divide by $2$ or multiply by $5$ .
|
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|
3,064,935 |
Let $\, f: \mathbb{R}^n \rightarrow [-\infty, +\infty]$ be a Lebesgue measurable function. I want to show that f is $\lambda_n$ - alomost everywhere equal to to a Borel measurable function $\, f'$ . I would very much appreciate your help. Best, KingDingeling
|
Define the digit sum of the positive integer $n$ as the sum $d(n)$ of its digits. The reduced digit sum $d^*(n)$ is obtained by iterating the computation of the digit sum. For instance $$
n=17254,\quad
d(n)=1+7+2+5+4=19,
\quad
d(d(n))=1+9=10,
\quad
d(d(d(n)))=1+0=1=d^*(n)
$$ Note that $d(n)\le n$ , equality holding if and only if $1\le n\le 9$ . Also, $1\le d^*(n)\le 9$ , because the process stops only when the digit sum obtained is a one-digit number. The main point is that $n-d(n)$ is divisible by $9$ : indeed, if $$
n=a_0+a_1\cdot10+a_2\cdot10^2+\dots+a_n\cdot10^n,
$$ then $$
n-d(n)=a_0(1-1)+a_1(10-1)+a_2(10^2-1)+\dots+a_n(10^n-1)
$$ and each factor $10^k-1$ is divisible by $9$ . This extends to the reduced digit sum, because we can write (in the example above) $$
n-d^*(n)=\bigl(n-d(n)\bigr)+\bigl(d(n)-d(d(n))\bigr)+\bigl(d(d(n))-d(d(d(n)))\bigr)
$$ and each parenthesized term is divisible by $9$ . This works the same when a different number of steps is necessary. Since the only one-digit number divisible by $9$ is $9$ itself, we can conclude that $d^*(n)=9$ if and only if $n$ is divisible by $9$ . Since $360$ is divisible by $9$ , its reduced digit sum is $9$ ; the same happenso for $180$ and so on. When you divide an even integer $n$ by $2$ , the quotient is divisible by $9$ if and only if $n$ is divisible by $9$ . What if the integer $n$ is odd? Well, the digits sum of $10n$ is the same as the digit sum of $n$ . So what you are actually doing when arriving at $45$ is actually \begin{align}
&45 \xrightarrow{\cdot10} 450 \xrightarrow{/2} 225 && d^*(225)=9 \\
&225 \xrightarrow{\cdot10} 2250 \xrightarrow{/2} 1225 && d^*(1225)=9
\end{align} and so on. Note that the first step can also be stated as $$
45 \xrightarrow{\cdot5} 225
$$ Under these operations divisibility by $9$ is preserved, because you divide by $2$ or multiply by $5$ .
|
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|
3,065,701 |
Can a cube (meaning $g(x) = f(x)^3 = f(x) \cdot f(x) \cdot f(x)$ ) of discontinuous function $f: D \to \mathbb{R}$ ( $D$ is subset of $\mathbb{R}$ ) be continuous? I think it can't, since $x^3$ is injective, but I am not able to prove it or find a counterexample.
|
Since $\phi : \mathbb{R} \to \mathbb{R}, \phi(x) = x^3$ , is a homeomorphism, you see that $f$ is continuous iff $\phi \circ f$ is continuous.
|
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|
3,067,148 |
A lily pad sits on a pond. It doubles in size every day. It takes 30
days for it to cover the pond. If you start with 8 lily pads instead,
how many days does it take to cover the pond? I think that the answer is $27$ , but I don't really think that makes sense intuitively. I think that, intuitively, the answer should be less than $30/4$ since it is increasing at an exponential rate.
|
Starting with 8 pcs. is like 3 days have gone by. So 27 days remain.
|
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|
3,068,013 |
In my probability book, it says that to count the total number of subsets of n elements is a process of $n$ stages with binary choice of either adding this element to the subset or not to add it. Therefore, the total number is $$2^n$$ But, for instance, we have 3 elements, according to this formula, there are 2 to the power of 3 elements, namely 8, which are $${\emptyset},A,B,C, AB, AC, BC, ABC$$ However, I have a hard time of imagining the process or N stages binary choice that form this many subsets. Can anyone explain/help me to understand it? I mean, ABC, if we are making the choice of A, put it in or do not put it in, exactly which subset are we choosing to put in or not? Thank you.
|
"Include A?" is stage 1 "Include B?" is stage 2 "Include C?" is stage 3.
|
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|
3,069,869 |
The question: The function p is defined by $-2|x+4|+10$ . Solve the equality $p(x) > -4$ Here were my steps to solving this: 1.) Subtract 10 from both sides -> $-2|x+4| > -14$ 2.) Divide both sides by -2 -> $|x+4|>7$ 3.) $x+4$ should therefore be 7 units or greater from zero on the number line, meaning either greater than 7 or less than -7: $x+4 > 7$ $x+4 < -7$ 4.) Subtract 4 from both sides: $x > 3$ $x < -11$ Graphing this, I see my signs are the wrong way round but I'm not quite sure where I've gone wrong.
|
$20$ is greater than $8$ , right? $$20 > 8$$ Now divide both sides by $-2$ : $$-10 > -4$$ Whoops! That's not right. This is because when you multiply or divide an inequality by a negative number, you must change the sense of the inequality: $>$ becomes $<$ , and $\le$ becomes $\ge$ etc: $$-10 < -4$$
|
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|
3,070,440 |
Through some calculation, I found that for all $r>0$ $$
\begin{array}{rcl}
{\displaystyle\int_{0}^{1}\left[\left(1 - x^{r}\right)^{1/r} - x\right]^{2}\,\mathrm{d}x} & {\displaystyle =} &
{\displaystyle{1 \over 3}}
\\
{\displaystyle\int_{0}^{1}\left[\left(1 - x^{r}\right)^{1/r} - x\right]^{4}\,\mathrm{d}x} & {\displaystyle =} &
{\displaystyle{1 \over 5}}
\\
{\displaystyle\int_{0}^{1}\left[\left(1 - x^{r}\right)^{1/r} - x\right]^{6}\,\mathrm{d}x} & {\displaystyle =} &
{\displaystyle{1 \over 7}}
\end{array}
$$ It seems like for $k \in \mathbb{N}$ $$
\int_{0}^{1}
\left[\left(1 - x^{r}\right)^{1/r} - x\right]^{2k}\mathrm{d}x = {1 \over 2k + 1}
$$ I want to prove this general form. Someone suggested to make the substitution $$y=(1-x^r)^{1/r}$$ Let $n=2k$ , and I rewrote the integral into $$\int_{0}^{1}((1-x^r)^{1/r}-x)^ndx=\int_{0}^{1}(y-x)^ndx$$ and tried to use the binomial formula: $$(y-x)^{n}=\sum _{k=0}^{n}{\binom {n}{k}}(-1)^{n-k}x^{n-k}y^{k}$$ The integral then becomes $$\begin{align}
\int_{0}^{1}(y-x)^ndx&=\int_{0}^{1}\sum _{k=0}^{n}{\binom {n}{k}}(-1)^{n-k}x^{n-k}y^{k}dx\\
&=\sum _{k=0}^{n}{\binom {n}{k}}(-1)^{n-k}\int_{0}^{1}x^{n-k}y^{k}dx\\
&=\sum _{k=0}^{n}{\binom {n}{k}}(-1)^{n-k}\int_{0}^{1}x^{n-k}(1-x^r)^{k/r}dx\\
\end{align}$$ Now I think I need to use Beta function: $$B(x,y) = \frac{(x-1)!(y-1)!}{(x+y-1)!}= \int_{0}^{1}u^{x-1}(1-u)^{y-1}du=\sum_{n=0}^{\infty}\frac{{\binom{n-y}{n}}}{x+n}$$ Am I on the right track? Are here any easier ways to prove the general form?
|
We present 3 different solutions. Solution 1 - slick substitution. We prove a more general statement: Proposition. Let $R \in (0, \infty]$ and let $\varphi : [0, R] \to [0, R]$ satisfy the following conditions: $\varphi$ is continuous on $[0, R]$ ; $\varphi(0) = R$ and $\varphi(R) = 0$ ; $\varphi$ is bijective and $\varphi^{-1} = \varphi$ . Then for any integrable function $f$ on $[0, R]$ , $$ \int_{0}^{R} f(|x-\varphi(x)|) \, \mathrm{d}x = \int_{0}^{R} f(x) \, \mathrm{d}x. $$ Proof. In case $\varphi$ is also continuously differentiable on $(0, R)$ , by the substitution $y = \varphi(x)$ , or equivalently, $x = \varphi(y)$ , $$ I
:= \int_{0}^{R} f( |x - \varphi(x)| ) \, \mathrm{d}x
= -\int_{0}^{R} f( |\varphi(y) - y| ) \varphi'(y) \, \mathrm{d}y. $$ Summing two integrals, \begin{align*}
2I
&= \int_{0}^{R} f( |x - \varphi(x)| ) (1 - \varphi'(x)) \, \mathrm{d}x \\
&= \int_{-R}^{R} f( |u| ) \, \mathrm{d}u = 2\int_{0}^{R} f(u) \, \mathrm{d}u, \tag{$u = x - \varphi(x)$}
\end{align*} proving the claim when $\varphi$ is continuously differentiable. This proof can be easily adapted to general $\varphi$ by using Stieltjes integral. ■ Now plug $\varphi(x) = (1-x^r)^{1/r}$ with $R = 1$ and $f(x) = x^n$ for positive even integer $n$ . Then $$ \int_{0}^{1} \left( (1-x^r)^{1/r} - x \right)^n \, \mathrm{d}x
= \int_{0}^{1} \left| x - (1-x^r)^{1/r} \right|^n \, \mathrm{d}x
= \int_{0}^{1} x^n \, \mathrm{d}x
= \frac{1}{n+1}. $$ Solution 2 - using beta function. Here is an alternative solution. Write $p = 1/r$ . Then using the substitution $x = u^p$ , \begin{align*}
\int_{0}^{1} \left( (1 - x^r)^{1/r} - x \right)^n \, \mathrm{d}x
&= \int_{0}^{1} \left( (1 - u)^{p} - u^p \right)^{n} pu^{p-1} \, \mathrm{d}u \\
&= \sum_{k=0}^{n} (-1)^k \binom{n}{k} p \int_{0}^{1} (1-u)^{p(n-k)} u^{p(k+1)-1} \, \mathrm{d}u \\
&= \sum_{k=0}^{n} (-1)^k \binom{n}{k} p \cdot \frac{(p(n-k))!(p(k+1)-1)!}{(p(n+1))!}
\end{align*} Here, $s! = \Gamma(s+1)$ . Now define $a_k = (pk)!/k!$ . Then the above sum simplifies to \begin{align*}
\int_{0}^{1} \left( (1 - x^r)^{1/r} - x \right)^n \, \mathrm{d}x
&= \frac{1}{(n+1)a_{n+1}} \sum_{k=0}^{n} (-1)^k a_{n-k}a_{k+1} \\
&= \frac{1}{(n+1)a_{n+1}} \left( a_0 a_{n+1} + \sum_{k=0}^{n-1} (-1)^k a_{n-k}a_{k+1} \right).
\end{align*} So it suffices to show that $\sum_{k=0}^{n-1} (-1)^k a_{n-k}a_{k+1} = 0$ . But by the substitution $l = n-1-k$ , we have $$ \sum_{k=0}^{n-1} (-1)^k a_{n-k}a_{k+1}
= - \sum_{l=0}^{n-1} (-1)^l a_{l+1}a_{n-l}. $$ (Here the parity of $n$ is used.) So the sum equals its negation, hence is zero as required. Solution 3 - using multivariate calculus. Let $\mathcal{C}_r$ denote the curve defined by $x^r + y^r = 1$ in the first quadrant, oriented to the right. Then $$ I(r) := \int_{0}^{1} \left( (1 -x^r)^{1/r} - x \right)^n \, \mathrm{d}x
= \int_{\mathcal{C}_r} ( y - x )^n \, \mathrm{d}x. $$ Notice that if $0 < r < s$ , then $\mathcal{C}_s$ lies above $\mathcal{C}_r$ , and so, the curve $\mathcal{C}_r - \mathcal{C}_s$ bounds some region, which we denote by $\mathcal{D}$ , counter-clockwise: $\hspace{10em}$ Then by Green's theorem, $$ I(r) - I(s)
= \int_{\partial \mathcal{D}} ( y - x )^n \, \mathrm{d}x
= - \iint_{\mathcal{D}} n (y - x)^{n-1} \, \mathrm{d}x\mathrm{d}y. $$ But since the region $\mathcal{D}$ is symmetric around $y = x$ and $n$ is even, interchanging the roles of $x$ and $y$ shows $$ \iint_{\mathcal{D}} n (y - x)^{n-1} \, \mathrm{d}x\mathrm{d}y
= \iint_{\mathcal{D}} n (x - y)^{n-1} \, \mathrm{d}x\mathrm{d}y
= - \iint_{\mathcal{D}} n (y - x)^{n-1} \, \mathrm{d}x\mathrm{d}y. $$ Therefore $I(r) = I(s)$ for any $ r < s$ , and in particular, letting $s \to \infty$ gives $$ I(r) = \int _{0}^{1} (1 - x)^n \, \mathrm{d}x = \frac{1}{n+1}. $$
|
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|
3,071,367 |
Equivalently, about variance? I realize it measures the spread of a distribution, but many other metrics could do the same (e.g., the average absolute deviation). What is its deeper significance? Does it have a particular geometric interpretation (in the sense, e.g., that the mean is the balancing point of a distribution)? any other intuitive interpretation that differentiates it from other possible measures of spread? What's so special about it that makes it act as a normalizing factor in all sorts of situations (for example, convert covariance to correlation)?
|
There's a very nice geometric interpretation. Random variables of finite mean form a vector space. Covariance is a useful inner product on that space. Oh, wait, that's not quite right: constant variables are orthogonal to themselves in this product, so it's only positive semi-definite. So, let me be more precise - on the quotient space formed by the equivalence relation "is a linear transformation of", covariance is a true inner product. (If quotient spaces are an unfamiliar concept, just focus on the vector space of zero-mean, finite-variance variables; it gets you the same outcome in this context.) Right, let's carry on. In the norm this inner product induces, standard deviation is a variable's length, while the correlation coefficient between two variables (their covariance divided by the product of their standard deviations) is the cosine of the "angle" between them. That the correlation coefficient is in $[-1,\,1]$ is then a restatement of the vector space's Cauchy-Schwarz inequality.
|
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|
3,071,381 |
Let $X,Y$ random variables. Every coninuous and not-decreasing monotone function $g:\mathbb{R}\to[0,1]$ fulfills $\mathbb{E}[g(Y)]\leq \mathbb{E}[g(X)]$ . Prove that $F_X(t)\leq F_Y(t)$ . I don't have any initial idea to solve it.
|
There's a very nice geometric interpretation. Random variables of finite mean form a vector space. Covariance is a useful inner product on that space. Oh, wait, that's not quite right: constant variables are orthogonal to themselves in this product, so it's only positive semi-definite. So, let me be more precise - on the quotient space formed by the equivalence relation "is a linear transformation of", covariance is a true inner product. (If quotient spaces are an unfamiliar concept, just focus on the vector space of zero-mean, finite-variance variables; it gets you the same outcome in this context.) Right, let's carry on. In the norm this inner product induces, standard deviation is a variable's length, while the correlation coefficient between two variables (their covariance divided by the product of their standard deviations) is the cosine of the "angle" between them. That the correlation coefficient is in $[-1,\,1]$ is then a restatement of the vector space's Cauchy-Schwarz inequality.
|
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|
3,077,308 |
I want to intuitively argue that there is no function with some properties, and find it tricky to explain it to someone who just understands that derivatives are representative of increase rates of a function. Here is the statement: There is no function $f(x)$ on $x \ge 0$ such that $f(0)=0$ , $f'(0)=0$ , $f(x)<x^2$ for $x>0$ and that the first, second and third derivative of $f(x)$ are strictly positive on $x>0$ . I appreciate any help!
|
I do not think it is true. Take for example $f(x) = 1 - x + \frac12x^2 - e^{-x}$ $f'(x) = - 1 + x + e^{-x}$ $f''(x) = 1 - e^{-x}$ $f'''(x) = e^{-x}$ Here is a graph: note that the black line for $x^2$ is above the dark blue line for $f(x)$ , while the green line for $f''(x)$ stays below $1$ while the second derivative of $x^2$ is $2$ , and the red line for $f'''(x)$ tends towards $0$ from above as $x$ increases
|
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|
3,077,641 |
I came across this Question where I have to find $$f^{(10)}$$ for the following function at $x = 0$ $$f(x) = e^x\sin x$$ I tried differentiating a few times to get a pattern but didn’t get one, can someone provide the solution.
|
Hint : As $\;\mathrm e^x\sin x=\operatorname{Im}\bigl(\mathrm e^{(1+i)x}\bigr)$ , you have to find first the real and imaginary parts of $(1+i)^{10}$ . Some details : There results from the above remark and linearity of differentiation that $\;(\mathrm e^x\sin x)'=\bigl(\operatorname{Im}(\mathrm e^{(1+i)x})\bigr)'= \operatorname{Im}\bigl((1+i)\mathrm e^{(1+i)x}\bigr)$ , hence $$\;(\mathrm e^x\sin x)''=\bigl(\operatorname{Im}((1+i)\mathrm e^{(1+i)x}))\bigr)'= \operatorname{Im}\bigl((1+i)^2\mathrm e^{(1+i)x}\bigr),$$ and more generally $$(\mathrm e^x\sin x)^{(k)}=\bigl(\operatorname{Im}(\mathrm e^{(1+i)x})\bigr)^{(k)}=\operatorname{Im}\bigl((1+i)^k(\mathrm e^{(1+i)x})\bigr).$$
|
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|
3,079,520 |
Here is a statement of the second part of the Fundamental Theorem of Calculus (FTC2), from a well-known calculus text (James Stewart, Calculus , 4th ed): If $f$ is continuous on $[a,b]$ , then $\int_a^b f(x) \, dx = F(b)-F(a)$ , where $F$ is any [emphasis mine] antiderivative of $f$ , that is, a function such that $F'=f$ . The following, however, seems to give a counterexample.* Can someone resolve this for me?: Let $f(x) = \frac{1}{4 \sin (x)+5}$ . $f$ is continuous on $[0, 2 \pi]$ : Consider two antiderivatives of $f$ , $F_1$ and $F_2$ : $$F_1(x) = \frac{x}{3}+\frac{2}{3} \tan^{-1}\left(\frac{\cos (x)}{\sin (x)+2}\right)$$ $$F_2(x)=\frac{1}{3} \left(\tan ^{-1}\left(2-\frac{3}{\tan \left(\frac{x}{2}\right)+2}\right)-\tan^{-1}\left(2-\frac{3}{\cot \left(\frac{x}{2}\right)+2}\right)\right).$$ Using Mathematica, I've confirmed that both $F_1'= f$ and $F_2'= f$ . According to my reading of the above statement of FTC(2), $\int_0^{2\pi} f (x) \, dx = F_1(2\pi)-F_1(0)= F_2(2\pi)-F_2(0)$ However, $F_1(2\pi)-F_1(0)=2\pi/3$ $F_2(2\pi)-F_2(0)=0$ Note from the plots below that $F_1$ is continuous on $[a,b]$ , while $F_2$ is not. Given all of this, it seems the sufficient condition for $\int_a^b f (x) \, dx = F(b)-F(a)$ is that the antiderivative be continuous on $[a,b]$ , not the integrand. $F_1 =$ $F_2=$ *I've taken this example function from a wolfram.com blog.
|
A differentiable function is continuous. If your "antiderivative" is not continuous, it is not an antiderivative of a continuous function.
|
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|
3,079,966 |
I suppose this is a question about mathematical convention. In a problem in Introduction to Probability by Bertsekas and Tsitsiklis, they ask the reader to prove an identity. But then their proof is mostly words: Problem 3.* Prove the identity $$A \cup \Bigg( \bigcap_{n=1}^\infty B_n \Bigg) = \bigcap_{n=1}^\infty\big(A \cup B_n\big).$$ Solution. If $x$ belongs to the set on the left, there are two possibilities. Either $x \in A$ , in which case $x$ belongs to all of
the sets $A \cup B_n$ , and therefore belongs to the set on the right.
Alternatively, $x$ belongs to all of the sets $B_n$ in which case, it
belongs to all of the sets $A \cup B_n$ , and therefore again belongs
to the set on the right. Conversely, if $x$ belongs to the set on the right, then it belongs to $A \cup B_n$ for all $n$ . If $x$ belongs to $A$ , then it belongs to
the set on the left. Otherwise, $x$ must belong to every set $B_n$ and
again belongs to the set on the left. In mathematics, why is this allowed? Can you say that this is more correct a proof that is, "Oh, it's obvious!" or "Just keep distributing $A$ over and over ad nauseum and you get the term on the right"? I'm not trolling. I'm genuinely curious as to how thorough one must be when using words as proof.
|
Exactly as thorough as you would have to be using any other kinds of symbols. It's just that vast messes of symbols are hellish for humans to read, but sentences aren't. Adding symbols to something doesn't make it more rigorous, less likely to be wrong, or really anything else. Symbols are useful for abbreviating in situations where this adds clarity, and making complex arguments easier to follow, but shouldn't be used where they do not help in this regard.
|
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|
3,085,568 |
The following was originally stated for n-tuples of elements from a scalar field, so most of the properties of "vectors" are easily established from the properties of the underlying scalar field. But the authors seem to want their development to be "self-reliant". For this reason I have replaced "n-tuple" with "vector". The equality relation for vectors has been established, as have the
associative and commutative laws of vector addition. The next property
of vector addition to be introduced is the neutral element: There exists a vector $\mathfrak{0}$ such that $\mathfrak{x}+\mathfrak{0}=\mathfrak{x}$ for every $\mathfrak{x}$ . It follows there can be only one neutral element, for if $\mathfrak{0}$ and $\mathfrak{0}^{\prime}$ were
two such elements we would have $\mathfrak{0}^{\prime}+\mathfrak{0}=\mathfrak{0}^{\prime}$ and $\mathfrak{0}+\mathfrak{0}^{\prime}=\mathfrak{0},$ so that by the commutative law of vector addition and the transitivity of vector equality we would have $\mathfrak{0}=\mathfrak{0}^{\prime}.$ Now suppose that for some $\mathfrak{x}$ we have $\mathfrak{x}+\mathfrak{z}=\mathfrak{x}.$ Do we have enough to prove that $\mathfrak{z}=\mathfrak{0}?$ I note in particular that the proof of the uniqueness of $\mathfrak{0}$ relies on the assumption that $\mathfrak{x}+\mathfrak{0}^{\prime}=\mathfrak{x}$ holds for all vectors, and thereby for $\mathfrak{x}=\mathfrak{0}$ . That assumption comes from the definition of $\mathfrak{0}$ satisfying $\mathfrak{x}+\mathfrak{0}=\mathfrak{x}$ for every vector, and the assumption that $\mathfrak{0}^\prime$ is also 'such an element'. Also note that the additive inverses have not yet been introduced.
|
No, this cannot be proved from just associativity, commutativity, and existence of a neutral element. For instance, consider the set $[0,1]$ with the binary operation $a*b=\min(a,b)$ . This operation is associative and commutative and $1$ is a neutral element. But for any $x,y$ with $x\leq y$ , we have $x*y=x$ , and $y$ is not necessarily the neutral element $1$ .
|
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|
3,085,587 |
Use the principle of mathematical induction to prove $$\left[\frac{1}{1}+ \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{n}\right] \leq \frac{n}{2} + 1~\text{for all}~n \in \mathbb{Z}^+$$ I have been trying to solve this however find myself getting stuck when I substitute $p_k$ back into the equation. This is my approach so far: (1) If $n= 1$ , LHS $= (1/1) = 1$ and RHS $= \frac{1}{2}+1 = \frac{3}{2}$ LHS $\leq$ RHS $p_1$ is true Assuming $p_k$ is true, $n = k$ $$\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{k} \leq \frac{k}{2}+1$$ If $p_{k+1}$ is true, $n = k+1$ $$\frac{1}{1} + \frac{1}{2}+ \frac{1}{3} + \ldots + \frac{1}{k} + \frac{1}{k+1} \leq \frac{k+1}{2} +1$$ Now LHS $\leq \frac{k}{2}+1+ \frac{1}{k+1}$ [using $p_k$ ] I don't really know what to do after this in terms of proving the sequence is less than or equal to $\frac{n}{2} + 1$ ?
|
No, this cannot be proved from just associativity, commutativity, and existence of a neutral element. For instance, consider the set $[0,1]$ with the binary operation $a*b=\min(a,b)$ . This operation is associative and commutative and $1$ is a neutral element. But for any $x,y$ with $x\leq y$ , we have $x*y=x$ , and $y$ is not necessarily the neutral element $1$ .
|
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|
3,087,356 |
This question is often asked as "why is 2 a prime number" and the only answers I can find are "The definition of prime numbers is written in such a way that 2 is prime". Sometimes, if the question was "Why is 2 a prime number even though it's even" the answer will include some explanation that being even is not such a special property, it just means a number is divisible by two and of course two is divisible by two, that's the number itself and three is divisible by three and it's still prime and so on. But there is something special about a number being even, and it 2 does have a special property that is not shared by any other prime number. It can be divided into equal subsets. The answer I'm looking for will answer the question in the title and also answer: If we change the definition of a prime number, what effects does that have on mathematics theories, proofs, etc. Are certain claims that were useful and important under the old definition no longer true? Specifically, What's the impact of changing the definition of prime to this: A prime is any integer that cannot be divided into smaller equal groups. N.B. this also changes the primeness of 1 but not any other number.
|
We define things in such a way that they have some practical use, and help us make sense of the world around us. This is exactly how words and concepts are created, and also evolve. For example, we decided to give a name to a class of objects in the sky with similar behavior: 'planets'. These objects form what a philosopher might call a 'natural' class of objects. Having labels for them allows us to talk and think about those objects more easily, helping us with explanations, predictions, doing science, and again making sense of things in general. But like I said, definitions can evolve: Pluto is no longer considered a 'planet', because after finding out more about our solar system we realized Pluto is in significant ways different from Neptune, Jupiter, Earth, etc. That is, by putting Pluto into different (though still related) class of 'dwarf-planets', we now look at it a little differently. The same holds for mathematical definitions. For example, we could define 'huppelflup numbers' to be exactly those numbers that can be divided by 17 or by 631 ... but there just isn't much practical use to such a definition, and so we don't. But the way we define prime numbers has lots of practical uses. For example, with the current definition, we get the nice, clean, result that every number has a unique prime factorization. And it's not just applications within mathematics that matter: prime numbers have tremendous importance for real life as well. Now, if we were to exclude $2$ as a prime, this would no longer be true. And a bunch of other results would likewise have to be stated in a much more cumbersome way. And by the way, this unique prime factorization theorem is exactly why mathematicians did exclude $1$ as a prime .. even though originally it was. So yes, you're right that it is not as if definitions are fixed until the end of time. And maybe at some point in the future we redefine the sets of primes again to also exclude $2$ , because doing so will have some other advantages. However, I wouldn't hold my breadth: the current definition is very nice.
|
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|
3,090,821 |
I am doing a project about irrational and transcendental numbers and I was wondering how could I construct a "new" transcendental number. I know that all Liouville numbers are transcendental so this could be a good place to start however I wish to try and create one not necessarily belonging to that group of numbers. In addition to this I am aware of Liouville's theorem surrounding transcendental numbers and have also found a proof to how that the theorem is true and holds for Liouville's constant. Once again I know about the Gelfond-Schneider theorem but the same idea applies. I would like to stay away from obvious substitutions i.e. $4^{\sqrt{17}}$ is transcendental but not really "interesting". Also any strange theorems I would like to try and prove for the sake of completion. I don't want to just assert something and just have it there. I would like to be as thorough as possible. Any help is greatly appreciated and thanks in advance
|
Take your favorite polynomial, $p(x) := x^2-x-1$ , and your favorite transcendental number $c := \pi$ . The number $p(c)$ is transcendental. The key to transcendental numbers is polynomials with integer or rational coefficients. If $p(c)$ was algebraic, then there exists polynomials $q(x)$ such that $q(p(c))=0$ but then the composition $r(x):=q(p(x))$ is also a polynomial and $r(c)=0$ which would imply that $c$ is algebraic. The contrapositive statement is now proven.
|
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|
3,091,706 |
I am learning about definite integrals and found the formula for finding an average of a function over a given interval: $$\frac{1}{b-a} \int_{a}^{b} f\left(x\right) dx$$ If we look at the average function for a set of numbers: $$\frac{1}{n} \sum_{i=1}^{n} x_i$$ It would seem as if the integral is essentially a summation of all values from $a$ to $b$ . Is it correct to think of it this way?
|
It's certainly helpful to think of integrals this way, though not strictly correct. An integral takes all the little slivers of area beneath a curve and sums them up into a bigger area - though there's, of course, a lot of technicality needed to think about it this way. There is an idea being obscured by your idea of an integral as a sum, however. In truth, a sum is a kind of integral and not vice versa - this is somewhat counterintuitive given that sums are much more familiar objects than integrals, but there's an elegant theory known as Lebesgue integration which basically makes the integral a tool which eats two pieces of information: We start with some indexing set and some way to "weigh" pieces of that set. We have some function on that set. Then, the Lebesgue integral spits out the weighted "sum" of that function. An integral in the most common sense arises when you say, "I have a function on the real numbers, and I want an interval to have weight equal to its length." A finite sum arises when you say "I have a function taking values $x_1,\ldots,x_n$ on the index set $\{1,\ldots,n\}$ . Each index gets weight $1$ " - and, of course, you can change those weights to get a weighted sum or you can extend the indexing set to every natural number to get an infinite sum. But, the basic thing to note is that there's a more general idea of integral that places summation as part of the theory of integration.
|
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|
3,091,832 |
This is an exercise from a sequence and series book that I am solving. I tried manipulating the number to make it easier to work with: $$111...1 = \frac{1}9(999...) = \frac{1}9(10^{55} - 1)$$ as the number of $1$ 's is $55$ . The exercise was under Geometric Progression and Geometric Mean. However, I am unable to think of a way to solve this problem using GP. How do I proceed from here?
|
The number is composite. We have \begin{align*}
\underbrace{11\ldots111}_{55 \text{ times }} = \frac{1}{9} \cdot (10^{55} - 1) \\
= \frac{1}{9} \cdot ((10^{5})^{11} - 1) \\
\end{align*} Also, note that $x^{m} - 1$ is divisible by $x - 1$ . Here, we can plug in $x = 10^{5}$ and $m = 11$ . As a result, we see that the quantity is divisible by $99999$ , meaning that the number must be divisible by $11111$ (and hence, composite).
|
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|
3,091,953 |
A spider needs a sock and a shoe for each of its eight legs. In how many ways can it put on the shoes and socks, if socks must be put on before the shoe? My attempt: If I consider its legs to be indistinguishable, then it's exactly the $8^{\text{th}}$ term of the Catalan sequence. However the legs are distinguishable. So the total number of ways equals $\frac1{8+1} \binom{16}{8} (8!)^2$ . Is this correct? Is there another way of doing it? Edit: All socks and shoes are distinguishable .
|
You can imagine doing this as writing a sequence, say $$3453228156467781$$ What does it mean? It means first put sock on leg $\color{red}{3}$ and on 4-th move put shoe on leg $\color{red}{3}$ then put sock on leg $\color{blue}{4}$ and on 11-th move put shoe on leg $\color{blue}{4}$ and so on... So for each leg you must choose a pair in this sequence. On smaller number in this pair put a sock and the other shoe. So we have $${16\choose 2}{14\choose 2}....{4\choose 2}{2\choose 2} = {16!\over 2!^8}$$
|
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|
3,092,340 |
Why is Euler's number $\mathtt 2.71828$ and not for example $\mathtt 3.7589$ ? I know that $e$ is the base of natural logarithms. I know about areas on hyperbola xy=1 and I know its formula: $$e =\sum_{n=0}^\infty \frac{1}{n!} \approx 2.71828$$ And I also know it has many other characterizations.
But, why is $e$ equal to that formula (which sum is approximately $\mathtt 2.71828$ )? I googled that many times and every time it ends in having " $e$ is the base of natural logarithms". I don't want to work out any equations using $e$ without understanding it perfectly. Summary: I'm looking for the origin of $e$ , if $\pi$ came from the radius of a circle with a unit diameter, then what is $e$ ???
|
$\sum\frac1{n!}$ is not that special. $\lim_{n\to\infty}\left(1+\frac1n\right)^n$ is not really special. $f'(x)=f(x)$ is a very simple differential equation, but unremarkable, really. $\ln (x)$ is only marginally nicer than other logarithms, in that its derivative is $\frac1x$ . The fact that a single number connects all of these (and many, many others) as intimately as $e$ does is nothing short of a miracle. Oh, and also $e$ happens to have the decimal expansion $2.718\ldots$
|
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|
3,092,659 |
I have a very basic problem. I am confused about "continuous function" term. What really is a continuous function? A function that is continuous for all of its domain or for all real numbers? Let's say: $\ln|x|$ - the graph clearly says it's continuous for all real numbers except for $0$ which is not part of the domain. So is this function continuous or not? I could say same about $\tan{x}$ or $\frac{x+1}{x}$ And also what about: $\ln{x}$ - the graph clearly says it's continuous for all of its domain: $(0; \infty)$ - so is this $f$ continuous or not? Thanks for clarification.
|
Mathematicians (but not all calculus books) mean "continuous at every point of its domain" when they say a function is "continuous." The functions $f(x) = 1/x$ and $f(x)=\ln x$ are continuous functions.
|
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|
3,093,400 |
So my question is this: $$V=\frac{4}{3}\pi r^3$$ And, $$\frac{dV}{dr}=4\pi r^2=SA$$ Is this a coincidence or are there some mathematical hoodoos that I'm unaware of? P.S. are there any more tags that I should use?
|
Start with a sphere of radius $r$ . Now let the radius of the sphere grow by some tiny amount $\Delta r$ . How much has the volume changed? By the definition of the derivative, it has changed by approximately $$
\Delta r \cdot V'(r)
$$ However, the added volume is basically a thin shell, and its volume is approximately equal to its surface area (the inner one, for convenience), multiplied by its thickness. This is $$
\Delta r\cdot SA(r)
$$ Thus we have $$
\Delta r \cdot V'(r)\approx \Delta r\cdot SA(r)\\
V'(r)\approx SA(r)
$$ Rigorous analysis of this setup will allow you to conclude that the approximation error above is small enough as $\Delta r$ becomes smaller, and thus that $V'(r)$ and $SA(r)$ are indeed equal.
|
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|
3,094,007 |
Anywhere I've looked, the definition of solution of equation is root(s) of that equation. But why does a textbook say that equation $y=7-3x$ has infinite number of solutions? Thanks
|
Equations have solutions, but not roots. Polynomials have roots, but not solutions. The equation $y=7-3x$ has infinitely many solutions. The equation $0=7-3x$ has only one solution. The polynomial $7-3x$ has only one root, which is the solution of the equation $0=7-3x$ .
|
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|
3,094,706 |
Continuity of function $f:X\to Y$ from topological space $X$ to $Y$ is defined by saying that for any open set $U_Y$ , $f^{-1}(U_Y)$ is also an open set. Intuitively, I find this weird. If we interpret “open set” informally as “a set whose elements are nearby each other” (indeed it is a set which is a neighbourhood of all its elements), then it makes intuitive sense to say that a continuous function $f$ is a function that does not “rip elements away from its neighbours”, i.e. if you input an open set $U_X$ (a set whose elements are “nearby each other”), then this should not produce a set where some elements are “not nearby each other”, i.e. it should produce an open set. So is there an intuitive explanation at this level of abstraction (i.e. without reference to metric spaces for example) of why we don’t define continuity as “for any open set $U_X$ , $f(U_X)$ is an open set”?
|
The problem with your intuition is that an "open set" is not "a set whose elements are nearby each other". For example, considering the real numbers with the standard topology, the set $(0, \infty)$ contains elements arbitrarily far away from each other, while $\{0\}$ contains elements extremely nearby each other. A better intuition is: an open set $X$ is a set such that if $x \in X$ , then all points that are close to $x$ are also in $X$ . This shows why the "forward definition" doesn't work: just because you are taking all points close to some $x$ , does not mean that you should map onto all points close to $f(x)$ -- it just means that you should hit only points close to $f(x)$ . But what does hitting only points close to $f(x)$ mean? It means that if you take all points $U$ close to $f(x)$ , then $f^{-1}(U)$ should include all points close to $x$ . If you try to make the ideas in the previous paragraph precise and formal, you end up with the ordinary definition of continuity. Edit: From the comments: I still find your intuition hard to reconcile with the idea of “a discontinuous function rips points apart”. Let us look at the converse, and take $f$ discontinuous. Informally, this means that there are $x$ , $y$ , which are close together, such that $f(x)$ and $f(y)$ are not close together. (Of course, to make this formal, you need to take one of $x$ or $y$ to be a sequence or even a net, etc.) My intuition of an open set says: let $X$ be an open set, then $x \in X$ if and only if $y \in X$ . Now let's see if the "forward definition of continuous" lets us prove that $f$ is discontinuous. Let's take any open set $X$ . If $x \notin X$ then also $y \notin X$ , and this doesn't seem to go anywhere. So let's look at open $X$ with $x, y \in X$ . Then $f(X)$ is also open, and therefore $f(x), f(y) \in f(X)$ -- but this is precisely not what we wanted to prove. Now let's apply my intuition to the "backward definition of continuous". Because $f(x), f(y)$ are far apart, there is an open set containing $f(x)$ but not $f(y)$ . Let's call it $Y$ . Then we have $x \in f^{-1}(Y)$ , but $y \notin f^{-1}(Y)$ . Thus $f^{-1}(Y)$ is not an open set, and $f$ is discontinuous.
|
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|
3,100,871 |
I think I have found out some property of an ellipse. Define the following: The point $P$ belongs to the ellipse. The tangent line $\ell$ goes through $P$ . The point $F$ is one of the foci . The line $m$ , through $F$ , is perpendicular to the line $\ell$ at point $I$ . $A$ and $B$ are the vertices of the ellipse (that is, the endpoints of the major axis). In this case, The $\angle AIB$ is always $\pi/2$ . (I think.....) But I didn't prove yet... Is it true? Could anybody give me some advice? Thanks in advance. Edit1 For jmerry Edit2 For jmerry Really thanks, makes me fun!
|
Yes, it's true. Some advice: the locus of all points $I$ such that $\angle AIB$ is a right angle is the circle with diameter $AB$ . So, then, if you can show that $OI=OA=OB$ where $O$ is the center of the ellipse, you'll have it. Next, add some more things to your diagram. The other focus (let's call it $G$ ) looks like a good place to start - after all, that lets us use nice facts like the segments from $P$ to the two focuses making the same angle to the tangent line $\ell$ , and that $PF+PG=AB$ . So now, how can you use those facts? Don't be afraid to extend lines and find new intersections. [Added in edit] All right, you've drawn some more things in, and some equal angles. Since we're looking at $I$ in particular, how about naming that point where $GP$ and $FI$ intersect? You've already labeled an angle there, so naming the point will make it easier to work with. (I called it $J$ in my diagram) Oh, and since we're interested in the length of $OI$ , that's another segment we should draw.
|
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|
3,102,218 |
Given a fraction: $$\frac{a}{b}$$ I now add a number $n$ to both numerator and denominator in the following fashion: $$\frac{a+n}{b+n}$$ The basic property is that the second fraction is suppose to closer to $1$ than the first one. My question is how can we prove that? What I have tried: I know $\frac{n}{n} = 1$ so now adding numbers $a$ and $b$ to it would actually "move it away" from $1$ . But I cannot understand why $\frac{a}{b}$ is actually farther away from $1$ than $\frac{a+n}{b+n}$ . Why is that? What does it mean to add a number to both the numerator and denominator?
|
There's a very simple way to see this. Just take the difference between the two fractions and 1. You want to show that this is smaller in modulus for the second fraction. You get $$ \frac{a}{b} - 1 = \frac{a-b}{b} $$ and $$ \frac{a+n}{b+n} -1 = \frac{a-b}{b+n} $$ So the second is smaller in modulus (provided $b$ and $n$ are positive, although I supposed it also works if both are negative) because it has same numerator and larger (modulus) denominator, QED.
|
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|
3,103,685 |
Zorn's Lemma states that if every chain $\mathcal{C}$ in a partially ordered set $\mathcal{S}$ has an upper bound in $\mathcal{S}$ , then there is at least one maximal element in $\mathcal{S}$ . Why can't we apply Zorn's Lemma in the following case? Let $\mathcal{S}$ be the set of all groups. Define a partial order $\preceq$ as follows: for $H, G \in \mathcal{S}$ , define $H \prec G$ if and only if $H$ is a subgroup of $G$ . Then every chain $\mathcal{C}=(G_{\alpha})_{\alpha \in A}$ in $\mathcal{S}$ has an upper bound $\cup_{\alpha \in A} G_{\alpha}$ in $\mathcal{S}$ . But certainly there is no maximal element in $\mathcal{S}$ . Could anyone tell me what is wrong with this argument?
|
There is no set of all groups, but Zorn's Lemma can be phrased for partially ordered classes as well, but we need to require that any chain has an upper bound, also proper class chains. However in that case it is easy to define a proper class chain which has no upper bound. Simply take for each ordinal $\alpha$ the free group generated by $\alpha$ . There are natural injections given by the natural injections between two ordinals. However this chain does not have an upper bound, since an upper bound would be a group, which is a set, that embeds all ordinals. This is a contradiction to Hartogs theorem, of course. (You can think about this in the following analog: an infinite chain of finite sets, or finite groups, will not have an upper bound which is also finite.)
|
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|
3,105,317 |
I have been having trouble with Hölder exponents. The definition of Hölder continuity tells me that a function $f$ between metric spaces must satisfy $d(f(x),f(y)) \leq C \cdot d(x,y)^\alpha$ for some exponent $\alpha > 0$ . The Wikipedia article however states that for an exponent $\alpha >1$ this condition implies that the function $f$ is constant. I've been mulling it over but just can't see why this is the case. Let's assume that $\alpha>1$ . Then I have two (interesting) cases: Case 1: $d(x,y) > 1$ . Here the Hölder condition tells me that the function values are allowed to be even further apart than the input values, which doesn't seem to enforce $f$ to be constant. Case 2: $d(x,y) < 1$ . This time the Hölder condition tells me that if the input values lie close together, then the function values have to be even tighter together. To me it is plausible that this is precisely what yields the continuity of Hölder-continuous functions, however again saying that $f$ needs to be constant seems to me to still be a strong conclusion. I've seen similar questions being posted, however they all make (at least indirect) use of some differentiability assumption on $f$ which I do not want to make use of. Can anyone enlighten me please? I appreciate your answers ;).
|
This is not true for general metric spaces - if you have a function whose domain is a two-point space, then this function is $\alpha$ -Hölder for every $\alpha>0$ . $\alpha$ -Hölder property for $\alpha>1$ implies the function being constant only in special spaces, like $\mathbb R^n$ . Let me just focus on functions $f:\mathbb R\to\mathbb R$ . You mention "[the proofs] all make (at least indirect) use of some differentiability assumption on $f$ ". This is not quite correct - you don't need to assume differentiability, since $\alpha$ -Hölder condition for $\alpha>1$ implies that the derivative exists - indeed, we just consider $$\left|\frac{f(x+h)-f(x)}{h}\right|\leq\frac{C|h|^\alpha}{|h|}\to 0,$$ so the limit exists and is zero everywhere. That way or another, we can prove that $f$ is constant directly as well: take any two points $x<y$ and let $x_0=x,x_1=x+\frac{y-x}{n},x_2=x+2\frac{y-x}{n},\dots,x_n=x+n\frac{y-x}{n}=y$ . Then we have $$|f(x)-f(y)|\leq|f(x_0)-f(x_1)|+|f(x_1)-f(x_2)|+\dots+|f(x_{n-1})-f(x_n)|\\
\leq C|x_0-x_1|^\alpha+C|x_1-x_2|^\alpha+\dots+C|x_{n-1}-x_n|^\alpha\\
=n\cdot C\left|\frac{x-y}{n}\right|^\alpha=n^{1-\alpha}C|x-y|^\alpha\to 0,$$ so that $f(x)=f(y)$ .
|
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3,105,330 |
Suppose that the inner product on $P_2(\mathbb{R})$ is defined by $$\langle f,g \rangle:= f(-1)g(-1)+f(0)g(0)+f(1)g(1).$$ Consider the operator $T \in B(P_2(\mathbb{R}))$ which is defined as $Tf=f'$ , so the derivative of $f$ . Find the adjoint of $T$ . I am trying to find the adjoint in the following way.
Since $\langle Tf, g\rangle = \langle f, T^*g\rangle$ then we have that $\langle f, T^*g\rangle = f'(-1)g(-1) + f'(0)g(0) + f'(1)g(1)$ . How do I continue now since I don't know how $T^*$ looks like? I appreciate your help.
|
This is not true for general metric spaces - if you have a function whose domain is a two-point space, then this function is $\alpha$ -Hölder for every $\alpha>0$ . $\alpha$ -Hölder property for $\alpha>1$ implies the function being constant only in special spaces, like $\mathbb R^n$ . Let me just focus on functions $f:\mathbb R\to\mathbb R$ . You mention "[the proofs] all make (at least indirect) use of some differentiability assumption on $f$ ". This is not quite correct - you don't need to assume differentiability, since $\alpha$ -Hölder condition for $\alpha>1$ implies that the derivative exists - indeed, we just consider $$\left|\frac{f(x+h)-f(x)}{h}\right|\leq\frac{C|h|^\alpha}{|h|}\to 0,$$ so the limit exists and is zero everywhere. That way or another, we can prove that $f$ is constant directly as well: take any two points $x<y$ and let $x_0=x,x_1=x+\frac{y-x}{n},x_2=x+2\frac{y-x}{n},\dots,x_n=x+n\frac{y-x}{n}=y$ . Then we have $$|f(x)-f(y)|\leq|f(x_0)-f(x_1)|+|f(x_1)-f(x_2)|+\dots+|f(x_{n-1})-f(x_n)|\\
\leq C|x_0-x_1|^\alpha+C|x_1-x_2|^\alpha+\dots+C|x_{n-1}-x_n|^\alpha\\
=n\cdot C\left|\frac{x-y}{n}\right|^\alpha=n^{1-\alpha}C|x-y|^\alpha\to 0,$$ so that $f(x)=f(y)$ .
|
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|
3,105,347 |
I am trying to learn some basic Scheme theory out of Eisenbud's book "Schemes: the Language of Modern Algebraic Geometry." I'm trying to understand how elements of a ring can be treated as functions over spec in the classical sense. Of course, the title is a little broad for this interpretation. However, this question arose as I was trying to solve another question in the book. My intuition tells me the answer is no, but I haven't been able to come up with a satisfying proof. Any help would be appreciated. Edit: in particular, I mean for $k'$ to be a proper subfield of $k$ , and the quotient of $A$ to be isomorphic to $k'$ as a ring.
|
This is not true for general metric spaces - if you have a function whose domain is a two-point space, then this function is $\alpha$ -Hölder for every $\alpha>0$ . $\alpha$ -Hölder property for $\alpha>1$ implies the function being constant only in special spaces, like $\mathbb R^n$ . Let me just focus on functions $f:\mathbb R\to\mathbb R$ . You mention "[the proofs] all make (at least indirect) use of some differentiability assumption on $f$ ". This is not quite correct - you don't need to assume differentiability, since $\alpha$ -Hölder condition for $\alpha>1$ implies that the derivative exists - indeed, we just consider $$\left|\frac{f(x+h)-f(x)}{h}\right|\leq\frac{C|h|^\alpha}{|h|}\to 0,$$ so the limit exists and is zero everywhere. That way or another, we can prove that $f$ is constant directly as well: take any two points $x<y$ and let $x_0=x,x_1=x+\frac{y-x}{n},x_2=x+2\frac{y-x}{n},\dots,x_n=x+n\frac{y-x}{n}=y$ . Then we have $$|f(x)-f(y)|\leq|f(x_0)-f(x_1)|+|f(x_1)-f(x_2)|+\dots+|f(x_{n-1})-f(x_n)|\\
\leq C|x_0-x_1|^\alpha+C|x_1-x_2|^\alpha+\dots+C|x_{n-1}-x_n|^\alpha\\
=n\cdot C\left|\frac{x-y}{n}\right|^\alpha=n^{1-\alpha}C|x-y|^\alpha\to 0,$$ so that $f(x)=f(y)$ .
|
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|
3,107,353 |
I computed the 10 first powers of 256 and I noticed that they all end by 6. 256^1 = 256
256^2 = 65536
256^3 = 16777216
256^4 = 4294967296
256^5 = 1099511627776
256^6 = 281474976710656
256^7 = 72057594037927936
256^8 = 18446744073709551616
256^9 = 4722366482869645213696
256^10 = 1208925819614629174706176 My intuition is that it is the same for all powers of 256 but I can't figure out how to prove it, any suggestion? My attempt was to show that $\forall n$ there is a $k$ such that $2^{8n} = k*10 + 6$ , $k$ and $n$ being non null integers. I tried to decompose the right member in powers of $2$ but that leaves me stuck.
|
$$(10x+6)(10y+6)=100xy+60 (x+y)+3\color {red}6. $$
|
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|
3,110,453 |
So much of the properties of compact sets are motivated by finite sets, to the point that thinking of compact sets as topologically finite sets may yield some deeper understanding. But finite sets have the intuitive property that every subset of a finite set is also compact(also finite), why is it that compact sets give this up? It is easy to state that they just do and provide the example $I=(0,1)$ and $K=[0,1]$ as an example, but the problem with that is it really only helps illuminate how compact sets work in the euclidean spaces. It isn't generally true in all spaces that open sets aren't compact, or that closed and bounded sets are compact. So there is a problem generalizing the ideas. The heart of my question is : Given a subset $E$ of a compact set $K$ why isn't it compact? Simple Answer : Because there is an open cover of $E$ which has no finite subcover The deeper question: Why?
|
One way to think of it is that compactness means that sequences (more generally, nets) cannot "run away". There are two types of "running away": Running away to infinity (failure of the set to be bounded, or some analog*. Failure of the sequence/net to have any limit point at all). Running away to a limit point which isn't in the set (failure of a set to be closed. The sequence/net has a limit point(s) in an ambient space, but that point is missing from the set in question). If I have a compact space $X$ , and I remove a point, $X-\{p\}$ may suddenly permit the second type of "running away". At first it seemed awkward to me to phrase these things in terms of sequences and nets, because sequences and nets are "discrete" objects describing a continuous thing. But one can always phrase everything in terms of nets/filters of open sets, not of points. That can make it seem a bit more natural. In any case, the basic point of a compact set is that it does not allow you to play a certain kind of game with infinity. *For example, a uniform space is compact iff totally bounded and Cauchy complete, which are exactly analogous to conditions 1 and 2 above.
|
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|
3,111,482 |
I am taking a basic analysis course. This is a general question that I often encounter in weekly homework. How should we start to attack this type of question: if the statement is true, prove it; if not true, give a counterexample? Yesterday evening I spent four hours looking for a counterexample, and finally figured that the statement was indeed true. I tried so hard to find a counterexample because I was not able to prove the statement at first. Are there any steps to follow in terms of this type of questions? In particular, sometimes a statement does not look like it is true. In this case, how can I quickly figure out whether I should try counterexamples or keep the efforts on proving the statement is true? The core problem is that this type of question is very time-consuming according to my homework experience. I desperately doubt I could figure out a correct answer during a timed exam next week.
|
One strategy is to first try to prove by contradiction that the statement is true. Such an effort will identify necessary conditions for a counterexample. If through such an analysis you realise you can also give sufficient conditions for a counterexample, and you can work out how to satisfy them, you'll have a counterexample. With any luck, in the event the statement is true you'll find a valid proof soon enough. (Once you do, it's worth checking whether it can be rewritten to not use contradiction; students new to proof-writing sometimes unnecessarily add a contradiction "wrapper" around a direct proof.) Note that if a counterexample exists in a textbook exercise, there will be a simple one that slightly complicates a situation that illustrates a weaker true claim. For example, if I asked you to prove or refute-by-counterexample the claim that all finite groups are Abelian, the hope is you'd quickly find this counterexample. It's slightly more complicated, though only slightly, than the case of a finite group with a single generator, which of course would be Abelian. So the hope is you'd think, "let's try to make a group with two generators; that might do it".
|
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|
3,112,970 |
$x,y$ are perpendicular if and only if $x\cdot y=0$ . Now, $||x+y||^2=(x+y)\cdot (x+y)=(x\cdot x)+(x\cdot y)+(y\cdot x)+(y\cdot y)$ . The middle two terms are zero if and only if $x,y$ are perpendicular. So, $||x+y||^2=(x\cdot x)+(y\cdot y)=||x||^2+||y||^2$ if and only if $x,y$ are perpendicular. ( I copied this ) I think this argument is circular because the property $x\cdot y=0 $ implies $x$ and $y$ are perpendicular comes from the Pythagorean theorem. Oh, it just came to mind that the property could be derived from the law of cosines. The law of cosines can be proved without the Pythagorean theorem, right, so the proof isn't circular? Another question : If the property comes from the Pythagorean theorem or cosine law, then how does the dot product give a condition for orthogonality for higher dimensions? Edit : The following quote by Poincare hepled me regarding the question: Mathematics is the art of giving the same name to different things.
|
I think the question mixes two quite different concepts together: proof and motivation. The motivation for defining the inner product, orthogonality, and length of vectors in $\mathbb R^n$ in the "usual" way (that is, $\langle x,y\rangle = x_1y_1 + x_2y_2 + \cdots + x_ny_n$ )
is presumably at least in part that by doing this we will be able to establish a property of $\mathbb R_n$ corresponding to the familiar Pythagorean Theorem from synthetic plane geometry.
The motivation is, indeed, circular in that we get the Pythagorean Theorem as one of the results of something we set up because we wanted the Pythagorean Theorem. But that's how many axiomatic systems are born. Someone wants to be able to work with mathematical objects in a certain way, so they come up with axioms and definitions that provide mathematical objects they can work with the way they wanted to. I would be surprised to learn that the classical axioms of Euclidean geometry (from which the original Pythagorean Theorem derives) were not created for the reason that they produced the kind of geometry that Euclid's contemporaries wanted to work with. Proof, on the other hand, consists of starting with a given set of axioms and definitions (with emphasis on the word "given," that is, they have no prior basis other than that we want to believe them), and showing that a certain result necessarily follows from those axioms and definitions without relying on any other facts that did not derive from those axioms and definitions.
In the proof of the "Pythagorean Theorem" in $\mathbb R^n,$ after the point at which the axioms were given, did any step of the proof rely on anything other than the stated axioms and definitions? The answer to that question would depend on how the axioms were stated.
If there is an axiom that says $x$ and $y$ are orthogonal if $\langle x,y\rangle = 0,$ then this fact does not logically "come from" the Pythagorean Theorem; it comes from the axioms.
|
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|
3,114,122 |
What is the best way to find determinant of the following matrix? $$A=\left(\begin{matrix}
1&ax&a^2+x^2\\1&ay&a^2+y^2\\ 1&az&a^2+z^2
\end{matrix}\right)$$ I thought it looks like a Vandermonde matrix, but not exactly. I can't use $|A+B|=|A|+|B|$ to form a Vandermonde matrix. Please suggest. Thanks.
|
\begin{align}
&|A|\\
&=\det\left(\begin{matrix}
1&ax&a^2+x^2\\1&ay&a^2+y^2\\ 1&az&a^2+z^2
\end{matrix}\right) \\
&=\begin{vmatrix}
1&ax&a^2\\1&ay&a^2\\ 1&az&a^2
\end{vmatrix}+\begin{vmatrix}
1&ax&x^2\\1&ay&y^2\\ 1&az&z^2
\end{vmatrix} \tag{multilinearity on 3rd column} \\
&=0+a\begin{vmatrix}
1&x&x^2\\1&y&y^2\\ 1&z&z^2
\end{vmatrix} \\
&= a (x-y)(y-z)(z-x)
\end{align}
|
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|
3,116,417 |
If $a_n$ is a sequence such that $$a_1 \leq a_2 \leq a_3 \leq \dotsb$$ and has the property that $a_{n+1}-a_n \to 0$ , then can we conclude that $a_n$ is convergent? I know that without the condition that the sequence is increasing, this is not true, as we could consider the sequence given in this answer to a similar question that does not require the sequence to be increasing. $$0, 1, \frac12, 0, \frac13, \frac23, 1, \frac34, \frac12, \frac14, 0, \frac15, \frac25, \frac35, \frac45, 1, \dotsc$$ This oscillates between $0$ and $1$ , while the difference of consecutive terms approaches $0$ since the difference is always of the form $\pm\frac1m$ and $m$ increases the further we go in this sequence. So how can we use the condition that $a_n$ is increasing to show that $a_n$ must converge? Or is this still not sufficient?
|
No. Just consider the case in which $a_n=1+\frac12+\frac13+\cdots+\frac1n$ . Note that then we have $$\lim_{n\to\infty}a_{n+1}-a_n=\lim_{n\to\infty}\frac1{n+1}=0.$$ But $a_n$ is the $n$ th partial sum of the harmonic series, and therefore $(a_n)_{n\in\Bbb N}$ diverges.
|
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|
3,117,052 |
I'm pretty sure that a zero vector is just a vector of length zero with direction, zero scalar is just the number zero, and that a zero function is any function that maps to zero. Not entirely sure what exactly a zero operator is however.
|
The zero vector is a vector, i.e. a member of whatever vector space is under consideration. It has the property that adding it to any vector $\bf v$ in the vector space leaves $\bf v$ unchanged. The zero scalar is a scalar, i.e. a member of the field that is part of the definition of the vector space (usually the real or complex numbers in an elementary linear algebra course). It has the property that multiplying
any vector $\bf v$ by it gives the zero vector of the second vector space. The zero operator is a linear operator, i.e. a linear map from a vector space to a vector space (possibly the same one). It has the property that it maps any member of the first vector space to the zero vector in the second vector space. The zero functional is a linear functional, i.e. a linear map from a vector space to the scalars. It has the property that it maps any member of the vector space to the zero scalar.
|
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|
3,117,260 |
So according to the commutative property for multiplication: $a \times b = b \times a$ However this does not hold for division $a \div b \neq b \div a$ Why is it that in the following case: $56 \times 100 \div 8 = 56 \div 8\times 100$ It seems like division is breaking the rule. There is something I am misunderstanding here. Is it because $a\times b\div c=a\div c\times b$ is allowed since $b\div c$ are not being rearranged so that $c\div b$ ? If this is the case are you allowed to rearrange values in equations so long as no values have the form $a \div b = b \div a$ and $a - b = b -a$ ?
|
Notice that you have always $\div 8$ , no matter the order of the other terms. You don't divide by a different number. It might help to think $\div c$ as a multiplication with $d=1/c$ . Then everything would look easier: $$a\times b\div c=a\times b\times d=a\times d\times b=a\div c\times b$$
|
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|
3,119,631 |
I know it is true that we have $$\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots}}}}=3$$ The argument is to break the nested radical into something like $$3 = \sqrt{9}=\sqrt{1+2\sqrt{16}}=\sqrt{1+2\sqrt{1+3\sqrt{25}}}=...=\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots}}}}$$ However, I am not convinced. I can do something like $$4 = \sqrt{16}=\sqrt{1+2\sqrt{56.25}}=\sqrt{1+2\sqrt{1+3\sqrt{\frac{48841}{144}}}}=...=\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots}}}}$$ Something must be wrong and the reason behind should be a misunderstanding of how we define infinite nested radical in the form of $$ \sqrt{a_{0}+a_{1}\sqrt{a_{2}+a_{3}\sqrt{a_{4}+a_{5}\sqrt{a_{6}+\cdots}}}} $$ I researched for a while but all I could find was computation tricks but not a strict definition. Really need help here. Thanks.
|
Introduction: The issue is what "..." really "represents." Typically we use it as a sort of shorthand, as if to say "look, I can't write infinitely many things down, just assume that the obvious pattern holds and goes on infinitely." This idea holds for all sorts of things - nested radicals, infinite sums, continued fractions, infinite sequences, etc. On Infinite Sums: A simple example: the sum of the reciprocals of squares: $$1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + ...$$ This is a well known summation. It is the Riemann zeta function $\zeta(s)$ at $s=2$ , and is known to evaluate to $\pi^2/6$ (proved by Euler and known as the Basel problem ). Another, easier-to-handle summation is the geometric sum $$1 + \frac 1 2 + \frac 1 4 + \frac 1 8 + ...$$ This is a geometric series where the ratio is $1/2$ - each summand is half the previous one. We know, too, that this evaluates to $2$ . Another geometric series you might see in proofs that $0.999... = 1$ is $$9 \left( \frac{1}{10} + \frac{1}{100} + \frac{1}{1,000} + \frac{1}{10,000} + ... \right)$$ which equals $1$ . In fact, any infinite geometric series, with first term $a$ and ratio $|r|<1$ can be evaluated by $$\sum_{n=0}^\infty ar^n = \frac{a}{1-r}$$ So a question arises - ignoring these "obvious" results (depending on your amount of mathematical knowledge), how would we know these converge to the given values? What, exactly, does it mean for a summation to converge to a number or equal a number? For finite sums this is no issue - if nothing else, we could add up each number manually, but we can't just add up every number from a set of infinitely-many numbers. Well, one could argue by common sense that, if the sequence converges to some number, the more and more terms you add up, the closer they'll get to that number. So we obtain one definition for the convergence of an infinite sum. Consider a sequence where the $n^{th}$ term is defined by the sum of the first $n$ terms in the sequence. To introduce some symbols, suppose we're trying to find the sum $$\sum_{k=1}^\infty x_k = x_1 + x_2 + x_3 + x_4 + ...$$ for whatever these $x_i$ 's are. Then define these so-called "partial sums" of this by a function $S(n)$ : $$S(n) = \sum_{k=1}^n x_k = x_1 + x_2 + ... + x_n$$ Then we get a sequence of sums: $$S(1), S(2), S(3), ...$$ or equivalently $$x_1 \;\;,\;\; x_1 + x_2\;\;,\;\; x_1 + x_2 + x_3\;\;,\;\; ...$$ Then we ask: what does $S(n)$ approach as $n$ grows without bound, if anything at all? (In calculus, we call this "the limit of the partial sums $S(n)$ as $n$ approaches infinity.") For the case of our first geometric sum, we immediately see the sequence of partial sums $$1, \frac{3}{2}, \frac{7}{4}, \frac{15}{8},...$$ Clearly, this suggests a pattern - and if you want to, you can go ahead and prove it, I won't do so here for brevity's sake. The pattern is that the $n^{th}$ term of the sequence is $$S(n) = \frac{2^{(n+1)}-1}{2^{n}}$$ We can then easily consider the limit of these partial sums: $$\lim_{n\to\infty} S(n) = \lim_{n\to\infty} \frac{2^{(n+1)}-1}{2^{n}} = \lim_{n\to\infty} \frac{2^{(n+1)}}{2^{n}} - \frac {1}{2^{n}} = \lim_{n\to\infty} 2 - \lim_{n\to\infty} \frac{1}{2^{n}}$$ Obviously, $1/2^{n} \to 0$ as $n$ grows without bound, and $2$ is not affected by $n$ , so we conclude $$\lim_{n\to\infty} S(n) = \lim_{n\to\infty} 2 - \lim_{n\to\infty} \frac{1}{2^n} = 2 - 0 = 2$$ And thus we say $$\sum_{k=0}^\infty \left(\frac 1 2 \right)^k = 1 + \frac 1 2 + \frac 1 4 + \frac 1 8 + ... = 2$$ because the partial sums approach $2$ . On Continued Fractions: That was a simple, "first" sort of example, but mathematicians essentially do the same thing in other contexts. I want to touch on one more such context before we deal with the radical case, just to nail that point home. In this case, it will be with continued fractions . One of the simpler such fractions is the one for $1$ : $$1 = \frac{1}{2-\frac{1}{2-\frac{1}{2-\frac{1}{...}}}}$$ As usual, the "..." denotes that this continues forever. But what it does it mean for this infinite expression to equal $1$ ? For this, we consider a more general analogue of the "partial sum" from before - a "convergent." We cut up the sequence at logical finite points, whatever those points being depending on the context. Then if the sequence of the convergents approaches a limit, we say they're equal. What are the convergents for a continued fraction? By convention, we cut off just before the start of the next fraction. That is, in the continued fraction for $1$ , we cut off at the $n^{th} \; 2$ for the $n^{th}$ convergent, and ignore what follows. So we get the sequence of convergents $$\frac{1}{2} , \frac{1}{2-\frac{1}{2}}, \frac{1}{2-\frac{1}{2-\frac{1}{2}}},...$$ Working out the numbers, we find the sequence to be $$\frac{1}{2},\frac{2}{3},\frac{3}{4},...$$ Again, we see a pattern! The $n^{th}$ term of the sequence is clearly of the form $$\frac{n-1}{n}$$ Let $C(n)$ be a function denoting the $n^{th}$ convergent. Then $C(1)=1/2,$ $C(2) = 2/3,$ $C(n)=(n-1)/n,$ and so on. So like before we consider the infinite limit: $$\lim_{n\to\infty} C(n) = \lim_{n\to\infty} \frac{n-1}{n} = \lim_{n\to\infty} 1 - \frac 1 n = \lim_{n\to\infty} 1 - \lim_{n\to\infty} \frac 1 n = 1 - 0 = 1$$ Thus we can conclude that the continued fraction equals $1$ , because its sequence of convergents equals $1$ ! On Infinite Radicals: So now, we touch on infinite nested radicals. They're messier to deal with but doable. One of the simpler examples of such radicals to contend with is $$2 = \sqrt{2 +\sqrt{2 +\sqrt{2 +\sqrt{2 +\sqrt{2 +...}}}}}$$ As with the previous two cases we see an infinite expression. We instinctively conclude by now: to logically define a limit for this expression - to assign it a value provided it even exists - we need to chop this up at finite points, defining a sequence of convergents $C(n)$ , and then find $C(n)$ as $n\to\infty$ . Nested radicals are a lot messier than the previous, but we manage. So first let the sequence of convergents be given by cutting off everything after the $n^{th} \; 2$ in the expression. Thus we get the sequence $$\sqrt 2 \;\;,\;\; \sqrt{2 + \sqrt{2}}\;\;,\;\; \sqrt{2+\sqrt{2+\sqrt{2}}}\;\;,\;\; \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}$$ Okay this isn't particularly nice already, but apparently there does exist, shockingly enough, a closed-form explicit expression for $C(n)$ : (from: S. Zimmerman, C. Ho) $$C(n) = 2\cos\left(\frac{\pi}{2^{n+1}}\right)$$ (I had to find that expression by Googling, I honestly didn't know that offhand. It can be proved by induction, as touched on in this MSE question .) So luckily, then, we can find the limit of $C(n)$ : $$\lim_{n\to\infty} C(n) = \lim_{n\to\infty} 2\cos\left(\frac{\pi}{2^{n+1}}\right)$$ It is probably obvious that the argument of the cosine function approaches $0$ as $n$ grows without bound, and thus $$\lim_{n\to\infty} C(n) = \lim_{n\to\infty} 2\cos\left(\frac{\pi}{2^{n+1}}\right) = 2\cos(0) = 2\cdot 1 = 2$$ Thus, since its convergents approach $2$ , we can conclude that $$2 = \sqrt{2 +\sqrt{2 +\sqrt{2 +\sqrt{2 +\sqrt{2 +...}}}}}$$ A Lengthy Conclusion: So, in short, how do we evaluate an infinite expression, be it radical, continued fraction, sum, or otherwise? We begin by truncating the expression at convenient finite places, creating a series of convergents, generalizations of the "partial sums" introduced in calculus. We then try to get a closed form or some other usable expression for the convergents $C(n)$ , and consider the value as $n\to\infty$ . If it converges to some value, we say that the expression is in fact equal to that value. If it doesn't, then the expression doesn't converge to any value. This doesn't mean each expression is "nice." Radical expressions in particular, in my experience, tend to be nasty as all hell, and I'm lucky I found that one closed form expression for the particularly easy radical I chose. This doesn't mean that other methods cannot be used to find the values, so long as there's some sort of logical justification for said method. For example, there is a justification for the formula for an infinite (and finite) geometric sum. We might have to circumvent the notion of partial sums entirely, or at least it might be convenient to do so. For example, with the Basel problem, Euler's proof focused on Maclaurin series, and none of this "convergent" stuff. (That proof is here plus other proofs of it!) Luckily, at least, this notion of convergents, even if it may not always be the best way to do it, lends itself to an easy way to check a solution to any such problem. Just find a bunch of the convergents - take as many as you need. If you somehow have multiple solutions, as you think with Ramanujan's radical, then you'll see the convergents get closer and closer to the "true" solution. (How many convergents you need to find depends on the situation and how close your proposed solutions are. It might be immediately obvious after $10$ iterations, or might not be until $10,000,000$ . This logic also relies on the assumption that there is only one solution to a given expression that is valid. Depending on the context, you might see cases where multiple solutions are valid but this "approaching by hand" method will only get you some of the solutions. This touches on the notion of "unstable" and "stable" solutions to dynamical systems - which I believe is the main context where such would pop up - but it's a bit overkill to discuss that for this post.) So I will conclude by showing, in this way, that the solution is $3$ to Ramanujan's radical. We begin with the radical itself: $$\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots}}}}=3$$ Let us begin by getting a series of convergents: $$\sqrt{1} \;\;,\;\; \sqrt{1 + 2\sqrt{1}} \;\;,\;\; \sqrt{1 + 2\sqrt{1 + 3\sqrt{1}}} \;\;,\;\;$$ Because the $\sqrt{1}$ isn't necessary, we just let it be $1$ . $$1 \;\;,\;\; \sqrt{1 + 2} \;\;,\;\; \sqrt{1 + 2\sqrt{1 + 3}} \;\;,\;\;$$ Okay so ... where to go from here? Honestly, my initial temptation was to just use a MATLAB script and evaluate it, but I can't think of even a recursive closed form for this that would be nice enough. So in any event, we just have to go by "hand" (and by hand I mean WolframAlpha). Let $C(n)$ be the $n^{th}$ convergent. Then $C(1) = 1$ $C(2) \approx 1.732$ $C(3) \approx 2.236$ $C(4) \approx 2.560$ $C(5) \approx 2.755$ $C(6) \approx 2.867$ $C(7) \approx 2.929$ $C(8) \approx 2.963$ $C(9) \approx 2.981$ $C(10) \approx 2.990$ To skip a few values because at this point the changes get minimal, I used a macro to make a quick code for $C(50)$ so I could put it into Microsoft Excel and got the approximate result $$C(50) \approx 2.999 \; 999 \; 999 \; 999 \; 99$$ So while not the most rigorous result, we can at least on an intuitive level feel like the convergents from Ramanujan's radical converge to $3$ , not $4$ or any other number. Neglecting that this is not an ironclad proof of the convergence, at least intuitively then we can feel like $$3 = \sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots}}}}$$ Whew! Hopefully that lengthy post was helpful to you! A late footnote, but Mathologer on YouTube did a video on this very topic, so his video would give a decent summary of all this stuff as well. Here's a link.
|
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3,119,656 |
I am trying to find the integral of: $$\int \frac{dx}{x\sqrt{x-1}}$$ if I do this $u$ -sub: $u = \sqrt{x-1}$ and $u^2 = x-1$ and $x = u^2 + 1$ and $\frac{dx}{du} = 2u$ and $dx = 2udu$ Then I get: $$\int \frac{2udu}{(u^2+1)u}$$ if I do a partial fraction of this I wind up stuck and I wonder why: $$\frac{2udu}{(u^2+1)u} = \frac{A}{u} + \frac{Bu + C}{u^2+1}$$ $$2u = Au^2 + A + Bu^2 + Cu$$ $$2u = u^2(A+B) + Cu + A$$ $$C = 2$$ $$A=0$$ $$B=0$$ But that doesn't seem right because then I'd get $\frac{2}{u^2+1}$ which is not equal to the original fraction. But if I do some cancelling: $$\frac{2udu}{(u^2+1)u} = \frac{2}{u^2+1} = 2 \frac{1}{u^2 + 1}$$ So then: $$\int \frac{2udu}{(u^2+1)u}$$ $$2 \int \frac{1}{u^2 + 1} du$$ $$ = 2\arctan(u)$$ $$2\arctan(\sqrt{x-1}) + C$$ Does not simplifying ruin partial fractions? EDIT ohhhhhh I get it. It's the same thing! Reduced or using partial fractions, I wind up with $\int \frac{2}{u^2+1}$ . Is that right?$
|
Introduction: The issue is what "..." really "represents." Typically we use it as a sort of shorthand, as if to say "look, I can't write infinitely many things down, just assume that the obvious pattern holds and goes on infinitely." This idea holds for all sorts of things - nested radicals, infinite sums, continued fractions, infinite sequences, etc. On Infinite Sums: A simple example: the sum of the reciprocals of squares: $$1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + ...$$ This is a well known summation. It is the Riemann zeta function $\zeta(s)$ at $s=2$ , and is known to evaluate to $\pi^2/6$ (proved by Euler and known as the Basel problem ). Another, easier-to-handle summation is the geometric sum $$1 + \frac 1 2 + \frac 1 4 + \frac 1 8 + ...$$ This is a geometric series where the ratio is $1/2$ - each summand is half the previous one. We know, too, that this evaluates to $2$ . Another geometric series you might see in proofs that $0.999... = 1$ is $$9 \left( \frac{1}{10} + \frac{1}{100} + \frac{1}{1,000} + \frac{1}{10,000} + ... \right)$$ which equals $1$ . In fact, any infinite geometric series, with first term $a$ and ratio $|r|<1$ can be evaluated by $$\sum_{n=0}^\infty ar^n = \frac{a}{1-r}$$ So a question arises - ignoring these "obvious" results (depending on your amount of mathematical knowledge), how would we know these converge to the given values? What, exactly, does it mean for a summation to converge to a number or equal a number? For finite sums this is no issue - if nothing else, we could add up each number manually, but we can't just add up every number from a set of infinitely-many numbers. Well, one could argue by common sense that, if the sequence converges to some number, the more and more terms you add up, the closer they'll get to that number. So we obtain one definition for the convergence of an infinite sum. Consider a sequence where the $n^{th}$ term is defined by the sum of the first $n$ terms in the sequence. To introduce some symbols, suppose we're trying to find the sum $$\sum_{k=1}^\infty x_k = x_1 + x_2 + x_3 + x_4 + ...$$ for whatever these $x_i$ 's are. Then define these so-called "partial sums" of this by a function $S(n)$ : $$S(n) = \sum_{k=1}^n x_k = x_1 + x_2 + ... + x_n$$ Then we get a sequence of sums: $$S(1), S(2), S(3), ...$$ or equivalently $$x_1 \;\;,\;\; x_1 + x_2\;\;,\;\; x_1 + x_2 + x_3\;\;,\;\; ...$$ Then we ask: what does $S(n)$ approach as $n$ grows without bound, if anything at all? (In calculus, we call this "the limit of the partial sums $S(n)$ as $n$ approaches infinity.") For the case of our first geometric sum, we immediately see the sequence of partial sums $$1, \frac{3}{2}, \frac{7}{4}, \frac{15}{8},...$$ Clearly, this suggests a pattern - and if you want to, you can go ahead and prove it, I won't do so here for brevity's sake. The pattern is that the $n^{th}$ term of the sequence is $$S(n) = \frac{2^{(n+1)}-1}{2^{n}}$$ We can then easily consider the limit of these partial sums: $$\lim_{n\to\infty} S(n) = \lim_{n\to\infty} \frac{2^{(n+1)}-1}{2^{n}} = \lim_{n\to\infty} \frac{2^{(n+1)}}{2^{n}} - \frac {1}{2^{n}} = \lim_{n\to\infty} 2 - \lim_{n\to\infty} \frac{1}{2^{n}}$$ Obviously, $1/2^{n} \to 0$ as $n$ grows without bound, and $2$ is not affected by $n$ , so we conclude $$\lim_{n\to\infty} S(n) = \lim_{n\to\infty} 2 - \lim_{n\to\infty} \frac{1}{2^n} = 2 - 0 = 2$$ And thus we say $$\sum_{k=0}^\infty \left(\frac 1 2 \right)^k = 1 + \frac 1 2 + \frac 1 4 + \frac 1 8 + ... = 2$$ because the partial sums approach $2$ . On Continued Fractions: That was a simple, "first" sort of example, but mathematicians essentially do the same thing in other contexts. I want to touch on one more such context before we deal with the radical case, just to nail that point home. In this case, it will be with continued fractions . One of the simpler such fractions is the one for $1$ : $$1 = \frac{1}{2-\frac{1}{2-\frac{1}{2-\frac{1}{...}}}}$$ As usual, the "..." denotes that this continues forever. But what it does it mean for this infinite expression to equal $1$ ? For this, we consider a more general analogue of the "partial sum" from before - a "convergent." We cut up the sequence at logical finite points, whatever those points being depending on the context. Then if the sequence of the convergents approaches a limit, we say they're equal. What are the convergents for a continued fraction? By convention, we cut off just before the start of the next fraction. That is, in the continued fraction for $1$ , we cut off at the $n^{th} \; 2$ for the $n^{th}$ convergent, and ignore what follows. So we get the sequence of convergents $$\frac{1}{2} , \frac{1}{2-\frac{1}{2}}, \frac{1}{2-\frac{1}{2-\frac{1}{2}}},...$$ Working out the numbers, we find the sequence to be $$\frac{1}{2},\frac{2}{3},\frac{3}{4},...$$ Again, we see a pattern! The $n^{th}$ term of the sequence is clearly of the form $$\frac{n-1}{n}$$ Let $C(n)$ be a function denoting the $n^{th}$ convergent. Then $C(1)=1/2,$ $C(2) = 2/3,$ $C(n)=(n-1)/n,$ and so on. So like before we consider the infinite limit: $$\lim_{n\to\infty} C(n) = \lim_{n\to\infty} \frac{n-1}{n} = \lim_{n\to\infty} 1 - \frac 1 n = \lim_{n\to\infty} 1 - \lim_{n\to\infty} \frac 1 n = 1 - 0 = 1$$ Thus we can conclude that the continued fraction equals $1$ , because its sequence of convergents equals $1$ ! On Infinite Radicals: So now, we touch on infinite nested radicals. They're messier to deal with but doable. One of the simpler examples of such radicals to contend with is $$2 = \sqrt{2 +\sqrt{2 +\sqrt{2 +\sqrt{2 +\sqrt{2 +...}}}}}$$ As with the previous two cases we see an infinite expression. We instinctively conclude by now: to logically define a limit for this expression - to assign it a value provided it even exists - we need to chop this up at finite points, defining a sequence of convergents $C(n)$ , and then find $C(n)$ as $n\to\infty$ . Nested radicals are a lot messier than the previous, but we manage. So first let the sequence of convergents be given by cutting off everything after the $n^{th} \; 2$ in the expression. Thus we get the sequence $$\sqrt 2 \;\;,\;\; \sqrt{2 + \sqrt{2}}\;\;,\;\; \sqrt{2+\sqrt{2+\sqrt{2}}}\;\;,\;\; \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}$$ Okay this isn't particularly nice already, but apparently there does exist, shockingly enough, a closed-form explicit expression for $C(n)$ : (from: S. Zimmerman, C. Ho) $$C(n) = 2\cos\left(\frac{\pi}{2^{n+1}}\right)$$ (I had to find that expression by Googling, I honestly didn't know that offhand. It can be proved by induction, as touched on in this MSE question .) So luckily, then, we can find the limit of $C(n)$ : $$\lim_{n\to\infty} C(n) = \lim_{n\to\infty} 2\cos\left(\frac{\pi}{2^{n+1}}\right)$$ It is probably obvious that the argument of the cosine function approaches $0$ as $n$ grows without bound, and thus $$\lim_{n\to\infty} C(n) = \lim_{n\to\infty} 2\cos\left(\frac{\pi}{2^{n+1}}\right) = 2\cos(0) = 2\cdot 1 = 2$$ Thus, since its convergents approach $2$ , we can conclude that $$2 = \sqrt{2 +\sqrt{2 +\sqrt{2 +\sqrt{2 +\sqrt{2 +...}}}}}$$ A Lengthy Conclusion: So, in short, how do we evaluate an infinite expression, be it radical, continued fraction, sum, or otherwise? We begin by truncating the expression at convenient finite places, creating a series of convergents, generalizations of the "partial sums" introduced in calculus. We then try to get a closed form or some other usable expression for the convergents $C(n)$ , and consider the value as $n\to\infty$ . If it converges to some value, we say that the expression is in fact equal to that value. If it doesn't, then the expression doesn't converge to any value. This doesn't mean each expression is "nice." Radical expressions in particular, in my experience, tend to be nasty as all hell, and I'm lucky I found that one closed form expression for the particularly easy radical I chose. This doesn't mean that other methods cannot be used to find the values, so long as there's some sort of logical justification for said method. For example, there is a justification for the formula for an infinite (and finite) geometric sum. We might have to circumvent the notion of partial sums entirely, or at least it might be convenient to do so. For example, with the Basel problem, Euler's proof focused on Maclaurin series, and none of this "convergent" stuff. (That proof is here plus other proofs of it!) Luckily, at least, this notion of convergents, even if it may not always be the best way to do it, lends itself to an easy way to check a solution to any such problem. Just find a bunch of the convergents - take as many as you need. If you somehow have multiple solutions, as you think with Ramanujan's radical, then you'll see the convergents get closer and closer to the "true" solution. (How many convergents you need to find depends on the situation and how close your proposed solutions are. It might be immediately obvious after $10$ iterations, or might not be until $10,000,000$ . This logic also relies on the assumption that there is only one solution to a given expression that is valid. Depending on the context, you might see cases where multiple solutions are valid but this "approaching by hand" method will only get you some of the solutions. This touches on the notion of "unstable" and "stable" solutions to dynamical systems - which I believe is the main context where such would pop up - but it's a bit overkill to discuss that for this post.) So I will conclude by showing, in this way, that the solution is $3$ to Ramanujan's radical. We begin with the radical itself: $$\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots}}}}=3$$ Let us begin by getting a series of convergents: $$\sqrt{1} \;\;,\;\; \sqrt{1 + 2\sqrt{1}} \;\;,\;\; \sqrt{1 + 2\sqrt{1 + 3\sqrt{1}}} \;\;,\;\;$$ Because the $\sqrt{1}$ isn't necessary, we just let it be $1$ . $$1 \;\;,\;\; \sqrt{1 + 2} \;\;,\;\; \sqrt{1 + 2\sqrt{1 + 3}} \;\;,\;\;$$ Okay so ... where to go from here? Honestly, my initial temptation was to just use a MATLAB script and evaluate it, but I can't think of even a recursive closed form for this that would be nice enough. So in any event, we just have to go by "hand" (and by hand I mean WolframAlpha). Let $C(n)$ be the $n^{th}$ convergent. Then $C(1) = 1$ $C(2) \approx 1.732$ $C(3) \approx 2.236$ $C(4) \approx 2.560$ $C(5) \approx 2.755$ $C(6) \approx 2.867$ $C(7) \approx 2.929$ $C(8) \approx 2.963$ $C(9) \approx 2.981$ $C(10) \approx 2.990$ To skip a few values because at this point the changes get minimal, I used a macro to make a quick code for $C(50)$ so I could put it into Microsoft Excel and got the approximate result $$C(50) \approx 2.999 \; 999 \; 999 \; 999 \; 99$$ So while not the most rigorous result, we can at least on an intuitive level feel like the convergents from Ramanujan's radical converge to $3$ , not $4$ or any other number. Neglecting that this is not an ironclad proof of the convergence, at least intuitively then we can feel like $$3 = \sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots}}}}$$ Whew! Hopefully that lengthy post was helpful to you! A late footnote, but Mathologer on YouTube did a video on this very topic, so his video would give a decent summary of all this stuff as well. Here's a link.
|
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|
3,119,664 |
I understand that a proportional relationship is $y=mx$ and a linear relationship is $y=mx+b$ . But if a line crosses $(0,0)$ , is it fair to say it is both proportional and linear assuming constant rate of change?
|
Introduction: The issue is what "..." really "represents." Typically we use it as a sort of shorthand, as if to say "look, I can't write infinitely many things down, just assume that the obvious pattern holds and goes on infinitely." This idea holds for all sorts of things - nested radicals, infinite sums, continued fractions, infinite sequences, etc. On Infinite Sums: A simple example: the sum of the reciprocals of squares: $$1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + ...$$ This is a well known summation. It is the Riemann zeta function $\zeta(s)$ at $s=2$ , and is known to evaluate to $\pi^2/6$ (proved by Euler and known as the Basel problem ). Another, easier-to-handle summation is the geometric sum $$1 + \frac 1 2 + \frac 1 4 + \frac 1 8 + ...$$ This is a geometric series where the ratio is $1/2$ - each summand is half the previous one. We know, too, that this evaluates to $2$ . Another geometric series you might see in proofs that $0.999... = 1$ is $$9 \left( \frac{1}{10} + \frac{1}{100} + \frac{1}{1,000} + \frac{1}{10,000} + ... \right)$$ which equals $1$ . In fact, any infinite geometric series, with first term $a$ and ratio $|r|<1$ can be evaluated by $$\sum_{n=0}^\infty ar^n = \frac{a}{1-r}$$ So a question arises - ignoring these "obvious" results (depending on your amount of mathematical knowledge), how would we know these converge to the given values? What, exactly, does it mean for a summation to converge to a number or equal a number? For finite sums this is no issue - if nothing else, we could add up each number manually, but we can't just add up every number from a set of infinitely-many numbers. Well, one could argue by common sense that, if the sequence converges to some number, the more and more terms you add up, the closer they'll get to that number. So we obtain one definition for the convergence of an infinite sum. Consider a sequence where the $n^{th}$ term is defined by the sum of the first $n$ terms in the sequence. To introduce some symbols, suppose we're trying to find the sum $$\sum_{k=1}^\infty x_k = x_1 + x_2 + x_3 + x_4 + ...$$ for whatever these $x_i$ 's are. Then define these so-called "partial sums" of this by a function $S(n)$ : $$S(n) = \sum_{k=1}^n x_k = x_1 + x_2 + ... + x_n$$ Then we get a sequence of sums: $$S(1), S(2), S(3), ...$$ or equivalently $$x_1 \;\;,\;\; x_1 + x_2\;\;,\;\; x_1 + x_2 + x_3\;\;,\;\; ...$$ Then we ask: what does $S(n)$ approach as $n$ grows without bound, if anything at all? (In calculus, we call this "the limit of the partial sums $S(n)$ as $n$ approaches infinity.") For the case of our first geometric sum, we immediately see the sequence of partial sums $$1, \frac{3}{2}, \frac{7}{4}, \frac{15}{8},...$$ Clearly, this suggests a pattern - and if you want to, you can go ahead and prove it, I won't do so here for brevity's sake. The pattern is that the $n^{th}$ term of the sequence is $$S(n) = \frac{2^{(n+1)}-1}{2^{n}}$$ We can then easily consider the limit of these partial sums: $$\lim_{n\to\infty} S(n) = \lim_{n\to\infty} \frac{2^{(n+1)}-1}{2^{n}} = \lim_{n\to\infty} \frac{2^{(n+1)}}{2^{n}} - \frac {1}{2^{n}} = \lim_{n\to\infty} 2 - \lim_{n\to\infty} \frac{1}{2^{n}}$$ Obviously, $1/2^{n} \to 0$ as $n$ grows without bound, and $2$ is not affected by $n$ , so we conclude $$\lim_{n\to\infty} S(n) = \lim_{n\to\infty} 2 - \lim_{n\to\infty} \frac{1}{2^n} = 2 - 0 = 2$$ And thus we say $$\sum_{k=0}^\infty \left(\frac 1 2 \right)^k = 1 + \frac 1 2 + \frac 1 4 + \frac 1 8 + ... = 2$$ because the partial sums approach $2$ . On Continued Fractions: That was a simple, "first" sort of example, but mathematicians essentially do the same thing in other contexts. I want to touch on one more such context before we deal with the radical case, just to nail that point home. In this case, it will be with continued fractions . One of the simpler such fractions is the one for $1$ : $$1 = \frac{1}{2-\frac{1}{2-\frac{1}{2-\frac{1}{...}}}}$$ As usual, the "..." denotes that this continues forever. But what it does it mean for this infinite expression to equal $1$ ? For this, we consider a more general analogue of the "partial sum" from before - a "convergent." We cut up the sequence at logical finite points, whatever those points being depending on the context. Then if the sequence of the convergents approaches a limit, we say they're equal. What are the convergents for a continued fraction? By convention, we cut off just before the start of the next fraction. That is, in the continued fraction for $1$ , we cut off at the $n^{th} \; 2$ for the $n^{th}$ convergent, and ignore what follows. So we get the sequence of convergents $$\frac{1}{2} , \frac{1}{2-\frac{1}{2}}, \frac{1}{2-\frac{1}{2-\frac{1}{2}}},...$$ Working out the numbers, we find the sequence to be $$\frac{1}{2},\frac{2}{3},\frac{3}{4},...$$ Again, we see a pattern! The $n^{th}$ term of the sequence is clearly of the form $$\frac{n-1}{n}$$ Let $C(n)$ be a function denoting the $n^{th}$ convergent. Then $C(1)=1/2,$ $C(2) = 2/3,$ $C(n)=(n-1)/n,$ and so on. So like before we consider the infinite limit: $$\lim_{n\to\infty} C(n) = \lim_{n\to\infty} \frac{n-1}{n} = \lim_{n\to\infty} 1 - \frac 1 n = \lim_{n\to\infty} 1 - \lim_{n\to\infty} \frac 1 n = 1 - 0 = 1$$ Thus we can conclude that the continued fraction equals $1$ , because its sequence of convergents equals $1$ ! On Infinite Radicals: So now, we touch on infinite nested radicals. They're messier to deal with but doable. One of the simpler examples of such radicals to contend with is $$2 = \sqrt{2 +\sqrt{2 +\sqrt{2 +\sqrt{2 +\sqrt{2 +...}}}}}$$ As with the previous two cases we see an infinite expression. We instinctively conclude by now: to logically define a limit for this expression - to assign it a value provided it even exists - we need to chop this up at finite points, defining a sequence of convergents $C(n)$ , and then find $C(n)$ as $n\to\infty$ . Nested radicals are a lot messier than the previous, but we manage. So first let the sequence of convergents be given by cutting off everything after the $n^{th} \; 2$ in the expression. Thus we get the sequence $$\sqrt 2 \;\;,\;\; \sqrt{2 + \sqrt{2}}\;\;,\;\; \sqrt{2+\sqrt{2+\sqrt{2}}}\;\;,\;\; \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}$$ Okay this isn't particularly nice already, but apparently there does exist, shockingly enough, a closed-form explicit expression for $C(n)$ : (from: S. Zimmerman, C. Ho) $$C(n) = 2\cos\left(\frac{\pi}{2^{n+1}}\right)$$ (I had to find that expression by Googling, I honestly didn't know that offhand. It can be proved by induction, as touched on in this MSE question .) So luckily, then, we can find the limit of $C(n)$ : $$\lim_{n\to\infty} C(n) = \lim_{n\to\infty} 2\cos\left(\frac{\pi}{2^{n+1}}\right)$$ It is probably obvious that the argument of the cosine function approaches $0$ as $n$ grows without bound, and thus $$\lim_{n\to\infty} C(n) = \lim_{n\to\infty} 2\cos\left(\frac{\pi}{2^{n+1}}\right) = 2\cos(0) = 2\cdot 1 = 2$$ Thus, since its convergents approach $2$ , we can conclude that $$2 = \sqrt{2 +\sqrt{2 +\sqrt{2 +\sqrt{2 +\sqrt{2 +...}}}}}$$ A Lengthy Conclusion: So, in short, how do we evaluate an infinite expression, be it radical, continued fraction, sum, or otherwise? We begin by truncating the expression at convenient finite places, creating a series of convergents, generalizations of the "partial sums" introduced in calculus. We then try to get a closed form or some other usable expression for the convergents $C(n)$ , and consider the value as $n\to\infty$ . If it converges to some value, we say that the expression is in fact equal to that value. If it doesn't, then the expression doesn't converge to any value. This doesn't mean each expression is "nice." Radical expressions in particular, in my experience, tend to be nasty as all hell, and I'm lucky I found that one closed form expression for the particularly easy radical I chose. This doesn't mean that other methods cannot be used to find the values, so long as there's some sort of logical justification for said method. For example, there is a justification for the formula for an infinite (and finite) geometric sum. We might have to circumvent the notion of partial sums entirely, or at least it might be convenient to do so. For example, with the Basel problem, Euler's proof focused on Maclaurin series, and none of this "convergent" stuff. (That proof is here plus other proofs of it!) Luckily, at least, this notion of convergents, even if it may not always be the best way to do it, lends itself to an easy way to check a solution to any such problem. Just find a bunch of the convergents - take as many as you need. If you somehow have multiple solutions, as you think with Ramanujan's radical, then you'll see the convergents get closer and closer to the "true" solution. (How many convergents you need to find depends on the situation and how close your proposed solutions are. It might be immediately obvious after $10$ iterations, or might not be until $10,000,000$ . This logic also relies on the assumption that there is only one solution to a given expression that is valid. Depending on the context, you might see cases where multiple solutions are valid but this "approaching by hand" method will only get you some of the solutions. This touches on the notion of "unstable" and "stable" solutions to dynamical systems - which I believe is the main context where such would pop up - but it's a bit overkill to discuss that for this post.) So I will conclude by showing, in this way, that the solution is $3$ to Ramanujan's radical. We begin with the radical itself: $$\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots}}}}=3$$ Let us begin by getting a series of convergents: $$\sqrt{1} \;\;,\;\; \sqrt{1 + 2\sqrt{1}} \;\;,\;\; \sqrt{1 + 2\sqrt{1 + 3\sqrt{1}}} \;\;,\;\;$$ Because the $\sqrt{1}$ isn't necessary, we just let it be $1$ . $$1 \;\;,\;\; \sqrt{1 + 2} \;\;,\;\; \sqrt{1 + 2\sqrt{1 + 3}} \;\;,\;\;$$ Okay so ... where to go from here? Honestly, my initial temptation was to just use a MATLAB script and evaluate it, but I can't think of even a recursive closed form for this that would be nice enough. So in any event, we just have to go by "hand" (and by hand I mean WolframAlpha). Let $C(n)$ be the $n^{th}$ convergent. Then $C(1) = 1$ $C(2) \approx 1.732$ $C(3) \approx 2.236$ $C(4) \approx 2.560$ $C(5) \approx 2.755$ $C(6) \approx 2.867$ $C(7) \approx 2.929$ $C(8) \approx 2.963$ $C(9) \approx 2.981$ $C(10) \approx 2.990$ To skip a few values because at this point the changes get minimal, I used a macro to make a quick code for $C(50)$ so I could put it into Microsoft Excel and got the approximate result $$C(50) \approx 2.999 \; 999 \; 999 \; 999 \; 99$$ So while not the most rigorous result, we can at least on an intuitive level feel like the convergents from Ramanujan's radical converge to $3$ , not $4$ or any other number. Neglecting that this is not an ironclad proof of the convergence, at least intuitively then we can feel like $$3 = \sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots}}}}$$ Whew! Hopefully that lengthy post was helpful to you! A late footnote, but Mathologer on YouTube did a video on this very topic, so his video would give a decent summary of all this stuff as well. Here's a link.
|
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|
3,121,694 |
Let $P_o$ be the primes excluding $2$ . $P_o \subset \mathbb{N}$ has the following property $Q$ : For any $a,b \in P_o$ , $a + b \not\in P_o$ . For any $a,b \in P_o$ , $ab \not\in P_o$ . So both addition and multiplication necessarily leave the set $P_o$ . $P_o$ has natural density $0$ . Q1 . Is there a set $S \subset \mathbb{N}$ with positive density
that satisfies property $Q$ ? Answered quickly by @JoséCarlosSantos: Yes .
Permit me then to add a new question: Q2 . What is largest density $S \subset \mathbb{N}$ that satisfies property $Q$ ? Santos's example has density $\frac{1}{3}$ .
|
New answer: Kurlberg, Lagarias and Pomerance, On sets of integers which are both sum-free and product-free ( arXiv:1201.1317) answers the question. The upper density of any such set is strictly less than 1/2, but can be arbitrarily close to 1/2. I don't see that they state this explicitly in the paper, but it follows pretty quickly from Theorem 1.3. Explicitly: Theorem 1.3 implies that for any $\varepsilon>0$ there is some $n$ and some subset $S\subset\mathbb{Z}/n\mathbb{Z}$ of residue classes that is sum-free and product-free consisting of at least $(\frac{1}{2}-\varepsilon)n$ classes. Then taking all integers in those residue classes gives a product-free sum-free set of integers of density at least $(\frac{1}{2}-\varepsilon)$ . Old answer: Andrew Treglow's talk On sum-free and solution-free sets of integers cites the following result of Deshouillers, Freiman, Sós and Temkin (1999): If $S\subseteq[n]$ is sum-free then at least one of the following holds: $\lvert S\rvert\le2n/5+1$ $S$ consists of odds $\lvert S\rvert\le\min(S)$ . Therefore, if the density of a sum-free product-free set $P$ of integers is greater than 2/5, then $P\cap[n]$ must fall in the second case for sufficiently large $n$ . (We can't be in the third case because $\min(P)<2n/5$ for sufficiently large $n$ .) So, the only way we could hope to do better than 2/5 is to use only odd numbers, and as a corollary the highest density we could hope for is 1/2. In fact, the proof of Remark 2.7 of Kurlberg, Lagarias and Pomerance, Product-free sets with high density carries over to the case of only odd numbers, showing that we cannot attain a density of 1/2. For completeness, we repeat the argument here with the appropriate modifications: Let $a$ denote the least element of $P$ , and let $P(x):=P\cap[1,x]$ . Since $P(x)\setminus{P(x/a)}$ lies in $(x/a,x]$ , $\lvert P(x)\rvert\le \lvert P(x/a)\rvert+\frac{x-\lfloor x/a\rfloor}{2}+1$ . Also, multiplying each member of $P(x/a)$ creates products in $[1,x]$ which cannot lie in $P$ , so we have $\lvert P(x)\rvert\le \frac{x}{2}+1-\lvert P(x/a)\rvert$ . Adding these two inequalities and dividing both sides by 2 gives $\lvert P(x)\rvert\le
\frac{x}{2}-\frac{\lfloor x/a\rfloor}{2}+2$ , which implies that the upper density of $P$ is at most $\frac{1}{2}-\frac{1}{2a}$ .
|
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|
3,121,707 |
For a matrix $A \in \mathbb R^{n \times n}$ , prove that $|tr(A)|\leq \sqrt{n}||A||_{F}$ , where $tr(A)$ denotes the trace of the matrix A, and $||.||_{F}$ denotes the Frobenius norm. We know that $|tr(A)|=|\sum_{i=1}^{n}a_{i,i}|$ , whereas, $||A||_{F}=\sqrt{\sum_{i=1}^{n}\sum_{j=1}^{n}|a_{i,j}|^{2}}$ . $\implies \sqrt{n}||A||_{F} = \sqrt{n(\sum_{i=1}^{n}\sum_{j=1}^{n}|a_{i,j}|^{2})} \geq \sqrt{n(\sum_{i=1}^{n}|a_{i,i}|^{2})} = \sqrt{\sum_{i=1}^{n}e_{i}^{2}}\sqrt{\sum_{i=1}^{n}|Ae_i|^{2}}$ , where $\{e_1,...,e_n\}$ forms the standard basis of $\mathbb R^n$ . But by Cauchy Schwarz Inequality, $\sqrt{\sum_{i=1}^{n}e_{i}^{2}}\sqrt{\sum_{i=1}^{n}|Ae_i|^{2}} \geq |\sum_{i=1}^{n} \langle e_i, Ae_i \rangle| = |tr(A)|$ . Hence, the proof. Is there any mistake in my proof? I will be grateful if someone could grade my proof and correct it if necessary. I am using this result to prove the Backward stability of Cholesky's method. Thanks.
|
New answer: Kurlberg, Lagarias and Pomerance, On sets of integers which are both sum-free and product-free ( arXiv:1201.1317) answers the question. The upper density of any such set is strictly less than 1/2, but can be arbitrarily close to 1/2. I don't see that they state this explicitly in the paper, but it follows pretty quickly from Theorem 1.3. Explicitly: Theorem 1.3 implies that for any $\varepsilon>0$ there is some $n$ and some subset $S\subset\mathbb{Z}/n\mathbb{Z}$ of residue classes that is sum-free and product-free consisting of at least $(\frac{1}{2}-\varepsilon)n$ classes. Then taking all integers in those residue classes gives a product-free sum-free set of integers of density at least $(\frac{1}{2}-\varepsilon)$ . Old answer: Andrew Treglow's talk On sum-free and solution-free sets of integers cites the following result of Deshouillers, Freiman, Sós and Temkin (1999): If $S\subseteq[n]$ is sum-free then at least one of the following holds: $\lvert S\rvert\le2n/5+1$ $S$ consists of odds $\lvert S\rvert\le\min(S)$ . Therefore, if the density of a sum-free product-free set $P$ of integers is greater than 2/5, then $P\cap[n]$ must fall in the second case for sufficiently large $n$ . (We can't be in the third case because $\min(P)<2n/5$ for sufficiently large $n$ .) So, the only way we could hope to do better than 2/5 is to use only odd numbers, and as a corollary the highest density we could hope for is 1/2. In fact, the proof of Remark 2.7 of Kurlberg, Lagarias and Pomerance, Product-free sets with high density carries over to the case of only odd numbers, showing that we cannot attain a density of 1/2. For completeness, we repeat the argument here with the appropriate modifications: Let $a$ denote the least element of $P$ , and let $P(x):=P\cap[1,x]$ . Since $P(x)\setminus{P(x/a)}$ lies in $(x/a,x]$ , $\lvert P(x)\rvert\le \lvert P(x/a)\rvert+\frac{x-\lfloor x/a\rfloor}{2}+1$ . Also, multiplying each member of $P(x/a)$ creates products in $[1,x]$ which cannot lie in $P$ , so we have $\lvert P(x)\rvert\le \frac{x}{2}+1-\lvert P(x/a)\rvert$ . Adding these two inequalities and dividing both sides by 2 gives $\lvert P(x)\rvert\le
\frac{x}{2}-\frac{\lfloor x/a\rfloor}{2}+2$ , which implies that the upper density of $P$ is at most $\frac{1}{2}-\frac{1}{2a}$ .
|
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|
3,122,554 |
Isn't the percentage symbol actually just a constant with the value $0.01$ ? As in $$
15\% = 15 \times \% = 15 \times 0.01 = 0.15.
$$ Isn't every unit actually just a constant? But why do we treat them in such a special way then?
|
Isn't the percentage symbol actually just a constant with the value $0.01$ ? No. If it were, all of the following would be valid constructs: $$
30+\%\,50=30.5\\
90\,\%\,\mathrm{cm}=0.9\,\mathrm{cm}\\
2-\%=1.99\\
\%^2=0.0001
$$ The percentage symbol is a unit . When converting between units, it's easy to treat them as constants that represent the conversion ratio, and multiply (for example, the $\mathrm{m}$ unit can be thought of as a constant equal to $100\,\mathrm{cm}$ , in $2\,\mathrm{m}=2(100\,\mathrm{cm})=200\,\mathrm{cm}$ ). But that isn't the same as saying they're "just constants", as they represent more than that. A unit is not just a ratio, it's a distance or a weight or an amount of time. This is less obvious with $\%$ because it's a dimensionless unit , representing something more abstract like "parts of a whole" rather than a physical property like mass or surface area. $1\,\%$ is "one one-hundredth of a thing", measuring an amount of something, anything, often something with its own units. A similarly dimensionless unit is the "degree", where $1^\circ$ is "one three-hundred-sixtieth of the way around". Another one is the "cycle", as in "one $\mathrm{Mhz}$ is one million cycles per second". Things like "wholes", "turns", and "cycles" are more abstract than inches or grams, but when applied they still represent tangible measurements, so they aren't any less powerful when treated as units. I mean, I guess every unit is actually just a constant, but why do we treat them in such a special way then? What then would you say the "constant" is that is represented by "inch", or "second", or "ounce"? Would these ideas not have clear numeric values if every unit were simply a constant? Again, a unit is not just a constant, it represents something more. I don't have exact vocabulary for this, but I would say a unit is an "amount" of a "dimension". The dimension can be time, space, energy, mass, etc. To even begin to treat a unit as a constant, we need to consider it in terms of a different unit in the same dimension . For example, the unit "millisecond" amounts to different constants depending on whether we think about it in terms of a second ( $0.001$ ), hour ( $2.77778\times10^{-7}$ ), microsecond ( $1000$ ), etc. This constant is not intrinsic to the units themselves, as it only arises when comparing to other units.
|
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|
3,127,366 |
I am reading this link on Wikipedia ; it states the following definition is given for a DAG. Definition: A DAG is a finite, directed graph with no directed cycles. Reading this definition believes me to think that the digraph below would be a DAG as there are no directed cycles here (there are cycles of the underlying graph but there are no directed cycles). However, all the pictures on Wikipedia show examples of DAGs with arrows pointing the same way. So, I think I am interpreting this definition wrong. In particular, why does the definition mention later on an equivalent definition is that it must have topological ordering such that "every edge is directed from earlier to later in the sequence"? Reading the definition above would lead me to believe that the graph above is a DAG, but then the equivalent definition would make me think otherwise.
|
The graph you show is a DAG. It is conventional to draw DAGs with all the arrows going in the roughly the same direction, because that usually gives a clearer intuition about what is going on in the graph. But remember that locations and directions are not part of the formal definition of a graph -- they're just incidental features of the particular drawing at the graph you're looking at, and it would be the same graph if you drew the vertices in different locations on the paper. (Even in your drawing, all the edges go in a broadly southeasterly direction -- or at least more southeast than northwest -- so you're actually following the convention). In particular, why does the definition mention later on an equivalent definition is that it must have topological ordering such that "every edge is directed from earlier to later in the sequence"? Because that is another way to define the same class of graphs, and sometimes (but not always) a more productive way to think about them. You should be able to prove that the finite directed graphs that have no directed cycles are exactly the same as the finite directed graphs that have a topological ordering.
|
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|
3,128,507 |
$$\int_0^{102}\left(\prod_{k=1}^{100}(x-k)\right)\left(\sum_{k=1}^{100}\frac1{x-k}\right)\,dx$$ I've tried solving this problem but only thing that comes to my mind is the manual integration by multiplication of the expressions which will literally take much longer than the allotted time for competitive exams Now this is a homework and exercises problem but I'd be glad if I could get some clues on how to solve this problem.
|
Hint : By the product rule you have the following result: $$\dfrac{\mathrm d}{\mathrm dx}\prod_{k=1}^{100}(x-k)=\left(\prod_{k=1}^{100}(x-k)\right)\left(\sum_{k=1}^{100}\dfrac{1}{(x-k)}\right)$$ Integrate both sides from $0$ to $102$ , use the Fundamental Theorem of Calculus and you'll be done in no time.
|
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|
3,130,579 |
$17!$ is equal to $$35568x428096y00$$ Both $x$ and $y$ , are digits. Find $x$ and $y$ . So, $$17!=2^{15}\times 3^6\times 5^3\times 7^2\times 11\times 13\times 17=(2^3\times 5^3)\times 2^{12}\times 3^6\times 7^2\times 11\times 13\times 17$$ If there`s a product of $(2\times 5)^3$ Then this number has $3$ zeros at the end, so $y=0$ How do I find the $x$ now?
|
HINT $17!$ is divisible by $9$ . What is an easy test for divisibility by 9, involving the digits of a number?
|
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|
3,131,186 |
I am not sure why "points exist" is not an axiom in geometry, given that the other axioms are likewise primitive and seemingly as obvious.
|
In the presentation I have most convenient to me (Hilbert's axioms), the axioms for plane geometry start with a trio of "axioms of incidence". One of those axioms is "There exist three non-collinear points". That axiom certainly includes the existence of points.
|
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|
3,134,219 |
More precisely, why is it that all rings are required by the axioms to have commutativity in addition, but are not held to the same axiom regarding multiplication? I know that we have commutative and non-commutative rings depending on whether or not they are commutative in multiplication, but I am wondering why it is that the axioms were defined that way, providing us with this option. I am using this list of axioms, from David Sharpe’s Rings and factorization : Definition 1.3.1. A ring is a non-empty set $R$ which satisfies the following axioms: (1) $R$ has a binary operation denoted by $+$ defined on it; (2) addition is associative, i.e. \begin{align}
a + \left(b+c\right) = \left(a+b\right) + c \text{ for all } a, b, c \in R
\end{align} (so that we can write $a+b+c$ without brackets); (3) addition is commutative, i.e. \begin{align}
a + b = b + a \text{ for all } a, b \in R;
\end{align} (4) there is an element denoted by $0$ in $R$ such that \begin{align}
0 + a = a \text{ for all } a \in R
\end{align} (there is only one such element because, if $0_1$ and $0_2$ are two such, then $0_1 = 0_1 + 0_2 = 0_2$ and they are the same -- we call $0$ the zero element of $R$ ); (5) for every $a \in R$ , there exists an element $-a \in R$ such that \begin{align}
\left(-a\right) + a = 0
\end{align} (there is only one such element for each $a$ , because if $b + a = 0$ and $c + a = 0$ , then \begin{align}
b = 0 + b = \left(c + a\right) + b = c + \left(a + b\right) = c + 0 = c;
\end{align} we call $-a$ the negative of $a$ ); (6) $R$ has a binary operation denoted by multiplication defined on it; (7) multiplication is associative, i.e. \begin{align}
a\left(bc\right) = \left(ab\right)c \text{ for all } a, b, c \in R;
\end{align} (8) multiplication is left and right distributive over addition, i.e. \begin{align}
a\left(b+c\right) = ab + ac,\ \left(a+b\right)c = ac + bc
\text{ for all } a, b, c \in R;
\end{align} (9) there is an element denoted by $1$ in $R$ such that $1 \neq 0$ and \begin{align}
1 \cdot a = a \cdot 1 = a \text{ for all } a \in R
\end{align} (as for the zero element, there is only one such element, and it is called the identity element of $R$ ).
|
The first rings that were considered were generally commutative; polynomial rings, then from work of Dedekind in number fields. The properties were then abstracted by Fraenkel and Noether, who still dealt mostly with commutative rings. However, it soon became apparent that there were too many instances where commutativity of multiplication did not hold. You had famously the quaternions, of course, but you also had matrices and, more generally, the endomorphism ring of an abelian group (where “multiplication” is composition of functions). So that we have two different, related, notions: commutative rings and noncommutative rings, just like we have noncommutative groups and commutative/abelian groups. Now, why do this with multiplication and not with addition? Well, if you take your definition of ring above, which includes a unity, but you drop condition (3) (that is, you require everything except you do not require that addition be commutative), it turns out that the other eight axioms force commutativity of addition. Indeed, suppose you have a structure $(R,+,\cdot,0,1)$ that satisfies axioms (1), (2), and (4)-(9) above. I claim that one can deduce (3). Indeed, let $a,b\in R$ . Then using distributivity on the left first, and distributivity on the right second, we have $$\begin{align*}
(1+1)(a+b) &= 1(a+b) + 1(a+b) = a+b+a+b\\
(1+1)(a+b) &= (1+1)a + (1+1)b = a+a+b+b.
\end{align*}$$ From this we get that $a+b+a+b = a+a+b+b$ . Now add the inverse of $a$ on the left and the inverse of $b$ on the right on both sides to get $$\begin{align*}
(-a) + a + b + a + b + (-b) &= 0+b+a+0 = b+a\\
(-a) + a + a + b + b + (-b) &= 0+a+b+0 = a+b
\end{align*}$$ Thus, we conclude that $a+b=b+a$ . That is, commutativity of addition is a consequence of the other eight axioms. The reason we include it is two-fold: one, is that it is much nicer to say that the first few axioms force $(R,+)$ to be a commutative/abelian group. The second is that it is also common to consider rings without unity, and if we do that, then it is no longer true that addition is forced to be commutative. To see this, note that if $(G,\cdot)$ is any group with identity element $e_G$ , and we define a multiplication on $G$ by letting $a*b=e_G$ for all $a,b\in G$ , then $(G,\cdot,*)$ satisfies axioms (1)-(8) given above. But if the original group is not commutative, then the “addition” in this ring is not commutative. So if we want to consider rings without unity, we do want to explicitly require addition to be commutative.
|
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|
3,134,563 |
One needs to choose six real numbers $x_1,x_2,\cdots,x_6$ such that the product of any five of them is equal to other number. The number of such choices is A) $3$ B) $33$ C) $63$ D) $93$ I believe this is not so hard problem but I got no clue to proceed. The work I did till now. Say the numbers be $a,b,c,d,e, abcde$ . Then, $b\cdot c\cdot d\cdot e\cdot abcde=a$ hence $bcde= +-1$ . Basically I couldn't even find a single occasion where such things occcur except all of these numbers being either $1$ or $-1$ . Are these all cases?
|
Here's an argument which extends to $n$ real numbers quite nicely! We can start by noting that $$x_1 (x_2x_3x_4x_5x_6) = x_1^2$$ And by commutativity, we get $x_i^2 = x_j^2$ which implies that all the magnitudes are equal. Now if $L$ is the magnitude, we also must have $L = L^5$ , and from this we conclude that $L=0$ or $1$ . Now we just have to go through the possibilities. If $L = 0$ , then we have all $0$ 's If $L = 1$ , then we have all $-1$ 's and $1$ 's. This configuration works iff the number of $-1$ 's is even, as this would imply that $$\frac{\prod_{i=1}^6 x_i}{1} = 1 \text{ and } \frac{\prod_{i=1}^6 x_i}{-1} = -1$$ Now we can count configurations. There will be $$\sum_{i=0}^3 \binom{6}{2i} = 2^5$$ possibilities. And finally, we have $1+32 = 33$ .
|
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|
3,134,607 |
I am trying to find the fixed point of the function $g(x) = e^{-x}$ . Wolfram Alpha tells me that this fixed point is approximately $x \approx 0,567$ . However, if I apply the Banach fixed point theorem, I can prove that $g(x)$ has a fixed point in the interval $[2,\infty)$ . I reasoned as follows: Banach fixed point theorem:
Let $X$ be a Banach space, $D \subseteq X$ a closed interval and $T:D \rightarrow D$ a contraction, which means that $T$ is Lipschitz continuous with Lipschitz constant $L < 1$ : \begin{equation}
\Vert T(u) - T(v) \Vert_X \leq L \Vert u-v \Vert_X \text{ } \forall u,v \in D.
\end{equation} Now $X = (\mathbb{R}, \Vert \cdot \Vert_1)$ is a Banach space and $D = [2,\infty)$ is a closed interval in $X$ . Now, I show that $g(x)$ is a contraction: without loss of generality, assume $x < y$ . Then \begin{equation}
\Vert g_1(x) - g_1(y) \Vert_1 = |e^{-x} - e^{-y}| = e^{-x} - e^{-y} = e^{-x}(1-e^{x-y}).
\end{equation} Since $e^a \geq 1+a \text{ } \forall a \in \mathbb{R}$ , I obtain \begin{equation}
\begin{split}
\Vert g_1(x) - g_1(y) \Vert \leq e^{-x}(1-(1+x-y)) = e^{-x}(y-x) \\
\leq e^{-2}(y-x) = e^{-2}|x-y| = e^{-2} \Vert x-y \Vert_1 = L \Vert x-y \Vert_1,
\end{split}
\end{equation} where $L = e^{-2}$ < 1. So the conditions of the Banach fixed point theorem are satisfied and $g(x)$ has a fixed point in the interval $[2, \infty)$ . However... This is not true!
Can anyone tell me what is going wrong here?
Thank you very much in advance!
|
$e^{-x}$ does not map $[2,\infty)$ into itself.
|
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|
3,134,611 |
If $T:\mathbb{R}^m \to \mathbb{R}^n$ is an onto linear transformation with $m>n$ , what is $\dim(\ker(T))$ ? Explain. I'm trying to understand this. A detailed explanation would be helpful. Thanks.
|
$e^{-x}$ does not map $[2,\infty)$ into itself.
|
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|
3,134,991 |
If nine coins are tossed, what is the probability that the number of heads is even? So there can either be 0 heads, 2 heads, 4 heads, 6 heads, or 8 heads. We have $n = 9$ trials, find the probability of each $k$ for $k = 0, 2, 4, 6, 8$ $n = 9, k = 0$ $$\binom{9}{0}\bigg(\frac{1}{2}\bigg)^0\bigg(\frac{1}{2}\bigg)^{9}$$ $n = 9, k = 2$ $$\binom{9}{2}\bigg(\frac{1}{2}\bigg)^2\bigg(\frac{1}{2}\bigg)^{7}$$ $n = 9, k = 4$ $$\binom{9}{4}\bigg(\frac{1}{2}\bigg)^4\bigg(\frac{1}{2}\bigg)^{5}$$ $n = 9, k = 6$ $$\binom{9}{6}\bigg(\frac{1}{2}\bigg)^6\bigg(\frac{1}{2}\bigg)^{3}$$ $n = 9, k = 8$ $$\binom{9}{8}\bigg(\frac{1}{2}\bigg)^8\bigg(\frac{1}{2}\bigg)^{1}$$ Add all of these up: $$=.64$$ so there's a 64% chance of probability?
|
The probability is $\frac{1}{2}$ because the last flip determines it.
|
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|
3,135,015 |
Do there exist functions $f,g$ in $L^1(\mathbf{R})$ such that the convolution $f \star g$ is (almost everywhere) equal to the indicator function of the interval $[0,1]$ ?
|
Yes, this works. Let's change the interval to $I=[-1,1]$ for convenience. Then we want to find $f,g\in L^1$ such that $$
\widehat{f}(t)\widehat{g}(t) = \widehat{\chi_I}=\frac{\sin t}{t} .
$$ Let's take $\widehat{f}(t) = |t|^{-1/2}$ for $|t|\ge 1$ and then of course $\widehat{g}$ such that $\widehat{f}\widehat{g}=\widehat{\chi_I}$ (so in particular, $\widehat{g}(t)=\textrm{sgn}(t)\,|t|^{-1/2}\sin t$ for $|t|\ge 1$ ), and then finally, I'm also going to insist that $\widehat{f}, \widehat{g}\in C^{\infty}$ . To show that $f,g\in L^1$ , let's first of all discuss the local behavior. This only depends on the large $|t|$ behavior of the Fourier transform, and $|t|^{-1/2}$ has FT $c|x|^{-1/2}$ , which is locally integrable (but not locally $L^2$ , which is just as well since it's easy to see that at least one of $f,g$ will fail to be in $L^2$ if $f*g=\chi_I$ ). The large $x$ asymptotics of $f(x)$ on the other hand depend on the smoothness of $\widehat{f}$ , and here we're clearly in great shape since $\widehat{f}''\simeq t^{-5/2}\in L^1$ , so $x^2f(x)\in C_0$ . The other function is of the same type, the extra factor $\sin t$ for large $t$ produces essentially a shift in $g$ , and $\textrm{sgn}(t)$ amounts to an extra Hilbert transform, which won't affect the local singularities here, so $g\in L^1$ also.
|
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|
3,135,016 |
If a $C^*$ -algebra $A$ has a tracial state $\tau$ , can we construct a nonzero representation $\pi: A\rightarrow B(H_{\tau})$ such that $\pi(ab)=\pi(ba),\forall a,b \in A$ through the GNS construction?
|
Yes, this works. Let's change the interval to $I=[-1,1]$ for convenience. Then we want to find $f,g\in L^1$ such that $$
\widehat{f}(t)\widehat{g}(t) = \widehat{\chi_I}=\frac{\sin t}{t} .
$$ Let's take $\widehat{f}(t) = |t|^{-1/2}$ for $|t|\ge 1$ and then of course $\widehat{g}$ such that $\widehat{f}\widehat{g}=\widehat{\chi_I}$ (so in particular, $\widehat{g}(t)=\textrm{sgn}(t)\,|t|^{-1/2}\sin t$ for $|t|\ge 1$ ), and then finally, I'm also going to insist that $\widehat{f}, \widehat{g}\in C^{\infty}$ . To show that $f,g\in L^1$ , let's first of all discuss the local behavior. This only depends on the large $|t|$ behavior of the Fourier transform, and $|t|^{-1/2}$ has FT $c|x|^{-1/2}$ , which is locally integrable (but not locally $L^2$ , which is just as well since it's easy to see that at least one of $f,g$ will fail to be in $L^2$ if $f*g=\chi_I$ ). The large $x$ asymptotics of $f(x)$ on the other hand depend on the smoothness of $\widehat{f}$ , and here we're clearly in great shape since $\widehat{f}''\simeq t^{-5/2}\in L^1$ , so $x^2f(x)\in C_0$ . The other function is of the same type, the extra factor $\sin t$ for large $t$ produces essentially a shift in $g$ , and $\textrm{sgn}(t)$ amounts to an extra Hilbert transform, which won't affect the local singularities here, so $g\in L^1$ also.
|
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|
3,135,036 |
Consider the plane $\mathbb{R}^2$ equipped with its standard metric $d$ given by $$d((x_1,y_1),(x_2,y_2)) = \sqrt{(x_1-x_2)^2 + (y_1 - y_2)^2}$$ Determine whether the following subsets of $\mathbb{R}^2$ are open
with respect to the metric $d$ . Justify your answers. a) $ A = \{(x,y) \in \mathbb{R}^2 | x \geq 0\}$ . I have been given the solution to this question but don't quite understand it. This is the solution I have been given: Let $r > 0$ .Denote the point $(0,0) \in \mathbb{R}^2$ by $0$ and consider the open ball $B_r(0) \subset \mathbb{R}^2$ . For the point $p_r = (-\frac{r}{2}, 0)$ we have $$d(p_r,0) = \sqrt{\frac{r^2}{4}} = \frac{r}{2}<r$$ and therefore $p_r \in B_r(0)\subset \mathbb{R}^2$ . Note that $0 \in A$ . If $A$ were open, then we would be able to find a radius $r>0$ with the property that $B_r(0)$ is never entirely contained in $A$ . Thus, $A$ is not an open set of $\mathbb{R}^2$ with respect to the metric $d$ . Looking at this solution I have a couple of questions. Why is the point $p_r = (-\frac{r}{2}, 0)$ used? I understand where the $d(p_r,0)$ comes from.
If someone could help me break it down a bit.
Thanks in advance.
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Yes, this works. Let's change the interval to $I=[-1,1]$ for convenience. Then we want to find $f,g\in L^1$ such that $$
\widehat{f}(t)\widehat{g}(t) = \widehat{\chi_I}=\frac{\sin t}{t} .
$$ Let's take $\widehat{f}(t) = |t|^{-1/2}$ for $|t|\ge 1$ and then of course $\widehat{g}$ such that $\widehat{f}\widehat{g}=\widehat{\chi_I}$ (so in particular, $\widehat{g}(t)=\textrm{sgn}(t)\,|t|^{-1/2}\sin t$ for $|t|\ge 1$ ), and then finally, I'm also going to insist that $\widehat{f}, \widehat{g}\in C^{\infty}$ . To show that $f,g\in L^1$ , let's first of all discuss the local behavior. This only depends on the large $|t|$ behavior of the Fourier transform, and $|t|^{-1/2}$ has FT $c|x|^{-1/2}$ , which is locally integrable (but not locally $L^2$ , which is just as well since it's easy to see that at least one of $f,g$ will fail to be in $L^2$ if $f*g=\chi_I$ ). The large $x$ asymptotics of $f(x)$ on the other hand depend on the smoothness of $\widehat{f}$ , and here we're clearly in great shape since $\widehat{f}''\simeq t^{-5/2}\in L^1$ , so $x^2f(x)\in C_0$ . The other function is of the same type, the extra factor $\sin t$ for large $t$ produces essentially a shift in $g$ , and $\textrm{sgn}(t)$ amounts to an extra Hilbert transform, which won't affect the local singularities here, so $g\in L^1$ also.
|
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|
3,135,798 |
Is there a math equivalent of the ternary conditional operator as used in programming? a = b + (c > 0 ? 1 : 2) The above means that if $c$ is greater than $0$ then $a = b + 1$ , otherwise $a = b + 2$ .
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From physics, I'm used to seeing the Kronecker delta , $$
{\delta}_{ij}
\equiv
\left\{
\begin{array}{lll}
1 &\text{if} & i=j \\
0 &\text{else}
\end{array}
\right. _{,}
$$ and I think people who work with it find the slightly generalized notation $$
{\delta}_{\left[\text{condition}\right]}
\equiv
\left\{
\begin{array}{lll}
1 &\text{if} & \left[\text{condition}\right] \\
0 &\text{else}
\end{array}
\right.
$$ to be pretty natural to them. So, I tend to use $\delta_{\left[\text{condition}\right]}$ for a lot of things. Just seems so simple and well-understood. Transforms: Basic Kronecker delta: To write the basic Kronecker delta in terms of the generalized Kronecker delta, it's just $$
\delta_{ij}
\Rightarrow
\delta_{i=j}
\,.$$ It's almost the same notation, and I think most folks can figure it out pretty easily without needing it explained. Conditional operator: The " conditional operator " or " ternary operator " for the simple case of ?1:0 : $$
\begin{array}{ccccc}
\boxed{
\begin{array}{l}
\texttt{if}~\left(\texttt{condition}\right) \\
\{ \\
~~~~\texttt{return 1;} \\
\} \\
\texttt{else} \\
\{ \\
~~~~\texttt{return 0;} \\
\}
\end{array}
~} &
\Rightarrow &
\boxed{~
\texttt{condition ? 1 : 0}
~} &
\Rightarrow &
\delta_{\left[\text{condition}\right]}
\end{array}
_{.}
$$ Then if you want a non-zero value for the false -case, you'd just add another Kronecker delta, $\delta_{\operatorname{NOT}\left(\left[\text{condition}\right]\right)} ,$ e.g. $\delta_{i \neq j} .$ Indicator function: @SiongThyeGoh's answer recommended using indicator function notation . I'd rewrite their example like $$
\begin{array}{ccccc}
\underbrace{a=b+1+\mathbb{1}_{(-\infty, 0]}(c)}
_{\text{their example}}
&
\Rightarrow &
\underbrace{a=b+1+ \delta_{c \in \left(-\infty, 0\right]}}
_{\text{direct translation}} &
\Rightarrow &
\underbrace{a=b+1+ \delta_{c \, {\small{\leq}} \, 0}}
_{\text{cleaner form}}
\end{array}
\,.
$$ Iverson bracket: Iverson bracket notation , as suggested in @FredH's answer , is very similar to the generalized Kronekcer delta. For example: $$
\delta_{i=j}
~~ \Rightarrow ~~
\left[i = j \right]
\,.$$ Dropping the $`` \delta "$ reduces backwards-compatibility withe the basic Kroncker delta, plus it weakens the signal about what the notation means, so it's probably not as good in general contexts right now. However, Iverson bracket notation should be easier to read and write, so when reinforcing the meaning of the notation isn't a big issue, it could be preferable. Note: " Conditional operator " rather than " ternary operator ". The conditional operator, condition ? trueValue : falseValue , has 3 arguments, making it an example of a ternary operator. By contrast, most other operators in programming tend to be unary operators (which have 1 argument) or binary operators (which have 2 arguments). Since the conditional operator is fairly unique in being a ternary operator, it's often been called " the ternary operator ", leading many to believe that that's its name. However, " conditional operator " is more specific and should generally be preferred.
|
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|
3,139,486 |
We refer to complex numbers as numbers. However we refer to vectors as arrays of numbers. There doesn’t seem to be anything that makes one more numeric than the other. Is this just a quirk of history and naming or is there something more fundamental?
|
They're called "numbers" for historical reasons, since the motivation in the development of the complex numbers was solving polynomial equations. They were viewed as natural extensions of the real numbers. It's somehow quite natural and satisfying to say "every polynomial equation can be solved by some (complex) number ". Is it more natural to regard $i$ as being a number which, when squared, is equal to $-1$ , or is it more natural to regard $i$ as being some non-number thingamajig which when squared is equal to $-1$ ? Clearly the former. "Higher" number systems, like quaternions, aren't really called numbers very often, for the simple fact that they are not as intimately connected with number theory and analysis in the same way that complex numbers are. Beyond these social conventions, I can't see any other reason. The word "number" doesn't have a strict or absolute definition in pure mathematics. $\mathbb{N}$ , $\mathbb{Z}$ , $\mathbb{Q}$ , $\mathbb{R}$ and $\mathbb{C}$ are technically just sets with a certain algebraic structure. I partially disagree with the other answers which claim that complex numbers are numbers simply by virtue of the fact that you can add and multiply them. Well, if that's the rationale, is every ring also a set of numbers?
|
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|
3,140,581 |
Is there a base $b$ such that: $$\log_b x = x $$ (The only one that comes to mind would be the invalid case of $\log_1 1 = 1 $ .) I'm fairly certain the answer is no , but I can't find a clear justification for it. (I don't have a strong mathematical background so an answer with the intuition would be much more helpful than any complex theorem proof.)
|
For a function to be a logarithm, it should satisfy the law of logarithms: $\log ab = \log a + \log b$ , for $a,b \gt 0$ .
If it were the identity function, this would become $ab = a + b$ , which clearly is not always true.
|
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|
3,141,819 |
I've seen a few conflicting pieces of information online. So far, I know that with real coefficients there will always be one real root. But how about with complex coefficients? At very least could you give me a counterexample? A cubic with no real roots.
|
One of the best things you can remember is that over a field (like the reals or complex numbers) roots come from linear factors. Use this to build your own examples: $f(z) =(z-i)^3$ . If you want three distinct complex roots, do something like $f(z) = (z-i)(z+i)(z-2i)$ .
|
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|
3,142,713 |
I've been self-studying from the amazing "Engineering Mathematics" by Stroud and Booth, and am currently learning about algebra, particularly logarithms. There is a question which I don't understand who they've solved. Namely, I'm supposed to express the following equations without logs: $$\ln A = \ln P + rn$$ The solution they provide is: $$A = Pe^{rn}$$ But I absolutely have no idea how they got to these solutions. (I managed to "decipher" some of the similar ones piece by piece by studying the rules of logarithms).
|
The basic idea behind all basic algebraic manipulations is that you are trying to isolate some variable or expression from the rest of the equation (that is, you are trying to "solve" for $A$ in this equation by putting it on one side of the equality by itself). For this particular example (and indeed, most questions involving logarithms), you will have to know that the logarithm is "invertible"; just like multiplying and dividing by the same non-zero number changes nothing, taking a logarithm and then an exponential of a positive number changes nothing. So, when we see $\ln(A)=\ln(P)+rn$ , we can "undo" the logarithm by taking an exponential. However, what we do to one side must also be done to the other, so we are left with the following after recalling our basic rules of exponentiation: $$
A=e^{\ln(A)}=e^{\ln(P)+rn}=e^{\ln(P)}\cdot e^{rn}=Pe^{rn}
$$
|
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|
3,149,368 |
Very short question. Could you please explain me why $$\sum_{i=0}^{n-1} a = na$$ with $a$ a constant?
I know that $$\sum_{i=1}^{n} a = na$$ but in my case the sum starts from zero and finishes for $(n-1)$ . Thanks.
|
In both cases – $$\sum_{i=0}^{n-1}a\quad \text { and }\quad\sum_{i=1}^{n}a$$ – there are exactly $n$ summands.
|
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|
3,149,428 |
My son gave me the following recurrence formula for $x_n$ ( $n\ge2$ ): $$(n+1)(n-2)x_{n+1}=n(n^2-n-1)x_n-(n-1)^3x_{n-1}\tag{1}$$ $$x_2=x_3=1$$ The task I got from him: The sequence has an interesting property, find it out. Make a conjecture and prove it. Obviously I had to start with a few values and calculating them by hand turned out to be difficult. So I used Mathematica and defined the sequence as follows: b[n_] := b[n] = n (n^2 - n - 1) a[n] - (n - 1)^3 a[n - 1];
a[n_] := a[n] = b[n - 1]/(n (n - 3));
a[2] = 1;
a[3] = 1; And I got the following results: $$ a_4=\frac{7}{4}, \ a_5=5, \ a_6=\frac{121}{6}, \ a_7=103, \ a_8=\frac{5041}{8}, \ a_9=\frac{40321}{9}, \\ a_{10}=\frac{362881}{10}, \ a_{11}=329891, \ a_{12}=\frac{39916801}{12}, \ a_{13}=36846277, \ a_{14}=\frac{6227020801}{14}\dots$$ Numbers don't make any sense but it's strange that the sequence produces integer values from time to time. It's not something that I expected from a pretty complex definition like (1). So I decided to find the values of $n$ producing integer values of $a_n$ . I did an experiment for $2\le n \le 100$ : table = Table[{i, a[i]}, {i, 2, 100}];
integers = Select[table, (IntegerQ[#[[2]]]) &];
itegerIndexes = Map[#[[1]] &, integers] ...and the output was: {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53,
59, 61, 67, 71, 73, 79, 83, 89, 97} Conjecture (pretty amazing, at least to me): $a_n$ is an integer if and only if $n$ is prime. Interesting primality test, isn't it? The trick is to prove that it's correct. I have started with the substitution: $$y_n=n x_n$$ ...which simplifies (1) a bit: $$(n-2)y_{n+1}=(n^2-n-1)y_n-(n-1)^2y_{n-1}$$ ...but I did not get much further (the next step, I guess, should be rearrangement).
|
The given difference equation can be solved in the following way. We have for $n\ge 3$ , $$
(n-2)(y_{n+1}-y_n) = (n-1)^2(y_n-y_{n-1}),\\
\frac{y_{n+1}-y_n}{n-1}=(n-1)\frac{y_{n}-y_{n-1}}{n-2}.
$$ If we let $\displaystyle z_n=\frac{y_{n}-y_{n-1}}{n-2}$ , it follows that $$
z_{n+1}=(n-1)z_n=(n-1)(n-2)z_{n-1}=\cdots =(n-1)!z_3=(n-1)!\frac{3x_3-2x_2}{1}=(n-1)!
$$ This gives $$
y_{n+1}-y_n=(n-1)(n-1)!=n!-(n-1)!,
$$ hence $y_n = nx_n = (n-1)!+c$ for some $c$ . Plugging $n=2$ yields $2=1!+c$ , thus we have that $$\displaystyle x_n =\frac{(n-1)!+1}{n}.$$
|
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|
3,149,432 |
I have three cones in $\mathbb{R}^3$ , explicitly defined by the equations: $$
(x-\alpha_x)^2+(y-\alpha_y)^2=(z-r_1)^2 \,, \\
(x-\beta_x)^2+(y-\beta_y)^2=(z-r_2)^2 \,, \\
(x-\gamma_x)^2+(y-\gamma_y)^2=(z-r_3)^2 \,. \tag{1}
$$ The parameters obey the conditions: $$
\alpha_x^2+\alpha_y^2 = \beta_x^2+\beta_y^2 = \gamma_x^2+\gamma_y^2 \leq 1 \,, \\
r_1+r_2+r_3 = 1 \,, \quad 0<r_1,r_2,r_3<1 \,.
$$ Side question 1: By demanding $z>c$ with $0<c<1$ , what are the conditions such that the cones have a unique point of intersection? What is the solution? What if I add more cones into the game? Assume now that a unique solution exists and let's denote it with $x_0, y_0, z_0$ . Let $\kappa \in [0,1]$ denote a parameter and consider the closely related problem $$
(x-\kappa\alpha_x)^2+(y-\kappa\alpha_y)^2=(z-r_1)^2 \,, \\
(x-\kappa\beta_x)^2+(y-\kappa\beta_y)^2=(z-r_2)^2 \,, \\
(x-\kappa\gamma_x)^2+(y-\kappa\gamma_y)^2=(z-r_3)^2 \,. \tag{2}
$$ That is, the projection on the $x-y$ plane of the vectors to the origins of the cones have been rescaled; or equivalently the origins of the cones are moved towards the $z$ axis by the same amount (the distance of the origins to the $z$ axis is the same for all three cones, which follows from the conditions on the parameters). This implies that in fact the origins of the cones have come closer to each other. Main Question : Assuming that solutions exist for all $\kappa$ (or at least for some subset of [0,1]), is there a way to relate the solutions of the second problem to the solutions of the first. That is, can I obtain $$
x_\kappa = f_1(\kappa, x_0, y_0,z_0) \,, \\
y_\kappa = f_2(\kappa, x_0, y_0,z_0) \,, \\
z_\kappa = f_3(\kappa, x_0, y_0,z_0) \,.
$$ Similarly, can one obtain the curve traced by the points of unique intersection with varying $\kappa$ ? Side Question 2: Are there any complications when I go to $\mathbb{R}^4$ or is it a direct generalisation? Regarding the main question, for some specific parameters I have numerically solved the problem and found a curve that looks like in the following graphic, which confirms my suspicion that at least in certain cases a curve like that exists.
|
The given difference equation can be solved in the following way. We have for $n\ge 3$ , $$
(n-2)(y_{n+1}-y_n) = (n-1)^2(y_n-y_{n-1}),\\
\frac{y_{n+1}-y_n}{n-1}=(n-1)\frac{y_{n}-y_{n-1}}{n-2}.
$$ If we let $\displaystyle z_n=\frac{y_{n}-y_{n-1}}{n-2}$ , it follows that $$
z_{n+1}=(n-1)z_n=(n-1)(n-2)z_{n-1}=\cdots =(n-1)!z_3=(n-1)!\frac{3x_3-2x_2}{1}=(n-1)!
$$ This gives $$
y_{n+1}-y_n=(n-1)(n-1)!=n!-(n-1)!,
$$ hence $y_n = nx_n = (n-1)!+c$ for some $c$ . Plugging $n=2$ yields $2=1!+c$ , thus we have that $$\displaystyle x_n =\frac{(n-1)!+1}{n}.$$
|
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|
3,151,152 |
Four married couples attend a party. Each person shakes hands with every other person, except their own spouse, exactly once. How many handshakes? My book gave the answer as $24$ . I do not understand why. I thought of it like this: You have four pairs of couples, so you can think of it as M1W2, M2W2, M3W3, M4W4, where
M is a man and W is a woman. M1 has to shake 6 other hands, excluding his wife. You have to do this 4 times for the other men, so you have $4\times 6$ handshakes, but in my answer, you are double counting. How do I approach this problem?
|
$8$ people. Each experiences handshakes with $6$ people. There are $6\times 8=48$ experiences of handshakes. Each handshake is experienced by two people so there $48$ experiences means $48\div 2=24$ handshakes.
|
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|
3,151,946 |
This is a problem from the AMC 8 (math contest): A certain calculator has only two keys $[+1]$ and $[\times 2]$ . When you press one of the keys, the calculator automatically displays the result. For instance, if the calculator originally displayed “ $9$ ” and you pressed $[+1]$ , it would display “ $10$ ”. If you then pressed $[\times 2]$ , it would display “ $20$ ”. Starting with the display “ $1$ ”, what is the fewest number of keystrokes you would need to reach “ $200$ ”? Intuitively I worked back from $200$ , dividing by $2$ until I reached an odd number, subtracting $1$ when I did, etc..to reach the correct answer of $9$ steps. However, I can't figure out how to convince myself beyond any doubt that it is the optimal solution. In other words, I can't prove it mathematically.
The best I can come up with is that beyond the first step from $1$ to $2$ , multiplication by $2$ is always going to yield a larger step than addition by $1$ and therefore I should take as many $[\times 2]$ steps as I can. This doesn't feel rigorous enough, though. EDIT: Just to be clear, I am asking for a proof or at least more rigorous explanation of why this is the optimal solution.
|
Look at what the operations $[+1]$ and $[\times 2]$ do to the binary expansion of a number: $[\times 2]$ appends a $0$ , and increases the length by one, leaving the total number of $1$ 's unchanged; if the final digit is $0$ , then $[+1]$ increases the number of $1$ 's by one, but doesn't change the length; if the final digit is $1$ , then $[+1]$ doesn't increase the total number of $1$ 's (it may in fact decrease it), and doesn't increase the total length by more than $1$ . Therefore, with a single key press: you can increase the length by one, but this won't increase the number of $1$ 's; you can increase the number of $1$ 's by one, but this won't increase the length. The binary expansion of $200$ is $200_{10}=11001000_2$ . This has three $1$ 's, and a length of eight. Starting from $1$ , we must increase the length by seven, and increase the number of $1$ 's by two. So this requires at least nine steps.
|
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|
3,152,962 |
How to know which line represents tangent to a curve $y=f(x)$ (in RED) ?From the diagram , I cannot decide which line to take as tangent , all seem to touch at a single point.
|
One of the defining points (no pun!) of a tangent is the idea that as you look at a smaller and smaller section of the curve, the curve starts to more and more looks like a straight line . That behaviour lets you do a huge number of things: You can estimate where the curve goes in places near that point (if tangent has a gradient of 2, then the curve at X + 0.000001 will probably be close to Y+0.000002, or whatever). You can approximate the curve by tiny sections of straight lines - this is the basis of simple calculus. You can take limits in several ways, and expect them to converge as you use narrower sections of the curve ? Many other things, all stemming from these. Not all curves behave like that, and yours doesn't. No matter how close you inspect the cusp of your curve (the pointy bit), it's never going to resemble anything like a straight line, on any scale. It'll always look like a cusp. That's the fundamental reason there isn't a tangent at that point. Because the curve just doesn't resemble a straight line, even in a microscopic close-up view, it doesn't have a gradient or tangent at that point, and it isn't differentiable at that point (much the same thing at a very simple level), and so on. That isn't in fact unusual. In fact more curves don't have a gradient than do - its just that we don't study curves at random, so you mainly look at curves that do - at least, at this level of maths! Other examples of curves that don't have a gradient at some or all points - the "step" curve (defined as y=0 if x<0, y=1 if x >= 0) a "curve" defined as y=1 if x contains a "1" when written in decimal, and y=0 otherwise. Because there are infinitely many numbers that simply don't have a 1 in their decimal expansion, slotted in between those that do! the curve y = 1/x at x=0 the " blancmange curve ", which doesn't have any breaks in it, and looks like it should be a nice straightforward curve, but actually "wobbles" so much at any microscopic level (however closely you look at it,and wherever you look!) that in the end, it doesn't have a gradient anywhere.
|
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|
3,153,739 |
I've read and heard in lectures that A way to prove that the Riemann hypothesis is true is to show that its negation is not provable. The argument (informally) usually goes like If a statement is false, then there must exist a counterexample showing its falsity. Hence, to prove any statement is false, one must have a constructive proof. Question : Why doesn't Godel's incompleteness theorem apply to false statements? That is, how do we know that all false statements are provably so?
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That is, how do we know that all false statements are provably so? This is simply wrong. There are both true and false statements that cannot be proven. What is true is that any sufficiently nice foundational system (i.e. one that has a proof verifier program and can reason about finite program runs) is $Σ_1$ -complete , meaning that it proves every true $Σ_1$ -sentence. Here, a $Σ_1$ -sentence is an arithmetical sentence (i.e. quantifies only over $\mathbb{N}$ ) that is equivalent to $∃k∈\mathbb{N}\ ( Q(k) )$ for some arithmetical property $Q$ that uses only bounded quantifiers. For example, "There is an even number that is not the sum of two primes." can be expressed as a $Σ_1$ -sentence. The " $Σ_1$ " stands for " $1$ unbounded existential". Similarly a $Π_1$ -sentence is an arithmetical sentence equivalent to one with only $1$ unbounded universal quantifier in Skolem normal form. In general, if you have a $Π_1$ -sentence $C ≡ ∀k∈\mathbb{N}\ ( Q(k) )$ , then $¬C$ is a $Σ_1$ -sentence. Thus if $C$ is false, $¬C$ is true and hence provable in any sufficiently nice foundational system by $Σ_1$ -completeness . This does not apply to all false sentences! It turns out that non-trivially RH (Riemann Hypothesis) is equivalent to a $Π_1$ -sentence , and hence by the above we know that if it is false then even PA (Peano Arithmetic) can disprove it. Also, I should add that no expert believes that it would be any easier to prove unprovability of RH over PA than to directly disprove RH, even if it is false in the first place. Godel's incompleteness theorem has completely nothing to do with $Σ_1$ -completeness. In fact, the generalized incompleteness theorem shows that any sufficiently nice foundational system (regardless of what underlying logic it uses) necessarily is either $Π_1$ -incomplete or proves $0=1$ . That is, if it is arithmetically consistent (i.e. does not prove $0=1$ ) then it also does not prove some true $Π_1$ -sentence. Moreover, we can find such a sentence uniformly and explicitly (as described in the linked post).
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3,161,647 |
I Just want to know the name for this if there is one because I don't think it satisifies any of the formal definitions for sets, n-tuples, sequences, combinations, permutations, or any other enumerated objects I can think of. For convenience, I will henceforth use the term $\mathbf{\ set^*}$ with an asterisk to refer to what I described in the title. As a quick example, let $\mathbf{A}$ and $\mathbf{B\ }$ be $\mathbf{\ set^*}$ 's where $$\mathbf{A = \{3,3,4,11,4,8\}}$$ $$\mathbf{B = \{4,3,4,8,11,3\}}$$ Then $\mathbf{A\ }$ and $\mathbf{\ B\ }$ are equal.
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If you're looking for something like a set which may have repeated elements, standard terms are multiset or bag . See multiset on wikipedia.
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3,164,020 |
Recently I was browsing math Wikipedia, and found that Harald Helfgott announced the complete proof of the weak Goldbach Conjecture in 2013, a proof which has been accepted widely by the math community, but according to Wikipedia hasn’t been accepted by any major journals. My first question: Has Helfgott’s proof been verified as of now? Why hasn’t it been published in a peer-reviewed journal yet (or has it and I’m just ignorant)? Secondly, I found that he announced his proof the same day Yitang Zhang announced his result of the 70,000,000-prime bound (a remarkable coincidence indeed). Zhang’s result got a lot of coverage, from Numberphile (who made like 5 videos about it compared to like 1 video about the Goldbach conjecture mentioning Helfgott in a passing comment), to science newspapers/magazines, to Terry Tao, James Maynard, and the polymath project. I mean, his work made it to the Annals of Mathematics in Princeton! Comparatively, I found very low coverage on Helfgott’s result, and it seems like people rank the importance of Zhang’s result above Helfgott’s in math ranking sites, such as https://www.mathnasium.com/top-10-mathematical-achievements-in-past-5-years , which explicitly gives the top spot to Zhang with “no surprise”. Also as I’ve mentioned before I don’t think he published in a major journal compared to Zhang who published in the Annals. Second question: Why did Helfgott’s proof produce less of a stir in the math community than Zhang’s work? Was Helfgott’s work not groundbreaking enough? (Is it perhaps because of the fact that Vinogradov had already proven the weak Goldbach Conjecture for sufficiently large numbers in 1937 and Helfgott “simply” lowered the bound, whereas Zhang’s work shrank the bound from infinity to a finite quantity? Still wouldn’t Helfgott’s work deserve publication in a peer-reviewed journal?)
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Update: I'm making most of the current version of the book publicly accessible. Comments and other feedback are much appreciated! Just a few remarks so as to keep everybody informed. (I came across this page
by chance while looking for something else.) As far as I know, nobody has found any serious issues with the proof.
(There was a rather annoying but non-threatening error that I found in section 11.2 and fixed myself, and of course some typos and slips here are there; none affect the overall strategy or the final result.) A manuscript containing the full proof was accepted for publication at
Annals of Mathematics Studies back in 2015. I was asked to rewrite matters
fairly substantially for expository reasons, though the extent of the revisions
was left up to my discretion. Publishing a lengthy proof (about 240 pages in its shortest complete version,
which was considered too terse by some) is never trivial. Publishing it in top
journals, where the backlog is often very large, is even more complicated.
(Many thanks are due to the editors of a top journal -- which does often
publish rather long articles -- for their candid description of complicated decisions in the
editorial process.) I was thus delighted when the manuscript was accepted for
publication in Annals of Mathematics Studies, which publishes book-length
research monographs. A very detailed referee report was certainly helpful; it was as detailed as
one could reasonably ask from a single author. At the same time, I felt that
it would be best for everybody if there were a second round of refereeing,
with individual referees taking care of separate chapters. So, I asked the publishers for such a second round, and they graciously accepted. One of the (first-round) referees had suggested that I treat the manuscript as a draft to be fairly thoroughly restructured, and that I add several introductory
chapters. While I found the request a little overwhelming at first, and while
the editors did not demand as much of me, I became convinced that the referee
was right, and set about the task. What follows is a long, still not quite finished story of a process that took
longer than expected, in part due to my commitments to other projects, in part
perhaps due to a certain perfectionism on my part, in part due to publishing
mishaps that you definitely do not want to hear about, and above all because
it became clear to me, not only that the proof had had fairly few
thorough readers, but that it would be worthwhile for it to have a
substantially wider readership. To expand on what has been said by other people who replied to or commented
on the original poster's question: knowing that ternary Goldbach holds for all
even integers $n\geq 4$ is not likely to have very many applications, though
it does have some. In that sense it may be seen as the end of a road. The
further use of the proof will reside mainly in the techniques that had to be
applied, developed and sharpened for its sake. For that matter, the same is
arguably true of Vinogradov's work -- it arguably brought the circle method to
its full maturity, after the foundational work of Hardy, Littlewood and
Ramanujan, besides showing the power that combinatorial identities can have in
work on the primes. From that perspective, it makes sense for the proof to be published as a book
that, say, a graduate student, or a specialist in a neighboring field,
can read with profit. Of course it is still fair and necessary to assume that the reader has taken the equivalent of a first graduate course in analytic number theory. In the current version, the first hundred pages are taken by an
introduction and by chapters on what can be called the basics of analytic
number theory from an explicit and computational viewpoint. Then come 40 pages on further
groundwork on the estimation of common sums in analytic number theory
- sums over primes, sums of $\mu(n)$ , sums of $\mu^2(n)/\phi(n)$ , etc. (I should
single out the contributions of O. Ramaré to the explicit understanding of
sums of $\mu(n)/n$ and $\mu^2(n)/\phi(n)$ as invaluable.) Then there
are close to 120 pages on improvements or generalizations on various versions
of the large sieve, their connection to the circle method, and also
on an upper-bound quadratic sieve. (This last subject got a little too
interesting at some point; I am glad my treatment is done!) Then comes an
explicit treatment of exponential sums, in some sense the core of the proof.
(The smoothing function used here has been changed from that in the original
version.) Then comes the truly complex-analytic part. I am editing that part a little
so that people who are not interested mainly in ternary Goldbach
will be able to take what they need on parabolic cylinder functions, the
saddle-point method or explicit formulas (explicit explicit formulas?). Then
comes the part where different smoothing functions have to be chosen - again,
I am currently editing so that others can readily pick up ideas that probably
have wider applicability. The calculations that are needed for the ternary
Goldbach problem and no other purpose take fewer than 20 pages at the end. I believe I can say the heavy part is mostly over; I am currently
doing some editing on the second half (or rather the last two fifths) of the
book while waiting to hear from several of the second-round referees I
requested myself. Of course I am also working on other things as well. All being said, I would not necessarily recommend any non-masochist friend to
write a book-length monograph in the future -- though some other people seem to
manage -- not just because the time things take seems to be quadratic on the
length of the text, which itself increases monotonically, but also because it
is frustrating that it is hard to post periodic updates (certainly
harder than for independent papers), in that always some part of the whole is
undergoing construction. At the same time, I hope to be happy with the end
result.
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3,164,182 |
I've done some research, and I'm hoping someone can check me. My question was this: Assume I have the function $f(x) = \frac{(x-3)(x+2)}{(x-3)}$ , so it has removable discontinuity at $x = 3$ . We remove that discontinuity with algebra: $f(x) = \frac{(x-3)(x+2)}{(x-3)} = (x+2)$ . BUT, the graph of the first function has a hole at $x = 3$ , and the graph of the second function is continuous everywhere. How can they be "equal" if one has a hole and the other does not? I think that this is the answer: Because the original function is undefined at the point $x = 3$ , we have to restrict the domain to $\mathbb{R} \setminus 3$ . And when we manipulate that function with algebra, the final result, $f(x) = (x + 2)$ is still using this restricted domain. So even though the function $f(x) = (x+2)$ would not have a hole if the domain were all of $\mathbb{R}$ , we are sort of "imposing" a hole at $x = 3$ by continuing to throw that point out of the domain. And then just to close the loop: Removing the removable discontinuity is useful because it allows us to "pretend" that we're working with a function that is everywhere continuous, which helps us easily find the limit. But the reality is that the function $f(x) = (x +2)$ is actually NOT continuous everywhere when we restrict the domain by throwing out the point 3. Or am I now taking things too far? Thanks in advance! EDIT: For anyone coming across this in the future, in addition to the excellent answers below, I also found this other question about the continuity of functions with removable discontinuities helpful.
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Two functions are typically defined to be equal if and only if they... Share the same domain Share the same codomain Take on the same values for each input. Thus, functions $f,g : S \to T$ for sets $S,T$ have $f=g$ if and only if $f(x) = g(x)$ for all $x$ in $S$ . For functions with holes, we typically restrict the domain by ensuring the values where the function is not defined at not included. For example, in the functions you have, you have $$f(x) = \frac{(x-3)(x+2)}{(x-3)} \;\;\;\;\; g(x) = x+2$$ Are these equal? Yes, and no. A function must be defined at all values of the domain. Thus, we can say $3$ is not in the domain of $f$ for sure. But we never specified otherwise the domains and codomains of these functions! Typically, unless stated otherwise, we often assume their domain to be $\Bbb R$ or $\Bbb C$ , minus whatever points are causing problems - and of course, in such cases, $f \neq g$ since $f(3)$ is not defined, and thus $f$ normally has domain $\Bbb R \setminus \{3\}$ and $g$ generally has domain $\Bbb R$ . But that restriction is not necessary. For example, we could define the functions to be $f,g : \Bbb R \setminus \Bbb Q \to \Bbb R$ . Notice that the domain of both functions are now all real numbers except rational numbers, i.e. the irrational numbers. This means $3$ is not in the domain of either function - and since that's the only "trouble spot," and the codomains are equal, and the values are equal at each point in the domain, $f=g$ here. Or even more simply: we could have $\Bbb R \setminus \{3\}$ be the domain of $f$ and $g$ and again have equality! The key point in all this is that, just because $f$ or $g$ do attain defined values for certain inputs, doesn't mean they have to be in the domain. In short, whether $f=g$ depends on your definitions of each. Under typical assumptions, $f \neq g$ in this case, but if we deviate from those assumptions even a little we don't necessarily have inequality.
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3,171,071 |
Ok I am in grade 9 and I am maybe too young for this. But I thought about this, why dividing by $0$ is impossible. Dividing by $0$ is possible would mean $1/0$ is possible, which would mean $0$ has a multiplicative inverse. So if we multiply a number by $0$ then by $1/0$ we get the same number. But that's impossible because all numbers multiplied by $0$ give $0$ therefore we can’t have an inverse for $0$ , as that gives us the initial number and thus division by $0$ is impossible Is this right?
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That's the most basic reason that division by $0$ is usually considered to be a Bad Thing, yes. Because if we did allow dividing by $0$ , we would have to give up at least of one of the following things (these are usually considered Very Nice): What $1$ means ( $1\cdot a = a$ for any $a$ ) What $0$ means ( $0 \cdot a = 0$ for any $a$ ) (actually a consequence of $0+a=a$ and $(a+b)\cdot c=a\cdot c+a\cdot b$ , two other Nice Things) What division means ( $\frac ab = c$ means $a = c\cdot b$ ) $0\neq1$
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3,174,227 |
Suppose that we consider English alphabets so we have 26 letters which 5 of them are vowels. I want to find all three letters sequences that contain at least one vowel. My first approach is $26^3-21^3=8315$ which is the number of all three lettres sequences minus the number of three letter sequences which do not contain vowels. Second approach: at least one vowel means one vowel or two vowels or three vowels so the answer is $(5\cdot21^2)+(5^2\cdot21)+5^3=2855$ . Why are these two answers different?
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Take the number of all three letter words, $26^3$ , and subtract from it the number that have only consonants, $21^3$ . Your second method does not address all possible sequences of vowels, such as vowel-consonant-vowel or vowel-vowel-consonant or consonant-vowel-vowel...
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3,174,538 |
What is the intuition behind short exact sequences of groups; in particular, what is the intuition behind group extensions? I'm sorry that the definitions below are a bit haphazard but they're how I learnt about them, chronologically. In Johnson's "Presentation $\color{red}{s}$ of Groups," page 100, there is the following . . . Definition 1: A diagram in a category $\mathfrak{C}$ , which consists of objects $\{A_n\mid n\in\Bbb Z\}$ and morphisms $$\partial_n: A_n\to A_{n+1}, n\in \Bbb Z,\tag{6}$$ is called a sequence in $\mathfrak{C}$ . Such a sequence is called exact if $$\operatorname{Im}\partial_n=\ker \partial_{n+1},\,\text{ for all }n\in \Bbb Z$$ [. . .] A short exact sequence in the category $\mathfrak{C}_{\Bbb R}$ of right $\Bbb R$ -modules is an exact sequence of the form $(6)$ with all but three consecutive terms equal to zero. [. . .] Also, ibid. , page 101, is this: It is fairly obvious that a sequence $$0\longrightarrow A\stackrel{\theta}{\longrightarrow}B\stackrel{\phi}{\longrightarrow}C\longrightarrow 0$$ is a short exact sequence if and only if the following conditions hold: $\theta$ is one-to-one, $\phi$ is onto, $\theta\phi=0$ , $\ker \phi\le\operatorname{Im}\theta$ . I'm reading Baumslag's "Topics in Combinatorial Group Theory" . Section III.2 on semidirect products starts with Let $$1\longrightarrow A\stackrel{\alpha}{\longrightarrow}E\stackrel{\beta}{\longrightarrow}Q\longrightarrow 1$$ be a short exact sequence of groups. We term $E$ an extension of $A$ by $Q$ . Thoughts: I'm aware that semidirect products can be seen as short exact sequences but this is not something I understand yet. My view of semidirect products is as if they are defined by a particular presentation and my go-to examples are the dihedral groups. Please help :)
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A short exact sequence $1\rightarrow A\rightarrow E\rightarrow Q\rightarrow1$ is really just a fancy way of saying " $E$ has a normal subgroup $A$ where $E/A\cong Q$ ". [The sequence also gives the isomorphism $\beta: E/A\rightarrow Q$ , while $\alpha$ corresponds to the embedding of the abstract group $A$ as a subgroup of $E$ .] Because you care about presentations: if $A$ has presentation $\langle \mathbf{x}\mid\mathbf{r}\rangle$ and $Q$ has presentation $\langle \mathbf{y}\mid\mathbf{s}\rangle$ then the group $E$ given by the above short exact sequence has presentation of the form: $$
\langle \mathbf{x, y}\mid SW_S^{-1} (S\in\mathbf{s}), \mathbf{r}, \mathbf{t}\rangle
$$ where $W_{S}\in F(\mathbf{x})$ for all $S\in\mathbf{s}$ , and $\mathbf{t}$ consists of words of the form $y^{-\epsilon}xy^{\epsilon}X^{-1}$ with $x\in\mathbf{x}$ , $y\in\mathbf{y}$ and $X\in F(\mathbf{x})$ . The intuition here is that relators in $\mathbf{t}$ ensure normality of $A$ , and so removing all the $x$ -terms makes sense. When they are removed you get the presentation $\langle \mathbf{y}\mid\mathbf{s}\rangle$ , because of the relators $SW_S^{-1}$ . I will leave you to work out where the maps $\alpha$ and $\beta$ fold in to this description. The above presentation justifies the term extension of $A$ by $Q$ : we've started with a presentation for $A$ , and then added in the presentation for $Q$ in a specific way to obtain a presentation for $E$ . For a worked example of the above (with some genuinely amazing applications, both in the paper and in subsequent research), look at the paper Rips, E. (1982), Subgroups of small Cancellation Groups . Bull. Lond. Math. Soc. 14: 45-47. doi: 10.1112/blms/14.1.45 It is an interesting question when a presentation of the above form does actually define a group extension. This was studied in the paper Pride, S., Harlander, J. & Baik, Y. (1998). The geometry of group extensions . J. Group Theory, 1(4), pp. 395-416. doi: 10.1515/jgth.1998.028
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3,175,604 |
Is there an easy example of one and the same space $X$ with two different metrics $d$ and $e$ such that one and the same sequence $\{x_n\}$ is a Cauchy sequence for both metrics, but converges only for one of them?
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Let $X=[0,\infty)$ , let $d(x,y)=\lvert x-y\rvert$ and let $$e(x,y)=\begin{cases}\lvert x-y\rvert&\text{ if }x,y\neq0\\\lvert x+1\rvert&\text{ if }x\neq0\text{ and }y=0\\\lvert y+1\rvert&\text{ if }x=0\text{ and }y\neq0\\0&\text{ if }x,y=0.\end{cases}$$ Then the sequence $\left(\frac1n\right)_{n\in\mathbb N}$ is a Cauchy sequence with respect to both metrics, but it converges only in $(X,d)$ .
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3,184,449 |
The obvious solution of Lattitude & Longitude doesn't work because it generates points more densely near the poles, and the other thing I came up with (Pick a random point in the unit cube, if it's in the unit sphere map it to the surface, and restart if it's outside) doesn't always find a point within a fixed number of tries.
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The Lambert cylindrical equal area projection maps the sphere to a cylinder, area to equal area. It is easy to generate a uniform distribution on a cylinder. Simply map it back to the sphere. For $(u_1,u_2)$ uniform on $[0,1]^2$ , either $\mathrm{lat}=\arcsin(2u_1-1),\mathrm{lon}=2\pi u_2$ or $z=2u_1-1,x=\sqrt{1-z^2}\cos(2\pi u_2),y=\sqrt{1-z^2}\sin(2\pi u_2)$
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3,185,630 |
You are given a function $f: \mathbb{R}\rightarrow \mathbb{R}$ . Every derivative $\frac{d^n}{dx^n}(f(x)), \,n >0$ of the function is continuous. Is there a function $g: \mathbb{R}\rightarrow \mathbb{R}$ , for which every derivative $\frac{d^n}{dx^n}(g(x)), \,n >0$ is also continuous, such that: $$\forall x\in[a,b]: \, g(x) = f(x)\land \, \exists x \notin [a,b]: f(x) \neq g(x),\, a \neq b$$ Thanks! Edit: I asked the question because I intuitively wondered if there would be functions, which could behave like "the same" in a given interval, but then behave differently so that they start diverging or at least stopped being the same the anymore. The answer to this question gave me a bigger understanding of real analysis. If you would like to know which made me think about such a problem: Although this is a vague formulation, generally, this question asks if two (completely) different things can develop (themselves) to be exactly equal in at least one part of there whole existence...
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Define the real functions $f$ and $g$ thus: $$
f(x) = \begin{cases} \exp\Big(-\frac{1}{(x - 1)^2}\Big)\ &\text{if } x > 1 \\
0\ &\text{if } x \in [-1, 1] \\
\exp\Big(-\frac{1}{(x + 1)^2}\Big)\ &\text{if } x < -1
\end{cases}
$$ and $$
g(x) = 0\quad \text{for all } x \in \mathbb{R}
$$ By construction, $f$ and $g$ are both $0$ on $[-1, 1]$ but they differ in value everywhere else. Obviously $g$ is continuously differentiable infinitely many times as it is a constant function. You can also check that $f$ is continuously differentiable infinitely many times at $x = -1$ and $x = 1$ by computing appropriate right and left hand limits of $\frac{d^nf(x)}{dx^n}$ ( $n \in \mathbb{Z}_+$ ) inductively. Let us compute for example $\lim\limits_{x \to 1^+}\frac{df(x)}{dx}$ : for $x > 1$ , \begin{align*}
0 < \frac{df(x)}{dx} &= \frac{2(x - 1)^{-3}}{\exp\big(\frac{1}{(x - 1)^2}\big)} \\
&= \frac{2y^3}{\exp\big(y^2\big)} \quad\text{ where }y = \frac{1}{x - 1} > 0 \\
&= \frac{2}{\frac{1}{y^3}\sum\limits_{k = 0}^\infty \frac{y^{2k}}{k!}} \quad\text{ divide by }y^3\text{ and use }e^z = \sum\limits_{k = 0}^\infty\frac{z^k}{k!} \\
&= \frac{2}{\frac{1}{y^3}\frac{y^0}{0!} + \frac{1}{y^3}\frac{y^2}{1!} + \frac{1}{y^3}\frac{y^4}{2!} + \sum\limits_{k = 3}^\infty \frac{y^{2k - 3}}{k!}} \\
&< \frac{2}{\frac{1}{y^3} + \frac{1}{y} + \frac{1}{2}y}
\end{align*} As $x \to 1^+$ , we get $y = \frac{1}{x - 1} \to \infty$ and the right hand fraction $\frac{2}{\frac{1}{y^3} + \frac{1}{y} + \frac{1}{2}y} \to 0$ . So, as $x \to 1^+$ , $\frac{df(x)}{dx} \to 0$ also by the Squeeze Theorem . That was a long calculation but take my word: it can be repeated inductively to show that $\lim\limits_{x \to 1+}\frac{d^nf(x)}{dx^n} = 0$ for all $n \in \mathbb{Z}_+!$ At all other points i.e. on $(-\infty, -1) \cup (-1, 1) \cup (1, \infty)$ , $f$ is infinitely differentiable because exponentials and constant functions are infinitely differentiable. Bonus Fact: Both $\frac{d^n f(x)}{dx^n}$ and $\frac{d^n g(x)}{dx^n}$ also have the same value $0$ on $[-1, 1]$ for all positive integers $n$ !
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3,186,480 |
I met the notation $ S=\{(a,b] ; a,b\in \mathbb R,a<b\}\cup\{\emptyset\} $ I know $S$ is a family of subsets ,a set of intervals, and from set theory $\emptyset$ is a subsets of every set then why in the notation : $ S=\{(a,b] ; a,b\in \mathbb R,a<b\}\cup\{\emptyset\} $ appear $\color{red}{\cup\{\emptyset\}}$ ?
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It is because the emptyset $\emptyset$ is a subset of every set, but not an element of every set.
It is $\emptyset\in S$ and you might want that to show, that the elements of $S$ define a topology. Or to be more clear it is $\{1\}\neq\{1,\emptyset\}$ . The set on the left has one element, the set on the right has two elements, with $\emptyset\in\{1,\emptyset\}$
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3,190,922 |
I was browsing through a Wikipedia article about the trigonometric identities, when I came across something that caught my attention, namely forgotten trigonometric functions. The versine (arguably the most basic of the functions), coversine, haversine and exsecant formulas had once been utilised for navigational purposes, prior to GPS tracking systems. However, recently, they have become less common in modern mathematics and beyond. Why is that? Here is a link to a PDF file describing all of these now-obsolete trig functions: The Forgotten Trigonometric Functions, or
How Trigonometry was used in the Ancient Art of Navigation (Before GPS!)
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Those functions are much less used than before for one reason: the advent of electronic computers. Before that, one had to rely either on tables or on slide rules . Tables were usually table of logarithms, and they included the logarithms of trigonometric functions as well. The trigonometric functions were then useful not only for geometric applications, but also to simplify algebraic calculations with logarithms. For instance, to compute $\log\sqrt{a^2+b^2}$ when $\log a$ and $\log b$ are known, you could find $\theta$ such that $\log\tan\theta=\log\frac ba=\log b-\log a$ , then $\log\sqrt{a^2+b^2}=\log a+\log\sqrt{1+\tan^2\theta}$ and $\log\sqrt{1+\tan^2\theta}=\log\frac{1}{\cos \theta}=-\log\cos\theta$ . There are many similar formulas. For geometric applications, sometimes versine and similar functions allow computing with greater precision while not adding too much computation. See for instance the haversine formula used to compute great circle distance (useful in navigation). The straightforward formula with arccosine has poor accuracy when the angle is small (the most common case), due to the fact that cosine is flat at $0$ . However the formula with haversine is more accurate. To achieve the same, you would have to use $\sin^2(\theta/2)$ everywhere, which require more computations (but it's still reasonable with logarithms). Therefore, navigation table like Nories's nautical tables have an haversine table. Note that tables of logarithms are usually accurate up to 5 digits (some larger tables had 7 digits, some very special ones had better precision but were difficult to use : more digits = much more space on paper). Slide rules have roughly 3 digits of precision. All of this is rendered pretty useless with calculators, which have usually around 15 digits of precision and compute fast enough that we don't have to worry about speeding things up with extra functions.
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3,193,067 |
Conic sections seem to get special attention in early math classes. My question is why do these cross sections of cones deserve more attention than those of, say, a rectangular prism, a cube, or some other 3D (or any dimensional) object? I have a couple of guesses: Studying a particular "simple" example can provide insight into the general idea (i.e. cross sections of higher dimensional objects). And conic sections are deemed simple. The applications of ellipses, parabolas, and hyperbolas are just so vast that their graphs and properties deserve special studying (e.g. elliptical orbits). I'd really appreciate some outside thoughts on this, even if it is just speculation. I've been giving cross sections some special study attention recently and have done a handful of google searches to try and understand why conic sections keep coming up (as can be seen in a lot of math curriculum). Thank you!
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One of the things that makes a cone simpler than a cube is that it is an “algebraic object” that can be defined by a simple polynomial identity ( $x^2 + y^2 - z^2 = 0$ ). Taking cross-sections preserves this algebraic nature (since an infinite plane is also algebraic) so we end up with a quadratic curve in two variables, which is a fairly nice object. In some sense these are the “simplest” possible shapes beyond straight lines. At the same time, you are probably aware that the shape of a quadratic curve varies drastically depending on where the minus signs live. The equation of a cone has enough minus signs that it is able to represent pretty much the entire spectrum of such curves, in contrast to, say, a sphere. Lastly, there is some extent to which we study these because they were studied classically by the ancient Greek geometers. This kind of speaks to the utility angle in that there would be more applications of things that are well studied. But the preceding two points show that there are objective (non-historical) reasons to consider conic sections interesting. They are simple enough to be studied very thoroughly, and this simplicity also increases the chances that they would emerge naturally in many situations.
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3,195,581 |
I conjecture that for irrational numbers, there is generally no pattern in the appearance of digits when you write out the decimal expansion to an arbitrary number of terms. So, all digits must be equally likely. I vaguely remember hearing this about $\pi$ . Is it true for all irrational numbers? If not, what about transcendental? If true, how might I go about proving this? For my attempt, I'm not quite sure how to approach this, all I have are some experimental results to validate the conjecture. I started with $\sqrt2$ . The occurrences of the various digits in the first 5,916 decimal terms are: 563, 581, 575, 579, 585, 608, 611, 565, 637, 612. And here are the occurrences of the decimals in the first 1993 digits of $\pi$ : 180, 212, 206, 188, 195, 204, 200, 196, 202, 210 Same for the first 9825 digits of $e$ : 955, 971, 993, 991, 968, 974, 1056, 990, 975, 952 It does seem that the percentage representations of each digit is very close to 10% in all cases. Edit: It's clear the conjecture is false (thanks for the answers). Still curious why all "naturally ocurring" irrational numbers (like the ones mentioned here) do appear to be normal. I know this is unproven, so feel free to provide conjectures.
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What you mention is not true for all irrational numbers, but for a special subset of them called Normal numbers . From the wiki article: While a general proof can be given that almost all real numbers are normal (in the sense that the set of exceptions has Lebesgue measure zero), this proof is not constructive and only very few specific numbers have been shown to be normal. And It is widely believed that the (computable) numbers $\sqrt{2}$ , $\pi$ , and $e$ are normal, but a proof remains elusive. Note that there are infinitely many irrational numbers that are not normal. In 1909, Borel introduced the concept of a Normal number and proved (with a few gaps resolved later) the following theorem: Almost all real numbers are normal, in the sense that the set of non-normal irrational numbers has Lebesgue measure zero . Some additional points of interest: (added with thanks to @leonbloy) The number of non-normal irrational numbers is uncountable Theorem 4 of this reference . There is a subset of normal numbers called Abnormal numbers and Absolutely Abnormal numbers which are uncountable. Abnormal numbers are not normal to a given base $b$ while Absolutely Abnormal numbers are not normal to any base $b \ge 2$ .
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3,197,429 |
I was given the following fact: there is a set $S$ of $11$ people, among which there are $4$ professors and $7$ students, $S=\{p_1, p_2, p_3,p_4, s_1, s_2,...,s_7\}$ We are requested to form from it a group of $5$ people, and we must have at least 3 professors. I find that the two answers I will expose should be equivalent, but are not, and I can't figure out why. Answer 1 The group of $5$ people must have at least $3$ professors. This means that three of the $5$ people will necessarily be a subset of $S_p$ , the subset of $S$ containing only the professors. There are $\binom{4}{3}$ subsets of $S_p$ , and therefore I have $\binom{4}{3}$ alternatives for the three professors that must be in the group. Now that I've made sure this $3$ professors are in the group, I have $11-3=8$ people left to choose from. The remaining two persons of the group can either be professors or students, so I can pick any of those $8$ . So for the two remaining places I have $\binom{8}{2}$ alternatives. At last, I have $\binom{4}{3} \binom{8}{2} = 112$ ways of forming a group of $5$ people in which there will definitely be at least $3$ professors. Answer 2 There are $4$ professors and, in my group of $5$ people, I must have at least $3$ of them. So I'll either have $3$ or $4$ professors. If I have $3$ professors, I'll choose them from the $4$ professors, and fill the remaining two places with $2$ of $7$ students. This is $\binom{4}{3} \binom{7}{2}$ . If on the other hand I have $4$ professors, I'll have $\binom{4}{4}$ alternatives for choosing them, and $\binom{7}{1}$ ways of choosing a student for the remaining last place. So at last there are $\binom{4}{3}\binom{7}{2}+\binom{4}{4}\binom{7}{1} = 91$ ways of making the group. Doubt As you can see, the answers are different. Answer $1$ says there are $112$ ways of making the group; answer two says $91$ . However, both reasonings seem okay to me and I can't see why should they differ nor where. Perhaps someone can clear this up for me.
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Your second solution is the correct one. Your first solution is incorrect because you overcount the scenarios where a professor is picked in the second step. The outcome where you pick the first three professors in the first step followed by the fourth professor in the second step: $\{p_1,p_2,p_3\},\{p_4,s_1\}$ is also counted where you picked the last three professors in the first step and the first professor in the second step: $\{p_2,p_3,p_4\},\{p_1,s_1\}$ . These outcomes should be considered the same however since in both scenarios you have the same five people selected. Be careful not to overcount things with multiplication principle. Objects selected in one step are treated differently than objects selected in a later step.
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3,198,918 |
When trying to solve a physics problem on decoupling a system of ODEs, I found myself needing to address the following problem: Let $A_n\in M_n(\mathbb R)$ be the matrix with all $1$ s above its main diagonal, all $-1$ s below its diagonal, and $0$ s everywhere else. Is $A_n$ always diagonalisable? If so, what is its diagonalisation (equivalently: what are its eigenvalues and corresponding eigenvectors)? For example, $$A_3=\begin{bmatrix}0&1&0\\-1&0&1\\0&-1&0\end{bmatrix},\quad A_5=\begin{bmatrix}0&1&0&0&0\\-1&0&1&0&0\\0&-1&0&1&0\\0&0&-1&0&1\\0&0&0&-1&0\end{bmatrix}.$$ Assuming my code is correct, Mathematica has been able to verify that $A_n$ is always diagonalisable up to $n=1000$ . If we use $\chi_n(t)\in\mathbb Z[t]$ to denote the characteristic polynomial of $A_n$ , a straightforward evaluation also shows that $$\chi_n(t)=-t\chi_{n-1}(t)+\chi_{n-2}(t)\tag{1}$$ for all $n\geq4$ . Furthermore, note that $A_n=-A_n^t$ so that, in the case where the dimension is even, $$\det(A_{2n}-\lambda I)=\det(A_{2n}^t-\lambda I)=\det(-A_{2n}-\lambda I)=\det(A_{2n}+\lambda I).$$ This implies that whenever $\lambda$ is an eigenvalue of $A_{2n}$ , so is $-\lambda$ . In other words, $\chi_{2n}(t)$ is always of the form $(t^2-\lambda _1^2)(t^2-\lambda_2^2)\dotsm(t^2-\lambda_n^2)$ for some $\lambda_i$ . And this is where I am stuck. In order for $A_n$ to be diagonalisable, we must have that all the eigenvalues are distinct, but trying to use the recurrence $(1)$ and strong induction, or trying to use the formula for the even case have not helped at all. It seems like the most probable line of attack would be to somehow show that $$\chi_{2n}'(t)=2t\sum_{k=1}^n\frac{\chi_{2n}(t)}{t^2-\lambda_k^2}$$ never shares a common zero with $\chi_{2n}$ (which would resolve the even case), though I don't see how to make this work. Note: I do not have any clue how to actually find the eigenvalues/eigenvectors even in the case where the $A_n$ are diagonalisable. As such even if someone cannot answer the second part of the question, but can prove that the $A_n$ are diagonalisable, I would appreciate that as an answer as well. Above I tried to look at the special case where the dimension is even, though of course the proof for all odd and even $n$ is more valuable. Even if this is not possible, for my purposes I just need an unbounded subset $S\subseteq\mathbb Z$ for which the conclusion is proven for $n\in S$ , so any such approach is welcome too. Thank you in advance!
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The matrix $A_n$ is a tridiagonal Toeplitz matrix with diagonal entries $\delta = 0$ and off-diagonal entries $\tau = 1$ and $\sigma = -1$ . Hence, we can use the formula in this paper to show that the eigenvalues are $$\lambda_k = 2i\cos\left(\dfrac{k\pi}{n+1}\right),$$ for $k = 1,\ldots,n$ , and the corresponding eigenvectors $v_1,\ldots,v_n$ have entries $$v_k[m] = i^m\sin\left(\dfrac{mk\pi}{n+1}\right).$$
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3,201,227 |
I am practicing for a math contest and I encountered the following problem that I don't know how to solve: For which value of $b$ is there only one intersection between the line $y = x + b$ and the parabola $y = x^2 + 5x + 3$ ? How do I solve it? Edit:
I made a mistake typing the problem. It should be $y = x^2 - 5x + 3$ instead of $y = x^2 + 5x + 3 $ the answer key says the answer is -6.
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Hint: $$x^2+5x+3=x+b\iff x^2+4x+3-b=0,$$ this quadratic equation should have exactly one solution therefore the discriminant $\Delta=?$
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3,203,028 |
What is the basic difference between canonical isomorphism and isomorphims? I need some basic analysis. As far as I consider on canonical isomorphism means a similarity between two geometric object having same kind of configuration and structure. While isomorphism means a map between two algebraic object or group or fields etc. But I am not satisfied with my own analysis. Can someone help me understanding these two definitions ?
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Great question. Canonical is more a term of art than a word with a strict mathematical definition. It's sometimes used as a synonym for “natural” or “obvious,” although natural is yet another idiom and obvious is in the eye of the beholder. You might think of it as meaning independent of any choices . It sounds like you're in an abstract algebra class now, but hopefully this linear algebra example will make sense. Let $V$ be a vector space over $\mathbb{R}$ of dimension $n$ . By choosing a basis $(v_1,\dots,v_n)$ , $V$ is isomorphic to $\mathbb{R}^n$ . What is the isomorphism? It takes $v\in V$ , decomposes it into $\alpha_1 v_1 + \dots + \alpha_n v_n$ , and assigns to $v$ the $n$ -tuple $(\alpha_1,\dots,\alpha_n)$ . Through the isomorphism to $\mathbb{R}^n$ , all vector spaces of the same dimension are isomorphic to each other, but not for any good reason, and not in any natural way. The isomorphism depends on the choice of basis. Let $V^*$ be the dual space to $V$ , that is, the vector space of linear functions $V \to \mathbb{R}$ . Once you choose a basis of $(v_1,\dots,v_n)$ , you can form a dual basis $(\lambda_1,\dots,\lambda_n)$ of $V^*$ , such that $\lambda_i(v_j) = \delta_{ij}$ . So $V$ and $V^*$ are isomorphic, but not canonically so. Now let $V^{**}$ be the dual space of $V^*$ . Elements of $V^{**}$ are linear functions from $V^*$ to $\mathbb{R}$ . One way to create such a map is to select $v \in V$ and send $\lambda \in V^*$ to $\lambda(v)$ . This association extends to a linear map $$
f \colon V \to V^{**},\ f(v)(\lambda) = \lambda(v)
$$ By a dimension count, this map has to be an isomorphism. And, we didn't have to choose a basis to create it. For this reason, we say that $V$ and $V^{**}$ are canonically isomorphic.
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3,203,041 |
We have an SI model with susceptible individuals $$\frac {dS}{dt} = bM\left(1 - \frac M K\right) - \mu S - \beta SI$$ and infected individuals $$ \frac {dI}{dt} = \beta SI - (\alpha+\mu)I.$$ In this case, show that the infection does not force the total popuation $M = S + I $ to die out. So I tried to find the equilibria of the system by expressing $\frac {dM}{dt}=\frac {dS}{dt} + \frac {dI}{dt}$ and also using $ S = M - I$ in the second equation to get $$\frac {dM}{dt} = bM\left(1 - \frac M K\right) - \mu M - \alpha I$$ $$ \frac {dI}{dt} = [\beta(M-I) - (\alpha+\mu)]I.$$ Both must be equal to $0$ so then I got equilibria $(0,0),(K ( 1 - \frac{\mu}{b}),0)$ and another quadratic equation. How do I proceed from here? Which case implies that the epidemic will not force the population to go extinct? Any help is appreciated.
|
Great question. Canonical is more a term of art than a word with a strict mathematical definition. It's sometimes used as a synonym for “natural” or “obvious,” although natural is yet another idiom and obvious is in the eye of the beholder. You might think of it as meaning independent of any choices . It sounds like you're in an abstract algebra class now, but hopefully this linear algebra example will make sense. Let $V$ be a vector space over $\mathbb{R}$ of dimension $n$ . By choosing a basis $(v_1,\dots,v_n)$ , $V$ is isomorphic to $\mathbb{R}^n$ . What is the isomorphism? It takes $v\in V$ , decomposes it into $\alpha_1 v_1 + \dots + \alpha_n v_n$ , and assigns to $v$ the $n$ -tuple $(\alpha_1,\dots,\alpha_n)$ . Through the isomorphism to $\mathbb{R}^n$ , all vector spaces of the same dimension are isomorphic to each other, but not for any good reason, and not in any natural way. The isomorphism depends on the choice of basis. Let $V^*$ be the dual space to $V$ , that is, the vector space of linear functions $V \to \mathbb{R}$ . Once you choose a basis of $(v_1,\dots,v_n)$ , you can form a dual basis $(\lambda_1,\dots,\lambda_n)$ of $V^*$ , such that $\lambda_i(v_j) = \delta_{ij}$ . So $V$ and $V^*$ are isomorphic, but not canonically so. Now let $V^{**}$ be the dual space of $V^*$ . Elements of $V^{**}$ are linear functions from $V^*$ to $\mathbb{R}$ . One way to create such a map is to select $v \in V$ and send $\lambda \in V^*$ to $\lambda(v)$ . This association extends to a linear map $$
f \colon V \to V^{**},\ f(v)(\lambda) = \lambda(v)
$$ By a dimension count, this map has to be an isomorphism. And, we didn't have to choose a basis to create it. For this reason, we say that $V$ and $V^{**}$ are canonically isomorphic.
|
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3,204,326 |
Relations defined by formulas such as " x has the same age as y" , " x comes from the same country as y " " a has the same image under function f as b " are obviously equivalence relations, due to the presence of the expression " same ... as". Are there many examples of equivalence relations that do not contain this " same ... as" expression and , consequently, that cannot immediately be recognized as equivalence relations ? Are there many examples of equivalence relations that , at first sight, for someone who reads their defining formula for the first time, do not at all look like equivalence relations? What I am looking for is relations such as " a is congruent to b ( modulo n) iff n divides a-b" in which one does not see any " same ... as" .
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As other answers point out it is always possible to phrase an equivalence relation as "has the same _ as" -- but sometimes the only natural way to do that is to start with the equivalence relation itself and say "same equivalence class". An important kind of equivalence relation have definitions of the shape "one thing can be reversibly made into the other by such-and-such kind of transformation": Let two closed curves in some topological space be related if they are homotopic. (They have the same homotopy class , but homotopy classes are themselves defined through this relation). Let two square matrices be related if they are similar . (Or congruent . Or variants of these where you require that the basis change is in some particular subgroup of $GL_n$ ). Let two elements of a group be related if they're conjugates. Let two sets be related if there exists a bijection between them. (They have the same cardinality , but cardinality is defined through this relation). Let two groups be related if they are isomorphic. (Or really any kind of thing you can speak of isomorphisms between). Let two polyhedra be related if one can cut one into a finite number of smaller polyhedra and reassemble them to produce the other. (This is actually the same relation as "the two polyhedra have the same volume and the same Dehn invariant ", but that is a somewhat deep result). Alternatively you can make an equivalence relation by taking the symmetric part of a larger preorder: Let two formulas of the propositional calculus be related if intuitionistic logic proves them to be equivalent. (With classical logic this would be the same as "they define the same truth function", but the situation for intuitionistic logic is not as simple). Let two infinite sequences of natural numbers be related if each of them is a subsequence of the other. (It feels plausible that one can puzzle out an equivalent characterization with a "has the same _ as" flavor that doesn't feel unnatural, but it's not immediately clear exactly what it would be). Let two sets of natural numbers be related if each is Turing reducible to the other. (They have the same Turing degree , but that is defined through this relation). Let two functions from naturals to naturals be related if each is Big Oh of the other as $n\to\infty$ . (They have the same asymptotic growth rate ). Let two sets be related if each of them admits an injection into the other. (This is the same as having the same cardinality, by the Cantor-Bernstein theorem. But that is not quite trivial). Let two groups be related if each of them admits an injective homomorphism into the other. (This is not the same relation as being isomorphic!) And here is a completely different approach: Let two real functions be related if they coincide on an open neighborhood of $0$ . (They have the same germ , but that is defined through this relation). Choose a free ultrafilter on $\mathbb N$ and let two sequences of real numbers be related if the set of indices where they agree is in the ultrafilter. (This example produces an ultrapower , which is used in non-standard analysis). Algebraic quotients are a bit of a corner case. You can define the equivalence relation as "generates the same coset as", but it is usually more natural to think of it as "the difference of the elements is in the chosen kernel".
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3,204,332 |
Let $s \geq 0$ be a real number. Show that there exists a natural number $n_0 \in \mathbb{N}$ such that $\forall n \in \mathbb{N},n \geq n_0,$ we have $$e^{-\sqrt{n!}}(n!)^s < \frac{1}{n^2}$$ By taking logarithms, we need to show that $$ -\sqrt{n!} + s \log{n!} + 2\log{n} < 0$$ for sufficiently large $n$ .
It would be sufficient to show that the limit of the sequence is a negative number; In order to do that, I thought of using the Stirling approximation, $n! \sim \sqrt{2\pi n}\left( \frac{n}{e} \right)^n:$ then it would suffice to show $$\lim_{n \to \infty} -\sqrt[4]{2\pi n}\left( \frac{n}{e} \right)^\frac{n}{2} + s\left(\frac{1}{2}(\log(2\pi) + \frac{1}{2}\log(n) + n(\log(n) - 1)\right)+ 2\log{n} < 0,$$ however this looks kind of scary. Intuitively, $\sqrt{n!}$ is a polynomial in $n!$ and so clearly grows much faster than any multiple of $\log{n!}.$ It also clearly exceeds $2\log{n}$ , but I don't know how to make this rigorous. Any ideas?
|
As other answers point out it is always possible to phrase an equivalence relation as "has the same _ as" -- but sometimes the only natural way to do that is to start with the equivalence relation itself and say "same equivalence class". An important kind of equivalence relation have definitions of the shape "one thing can be reversibly made into the other by such-and-such kind of transformation": Let two closed curves in some topological space be related if they are homotopic. (They have the same homotopy class , but homotopy classes are themselves defined through this relation). Let two square matrices be related if they are similar . (Or congruent . Or variants of these where you require that the basis change is in some particular subgroup of $GL_n$ ). Let two elements of a group be related if they're conjugates. Let two sets be related if there exists a bijection between them. (They have the same cardinality , but cardinality is defined through this relation). Let two groups be related if they are isomorphic. (Or really any kind of thing you can speak of isomorphisms between). Let two polyhedra be related if one can cut one into a finite number of smaller polyhedra and reassemble them to produce the other. (This is actually the same relation as "the two polyhedra have the same volume and the same Dehn invariant ", but that is a somewhat deep result). Alternatively you can make an equivalence relation by taking the symmetric part of a larger preorder: Let two formulas of the propositional calculus be related if intuitionistic logic proves them to be equivalent. (With classical logic this would be the same as "they define the same truth function", but the situation for intuitionistic logic is not as simple). Let two infinite sequences of natural numbers be related if each of them is a subsequence of the other. (It feels plausible that one can puzzle out an equivalent characterization with a "has the same _ as" flavor that doesn't feel unnatural, but it's not immediately clear exactly what it would be). Let two sets of natural numbers be related if each is Turing reducible to the other. (They have the same Turing degree , but that is defined through this relation). Let two functions from naturals to naturals be related if each is Big Oh of the other as $n\to\infty$ . (They have the same asymptotic growth rate ). Let two sets be related if each of them admits an injection into the other. (This is the same as having the same cardinality, by the Cantor-Bernstein theorem. But that is not quite trivial). Let two groups be related if each of them admits an injective homomorphism into the other. (This is not the same relation as being isomorphic!) And here is a completely different approach: Let two real functions be related if they coincide on an open neighborhood of $0$ . (They have the same germ , but that is defined through this relation). Choose a free ultrafilter on $\mathbb N$ and let two sequences of real numbers be related if the set of indices where they agree is in the ultrafilter. (This example produces an ultrapower , which is used in non-standard analysis). Algebraic quotients are a bit of a corner case. You can define the equivalence relation as "generates the same coset as", but it is usually more natural to think of it as "the difference of the elements is in the chosen kernel".
|
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