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2,719,600 |
Suppose we have events $A$ and $B$. We want to write the probability that exactly one of the events $A,B$ occurs in terms of $P(A),P(B)$ and $P(A \cap B)$ only My thought: Since I want only one occurring, $A$ or $B$, we must find $P(A \cup B)$ which equals $P(A) + P(B) - P(A \cap B)$.. However, on my answer sheet it says the answer is $P(A) + P(B) - 2P(A \cap B )$. Am I missing something?
|
$P(A\cup B)$ includes those outcomes where both events $A$ and $B$ occur. We want to exclude this case, because if they both occur it is not exactly one occurring. This undesirable case is exactly $P(A\cap B)$, so we need to subtract it, as $$P(A\cup B)-P(A\cap B)$$
|
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|
2,719,614 |
Question: suppose $f : [a,b] \rightarrow R$ is a differentiable and $c \in [a,b]$. Then show there exists a sequence $\{x_n\}$ converging to $c$, $x_n \not = c$ for all $n$, such that $f^\prime (c) = \lim_{n \to \infty} f^\prime (x_n)$. Do note this does not imply that $f^\prime$ is continuous (why?). Attempt: Let $x_n \in [a,b]$ Then, $f(x_n)$ is differentiable. Then, $f^\prime (c) = f^\prime(\lim_{n \to \infty}x_n)=\lim_{n \to \infty}f^\prime(x_n)$. I don't understand why this does not imply continuity because if $f(c)=\lim_{n \to \infty}f(x_n)$, $f$ is continuous at $c$. Isn't it the case for the derivative function? Thank you in advance. Edit: I agree that my attempt above assumes that $f^\prime$ is continuous, which might not be true. So, could you give me some hint to solve this problem without using that assumption? Let $x_n \in [a,b]$. Then, $f(x_n)$ is differentiable. Then, $\lim_{x_n \to c} \frac {f(x_n)-f(c)}{x_n-c}=f^\prime (c)$.I don't know how to proceed from here.
|
$P(A\cup B)$ includes those outcomes where both events $A$ and $B$ occur. We want to exclude this case, because if they both occur it is not exactly one occurring. This undesirable case is exactly $P(A\cap B)$, so we need to subtract it, as $$P(A\cup B)-P(A\cap B)$$
|
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|
2,719,618 |
In my Complex Analysis classes I was studying the Riemann's zeta function. At some point my teacher was able to demonstrate the product formula in this way: "Suppose M and N are positive integers with M>N. Observe now that, by the fundamental theorem of arithmetc, any positive integer $ n \le N$ can be written uniquely as a product of primes, and that each prime that occurs in the product must be less than or equal to N and repeated less than M times. Therefore:" $$\zeta(s)\le \prod_{p \le N} (1+p^{-s}+p^{-2s}+...+p^{-Ms}) \mbox{
(1)}$$, p primes. Letting N tend to infinity, we have: $$\zeta(s)\le \prod_{p}\frac{1}{1-p^{-s}}$$
And then, using a similar argument with the fundamental theorem of arithmetic: $$\prod_{p \le N} (1+p^{-s}+p^{-2s}+...+p^{-Ms}) \le \zeta(s) \mbox{
(2)}$$
And conclude that: $$\zeta(s)=\prod_{s} (\frac{1}{1-p^{-s}})$$ Assuming $s$ real But I'm not understanding how to get the inequalities (1) and (2). Can someone explain it to me?
|
$P(A\cup B)$ includes those outcomes where both events $A$ and $B$ occur. We want to exclude this case, because if they both occur it is not exactly one occurring. This undesirable case is exactly $P(A\cap B)$, so we need to subtract it, as $$P(A\cup B)-P(A\cap B)$$
|
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|
2,720,550 |
Suppose you have the function: $$f(x) = e^x \cos{x} $$
and you need to find the 3rd degree Taylor Series representation. The way I have been taught to do this is to express each separate function as a power series and multiply as necessary for the 3rd degree. For example for
$$ \cos x =\sum_{n=0}^\infty (-1)^n\frac{ x^{2n}}{(2n)!} = 1-\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots \text{ and } e^x =\sum_{n=0}^\infty \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!}+\cdots $$ multiply the terms on the right of each until you get the 3rd degree. Logically, I am happy. However, I have not seen a theorem or any rule that says you can just multiply series in this way. Doing it this way, is there a guarantee that I will always get the power series representation of $f(x)$? Additionally, if instead of multiplying, functions were being added? Would the above hold true - take the series of each function and add up the necessary terms?
|
Taylor's theorem allows you to use the Big O notation :
$$\cos(x)= 1-\frac{x^2}{2!}+O(x^4)\quad\mbox{and}\quad e^x=1 + x + \frac{x^2}{2!} + \frac{x^3}{3!}+O(x^4).$$
Therefore
$$e^x\cos(x)=1 + x + \frac{x^2}{2!} + \frac{x^3}{3!}+O(x^4)-\frac{x^2}{2!}(1+x+O(x^2))=1+x-\frac{x^3}{3}+O(x^4).$$
|
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2,723,860 |
I know, I know, there are tons of questions on this -- I've read them all, it feels like. I don't understand why $(F \implies F) \equiv T$ and $(F \implies T) \equiv T$. One of the best examples I saw was showing how if you start out with a false premise like $3=5$ then you can derive all sorts of statements that are true like $8=8$ but also false like $6=10$, hence $F \implies T$ is true but so is $F \implies F$. But for me examples don't always do it for me because how do I know if the relationship always holds even outside the example? Sometimes examples aren't sufficiently generalized. Sometimes people say "Well ($p \implies q$) is equivalent to $\lnot p \lor q$ so you can prove it that way!" except we arrived at that representation from the truth table in the first place from disjunctive normal form so the argument is circular and I don't find it convincing. Sometimes people will use analogies like "Well assume we relabeled those two "vacuous cases" three other ways, $F/F, F/T, T/F$ -- see how the end results make no sense?" Sure but T/T makes no sense to me either so I don't see why this is a good argument. Just because the other three are silly doesn't tell me why T/T is not silly. Other times I see "Well it's just defined that way because it's useful"... with no examples of how it's indeed useful and why we couldn't make do with some other definition. Then this leads to the inevitable counter-responders who insist it's not mere definition of convenience but a consequence of other rules in the system and so on, adding to the confusion. So I'm hoping to skip all that: Is there some other way to show without a doubt that $(F \implies q) \equiv T$?
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I've never been satisfied with the definition of the material implication in the context of propositional logic alone. The only really important things in the context of propositional logic are that $T \Rightarrow T$ is true and $T \Rightarrow F$ is false. It feels like the truth values of $F \Rightarrow T$ and $F \Rightarrow F$ are just not specified by our intuition about implication. After all, why should "if the sky is green, then clouds are red" be true? But in predicate logic, things are different. In predicate logic, we'd like to be able to say $\forall x (P(x) \Rightarrow Q(x))$ and have the $x$ 's for which $P(x)$ is false not interfere with the truth of the statement. For example, consider "among all integers, all multiples of $4$ are even". That statement is true even though $1$ is not even (an instance of $F \Rightarrow F$ ). It's also true even though $2$ is even despite not being a multiple of $4$ (an instance of $F \Rightarrow T$ ). But now in classical logic, every proposition has a single truth value. Thus the only way to define $\forall x R(x)$ is "for every $x$ , $R(x)$ is true". We can't define it in some other way, like "for every $x$ , either $R(x)$ is true or $R(x)$ is too nonsensical to have a truth value". Thus we are stuck defining $F \Rightarrow T$ and $F \Rightarrow F$ to both be true, if $\forall x (P(x) \Rightarrow Q(x))$ is going to behave the way we want. In a different system of logic, we might do things differently. But in classical logic, "every proposition has a truth value" is basically an axiom.
|
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2,727,733 |
I'm looking for a way to look at a triangle, and perhaps visualize a few extra lines, and be able to see that the interior angles sum to $180^\circ$. I can visualize that supplementary angles sum to $180^\circ$. I'd like to be able to see the interior angle sum similarly... I can see that the exterior angles must sum to $360^\circ$, because if you walked around the perimeter, you would turn around exactly once (though I can tell this is true, I don't really see it). I also saw a proof on KA, where the exterior angles were superimposed, to show they summed to $360^{\circ}$ (though I'm not 100% comfortable with this one). Finally, for $a$, $b$, and $c$ exterior angles $a+b+c=360$: \begin{align}
(180-a) + (180-b) + (180-c) & = 3\times 180 - (a+b+c) \\
& = 3\times 180 - 360 \\
& = 180 \\
\end{align} But I find this algebra hard to see visually/geometrically. Is there proof that enables one to directly see that the interior angles of triangle sum to $180^\circ$? A couple of secondary questions: am I visually deficient in my ability to imagine? or, am I asking too much of a proof, that I be able to see it, and that beimg able to tell that it is true should be enough...?
|
Since that fact about the angle sum is equivalent to the parallel postulate, any visualization is likely to include a pair of parallel lines. Here's one from wikipedia :
|
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2,728,317 |
As I know when you move to "bigger" number systems (such as from complex to quaternions) you lose some properties (e.g. moving from complex to quaternions requires loss of commutativity), but does it hold when you move for example from naturals to integers or from reals to complex and what properties do you lose?
|
The most important ones as I see it: Naturals to integers: lose well-orderedness, gain "abelian group" (and, indeed, "ring"). Integers to rationals: lose discreteness, gain "field". Rationals to reals: lose countability, gain "Cauchy-complete". Reals to complexes: lose a compatible total order, gain the Fundamental Theorem of Algebra.
|
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2,728,646 |
I am failing to understand why the integral is defined as: $$\int_a^b f(x) dx = \lim_{n\to\infty}\sum_{i=1}^n f(x_i^*)\Delta x$$ instead of: $$\int_a^b f(x)dx=\sum_{i=1}^\infty f(x_i^*)\Delta x$$ Is the former just popular preference or is there something I am not conceptually understanding here?
|
Let's cut $[a, b]$ up into infinitely many equally spaced slices. So $x_1 = a$, obviously. What's $x_2$? What's $\Delta x$, if not zero? In general, there isn't a "nice" way to cut a finite interval into infinitely many sampling intervals. Ultimately, the Riemann integral samples a bunch of function values in a fairly uniform way (meaning one from each interval of length $\Delta x$) and averages them. It doesn't make sense to make an infinite uniform sampling of a bounded interval.
|
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2,728,652 |
Is it true that any circle in $\mathbb{R^2}$ contains a point with rational coordinates? what about any simple closed curve? If it is, could you please help me with the proof?
|
Let's cut $[a, b]$ up into infinitely many equally spaced slices. So $x_1 = a$, obviously. What's $x_2$? What's $\Delta x$, if not zero? In general, there isn't a "nice" way to cut a finite interval into infinitely many sampling intervals. Ultimately, the Riemann integral samples a bunch of function values in a fairly uniform way (meaning one from each interval of length $\Delta x$) and averages them. It doesn't make sense to make an infinite uniform sampling of a bounded interval.
|
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2,730,687 |
So - I am no math genius but I do have shower thoughts. And there is one thought about normal distribution that I just couldn't let go. I converted it into a little story to visualize it a little better. Let's see if it makes sense and if it really is a paradox I came up with. Here is the story: A man is in court. He is said to have murdered someone. There is evidence that stats that he did it - but chances are it is all a coincidence. The judge comes up with a simple solution:
"Tomorrow at 8am on market square - you are to toss a coin. Head and your head comes off - tails and you go home a free man. Let the gods decide whether you are to die or not." The man gladly accepts this offer. You must know - even though it is the middle ages, he is a mathematican - not one to believe in gods. And he also knows probabilites and thinks he has a way of how to manipulate those. The man takes his fate deciding coin home with him and begins tossing it all night. The morning comes and everyone is waiting on market square. It is 8 am sharp and the man, as promised shows up with his coin in his hand. He is very confident, because he knows - his chances of dying are at about 0.1 %.
In front of everybody, he tosses the coin and: tails. Then man is free to go. Not even the tiniest bit nervous about his fate. How was that possible? He must have known that his chances where 50 - 50 (assuming the coin cannot land on its edge and will always be tossed and flipped randomly). Well, here is the thing that I cannot explain: Last night, the man was home - as I said - flipping his coin over and over again. Since this is a normal distribution, in within the first 10 tosses, the coin showed head 5 times, and tails 5 times. But, after many, many tosses - the coin finally showed head 9 times in a row. This happening comes with a likelihood of 0.2% (according to one of those tree-diagrams). Now - for the 10th time, the chances of head again would be only 0.1% percent if I am not mistaken. Now - in my eyes: All the man had to do was to NOT throw that coin again until his fate was about to be decided - because heads again? That would be insanly unlikely - wouldn't it be? So, that is my paradox. A random coin toss cannot be manipulated only by waiting for it to be unlikely to show a certain outcome over and over again - or can it? Thanks for reading my little story :) I hope you guys understand what I am trying to convey here :)
|
Something to think about: Since the coin flips are independent, and assuming the coin is fair, the probability that ten coin flips land heads is: $$P(H)\cdot P(H)\cdot P(H)\cdot P(H)\cdot P(H)\cdot P(H)\cdot P(H)\cdot P(H) \cdot P(H)\cdot P(H)=(0.5)^{10}$$ The probability that nine coin flips land heads and the tenth lands tails is: $$P(H)\cdot P(H)\cdot P(H)\cdot P(H)\cdot P(H)\cdot P(H)\cdot P(H)\cdot P(H) \cdot P(H)\cdot P(T)=(0.5)^{10}$$ The probabilities are the same! So he had equal chances of dying or not dying.
|
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2,733,613 |
Why, when we perform Fourier transforms/decompositions, do we use sine/cosine waves (or more generally complex exponentials) and not other periodic functions? I understand that they form a complete basis set of functions (although I don't understand rigourously why), but surely other period functions do too? Is it purely because sine/cosine/complex exponentials are convenient to deal with, or is there a deeper reason? I get the relationship these have to circles and progressing around them at a constant rate, and how that is nice and conceptually pleasing, but does it have deeper significance?
|
The Fourier basis functions $e^{i \omega x}$ are eigenfunctions of the shift operator $S_h$ that maps a function $f(x)$ to the function $f(x - h)$:
$$
e^{i \omega (x-h)} = e^{-i\omega h} e^{i \omega x}
$$
for all $x \in \mathbb R$. All of the incarnations of the Fourier transform (such as Fourier series and the discrete Fourier transform) can be understood as changing basis to a basis of eigenvectors for a shift operator. It is possible to consider other operators, which have different eigenfunctions leading to different transforms. But this shift operator is so simple and fundamental that it's not surprising the Fourier transform turns out to be particularly useful.
|
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2,733,773 |
Let $A=(a_{ij})$ be an $n\times n$ matrix. Suppose that $A^2$ is diagonal? Must $A$ be diagonal. In other words, is it true that $$A^{2}\;\text{is diagonal}\;\Longrightarrow a_{ij}=0,\;i\neq j\;\;?$$
|
As a simplified case, consider $n=2$ and let $$A=\begin{pmatrix}a&b\\c&d\end{pmatrix}\implies A^2=\begin{pmatrix}a^2+bc&b(a+d)\\c(a+d)&d^2+bc\end{pmatrix}$$
so $$A^2\,\,\text{diagonal}\iff b(a+d)=c(a+d)=0\implies\left\{\begin{matrix}
b=c=0 \implies A\,\,\text{diagonal}\\
a+d=0\implies a=-d
\end{matrix}\right.$$ Hence a counterexample would be any matrix of the form $$A=\begin{pmatrix}-d&b\\c&d\end{pmatrix}$$ with $b \ne 0$ or $c \ne 0$ or both.
|
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2,733,802 |
I've done quite a bit of research, but I always fail right before the end.
I know the integral of a circle, I know the substitution $$\theta = \arcsin\left(\frac x r\right)$$ However, I never get the right results. My question is: How do I find the area under a semicircle from $-r$ to $x_1; x_1 \in[-r;+r]$? Also if I want to find the are under a semicircle offset in $$y= f(x) = \sqrt{r^2 - x^2} - d$$ I would simply have to use the above formular and subtract $(x + r) * d$?
|
As a simplified case, consider $n=2$ and let $$A=\begin{pmatrix}a&b\\c&d\end{pmatrix}\implies A^2=\begin{pmatrix}a^2+bc&b(a+d)\\c(a+d)&d^2+bc\end{pmatrix}$$
so $$A^2\,\,\text{diagonal}\iff b(a+d)=c(a+d)=0\implies\left\{\begin{matrix}
b=c=0 \implies A\,\,\text{diagonal}\\
a+d=0\implies a=-d
\end{matrix}\right.$$ Hence a counterexample would be any matrix of the form $$A=\begin{pmatrix}-d&b\\c&d\end{pmatrix}$$ with $b \ne 0$ or $c \ne 0$ or both.
|
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2,733,823 |
Task: Does the exact differential equation
$$
x\,dx+y\,dy+x\,dy-y\,dx=0
$$
have an integrating factor of the form $α=α(z)$, $z=x^2+y^2$. In the provided solution there is a step where in the original equation (for the integrating factor)
$$\frac{\partial{\ln\alpha(x,y)}}{\partial y} M - \frac{\partial{\ln\alpha(x,y)}}{\partial x} N = \frac{\partial N }{ \partial x} - \frac{\partial M}{ \partial y}$$ we choose
$z = x^2 + y^2$
and say $\alpha = \alpha(z)$ and so original equation takes the form $$2(My-Nx)\frac{\partial\ln\alpha}{\partial z} = \frac{\partial N }{ \partial x} - \frac{\partial M}{ \partial y}$$ It says that when we pick the function $\alpha$ as a function of $(x^2+y^2)$ equation comes to this form. I'm trying to understand it and prove it but I couldn't.
|
As a simplified case, consider $n=2$ and let $$A=\begin{pmatrix}a&b\\c&d\end{pmatrix}\implies A^2=\begin{pmatrix}a^2+bc&b(a+d)\\c(a+d)&d^2+bc\end{pmatrix}$$
so $$A^2\,\,\text{diagonal}\iff b(a+d)=c(a+d)=0\implies\left\{\begin{matrix}
b=c=0 \implies A\,\,\text{diagonal}\\
a+d=0\implies a=-d
\end{matrix}\right.$$ Hence a counterexample would be any matrix of the form $$A=\begin{pmatrix}-d&b\\c&d\end{pmatrix}$$ with $b \ne 0$ or $c \ne 0$ or both.
|
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2,738,115 |
The second step in proof by induction is to: Prove that if the statement is true for some integer $n=k$, where
$k\ge n_0$ then it is also true for the next larger integer, $n=k+1$ My question is about the "if"-statement. Can we just assume that indeed the statement is true? If we assume it, then the proof works... but isn't that similar to the following "proof": Let $N$ be the largest positive integer. Since $1$ is a positive integer, we must have $N\ge1$. Since $N^2$ is a positive integer, it cannot exceed the largest positive integer. Therefore, $N^2\le N$ and so $N^2-N\le0$. Thus, $N(N-1)\le0$ and we must have $N-1\le0$. Therefore, $N\le1$. Since also $N\ge1$, we have $N=1$. Therefore, $1$ is the largest positive integer. The only thing that is wrong with this "proof" is that we falsely assume there actually exists a largest positive integer. So both in the above case and in proof by induction we do an assumption. In the second case the assumption leads to a false conclusion. What is the difference with proof by induction? Why is doing the assumption that the hypothesis is actually true valid here and why doesn't it lead to a similar contradiction? EDIT: the "proof" above is not mine, it is taken from Calculus a Complete Course 8th edition as an example of why existence proofs are important.
|
Indeed, a fallacious conclusion may be reached by valid deduction from an unjustifiable premise. This is why the first step of induction is to prove that the predicate is justified for the base case; to ensure that we do not do that. If $\mathcal P(0)$ is proven and for all natural numbers $n$ we can show that $\mathcal P(n)\to\mathcal P(n+1)$ is true, then we may successively prove $\mathcal P(1)$ , $\mathcal P(2)$ , $\mathcal P(3)$ , and so forth, by iterative applications of modus ponens . $${\text{We may soundly prove }\mathcal P(1)\text{ from having proven }\mathcal P(0)\textit{ and }\mathcal P(0)\to\mathcal P(1).\\\text{We may soundly prove }\mathcal P(2)\text{ from having proven }\mathcal P(1)\textit{ and }\mathcal P(1)\to\mathcal P(2).\\\text{We may soundly prove }\mathcal P(3)\text{ from having proven }\mathcal P(2)\textit{ and }\mathcal P(2)\to\mathcal P(3).\\\text{And so forth, and so on,}\textit{ et cetera,}\ldots}$$
|
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2,738,360 |
It seems to me an equation, in an abstract sense, must always involve some varying quantities where the varying quantities belong in some space (set, algebraic structure, what have you). In order to make precise the phrase, "vary quantities", it seems to me that one must have a mechanism for evaluation of each side of the equation. Ultimately, I think solving an equation must always be, in essence, finding the pre-image of a mapping. If the pre-image is empty then there are no solutions. Consider the equation over $\mathbb{C}$
$$x^2-3x+2=0$$ We have that $x^2-3x+2$ is a polynomial and this polynomial induces a natural map from to $\mathbb{C}$ to $\mathbb{C}$ called evaluation. The equation is really asking for the pre-image of $0$ of this map. Consider the functional equation $$f(x+y) + f(x-y) = 0$$
where we are looking for solutions that are functions from $\mathbb{R}$ to $\mathbb{R}$. I think ultimately that this equation can be thought of in terms of the pre-image of the map $G$ that takes functions like $f$ and maps them to the function from $\mathbb{R^2}$ to $\mathbb{R}$ by sending $f$ to the function of two variables $f(x+y)+f(x-y)$. And the equation is really asking for the pre-image of the zero function under this map. Is it correct to view all equations in this manner? That is finding the solutions must always be equivalent to finding the pre-image of some element of some mapping? Think of basic equations one finds in college algebra books. I tell my students that we take the given equation and apply solution preserving operations to it to transform the equation into a simpler one. The goal is to ultimately end up with a simpler equation whose solutions we can find by inspection. For instance, $$3x - 2 = 5 \implies 3x = 7 \implies x = \frac{7}{3}$$ At each step we transform the equation to a simpler one whose solution set is the same. We can solve the last equation by inspection. Ultimately, isn't this how all equations are solved? We transform the equation to simpler equation(s) and end up with an equation that can be solved by inspection.
|
One approach: An equation is a predicate , $P(x)$, of the form $s(x) = t(x)$ where $x$ is a free variable (or vector of free variables) and $s(x), t(x)$ are terms -- expressions which evaluate to elements of the universe (e.g. real numbers if you are doing mathematics over the reals) when values are substituted for variables. This means that $P(x)$ is something which evaluates to either $true$ or $false$ when $x$ is replaced by members of the universe. In the 1-variable case it can be thought of as a function of the form $$P(x): U \mapsto \{true,false\}$$ where $U$ is the domain of discourse. To solve an equation is to determine $P^{-1}(true)$, the set of all values which make the predicate true.
|
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|
2,739,592 |
I self-realized an interesting property today that all numbers $(a,b)$ belonging to the infinite set $$\{(a,b): a=(2l+1)^2, b=(2k+1)^2;\ l,k \in N;\ l,k\geq1\}$$ have their AM and GM both integers. Now I wonder if there exist distinct real numbers $(a,b)$ such that their arithmetic mean, geometric mean and harmonic mean (AM, GM, HM) all three are integers. Also, I wonder if a stronger result for $(a,b)$ both being integers exists. I tried proving it, but I did not find it easy. For the AM, it is easy to assume a real $a$ and an AM $m_1$ such that the second real $b$ equals $2m_1-a$. For the GM, we get a condition that $m_2=\sqrt{(2m_1-a)a}$. If $m_2$ is an integer, then… what? I am not sure exactly how we can restrict the possible values of $a$ and $m_1$ in this manner.
|
Expanding on Christian Blatter's answer. There are a few key points. The arithmetic mean of two rational numbers is always rational. The harmonic mean of two non-zero rational numbers is always rational. The geometric mean of two squared positive integers is always an integer. For all three types of mean if we multiply every input by a positive real value we also multiply the result by that same value. These key points lead to a strategy for finding numbers whose am, gm and hm are all integers. pick a pair of integers whose GM is an integer. calculate the AM and HM multiply through by the denominators of the AM and HM. Now to work this through, pick any two distinct positive integers $x$ and $y$. $$\mathrm{GM}(x^2,y^2) = xy$$
$$\mathrm{AM}(x^2,y^2) = \frac{x^2+y^2}{2}$$
$$\mathrm{HM}(x^2,y^2) = \frac{2x^2y^2}{x^2+y^2}$$ Let $t = 2(x^2 + y^2)$ Let $a=tx^2$ Let $b=ty^2$. Since only addition, multiplication and squaring of positive integers is involved it is clear that $t$, $a$ and $b$ are all positive integers. It is also clear that a and b are distinct. $$\mathrm{GM}(a,b) = txy$$
$$\mathrm{AM}(a,b) = t\frac{x^2+y^2}{2} = (x^2+y^2)^2$$
$$\mathrm{HM}(a,b) = t\frac{2x^2y^2}{x^2+y^2} = 4x^2y^2$$ Again since all these values can be calculated merely by adding, multiplying and squaring positive integers they are all positive integers. Lets plug in some numbers, for example $x=1$ and $y=2$ $$t = 10$$
$$a = 10$$
$$b = 40$$
$$\mathrm{GM}(10,40) = 20$$
$$\mathrm{AM}(10,40) = 25$$
$$\mathrm{HM}(10,40) = 16$$ Indeed we can extend this techiquie to find an arbitary size list of integers the AM, GM, and HM of any subset of which are integers. Just start with integers of the form $x^{n!}$ so the GMs are all integers. Then work out the AMs and HMs and multiply through.
|
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|
2,740,316 |
One day I was reading an article on the infinitude of prime numbers in the Proof Wiki . The article introduced a proof that used only topology to prove the infinitude of primes, and I found it very interesting and satisfying. I'm wondering, if there are similar proofs that use topology where it's not obvious that it can be applied. I'm sure that seeing such proofs could also strengthen my intuition with topology. So my question is: "Which theorems, not directly linked with topology, have interesting proofs that use topology?". Thanks in advance!
|
One of the richest sources for this is geometric group theory , and especially the parts of it that construct topological spaces corresponding to particular groups (schreier graphs, cayley complexes, BG spaces etc.) For example, the proof that any subgroup of a free group is free (Nielsen Schreier) epitomizes how topological arguments greatly simplify some parts of group theory. There are proof methods for theorems from discrete geometry that utilize methods from equivariant topology (such as the ham-sandwich theorem via Borsuk ulam .) These use the CS/TM paradigm, which studies maps between the space of "geometric configurations" to the "Space of viable solutions." See here for a general reference and wikipedia for a shorter exposition. There are some other miscellaneous examples that I know of: Bezout's identity via separating curves (assumes the euclidian algorithm in disguise.) Fundamental Theorem of Algebra (resp. existence of eigenvalues) via lefschetz fixed point theorem the use of "Denseness arguments" in algebraic geometry (such as cayley-hamilton ).
|
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|
2,741,229 |
I have searched a lot, but i haven't found any proof about that statement. I have checked the proof of If $f$ is differentiable, then $f$ is continuous but it's not the same argument I think. Also, I want to know what's your opinion about the statement If derivative of $f$ is not continuous, then $f$ is not continuous
|
If $f$ is differentiable, then $f$ is continuous. The continuity of $f'$ is irrelevant here. In particular, even if $f'$ is discontinuous, $f$ is continuous .
|
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|
2,742,126 |
A definition in classical geometry (for example, Birkhoff's formulation, but I suppose it could be all of them) is that a line is always considered to be parallel to itself. I understand this is probably for convenience, but in my mind since two distinct lines are parallel if they have no points in common and a line has infinitely many points in common with itself. Perhaps the idea is to ease the definition that two (non-parallel) lines intersect at one and only one point? Q: What's the purpose/what inconvenience would be caused if we didn't have that definition?
|
The idea is that you want "parallel" to define equivalence classes (called "pencils", cf. Coxeter, Projective Geometry , and Artin, Geometric Algebra ), which require the defining relationship to be an equivalence relationship: reflexive, symmetric, and transitive. Those classes then have some nifty uses, like defining projective space by adding a point at infinity for each pencil (which is the considered to be on each of those lines) and a line at infinity for each class of parallel planes (this line containing all the points at infinity corresponding to the pencils of lines in that class of planes). Also, you were already going to have to rethink the definition of "parallel" as having no points in common, if you are going to do solid geometry. Parallel lines also need to be coplanar...i.e., there need to be two other lines that intersect each other and that each intersect the parallel lines (five distinct points of intersection). Lines that are not coplanar are called "skew" not "parallel".
|
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|
2,743,302 |
Prove that any real number $r$ can be expressed as the sum of two irrational numbers $x$ and $y$. Progress: I have a specific example for any rational number $r$: $x = r-\pi$ and $y = \pi$ (or replace $\pi$ with any irrational number.) However, I can't seem to find a way to prove this in general for irrational $r$. Any help is greatly appreciated.
|
If $r$ is irrational, then $r=\frac{r}2+\frac{r}2$. To tell the truth, for $r$ rational we also have $r=\frac{r}2+\frac{r}2$, but it is not useful in this case. :-)
|
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|
2,743,440 |
Let the function $f:\mathbb{R}\rightarrow\mathbb{R}$ be differentiable at $x=0$. Prove that $\lim_{x\rightarrow 0}\frac{f(x^2)-f(0)}{x}=0$. The result is pretty obvious to me but I am having a difficult time arguing it precise enough for a proof. What I have so far is of course that since $f$ is differentiable;
$$f'(0)=\lim_{x\rightarrow 0}\frac{f(x)-f(0)}{x}$$
exists. Any help would be greatly appreciated.
|
HINT: $$\frac{f(x^2)-f(0)}{x}=\left(\frac{f(x^2)-f(0)}{x^2}\right)x$$
|
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|
2,744,850 |
The goal:
Prove that there is no integer $k$ such that ${(k+1)^2\over{k^2}}=2$. My proof:
If ${(k+1)^2\over{k^2}}=2$, then ${{k+1}\over{k}}=\sqrt2$, and if $k$ is an integer, $k+1$ is also an integer. This implies that $\sqrt2$ is a rational number, which is also provably false .
|
How about solving the equation: $(k+1)^2=2k^2$ for $k\ne 0$? $$(k+1)^2=2k^2$$ $$k^2+2k+1=2k^2$$ $$k^2-2k-1=0$$ $$k^2-2k+1=2$$ $$(k-1)^2-(\sqrt{2})^2=0$$ $$(k-1-\sqrt{2})(k-1+\sqrt{2})=0$$ Then $k=1+\sqrt{2}$ or $k=1-\sqrt{2}$, neither of them are integers, because you can prove that $\sqrt{2}$ is an irrational number (but knowing that it is not an integer is enough here). For your method, the only minor mistake you made is: $${(k+1)^2\over{k^2}}=2\Leftrightarrow {{k+1}\over{k}}=\pm \sqrt2.$$
|
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|
2,744,878 |
I am so confusing about get the tangent equation for a particular curve, I read many topics about tangent equation but i got many questions to need to answers
I read that to get line equation i have to follow that
1- https://i.stack.imgur.com/cwibt.png 2- https://i.stack.imgur.com/Q0raq.png So how can i use the same approach to get tangent for specific curve equation like helix and so on ... i don't want equation or example ... i just want to understand by graph how can i get it ..
For example tangent equation for K(t)=(sin3t cost, sin3t sint, 0) at point pi/3
Thanks
|
How about solving the equation: $(k+1)^2=2k^2$ for $k\ne 0$? $$(k+1)^2=2k^2$$ $$k^2+2k+1=2k^2$$ $$k^2-2k-1=0$$ $$k^2-2k+1=2$$ $$(k-1)^2-(\sqrt{2})^2=0$$ $$(k-1-\sqrt{2})(k-1+\sqrt{2})=0$$ Then $k=1+\sqrt{2}$ or $k=1-\sqrt{2}$, neither of them are integers, because you can prove that $\sqrt{2}$ is an irrational number (but knowing that it is not an integer is enough here). For your method, the only minor mistake you made is: $${(k+1)^2\over{k^2}}=2\Leftrightarrow {{k+1}\over{k}}=\pm \sqrt2.$$
|
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|
2,746,430 |
A common statement of The Fundamental Theorem of Arithmetic goes: Every integer greater than $1$ can be expressed as a product of powers of distinct prime numbers uniquely up to a reordering of the factors . Now the statement makes a point of mentioning that factorization is unique up to reordering of the factors, saying basically that we don't have to worry about it because multiplication in the integers is commutative. But why not specify that it's also unique up to the choice in which order we multiply the factors? I.e, that we don't have to worry about it because multiplication in the integers is associative too? If we insist on multiplication being a binary operation, then we need to define some grouping when we have a product of more than two integers. Shouldn't there be a clause in the Fundamental Theorem that indicates, for example, that $30 = (2\times (3 \times 5))$ and $30 = ((2\times 3) \times 5)$ are not distinct factorizations? It should be noted that some answers to this question were merged from another question , so they may not be completely consistent with this question exactly as it's stated.
|
"Uniquely" means that there is exactly one way to write an integer as a $k$-ary product of primes (up to permutation of the factors). Since thanks to associativity, all placements of parentheses give the same product, it does not matter which of the concatenations of binary operations one uses for the definition of the $k$-ary product. One symmetric way to think about it, is to define it as an equivalence class of all these expressions. If you insist on Polish notation, then we get, say, $30=*_3 2\ 3\ 5 $ where $*_3$ denotes the ternary multiplication operator.
|
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|
2,749,422 |
From a deck of $52$ cards, cards are picked one by one, randomly and without replacement. What is the probability that no club is extracted before the ace of spades? I think using total probability for solve this $$P(B)=P(A_1)P(B\mid A_1)+\ldots+P(A_n)P(B\mid A_n)$$ But I am not sure how to solve this. Can someone help me?
|
The event that you find $\spadesuit A$ before any $\clubsuit$ is entirely determined by the order in which the $14$ cards $\spadesuit A,\clubsuit A,\clubsuit 2,...,\clubsuit K$ appear in the deck. There are $14!$ possible orderings of these $14$ cards, and each of these orderings are equally likely. How many of these orderings have $\spadesuit A$ appearing first? The first card must be $\spadesuit A$, there are $13$ choices for the second card, $12$ for the third, and so on, so there are $13!$ such orderings. Therefore, the probability is $13!/14!=\boxed{1/14}$. Put even more simply: of the fourteen cards $\spadesuit A,\clubsuit A,\clubsuit 2,...,\clubsuit K$, each is equally likely to appear earliest in the deck, so the probability that you find $\spadesuit A$ first is $1/14.$ Added Later: There is also a way to solve this using the law of total probability. We may as well stop dealing cards once the $\spadesuit A$ or any $\clubsuit$ shows up. Let $E_n$ be the event that exactly $n$ cards are dealt. Then
$$
P(\spadesuit A\text{ first})=\sum_{n=1}^{39}P(\spadesuit A\text{ first }|E_n)P(E_n)
$$
Now, given that the experiment ends on the $n^{th}$ card, we know that the $n^{th}$ card is one of $\spadesuit A,\clubsuit A,\clubsuit 2,...,\clubsuit K$, and none of the previous cards are. Each of these is equally likely (due to the symmetry among the 52 cards), so $P(\spadesuit A\text{ first }|E_n)=1/14$. Therefore,
$$
P(\spadesuit A\text{ first})=\sum_{n=1}^{39}\frac1{14}P(E_n)=\frac1{14}\sum_{n=1}^{39}P(E_n)=\frac1{14}\cdot 1,
$$
using the fact that the events $E_n$ are mutually exclusive and exhaustive. I offer one final method which is more direct, but leads to a summation which is difficult to simplify. Let $F_n$ be the event that the $n^{th}$ card is the $\spadesuit A$. Then
$$
P(\spadesuit A\text{ first})=\sum_{n=1}^{52}P(\spadesuit A\text{ first }|F_n)P(F_n)=\sum_{n=1}^{52}\frac{\binom{52-n}{13}}{\binom{51}{13}}\cdot\frac1{52}
$$
You can simplify this to $1/14$ using the hockey-stick identity: $$\sum_{n=1}^{52}\binom{52-n}{13}=\sum_{m=13}^{51}\binom{m}{13}=\binom{52}{14}$$
|
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|
2,750,595 |
This is a simple integral. $$
\int \frac{1}{3x}dx
$$ with an equally simple solution of $$
\frac{1}{3}\ln|x| +c
$$ My question is that if you chose to use u-substitution and used u = 3x, the solution appears to work out as follow: $$
\int \frac{1}{u} \frac {du}{3}
$$
$$
\frac{1}{3} \ln|3x|+c
$$ which seems correct as well. Is this in fact correct? The 2 graphs appear nothing alike.
|
It's the same since you have a constant of integration..
$$ \frac{1}{3} \ln|3x|+c= \frac 13\ln |x|+ \underbrace{\frac{1}{3} \ln|3|+c}_{ \text { is a constant } }=\frac{1}{3} \ln|x|+K$$
|
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|
2,755,987 |
Show that $0,2,4$ are the eigenvalues for the matrix $A$:
$$A=\pmatrix{
2 & -1 & -1 & 0 \\
-1 & 3 & -1 & -1 \\
-1 & -1 & 3 & -1 \\
0 & -1 & -1 & 2 \\
}$$
and conclude that $0,2,4$ are the only eigenvalues for $A$. I know that you can find the eigenvalues by finding the $\det(A-\lambda \cdot I)$, but it seems to me that the computation will be rather difficult to compute as it is a $4 \times 4$ matrix. My question: is there an easier method to calculate the eigenvalues of $A$?
And if I have to conclude that these are the only eigenvalues, is there a theorem that argues how many eigenvalues a matrix can have?
|
$A$ has zero row sums. Therefore the all-one vector $\mathbf e$ is an eigenvector of $A$ for the eigenvalue $0$. Since $A$ is also real symmetric, $\mathbf e$ can be extended to an orthogonal eigenbasis $\{\mathbf u,\mathbf v,\mathbf w,\mathbf e\}$ of $A$. But this is also an eigenbasis of $A+\mathbf e\mathbf e^\top$. Hence the spectrum of $A$ is $\{a,b,c,0\}$ if and only if the spectrum of $A+\mathbf e\mathbf e^\top$ is $\{a,b,c,\|\mathbf e\|^2\}=\{a,b,c,4\}$. It is easy to see that four eigenvalues of
$$
A+\mathbf e\mathbf e^\top=\pmatrix{3&0&0&1\\ 0&4&0&0\\ 0&0&4&0\\ 1&0&0&3}
$$
are $2,4,4,4$. Therefore the eigenvalues of $A$ are $2,4,4,0$.
|
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|
2,758,249 |
I have trouble understanding this concept. Why is it necessary to prove that addition or multiplication is well defined in equivalence classes? My understanding of equivalence classes is that it must be reflexive, symmetric and transitive. Doesn't proving it automatically imply that addition and multiplication can be done? Why the additional need to prove that it is 'well defined'? Apologies if this question is too trivial; my understanding of this topic is limited.
|
Consider this equivalence relation on $\mathbb N$:
$$a \sim b \quad\text{iff}\quad \lfloor a/10 \rfloor = \lfloor b/10 \rfloor $$
which says that two naturals are related if they differ only in their last digit. This is a perfectly good equivalence relation, but we can't extend addition to equivalence classes by the same rule that works for modular arithmetic: $$ [a]_\sim + [b]_\sim = [a+b]_\sim $$ The problem is that the sum of two equivalence classes now depend on which representatives we use to define their sum. For example, the rule seems to imply that
$$ [11]_\sim + [32]_\sim = [43]_\sim \\
[17]_\sim + [35]_\sim = [52]_\sim $$
However $[11]_\sim$ is the same equivalence class as $[17]_\sim$, and $[32]_\sim = [35]_\sim$, but $[43]_\sim$ is not the same as $[52]_\sim$. So the definition doesn't really tell us which of the equivalence classes should be the result of
$$ \{10,11,\ldots,19\} + \{30,31,\ldots,39\} $$ Checking that addition on the equivalence classes "is well-defined" means convincing oneself that this situation does not occur for the equivalence relation you're looking at.
|
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|
2,758,250 |
Determine whether the following fields are Galois over $\mathbb{Q}$ . (a) $\mathbb{Q}(\omega )$ , where $\omega = exp(2\pi i/3)$ . (b) $\mathbb{Q}(\sqrt[4]{2})$ . (c) $\mathbb{Q}(\sqrt{5}, \sqrt{7})$ . I have already shown that (a) is Galois over $\mathbb{Q}$ , and (b) is not. For (c), I know that $\mathbb{Q}(\sqrt{5},\sqrt{7}) = \mathbb{Q}(\sqrt{5} + \sqrt{7})$ , but I couldn't conclude. Any hint? PS: Use only results of automorphism and Galois extensions. Definition. Let $K$ be an algebraic extension of $F$ . Then $K$ is Galois over $F$ if $F = \mathcal{F}(Gal(K/F))$ . $\mathcal{F}$ is the fixed field. Proposition. Let $K$ be a field extension of $F$ . Then $K/F$ is Galois if and only if $|Gal(K/F)|=[K:F]$ .
|
Consider this equivalence relation on $\mathbb N$:
$$a \sim b \quad\text{iff}\quad \lfloor a/10 \rfloor = \lfloor b/10 \rfloor $$
which says that two naturals are related if they differ only in their last digit. This is a perfectly good equivalence relation, but we can't extend addition to equivalence classes by the same rule that works for modular arithmetic: $$ [a]_\sim + [b]_\sim = [a+b]_\sim $$ The problem is that the sum of two equivalence classes now depend on which representatives we use to define their sum. For example, the rule seems to imply that
$$ [11]_\sim + [32]_\sim = [43]_\sim \\
[17]_\sim + [35]_\sim = [52]_\sim $$
However $[11]_\sim$ is the same equivalence class as $[17]_\sim$, and $[32]_\sim = [35]_\sim$, but $[43]_\sim$ is not the same as $[52]_\sim$. So the definition doesn't really tell us which of the equivalence classes should be the result of
$$ \{10,11,\ldots,19\} + \{30,31,\ldots,39\} $$ Checking that addition on the equivalence classes "is well-defined" means convincing oneself that this situation does not occur for the equivalence relation you're looking at.
|
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|
2,762,306 |
Which is greater between $$\sqrt[8]{8!}$$ and $$\sqrt[9]{9!}$$? I want to know if my proof is correct... \begin{align}
\sqrt[8]{8!} &< \sqrt[9]{9!} \\
(8!)^{(1/8)} &< (9!)^{(1/9)} \\
(8!)^{(1/8)} - (9!)^{(1/9)} &< 0 \\
(8!)^{(9/72)} - (9!)^{8/72} &< 0 \\
(9!)^{8/72} \left(\left( \frac{8!}{9!} \right)^{(1/72)} - 1\right) &< 0 \\
\left(\frac{8!}{9!}\right)^{(1/72)} - 1 &< 0 \\
\left(\frac{8!}{9!}\right)^{(1/72)} &< 1 \\
\left(\left(\frac{8!}{9!}\right)^{(1/72)}\right)^{72} &< 1^{72} \\
\frac{8!}{9!} < 1 \\
\frac{1}{9} < 1 \\
\end{align} if it is not correct how it would be?
|
$$(\sqrt[8]{8!})^ {72}= (8!)^9 = (8!) (8!)^8 $$ $$(\sqrt[9]{9!})^ {72} = (9!)^8 = (9\times 8!)^8 = 9^8 (8!)^8$$ The second one, wins hands down.
|
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|
2,762,317 |
Let $\beta>0$, $x\ge 0$, $\alpha\in\mathbb{C}$. Compute $$\int_{-\infty}^\infty\dfrac{k^2+\beta^2+\beta}{(k^2+\alpha^2)(k^2+\beta^2)}\cos(kx)dk$$ Let $$f(k)=\dfrac{k^2+\beta^2+\beta}{(k^2+\alpha^2)(k^2+\beta^2)}$$ If $\alpha^2\in\mathbb{R}$, the integral equals $$\Re\left(\lim_{R\to\infty}\left(\int_Cf(k)e^{ikx}dk-\int_{C_R}f(k)e^{ikx}dk\right)\right)$$ where $C_R=\{|k|=R,\arg k\in(0,\pi)\}$ and $C=C_R\cup[-R,R]$. The integral over $C$ can be computed using Cauchy's residue theorem, with special cases $\alpha^2=\beta^2$ and $\alpha^2=\beta^2+\beta$. For all $k\in C_R$, $|e^{ikx}|\le 1$ since $\Im(k),x\ge 0$ and $|f(k)|=\Theta(R^{-2})$. $C_R$ has length $\pi R$, so the integral over $C_R$ has magnitude at most $\Theta(R^{-1})$, which tends to $0$ as $R\to\infty$, so $$\lim_{R\to\infty}\int_{C_R}f(k)e^{ikx}dk=0$$ so we can compute the integral when $\alpha^2\in\mathbb{R}$. If $\alpha^2\not\in\mathbb{R}$, $f(k)\Re(e^{ikx})\neq \Re(f(k)e^{ikx})$, so this approach doesn't work. What do we do then? This integral arises in the Fourier cosine transform solution to $y''-\alpha^2y=e^{-\beta x}$ for all $x\ge 0$ with $y'(0)=1$ and $y$ bounded as $x\to\infty$.
|
$$(\sqrt[8]{8!})^ {72}= (8!)^9 = (8!) (8!)^8 $$ $$(\sqrt[9]{9!})^ {72} = (9!)^8 = (9\times 8!)^8 = 9^8 (8!)^8$$ The second one, wins hands down.
|
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|
2,763,341 |
When in high school I used to see mathematical objects as ideal objects whose existence is independent of us. But when I learned set theory, I discovered that all mathematical objects I was studying were sets, for example: $ 0 = \emptyset $ What does it mean to say that the number $ 1 $ is the singleton set of the empty set? Thank you all for the answers, it is helping me a lot. Obs. This question received far more attention than I expected it to do.
After reading all the answers and reflecting on it for a while I've come to the following conclusions (and I would appreciate if you could add something to it or correct me):
Let's take a familiar example, the ordered pair. We have an intuitive, naive notion of this 'concept' and of its fundamental properties like, for example, it has two components and $ (x, y) = (a, b) $ iff $ x = a $ and $ y = b $ But mathematicians find it more convenient (and I agree) to define this concept in terms of set theory by saying that $ (x, y) $ is a shortcut for the set { {x}, {x, y} } and then proving the properties of the ordered pair. I.e. showing that this set-theoretical ordered pair has all the properties one expect the 'ideal' ordered pair to have. Secondly, mathematicians don't really care much about these issues.
|
Warning: personal opinion ahead! It depends upon what the meaning of the word "is" is. One way to think about it - not necessarily historically correct - is the following: that the axioms of set theory are intricate enough - or, if you prefer, describe structures (their models ) which are intricate enough - to "implement" all of mathematics. This is analogous to the relation between algorithms (as clear-but-informal descriptions of processes) and programs (their actual implementation), with the observation that the same algorithm can be implemented in different ways. It's worth noting that this "in different ways" has multiple senses: there will be many ways to write a program which carries out a specific algorithm in a given programming language, and there are also lots of programming languages. Correspondingly, we have: There are lots of ways to implement (say) basic arithmetic inside ZFC; the von Neumann approach is just the standard one. There are also different theories which similarly are intricate enough to implement all of mathematics. Asking what $1$ "is" is an ontological question, but set theory doesn't need to be thought of ontologically - the pragmatic approach ("how can we formally and precisely implement mathematics?") is sufficient. I'm not claiming that this is the universal view; for example, one can also argue (although I don't) that the cumulative hierarchy (= the sets "built from" the emptyset) consists of all the mathematical objects which are guaranteed to exist, in a Platonic sense. But I think the view above probably more faithfully reflects the general attitude of the mathematical community, and is certainly how I approach the question.
|
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|
2,763,381 |
How do I compute $$\sum_{r=1}^{\infty} \frac{8r}{4r^4 +1}$$ Calculating first few terms tells me that the sum converges to 2.
I have also tried squeezing the term.
|
Partial fraction expansion gives us$$\begin{align*} & \sum\limits_{r=1}^n\frac {8r}{4r^4+1}=\sum\limits_{r=1}^n\frac 2{2r^2-2r+1}-\sum\limits_{r=1}^n\frac 2{2r^2+2r+1}\\ & =\left[2+\frac 2{5}+\frac 2{13}+\cdots+\frac 2{2n^2-2n+1}\right]-\left[\frac 2{5}+\frac 2{13}+\cdots+\frac 2{2n^2+2n+1}\right]\end{align*}$$Notice how all but the last fraction in the second sum cancels out with the fractions in the first sum. Continuing on indefinitely until the end gives us$$\sum\limits_{r=1}^n\frac {8r}{4r^4+1}=2-\frac 2{2n^2+2n+1}$$As $n\to\infty$, the fraction tends to zero, so your sum equals$$\sum\limits_{r\geq1}\frac {8r}{4r^4+1}=2$$ EDIT: To find the partial fraction decomposition, we first factor the denominator as a product of two quadratics. This can be done by adding and subtracting $4r^2$ so the quartic factors as$$4r^4+1=(2r^2+2r+1)(2r^2-2r+1)$$Now, the decomposition is set up as$$\frac {8r}{(2r^2+2r+1)(2r^2-2r+1)}=\frac {Ar+B}{2r^2-2r+1}+\frac {Cr+D}{2r^2+2r+1}$$
|
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|
2,763,383 |
Let $(0,0), (2,2), (4,0)$ be the vertices of a triangle. Then find the line $l$ that passes through $(1,-1)$ and divides the triangle in half (equal area).
I think can solve this problem by Letting $P$ and $Q$ be the intersection of the line and the triangle and $m$ be the slope of $l$.
$$
\frac{y-1}{x+1}=m\\
$$ Then the height of the $\triangle APQ $ is $$
\frac{|mx-y+m+1|}{\sqrt{m^2+1}}=\frac{|5m+1|}{\sqrt{m^2+1}}
$$
and then finding the distance between $P$ and $Q$... Finding the half area in terms of $m$...But is there more clean and easy way to do this? Am I missing something obvious?
|
Partial fraction expansion gives us$$\begin{align*} & \sum\limits_{r=1}^n\frac {8r}{4r^4+1}=\sum\limits_{r=1}^n\frac 2{2r^2-2r+1}-\sum\limits_{r=1}^n\frac 2{2r^2+2r+1}\\ & =\left[2+\frac 2{5}+\frac 2{13}+\cdots+\frac 2{2n^2-2n+1}\right]-\left[\frac 2{5}+\frac 2{13}+\cdots+\frac 2{2n^2+2n+1}\right]\end{align*}$$Notice how all but the last fraction in the second sum cancels out with the fractions in the first sum. Continuing on indefinitely until the end gives us$$\sum\limits_{r=1}^n\frac {8r}{4r^4+1}=2-\frac 2{2n^2+2n+1}$$As $n\to\infty$, the fraction tends to zero, so your sum equals$$\sum\limits_{r\geq1}\frac {8r}{4r^4+1}=2$$ EDIT: To find the partial fraction decomposition, we first factor the denominator as a product of two quadratics. This can be done by adding and subtracting $4r^2$ so the quartic factors as$$4r^4+1=(2r^2+2r+1)(2r^2-2r+1)$$Now, the decomposition is set up as$$\frac {8r}{(2r^2+2r+1)(2r^2-2r+1)}=\frac {Ar+B}{2r^2-2r+1}+\frac {Cr+D}{2r^2+2r+1}$$
|
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|
2,764,985 |
Why is the difference between these two functions a constant? $$f(x)=\frac{2x^2-x}{x^2-x+1}$$
$$g(x)=\frac{x-2}{x^2-x+1}$$ Since the denominators are equal and the numerators differ in degree I would never have thought the difference of these functions would be a constant. Of course I can calculate it is true: the difference is $2$, but my intuition is still completely off here. So, who can provide some intuitive explanation of what is going on here? Perhaps using a graph of some kind that shows what's special in this particular case? Thanks! BACKGROUND: The background of this question is that I tried to find this integral: $$\int\frac{x dx}{(x^2-x+1)^2}$$ As a solution I found: $$\frac{2}{3\sqrt{3}}\arctan\left(\frac{2x-1}{\sqrt{3}}\right)+\frac{2x^2-x}{3\left(x^2-x+1\right)}+C$$ Whereas my calculusbook gave as the solution: $$\frac{2}{3\sqrt{3}}\arctan\left(\frac{2x-1}{\sqrt{3}}\right)+\frac{x-2}{3\left(x^2-x+1\right)}+C$$ I thought I made a mistake but as it turned out, their difference was constant, so both are valid solutions.
|
It is just a bit of clever disguise. Take any polynomial $p(x)$ with leading term $a_n x^n$. Now consider
$$\frac{p(x)}{p(x)}$$
This is clearly the constant $1$ (except at zeroes of $p(x)$). Now separate the leading term:
$$\frac{a_n x^n}{p(x)} + \frac{p(x) - a_n x^n}{p(x)}$$ and re-write to create the difference: $$\frac{a_n x^n}{p(x)} - \frac{a_n x^n - p(x)}{p(x)}$$ Obviously the same thing and hence obviously still $1$ but the first has a degree $n$ polynomial as its numerator and the second a degree $n - 1$ or less polynomial. Similarly, you could split $p(x)$ in many other ways.
|
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|
2,769,357 |
In an exam with $12$ yes/no questions with $8$ correct needed to pass, is it better to answer randomly or answer exactly $6$ times yes and 6 times no, given that the answer 'yes' is correct for exactly $6$ questions? I have calculated the probability of passing by guessing randomly and it is $$\sum_{k=8}^{12} {{12}\choose{k}}0.5^k0.5^{n-k}=0.194$$ Now given that the answer 'yes' is right exactly $6$ times, is it better to guess 'yes' and 'no' $6$ times each? My idea is that it can be modelled by drawing balls without replacement. The balls we draw are the correct answers to the questions. Looking at the first question, we still know that there are $6$ yes and no's that are correct. The chance that a yes is right is $\frac{6}{12}$ and the chance that a no is right is also $\frac{6}{12}$ . Of course the probability in the next question depends on what the first right answer was. If yes was right, yes will be right with a probability of $5/11$ and a no is right with the chance $6/11$ . If no was right, the probabilities would change places. Now that we have to make the choice $12$ times and make the distinction which one was right, we get $2^{12}$ paths total. We cannot know what the correct answers to the previous questions were. So we are drawing $12$ balls at once, but from what urn? It cannot contain $24$ balls with $12$ yes and $12$ no's. Is this model even correct? Is there a more elegant way to approach that? I am asking for hints, not solutions, as I'm feeling stuck. Thank you. Edit : After giving @David K's answer more thought, I noticed that the question can be described by the hypergeometric distribution , which yields the desired result.
|
We are given the fact that there are $12$ questions, that $6$ have the correct answer "yes" and $6$ have the correct answer "no." There are $\binom{12}{6} = 924$ different sequences of $6$ "yes" answers and $6$ "no" answers.
If we know nothing that will give us a better chance of answering any question
correctly than sheer luck, the most reasonable assumption is that every possible sequence of answers is equally likely, that is, each one has
$\frac{1}{924}$ chance to occur. So guess "yes" $6$ times and "no" $6$ times. I do not care how you do that:
you may guess "yes" for the first $6$, or flip a coin and answer "yes" for heads and "no" for tails until you have used up either the $6$ "yeses" or the $6$ "noes" and the rest of your answers are forced, or you can put $6$ balls labeled "yes" and $6$ labeled "no" in an urn, draw them one at a time, and answer the questions in that sequence. No matter what you do, you end up with some sequence of "yes" $6$ times and "no" $6$ times. You get $12$ correct if and only if the sequence of correct answers is exactly the same as your sequence.
That probability is $\frac{1}{924}.$ There is no way for you to get $11$ correct. You get $10$ correct if and only if the correct answers are "yes" on $5$ of your "yes" answers and "no" on your other "yes" answers.
The number of ways this can happen is the number of ways to choose $5$ correct answers from your $6$ "yes" answers, times the number of ways to choose $5$ correct answers from your $6$ "no" answers:
$\binom 65 \times \binom 65 = 36.$ There is no way for you to get $9$ correct. You get $8$ correct if and only if the correct answers are "yes" on $4$ of your "yes" answers and "no" on your other "yes" answers.
The number of ways this can happen is the number of ways to choose $4$ correct answers from your $6$ "yes" answers, times the number of ways to choose $4$ correct answers from your $6$ "no" answers:
$\binom 64 \times \binom 64 = 225.$ In any other case you fail. So the chance to pass is
$$
\frac{1 + 36 + 225}{924} = \frac{131}{462} \approx 0.283550,
$$
which is much better than the chance of passing if you simply toss a coin for each individual question
but not nearly as good as getting $4$ or more heads in $6$ coin tosses. Just to check, we can compute the chance of failing in the same way:
$6$ answers correct ($3$ "yes" and $3$ "no"), $4$ answers correct,
$2$ correct, $0$ correct. This probability comes to
$$
\frac{\binom 63^2 + \binom 62^2 + \binom 61^2 + 1}{924}
= \frac{400 + 225 + 36 + 1}{924}
= \frac{331}{462} \approx 0.716450,
$$
which is the value needed to confirm the answer above.
|
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|
2,770,854 |
In college and later the first year of university(Engineering), I was taught that you can multiply the constant of integration by a constant value and it doesn't change, like in these examples: $$
y = \frac{1}{5}\int dx = \frac{x+C}{5} = \frac{x}{5} + C
$$ $$
y = e^{\int dx} = e^{x+C} = Ce^x
$$ I get what's happening here, but is it considered bad form to do this? Should I create another constant, say $K$, and let this equal (in the first example) $\frac{C}{5}$ so I can say that $y=\frac{x}{5}+K$, or is it just accepted that it's slightly iffy but everyone understands what you've done?
|
It is considered bad form. The usual approach, at least in a final presentation (as opposed to when you're actually doing the calculations), is to know how many different $C$s you need, and use either $C',C''$, etc. until you get to the final one, which is just $C$ (if there aren't too many, $C'''''$ is a bit ridiculous), or use indices: $C_1,C_2,$ etc. If what you're writing is only meant for your eyes, you can really do whatever you feel like; mathematical notation conventions are there to facilitate communication between people, not to put restrictions on what people write down as their own personal notes. That being said, eliminating a potential source of confusion at nearly no cost of writing speed or cognitive load sounds like a good thing to me, so I would suggest you use something like this in those cases as well.
|
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|
2,772,612 |
Let us roll a fair die $4$ independent times, and denote the outcomes as $X_1, X_2, X_3$ and $X_4$. What is the probability of $X_1+X_2+ X_3+X_4 > X_1X_2 X_3X_4$? My try:
I could get the answer for $2$ rolling case by enumerating possibilities, but couldn't get this larger problem. Can someone help me with this? Thanks in advance for any help!
|
Case work: If 3 or 4 of the rolls come up 1, it's straightforward to see that the desired inequality holds. If 0 or 1 of the rolls come up 1, then we can show that the inequality never holds. (The product of three integers, all $>1$, is always at least 2 greater than their sum.) The remaining case is when exactly 2 of the rolls come up 1. Empirically, you can find that the only case when the inequality is satisfied is when the rolls are some permutation of $3,2,1,1$ or $2,2,1,1$. There is only 1 combination of all 1's, $5\times 4=20$ combinations with three 1's, $4\times 3=12$ permutations of $3,2,1,1$, and $\binom{4}{2}=6$ permutations of $2,2,1,1$. This adds up to a total of 39 rolls that satisfy the inequality. There are $6^4=1296$ total rolls, so the probability that the inequality holds is $39/1296 \approx .03009$.
|
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|
2,773,097 |
I knew that for $ x^2 + y^2 = 1$ the x and y can be expressed by introducing one more variable where $\ m=y/(x+1) $, then $\ x= 2m/(1+m^2) $ and $\ y= (1-m^2)/(1+m^2) $. What about $\ x^2 + 3y^2 = 7 $, should I divide the equation by 7 in order to get the 1 at the right hand side ? Then how to deal with the $\ 3y^2 $ ? Thank you!
|
The key here is to understand where the substitution $$m = \frac{y}{x + 1}$$ comes from. Geometrically, this variable represents the slope of the line between a point $(x, y)$ and the point $(-1, 0)$. What you are then doing is considering a rational slope $m$, taking the line $L$ through $(-1,0)$ of slope $m$, and solving for the second intersection point of this line with the conic $x^2+y^2=1$ (the first intersection point being $(-1,0)$). This same approach works for any conic, as long as you have a single rational point on the conic to play the role of $(-1,0)$. In the case of $x^2+3y^2=7$, for instance, you could take $(2,1)$ as your initial point. So then you would define $$m = \frac{y - 1}{x - 2}$$ and solve for $(x,y)$ as the second point where the line $L$ through $(2,1)$ of slope $m$ intersects the conic $x^2+3y^2=7$.
|
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|
2,775,198 |
Suppose I have a $n\times n$ matrix $A$. Can I, by using only pre- and post-multiplication by permutation matrices, permute all the elements of $A$? That is, there should be no binding conditions, like $a_{11}$ will always be to the left of $a_{n1}$, etc. This seems to be intuitively obvious. What I think is that I can write the matrix as an $n^2$-dimensional vector, then I can permute all entries by multiplying by a suitable permutation matrix, and then re-form a matrix with the permuted vector.
|
It is not generally possible to do so. For a concrete example, we know that there can exist no permutation matrices $P,Q$ such that $$
P\pmatrix{1&2\\2&1}Q = \pmatrix{2&1\\2&1}
$$ If such a $P$ and $Q$ existed, then both matrices would necessarily have the same rank.
|
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|
2,776,158 |
Problem: A vertex of one square is pegged to the centre of an identical square, and the overlapping area is blue. One of the squares is then rotated about the vertex and the resulting overlap is red. Which area is greater? Let the area of each large square be exactly $1$ unit squared. Then, the area of the blue square is exactly $1/4$ units squared. The same would apply to the red area if you were to rotate the square $k\cdot 45$ degrees for a natural number $k$ . Thus, I am assuming that no area is greater, and that it is a trick question $-$ although the red area might appear to be greater than the blue area, they are still the same: $1/4$ . But how can it be proven ? I know the area of a triangle with a base $b$ and a height $h\perp b$ is $bh\div 2$ . Since the area of each square is exactly $1$ unit squared, then each side would also have a length of $1$ . Therefore, the height of the red triangle area is $1/2$ , and so $$\text{Red Area} = \frac{b\left(\frac 12\right)}{2} = \frac{b}{4}.$$ According to the diagram, the square has not rotated a complete $45$ degrees, so $b < 1$ . It follows, then, that $$\begin{align} \text{Red Area} &< \frac 14 \\ \Leftrightarrow \text{Red Area} &< \text{Blue Area}.\end{align}$$ Assertion: To conclude, the $\color{blue}{\text{blue}}$ area is greater than the $\color{red}{\text{red}}$ area. Is this true? If so, is there another way of proving the assertion? Thanks to users who commented below, I did not take account of the fact that the red area is not a triangle $-$ it does not have three sides! This now leads back to my original question on whether my hypothesis was correct. This question is very similar to this post . Source: The Golden Ratio (why it is so irrational) $-$ Numberphile from $14$ : $02$ .
|
The four numbered areas are congruent. [Added later] The figure below is from a suggested edit by @TomZych, and it shows the congruent parts more clearly. Given all the upvotes to the (probably tongue-in-cheek) comment “This answer also deserves the tick for artistic reasons,” I’m leaving my original “artistic” figure but also adding Tom’s improved version to my answer.
|
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|
2,776,864 |
I' m trying to prove that $$A=\begin{pmatrix}
4 & 2 & 0 & 0 & 0 \\
2 & 5 & 2 & 0 & 0 \\
0 & 2 & 5 & 2 & 0 \\
0 & 0 & 2 & 5 & 2 \\
0 & 0 & 0 & 2 & 5 \\
\end{pmatrix}$$ admits a Cholesky decomposition . $A$ is symmetric, so it admits a Cholesky decomposition iff it is positive definite . The only methods I know for checking this are: $X^tAX > 0, \quad \forall X \in \mathbb{K}^n- \{0\}$. If $\lambda$ is an eigenvalue of $A$, then $\lambda>0.$ I have failed to prove it using 1 and 2 is taking me so much time. Is there any easier way to do this, given that $A$ is tridiagonal ?
|
Notice $A$ can be rewritten as a sum of 5 matrices.
$$A =
\left[\begin{smallmatrix}
2 & 0 & 0 & 0 & 0\\
0 & 1 & 0 & 0 & 0\\
0 & 0 & 1 & 0 & 0\\
0 & 0 & 0 & 1 & 0\\
0 & 0 & 0 & 0 & 3
\end{smallmatrix}\right] +
\left[\begin{smallmatrix}
2 & 2 & 0 & 0 & 0\\
2 & 2 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0
\end{smallmatrix}\right] +
\left[\begin{smallmatrix}
0 & 0 & 0 & 0 & 0\\
0 & 2 & 2 & 0 & 0\\
0 & 2 & 2 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0
\end{smallmatrix}\right] +
\left[\begin{smallmatrix}
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 2 & 2 & 0\\
0 & 0 & 2 & 2 & 0\\
0 & 0 & 0 & 0 & 0
\end{smallmatrix}\right] +
\left[\begin{smallmatrix}
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 2 & 2\\
0 & 0 & 0 & 2 & 2
\end{smallmatrix}\right]
$$
The first matrix
is diagonal with positive entries on diagonals, so it is positive definite.
The remaining four matrices are clearly positive semi-definite. Being a sum
of a positive definite matrix with a bunch of positive semi-definite matrices,
$A$ itself is positive definite.
|
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|
2,778,826 |
The sequence in question is
$$
a_n = \sqrt{n^2+4}-n\,.
$$
I can see that it is strictly decreasing by finding the derivative and observing that it is negative on the entire range containing the relevant values of $n$. But I feel like there must be a simpler algebraic reason that I'm just not seeing.
|
Notice that we have $$\sqrt{n^2+4}-n = \frac{4}{\sqrt{n^2+4}+n}$$ hence as $n$ increases, the expression decreases as the denominator is positive and increases while the numerator is a positive constant.
|
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|
2,780,012 |
The free group over a set $S$ only includes finitely-long words made up of letters from $S$ and their inverses. It seems natural to me to also allow infinitely long words. While this would obviously be impossible to operationalize on a computer, in principle these infinitely-long group elements seem perfectly well-defined. Is there some fundamental logical problem with including infinitely long words? If not, has this concept ever been studied? (I know that such a generalization would no longer count as being "generated" by $S$ - since by definition every group element needs to be reachable by a finite number of generator multiplications - so you'd need to find a new name.)
|
This doesn't work if you just do it naively. The most naive definition of an "infinite word" would be an infinite string $s_1s_2s_3\dots$ where each $s_n$ is an element of $S$ or the formal inverse of an element of $S$. This fails horribly, since such strings are not closed under composition or inverses. For instance, the inverse of such an infinite string would be infinitely long on the left, instead of on the right. And if you concatenate two such infinite strings, you would get a string of the form $(s_1s_2s_3\dots)(t_1t_2t_3\dots)$ where the "letters" in the word are now arranged in a sequence indexed by the ordinal $\omega+\omega$, instead of just by the natural numbers. So, to make sense of this idea, you need to allow a more exotic variety of "infinite words" that can have many different infinite totally ordered sets as their index sets. Another important obstacle is the Eilenberg swindle . Namely, let $s\in S$ (or more generally let $s$ be any word) and consider the infinite word $w=sss\dots$. Then any reasonable definition of a "group of infinite words" should have $sw=w$, which then implies $s=1$! So if you want your group to be nontrivial, you need to impose some restriction that disallows words of this type. However, the news is not all bad! There does exist at least one interesting construction of a "free group with infinite words" (there are probably others too; I don't know the literature on this subject). You can find the details in Section 3 of the nice paper The Combinatorial Structure of the Hawaiian Earring Group by J. W. Cannon and G. R. Conner. Specifically, Cannon and Conner define a "transfinite word" on a set $S$ to be a map $f:I\to S\cup S^{-1}$ where $I$ is a totally ordered set, $S^{-1}$ is the set of formal inverses of elements of $S$, and each fiber of $f$ is finite. So, you can have a word indexed by any totally ordered set, as long as each element of $S$ only appears in it only finitely many times (this finiteness condition avoids Eilenberg swindles). Two words $f:I\to S\cup S^{-1}$ and $g:J\to S\cup S^{-1}$ are identified if there is an order-isomorphism $I\cong J$ which turns $f$ into $g$. They then define an equivalence relation on such words using a sort of cancellation, prove that each word is equivalent to a unique "reduced" word, and use this to define a "big free group" on $S$ consisting of transfinite words modulo equivalence (or equivalently, reduced transfinite words). When $S$ is countably infinite, this group is isomorphic to the fundamental group of the Hawaiian earring . Sadly, this construction is only really of interest when $S$ is infinite, since when $S$ is finite all words must be finite and you just get the ordinary free group.
|
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|
2,780,138 |
In Season 5 Episode 16 of Agents of Shield, one of the characters decides to prove she can't die by pouring three glasses of water and one of poison; she then randomly drinks three of the four cups. I was wondering how to compute the probability of her drinking the one with poison. I thought to label the four cups $\alpha, \beta, \gamma, \delta$ with events $A = \{\alpha \text{ is water}\}, \ a = \{\alpha \text{ is poison}\}$ $B = \{\beta \text{ is water}\},\ b = \{\beta \text{ is poison}\}$ $C = \{\gamma \text{ is water}\},\ c = \{\gamma \text{ is poison}\}$ $D = \{\delta \text{ is water}\},\ d = \{\delta \text{ is poison}\}$ If she were to drink in order, then I would calculate $P(a) = {1}/{4}$.
Next $$P(b|A) = \frac{P(A|b)P(b)}{P(A)}$$ Next $P(c|A \cap B)$, which I'm not completely sure how to calculate. My doubt is that I shouldn't order the cups because that assumes $\delta$ is the poisoned cup. I am also unsure how I would calculate the conditional probabilities (I know about Bayes theorem, I mean more what numbers to put in the particular case). Thank you for you help.
|
The probability of not being poisoned is exactly the same as the following problem: You choose one cup and drink from the other three. What is the probability of choosing the poisoned cup (and not being poisoned)? That probability is 1/4. Therefore, the probability of being poisoned is 3/4.
|
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|
2,780,582 |
I was solving a question from the Regional Math Olympiad (RMO) 2014. Find all positive real numbers $x,y,z$ such that $$2x-2y+\frac1z=\frac1{2014},\quad2y-2z+\frac1x=\frac1{2014},\quad2z-2x+\frac1y=\frac1{2014}$$ Here's my solution: These expressions are cyclic. Therefore all solution sets must be unordered. This implies that $x=y=z$ . Thus, $x=2014$ and the solution is $$x=2014\quad y=2014\quad z=2014$$ Here's the official solution: Adding the three equations, we get $$\frac1x+\frac1y+\frac1z=\frac3{2014}$$ We can also write them as $$2xz-2yz+1=\frac z{2014},\quad2xy-2xz+1=\frac x{2014},\quad2yz-2xy+1=\frac y{2014}$$ Adding these, we get $$x+y+z=3\times2014$$ Therefore, $$\left(\frac1x+\frac1y+\frac1z\right)(x+y+z)=9$$ Using $\text{AM-GM}$ inequality, we therefore obtain $$9=\left(\frac1x+\frac1y+\frac1z\right)(x+y+z)\ge9\times(xyz)^{\frac13}\left({1\over xyz}\right)^{\frac13}=9$$ Hence equality holds and we conclude that $x=y=z$ . Thus we conclude $$x=2014\quad y=2014\quad z=2014$$ What I wonder is if there is something wrong with my approach. If yes, what is it? If no, then why is the official solution so long winded?
|
Consider the system of equations $xy + z = 1, \quad yz + x = 1, \quad zx + y = 1$ These equations are related by cyclic permutations of $(x,y,z)$, but they are satisified by $(1,1,0)$ (and its cyclic permutations) when $x$, $y$ and $z$ are not all equal. There are also solutions where $x=y=z=\frac{\pm \sqrt{5}-1}{2}$, but these are not the only solutions.
|
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|
2,782,717 |
I know how basic operations are performed on matrices, I can do transformations, find inverses, etc. But now that I think about it, I actually don't "understand" or know what I've been doing all this time. Our teacher made us memorise some rules and I've been following it like a machine. So what exactly is a matrix? And what is a determinant? What do they represent? Is there a geometrical interpretation? How are they used? Or, rather, what for are they used? How do I understand the "properties" of matrix? I just don't wanna mindlessly cram all those properties, I want to understand them better. Any links, which would improve my understanding towards determinants and matrices? Please use simpler words. Thanks :)
|
A matrix is a compact but general way to represent any linear transform. (Linearity means that the image of a sum is the sum of the images.) Examples of linear transforms are rotations, scalings, projections. They map points/lines/planes to point/lines /planes. So a linear transform can be represented by an array of coefficients. The size of the matrix tells you the number of dimension of the domain and the image spaces. The composition of two linear transforms corresponds to the product of their matrices. The inverse of a linear transform corresponds to the matrix inverse. A determinant measures the volume of the image of a unit cube by the transformation; it is a single number. (When the number of dimensions of the domain and image differ, this volume is zero, so that such "determinants" are never considered.) For instance, a rotation preserves the volumes, so that the determinant of a rotation matrix is always 1. When a determinant is zero, the linear transform is "singular", which means that it loses some dimensions (the transformed volume is flat), and cannot be inverted. The determinants are a fundamental tool in the resolution of systems of linear equations. As you will later learn, a linear transformation can be decomposed in a pure rotation, a pure (anisotropic) scaling and another pure rotation. Only the scaling deforms the volumes, and the determinant of the transform is the product of the scaling coefficients.
|
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|
2,785,718 |
For example: $12$ reversed is $21$ and $12$ + $9$ = $21$. $17$ with the two values swapped is $71$, and $17$ + $9$ + $9$ + $9$ + $9$ + $9$ + $9$ = $71$. Take the number $123$ and add $9$ a total of $22$ times you get $321$, which is the reverse. It seems to work for every number. Why is this the case? Is it just a addition thing?
|
Any number is congruent modulo $9$ to the sum of its digits:
$$a_na_{n-1}\ldots a_1a_0\equiv a_n+a_{n-1}+\ldots +a_1+a_0\pmod 9$$
If you reverse the digits, the sum is unchanged. Therefore $$a_0a_1\ldots a_{n-1}a_n\equiv a_na_{n-1}\ldots a_1a_0\pmod 9$$
which is equivalent to your statement. Note a stronger statement implied by the same reasoning: By adding or subtracting a suitable amount of $9$'s, you can reach any permutation of the digits.
|
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|
2,786,600 |
Is it possible to revert the softmax function in order to obtain the original values $x_i$ ? $$S_i=\frac{e^{x_i}}{\sum e^{x_i}} $$ In case of 3 input variables this problem boils down to finding $a$ , $b$ , $c$ given $x$ , $y$ and $z$ : \begin{cases}
\frac{a}{a+b+c} &= x \\
\frac{b}{a+b+c} &= y \\
\frac{c}{a+b+c} &= z
\end{cases} Is this problem solvable?
|
Note that in your three equations you must have $x+y+z=1$.
The general solution to your three equations are $a=kx$, $b=ky$, and $c=kz$ where $k$ is any scalar. So if you want to recover $x_i$ from $S_i$, you would note $\sum_i S_i = 1$ which gives the solution $x_i = \log (S_i) + c$ for all $i$, for some constant $c$.
|
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|
2,787,356 |
Call a function $f : \mathcal{N} \rightarrow \mathcal{N}$ repetitive if for every finite sequence of natural numbers $(a_1, a_2, \cdots,a_n)$ there exists a number $k \in \mathcal{N}$ satisfying $f(k) = a_1$, $f(k+1) = a_2$, ... , $f(k + n -1) = a_n$. Show that if $f$ is repetitive, then for any combination of $a_i$'s there are actually infinitely many numbers $k$ with this property. I've been trying to figure out a proof for this for a while now. I'm feeling that the fact that the number $k$ exists for every sequence allows us to manipulate other sequences or something like that and get infinitely many numbers that way. But I haven't been able to actually find a way to construct infinitely many numbers $k$ for a random sequence.
|
There's a $k$ with $f(k)=a_1,\ldots,f(k+n-1)=a_n$, $f(k+n)=1$. There's a $k'$ with $f(k')=a_1,\ldots,f(k'+n-1)=a_n$, $f(k'+n)=2$ etc.
|
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|
2,789,366 |
What is the name of the number that is being added to another number? I am writting a paper, and I want to know how to call the number that you add to another number (for example: 2+4, how can I call the four without saying the word "4"). Thanks.
|
Either addend or summand is correct. If you are adding more than two numbers, it might be more appropriate to use the term addend or summand to refer to any term being added, e.g.
$$\text{summand} + \text{summand} + \dotsb + \text{summand} = \text{sum (or total)} $$
or
$$\text{addend} + \text{addend} + \dotsb + \text{addend} = \text{sum (or total)}. $$
If you need to refer to a specific term, the phrase "the $n$-th summand (or addend)" gets the job done. In a case when you are adding exactly two numbers, you can use augend and addend , to refer to the first and second terms in a summation, i.e. you can write
$$\text{augend} + \text{addend} = \text{sum}.$$
That being said, this use of the term "addend" is not unambiguous: for example, Mathworld defines "summand" as a synonym of addend, with augend referring to the first summand (or addend) in any sum. Hence it might be inadvisable to use "addend" to refer to only the second term in a sum, as the term "addend" is not unambiguously understood in this way; it could refer to any term in a sum. Moreover, the term "augend" is somewhat esoteric, and its use will likely draw requests for clarification.
|
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|
2,789,484 |
1. THE CONTEXT OF THE PROBLEM This question came to me when I was exploring complex exponents. The key identity to computing expressions with complex exponents is the Euler's identity: $$e^{i\theta}=\cos\theta+i\sin\theta$$ This enables us to compute, for example what $2^i$ is by some algebraic manipulation. The computation is as follows: $$2^i=e^{\ln2^i}=e^{i\ln2}=\cos\ln2+i\sin\ln2\;\approx\;0.769+0.639i$$ 2. THE PROBLEM A question arises in my head. We compute sines and cosines with radian values. And since $\ln2\approx0.693$ then $\sin\ln 2$ will be the $y$ coordinate of the unit circle when the angle is $0.693$ radians . But if we use turns instead of radians, then $\sin\ln 2$ will be the $y$ coordinate of the unit circle when the angle is $0.693$ turns . So the value of sine when using turns or radians is different. Similarly, the value of cosine is different. But that creates a problem. When using radians , $2^i$ computes to about $0.769+0.639i$. But when using turns , $2^i$ computes to about $0.994+0.110i$. 3. POSSIBLE IMPLICATIONS The problem above illustrates that $2^i$ and any expression with complex exponent is a generalisation that directly depends on what units we use for angles. There are only two possibilities of "the state of truth" that this fact implies. Either complex exponents are a concept completely made up by humans which would mean that expressions like these are really undefined to the code of the universe (we only make it defined because we think that radians are the true angle units), or there must be a unit that is the truest unit for measuring angles, whether that would be radians or something else. If the first statement is true, then this would mean that we can accept $2^i$ to be $0.769+0.639i$. After all, this concept would be defined by radians only because we defined it like this. However, if the second statement is true, then there are even more questions to be asked. If there is a "truest" unit for measuring angles then what is it? Perhaps radians are really the truest unit, meaning that $2^i$ is unambiguously $0.769+0.639i$, but if so, what justifies this fact? What makes radians truer than turns?
|
The actual definition of the complex exponential map is equivalently via an ODE ($f'=f$, $f(0)=1$) or via power series ($f(z)= \sum_{k\geq 0}\frac{z^k}{k!}$). Similarly, cosine and sine functions are defined via power series above all. If you are to change the way you measure angles and therefore define a new, distorted version of sine and cosine, then the most natural thing to say would be that the Euler formula simply doesn't hold in your new settings. Or you could use it as a new definition for a distorted exponential map, but then what is more important for you: the fact that Euler holds, or the fact that $\exp$ satisfies the above ODE? All of mathematics rely on the fact that we agree on axioms, language and definitions. How natural these are depend on how beautiful the resulting theorems seem to the trained eye.
|
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|
2,790,867 |
You have a standard deck of cards and randomly take one card away without looking at it and set it aside. What is the probability that you draw the Jack of Hearts from the pile now containing 51 cards? I'm confused by this question because if the card you removed from the pile was the Jack of Hearts then the probability would be zero so I'm not sure how to calculate it. Edit: I asked this question about a year ago because I was struggling to get an intuitive understanding of an important concept in probability, and the comments and answers were really helpful for me (especially the one about "no new information being added so the probability doesn't change").
|
The Hard Way The probability that the first card is not the Jack of Hearts is $\frac {51}{52}$ so the probability that the first card is not the Jack of Hearts and the second card is the Jack of Hearts is $\frac {51}{52}\times \frac 1{51}$. The probability that the first card is the Jack of Hearts is $\frac 1{52}$ so the probability that the first card is the Jack of Hearts and the second card is the Jack of Hearts is $\frac 1{52}\times 0$. So the total probability that the second card is Jack of Hearts is: The probability that the second card is after the first card is not + The probability that the second card is after the first card already was
$$= \frac {51}{52}\times \frac 1{51} + \frac 1{52}\times 0 = \frac 1{52} + 0 = \frac 1{52}$$ That was the hard way. The Easy Way The probability that any specific card is any specific value is $\frac 1{52}$. It doesn't matter if it is the first card, the last card, or the 13th card. So the probability that the second card is the Jack of Hearts is $\frac 1{52}$. Picking the first card and not looking at, just going directly to the second card, putting the second card in an envelope and setting the rest of the cards on fire, won't make any difference; all that matters is the second card has a one in $52$ chance of being the Jack of Hearts. Any thing else just wouldn't make any sense. The thing is throwing in red herrings like "what about the first card?" doesn't change things and if you actually do try to take everything into account, the result, albeit complicated, will come out to be the same.
|
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|
2,791,002 |
Why are the curves of exponential, log, and parabolic functions all smooth, even though the gradient is being changed at every point? Shouldn't it be much more choppy? By the way, if possible, can this be explained intuitively (not too rigorously), and without Calculus? Because I want to understand this, but I haven't learnt Calculus yet.
|
"At every point" The gradient might change "at every point", but you need to remember that those points can be arbitrarily close to each other (see "real numbers" ): When you reduce the distance between the sampling points for $e^x, x^2, \ln(x), \sin(x)$, the gradient changes also become smaller. After a few iterations, the screen resolution isn't high enough to show any change anymore and the curves look smooth. On the other hand, the $|x|$ curve ( absolute value , the green curve on the graph) doesn't change anymore as soon as $x = 0$ is plotted: there's an abrupt change of gradient around $x = 0$, even at a very high resolution. At $x = 0$, it's not possible to define a gradient for this curve. If you zoom infinitely on the smooth curves, they will look just like straight lines. If you zoom on $|x|$ at $x=0$, you'll always see the sharp corner: However small $\varepsilon$ is, going from $x=-\varepsilon$ to $x = \varepsilon$ will change the gradient of $|x|$ from $-1$ to $1$.
|
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|
2,792,646 |
I get that if all the other options are proven wrong, then the only option left must be the correct option. But why does this work? What is the logic behind it? It just doesn't click with me for some reason, although I know I should get it because I "understand" it, at least at a cursory level. Update: "Exhaustion" changed to "elimination" in the title to avoid confusion.
|
To reword my comment, consider this. There are five cats in a room, and you know that exactly one of them is black. If you show that four of them are not black, then you can reasonably conclude that since one must be black , it has to be the final one. Otherwise, if the last one was also not black, then it would have been false to say that there is exactly one black one in the first case. My first experience with this type of proof was a group theory proof. I had a group of order $8$, and I needed to show that it belonged to a particular isomorphism class (that is, that it was isomorphic to one of the 'standard' order $8$ groups). Since it was a group of order $8$, it must be isomorphic to one of the five 'standard' order $8$ groups. Rather than construct the isomorphism explicitly, I could show that it was not isomorphic to four of the 'standard' order $8$ groups easily, which meant that it had to be isomorphic to the final 'standard' group of order $8$. If it were not, then either it wasn't an order $8$ group to start with, which would be false, or it would be an entirely new group structure of order $8$, which is also false. Thus it had to be isomorphic to the final case. Some other (fairly trivial) examples include: Let $n$ be an integer, and we clarify that $0$ is even. Either $n$ is odd or $n$ is even. If you show that it's not divisible by $2$, you've shown that it's not even ($0$ is divisible by $2$), so it must be odd. This is how you typically check if something is even or odd. Let $a$ be a real number. Then it's either positive, negative, or zero. If you show that it's neither negative nor zero, it must be positive. Similarly, this is why the phrase "$a$ is non-negative" is equivalent to "either $a$ is zero or $a$ is positive". Let $x$ be a positive integer. Then modulo $4$, its remainder is one of $0$, $1$, $2$, or $3$. If you show that its remainder is none of $0$, $1$, or $2$, then its remainder must be $3$. Let $G$ be a cyclic group of order $4$ with elements $a, b, c, d$. It is known that the elements must have order $1$, $2$, or $4$. Thus if you show that, say, the element $a$ does not have order $1$ or order $2$ (showing that neither itself nor its square are the identity), it must have order $4$ and it would be a generator. See here for an example. I'll stress that the proof by elimination method is unlikely the best proof method for a lot of the above examples, but it would still work if you know that your object has to satisfy one of the cases you were going to check. A simple counterexample is to let $a$ be a real number, show that it's not divisible by $2$, and conclude then that it must be odd. This is clearly false since real numbers cannot be partitioned into only even and odd numbers, so it is true that it could be neither.
|
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|
2,793,741 |
I actually have to find the number of real roots of $m^3+4m+2=0$ for a conic sections question in which $m$ is the slope of the normal. The answer is that there is only one real value of $m$, and therefore only one normal. How did they get this? How can I find out how many real roots such a cubic equation has without spending too much time on it?
|
The derivative of this function, $3m^2+4$, is always positive, so the function is always increasing. An increasing function on the real line cannot have more than one zero.
|
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|
2,793,752 |
Are there any degree 7 polynomials over $\mathbb{Q}$ having Galois group $S_7$? If so, is there one for which this is easy to check with pencil and paper? I know that for degree 3 polynomials, the Galois group will be $S_3$ if the discriminant is not a square in $\mathbb{Q}$, but I don't think the same holds for a degree 7.
|
The derivative of this function, $3m^2+4$, is always positive, so the function is always increasing. An increasing function on the real line cannot have more than one zero.
|
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|
2,794,903 |
How can you correctly reason that this directed graph is acyclic? I can only visually say that this graph is acyclic because there is not a single path in the graph where the starting vertex is equal to the ending vertex. But is this actually a reason you can say if someone asks you? Is there maybe some kind of rule / formula where you can determine this by comparing the amount of vertices with the amount of edges? We have here $6$ vertices and $10$ edges. I hope you can give me some help because if someone asks me why this directed graph or generally a directed graph is acyclic, I would go like that here and I don't feel good about it.
|
Step 1. Since $A$ has no indegree it can't be part of any cycle. So remove it. We have now graph $G_1$. Step 2. Since $C$ has no indegree it can't be part of any cycle (in this new graph $G_1$). So remove it. We get $G_2$ Step 3. Now in $G_2$ nodes $B$ and $D$ have no indegree so remove them.
|
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|
2,795,029 |
I was programming and I realized that the last digit of all the integer numbers squared end in $ 0, 1, 4, 5, 6,$ or $ 9 $. And in addition, the numbers that end in $ 1, 4, 9, 6 $ are repeated twice as many times as the numbers that end in $ 0, 5$ I checked the numbers from $1$ to $1000$, and the results are: $1.$ The numbers on the left are the last digit of each digit squared. $2.$ The numbers on the right are the number of times that the last digit is repeated. $$
\begin{array}{cc}
0: &100, \\
1: &200, \\
4: &200, \\
5: &100, \\
6: &200, \\
9: &200
\end{array}
$$ So, why does this happen? What is the property that all integers have?
|
The answer to this question is a bit less profound than you might hope. To see why, first note that the last digit of the square of any natural number only depends on the number's last digit - any other digits represent powers of 10 and do not make any difference to the last digit of the square. So the problem amounts to working out the last digit of the squares of single digit numbers (and 10, if we don't consider 0 a natural number). They are: 1 4 9 6 5 6 9 4 1 0 The relative frequencies of these last digits here explain why they take up the proportions of square numbers that you observe.
|
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|
2,795,040 |
I go to the gym every Mon/Wed/Fri while a friend of mine goes every 3 days regardless of the day. A typical two weeks could look like: Su | M | T | W | R | F | Sa | Su | M | T | W | R | F | Sa
ME: X X X X X X
HIM: X X X X To my surprise, I've observed that I see him exactly once a week. I decided to look at it from a math perspective. It's clear the number of times a week I see him has an upper bound of 1. If I see him Mon, he'll come in Thurs and I won't see him Wed/Fri. The same logic can be used for if I see him Wed/Fri. However, I'm not sure why the lower bound here is 1. My intuition tells me it has something to do with modulus 7, since the days he comes to the gym are cyclical (this would also make some kind of sense since I'm coming to the gym every two days 3 times while he comes every 3 days and his cycle repeats every 3 weeks) but I can't make the mathematical leap. I also tried to generalize. What if I come in for 3 days that are each 1 apart? Then it's obvious I'll see him exactly once. But what if I come in 3 days that are each 4 days apart? I can bruteforce this to find that I'm right, but intuitively, why are our days lining up exactly once a week?
|
Let's think of numbers $0, 1, \ldots, 6, \ldots$ as the days starting from some Sunday. Then every Monday, Wednesday, and Friday, means $\{7n + 1, 7n + 3, 7n + 5\}$. "Every three days" is all days of the form $3n + k$, where $k \in \{0, 1, 2\}$. Equivalently, these are all days that are $\equiv k \pmod 3$. Taking $\{\rm M, W, F\}$ modulo $3$, we get $\{n+1, n, n+2\}$, which is the same as $\{0,1,2\}$, thus one of Monday, Wednesday, Friday is $k \pmod{3}$. Thus for any $k$, there is a day that matches both sets.
|
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|
2,795,340 |
Suppose we have a basis $B$ for an endomorphism $f$ that has eigenvalues $\lambda_{1},\dots,\lambda_{k}$. Do these eigenvalues change or stay the same if we change to another basis $B'$?
|
No, eigenvalues are invariant to the change of basis, only the representation of the eigenvectors by the vector coordinates in the new basis changes. Indeed suppose that $$Ax=\lambda x$$ and let consider the change of basis $x=My$ then $$Ax=\lambda x\implies AMy=\lambda My\implies M^{-1}AMy=\lambda y \implies By=\lambda y$$
|
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|
2,795,884 |
I recently posted a tweet claiming I had encountered a real life Monty Hall dilemma . Based on the resulting discussion, I'm not sure I have. The Scenario I have 3 tacos (A,B,C) where tacos A and C are filled with beans, and taco B is filled with steak. I have no foreknowledge of the filling of any tacos. My wife only knows that taco C is filled with beans. My wife and I both know that I want steak. After I pick taco A, my wife informs me taco C is filled with beans. I switch my pick from taco A to taco B, thinking the logic behind the Monty Hall problem is relevant to my choice. Edit for clarity Timing: The contents of taco C were not revealed to me until after I had made my selection of taco A. My knowledge of what my wife knew: When she told me the contents of taco C, I knew that she had previously opened taco C. I also knew that she had no other knowledge of the contents of the other tacos. Questions Even though my wife does not know the fillings of all the tacos, does her revealing that taco C is definitively not the taco I want after I've made my initial selection satisfy the logic for me switching (from A to B) if I thought it would give me a 66.6% chance of getting the steak taco? If this is not a Monty Hall situation, is there any benefit in me switching?
|
No, this is not a Monty Hall problem. If your wife only knew the contents of #3, and was going to reveal it regardless, then the odds were always 50/50/0. The information never changed. It was just delayed until after your original choice. Essentially, you NEVER had the chance to pick #3, as she would have immediately told you it was wrong. (In this case, she is on your team, and essentially part of the player). #3 would be eliminated regardless: "No, not that one!" Imagine you had picked #3. Monty Hall never said, "You picked a goat. Want to switch?" If he did, the odds would immediately become 50/50, which is what we have here. Monty always reveals the worst half of the 2/3 you didn't select, leaving the player at 33/67/0.
|
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|
2,797,360 |
There are 4 basic combinatoric formulas when picking $k$ elements among $n$ We have repetition is allowed or not allowed, and order matters or does not matter. When order matters and repetition is not allowed we call it a permutation. When order does not matter and repetition is not allowed we call it a combination. What are the names of the missing two and what are the formulas for each?
|
We have the following cases for the number of subsets of size $k$ chosen from a set of $n$ distinct elements: replacement and ordered, "permutation with repetition" $$n^k$$ no replacement and ordered, "k-permutations of n" $$\frac{n!}{(n-k)!}$$ no replacement and unordered, "combinations" $$\binom{n}k$$ replacement and unordered, "combination with repetitions" $$\binom{n+k-1}k$$
|
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|
2,798,387 |
Occasionally when I'm bored, I'll play a game: Pick a random positive integer $X$. Add $+1$, $0$, $-1$ to make it divisible by $3$. $^\dagger$ Divide by $3$ to create a new $X$. Repeat steps $2$ and $3$ until you reach $1$. $^\dagger$ Keep track of how much you've added and subtracted. That is your "score". Assuming a random starting number $X$ (rather than a pseudo-random one), is there an equal chance that this game results in a positive or negative score? (I assume that this can be done with induction, hence the tag, but I'm fine with any proof)
|
If you write $X$ in base $3$ your game does the following:
Look at the ones bit of $X$. If it is $0$ do nothing. If it is $1$, subtract $1$ to make it $0$. If it is $2$, add $1$ to make it zero and carry $1$. Then erase the $0$ in the ones place. Repeat. This is all fair until you get to the last place. In that place you can never subtract. If it is a $2$ you add $1$. It could either have started as $2$ or have become $2$ because of a carry from below. For most ranges that $X$ could be chosen from, this will introduce a bias upwards.
|
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|
2,800,013 |
I'm learning math. I've recently thought more about the proof by contradiction technique, and I have a question that I would like cleared up. Let me set the stage. Suppose I am trying to prove a theorem. Theorem: If A and $\neg$B, then $\neg$C. Proof (contradiction): Let us suppose that A is true and $\neg$B is true. Let us assume that C is true ($\neg$C is false). [blah blah blah] From this, we arrive at a contradiction because we see that B is true ($\neg$B is false), but we know that $\neg$B is true (because we assumed it to be true). Thus, since assuming that C is true lead us to a contradiction, it must be the case that C is false ($\neg$C is true). QED. My issue with this : why is it that C leading to a contradiction must mean that $\neg$C is true? What if $\neg$C also leads to a contradiction? In that case, doesn't a proof by contradiction not prove anything? Why can we be sure that C leading to a contradiction must mean that $\neg$C doesn't lead to a contradiction? I'm sorry if this question has already been asked. I searched for a bit before asking to see if anyone had this same specific question, but most results just asked why a proof by contradiction works in general without any clear question.
|
If both $C$ and $\neg C$ lead to a contradiction, then you must be working with an inconsistent set of assumptions ... from which anything can be inferred ... including $\neg C$. As such, $\neg C$ can still be concluded given that assumption $C$ leads to a contradiction. So, regardless of whether $\neg C$ also leads to a contradiction or whether it does not, we can conclude $\neg C$ once assumption $C$ leads to a contradiction. The thing to remember is that when in logic we say that we can 'conclude' something, we mean that that something follows from the assumptions ... not that that something is in fact true. I think that's the source of your confusion. You seem to be saying: "OK, if $C$ leads to a contradiction, then we want to say that $\neg C$ is true ... But wait! What if $\neg C$ leads to a contradiction as well .. wouldn't that mean that $\neg C$ cannot be true either? So, how can we say $\neg C$ is true?!". But it's not that $\neg C$ is true .. it's just that it logically follows from the assumptions. That is: if the assumptions are all true, then $\neg C$ will be true as well. Well, are they? .. and is it? Funny thing is, as logicians, we don't really care :)
|
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|
2,800,484 |
Let $A$ be an $n \times n$ real matrix with the following property: All the conjugates of $A$ have only zeros on the diagonal. Does $A=0$? (By conjugates, I mean all the matrices similar to it, over $\mathbb{R}$, that is I require the conjugating matrix to be real). Of course, if $A$ is diagonalizable, then clearly it must be zero. The only idea I have is to use the Jordan form for real matrices , but after some thought I am not sure this is a good approach.
|
I think it is true: Suppose $A$ is nonzero. Then we find nonzero $v,w\in \mathbb R^n$ with $Av=w$. If $w$ and $v$ are linearly dependent, extend $v$ to a basis, then $A$ written in that basis will have a nonzero entry in the diagonal. If they are independent, then so are $v$ and $v+w$. Extending $\{v,v+w\}$ to a basis will then yield a nonzero diagonal element in $A$ written in this basis.
|
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|
2,804,558 |
This is something of a pedagogical question. Most of us have scratched our heads at some point as to why the definition of "$f$ is continuous" requires that the pre image of any open set under $f$ is open, then done a bunch of examples, then convinced ourselves. I think our confusion comes because when we first encounter "continuity" we are generally expecting a statement about what $f$ does to the images of sets, and get frustrated when we see a statement about what $f$ guarantees about the preimages of sets. I think that we might be able to restore this by instead looking at subspaces of the topological space and whether they are connected according to their subspace topology. Let's call a subset of a topological space which is connected according to its subspace topology a "connected subset". Connectedness is inherently a negative notion ("there does not exist two open sets...") and thus one is naturally induced to look at a contrapositive -- at images rather than preimages. So I think there might be a way to start from a more-intuitive definition, for example: "$f: X \to Y$ is continuous iff $X$ and $Y$ are topological spaces and the image of any connected subset of $X$ is a connected subset of $Y$." This would seem immediately intuitive to a student, I feel. And if we start from there, then immediately the contrapositive accomplishes the reversal: "$f: X \to Y$ is continuous iff the preimage of any disconnected subset of $Y$ is a disconnected subset of $X$." Now it's clear that the latter is necessary for $f$ to be continuous according to the conventional definition. If $f$ is conventionally-continuous then when we factor a disconnected subset of $Y$ into two disjoint open subsets, the preimage of each subset is open and the preimages of two disjoint sets must also be disjoint, proving that the subset of $X$ was disconnected, too. However I am struggling with rigorously proving sufficiency and thereby deriving the conventional definition from the intuitive one. The existential quantifiers of disconnectedness seem to really get in my way: for example even if I could somehow state "for every nonempty open set there exists some other disjoint nonempty open set" (which is false for the trivial topology but maybe that's not such an issue) the above definition seemingly does not guarantee that the corresponding preimages of the two are open -- maybe this partition of the $Y$-subset has very little to do with the partition of the $X$-subset, for example. So it's not clear to me how to push past these weird existentials and come to "the preimage of any open set is open" again. Am I missing some obvious counterexample that shows this whole enterprise is a mistake, or clarifies some extra assumptions I'll need in my definition to make it work? If not, what technique might I use to finish the sufficiency side of the proof?
|
The short answer is that there are functions which are not conventionally continuous, but which are connected-set-contininuous. For example, one can use any $f:\mathbb{R}\rightarrow \mathbb{R}$ with the following property: If $X:=(a,b)$ is any non-empty open interval, then $f(X) =\mathbb{R}$. One example of such an $f$ is Conway's base 13 function . Such a function cannot be continuous because it is unbounded on every non-trivial closed interval (because a non-trivial closed interval contains a non-trivial open interval as a subset). On the other hand, the connected subsets of $\mathbb{R}$ are points or are intervals (with or without some end points). Of course, the image of a point is connected. The image of an interval is also connected: every interval has an open subinterval, and then $f$ applied to the open subinterval is already all of $\mathbb{R}$, which is connected.
|
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|
2,804,568 |
$D$ and $E$ are points on sides $AB$ and $AC$ of a triangle $ABC$ such that $DE$ is parallel to $BC$ and tangent to the incircle $I$. Prove that $$DE\leq\frac{AB+BC+AC}{8}$$ This looks like $AM-GM$, but I don't have a cube on $DE$. I tried using the triangle inequality like this: I rotated $ABC$ by $180^\circ$ around the incentre, then applied the triangle inequality to get $$AD+AE+B'E+B'K+CK+CL\gt DE+EK+KL$$ Now, if this relation is true it should apply for all of $DE, EK,KL$. I'm not sure where to go from here. Please help. Hints will be appreciated. I just saw this , and it seems to be related, but different.
|
The short answer is that there are functions which are not conventionally continuous, but which are connected-set-contininuous. For example, one can use any $f:\mathbb{R}\rightarrow \mathbb{R}$ with the following property: If $X:=(a,b)$ is any non-empty open interval, then $f(X) =\mathbb{R}$. One example of such an $f$ is Conway's base 13 function . Such a function cannot be continuous because it is unbounded on every non-trivial closed interval (because a non-trivial closed interval contains a non-trivial open interval as a subset). On the other hand, the connected subsets of $\mathbb{R}$ are points or are intervals (with or without some end points). Of course, the image of a point is connected. The image of an interval is also connected: every interval has an open subinterval, and then $f$ applied to the open subinterval is already all of $\mathbb{R}$, which is connected.
|
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|
2,805,639 |
I imagine this problem is a common one, however without having any source to refer to I don't know its usual name, and am having trouble finding an answer. I am taking the definition: A cover $C$ of a set $S$ is a set such that $\cup C = S$ I want to know if every cover has a minimal subcover. Thanks.
|
How about $S=\Bbb R$ and the cover composed of the intervals $(-n,n)$?
Any subcover of this cover remains a subcover if you omit one of its elements.
|
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|
2,806,389 |
My knowledge of geometry is just a little bit above high school level and I know absolutely nothing about topology. So, what is the point of this meme? (Original unedited webcomic: “Juncrow” by False Knees )
|
The topologist bird is stating some nice simple topological concepts. The geometer bird comes along and interrupts the topologist bird, squawking loudly over the top with something much longer, more complex-looking and uglier, involving iterated partial derivatives, multiple nested subscripts and superscripts, and so on. The topologist bird looks pissed off. The meme was probably made by a topologist, in a frustrated attempt to convey "why do geometers always overcomplicate things?" or "why is geometry so ugly?" or "why do geometers always talk over topologists?" or something along those lines. (As in most memes, this probably isn't an entirely fair reflection of reality.) I imagine a geometer could make a meme in reverse that implied "why do topologists always take nice things like doughnuts and turn them into coffee cups and start calculating their cohomology?". I don't think there's anything deep going on here. The contents of the speech bubbles don't actually mean much as sentences. They're just some common expressions that you find in topology and geometry respectively, thrown together.
|
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|
2,806,391 |
The rate of decay of a radioactive source falls from $3000$ counts per minute to $2000$ counts per minute 5 minutes later. From this information, determine the half life of the substance. The answers for this question say to use the formula $N = Ae^{kt}$, using: $A = 3000, N = 2000 \text { and } t = 50$ (to find $k$) I don't understand how $3000$ can be used as an 'initial value', when it represents a rate of change? Wouldn't this be equal, then, to the derivative of the equation?
|
The topologist bird is stating some nice simple topological concepts. The geometer bird comes along and interrupts the topologist bird, squawking loudly over the top with something much longer, more complex-looking and uglier, involving iterated partial derivatives, multiple nested subscripts and superscripts, and so on. The topologist bird looks pissed off. The meme was probably made by a topologist, in a frustrated attempt to convey "why do geometers always overcomplicate things?" or "why is geometry so ugly?" or "why do geometers always talk over topologists?" or something along those lines. (As in most memes, this probably isn't an entirely fair reflection of reality.) I imagine a geometer could make a meme in reverse that implied "why do topologists always take nice things like doughnuts and turn them into coffee cups and start calculating their cohomology?". I don't think there's anything deep going on here. The contents of the speech bubbles don't actually mean much as sentences. They're just some common expressions that you find in topology and geometry respectively, thrown together.
|
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|
2,806,508 |
In my linear algebra class, I get the answer and everything right but lose points on the mathematical writing, symbols, notations etc ... Is there a comprehensive book or document that provides extensive explanation and examples on how to answer questions in the proper mathematical way?
|
You mention mathematical writing and symbols/notation. Though I’ve never taken a linear algebra course, here are some relevant tips: Make sure every line of work has a meaningful relation (symbol) in it. This helps you label your answers. If you define your own notations, explain that. Aesthetically organize your work. Do not be afraid to blend English (or whatever language you write in) with mathematics. Use an ample and appropriate vocabulary of mathematical jargon. I will stick to calculus examples, because it is something I know and that most readers can probably follow. 1. Meaningful Relations & Labeled Answers This practice is the easiest to implement and comes directly from an instructor of my own, so I’ve put it first and foremost. I don’t have the clearest explanation, but connecting mathematical phrases without a relation symbol is illogical. Differentiate $\newcommand{\b}{\boldsymbol} \b{x^2+e}$ with respect to $\b x$. $\newcommand{\r}{\color{red}}\qquad\qquad\quad\r{ ↳ 2x}$ This is a bad answer because there no is meaningful relation specified—namely equality. You need that equals sign! Differentiate $\b{x^2+e}$ with respect to $\b x$. $$\newcommand{\g}{\color{green}} \g{\frac{d}{dx}\left(x^2+e\right)=2x}$$ 2. Explain Notations You can’t just assume your grader will intuitively follow your approach to problem solving. This is especially true with notation, where no absolutes exist! As a result you need to explain the meaning of any notation you make up (that wasn’t given in the problem). Prove that $\b{\displaystyle\frac{d}{dx}\int_a^x f(t) \, dt = f(x)}$ for constant $\b a$. $$\r{\frac{d}{dx}\bigl(F(x) - F(a)\bigr) = F’(x)-(F(a))’=f(x)}$$ The problem here is that the meaning of $F$ has not been explained explicitly. (Yes, you can intuit the meaning here , but that’s beside the point.) Also, the final value $f(x)$ has not been attached to the initial integral, so since there is no appropriate labeling, the whole thing is irrelevant! A better response might go something like: Prove that $\b{\displaystyle\frac{d}{dx}\int_a^x f(t) \, dt = f(x)}$ for constant $\b a$. $\g{\text{Taking $F’(t)=f(t)$},}$ $$\begin{align}
\g{\frac{d}{dx}\int_a^x f(t)\,dt} &\g= \g{\frac{d}{dx}\left(F(t)\bigr|_a^x\right)} \\[2ex]
&\g= \g{\frac{d}{dx}\bigl(F(x)-F(a)\bigr)} \\[2ex]
&\g= \g{F’(x) - \frac{d\,F(a)}{dx}} \\[2ex]
&\g= \g{F’(x)-0} \\[2ex]
&\g= \g{f(x)}
\end{align}$$ $\g{\text{since $F(a)$ evaluates as a constant.}}$ The above example also demonstrates point 3 by spacing lines of work appropriately and point 4 with its last corollary. 3. Aesthetically Organize Your Work Do not be afraid to use a bunch of paper! (I promise that deforestation of luscious ecosystems is not caused by mathematicians but rather greedy capitalists purchasing plots of the Amazon on which they implement slash-and-burn agriculture and which they abandon after one season of soy.) This is how I record my answers on math tests: This setup allows me to use ink (which is more visible), switch back and forth between questions, et cetera, because I merely cross out erroneous work with a single line and get more paper when I need it. If you’re doing stuff with matrices, you might invest in graph paper. DON’T try to solve an entire problem in the blank space under the prompt. That’s a mess! Also, be sure to align your writing appropriately. 4. Blend In Spoken Language Sometimes, a blend between spoken language and mathematical notation is advantageous as a more natural way to communicate yourself. You want your math and your language to supplement one another. What is the definition of a limit? $$\r{\lim_{x\to c}f(x)=L \iff \forall\epsilon>0:\exists\delta>0:\lvert x-c\rvert<\delta\Rightarrow\lvert f(x)-L\rvert<\epsilon}$$ If you were a grader, would you (a) spend time sorting through that mess and (b) be satisfied that you the student have an understanding of whatever that mess is? What is the definition of a limit? $\newcommand{\G}[1]{\color{green}{\text{#1}}} \G{Let $f(x)$ be a function,}$
$\G{and let $c$}$
$\G{be an $x$-value}$
$\G{to approach.}$
$\G{It is}$
$\G{defined that}$
$\G{$\lim_{x\to c}f(x)=L$}$
$\G{if the values}$
$\G{of $f(x)$ get}$
$\G{arbitrarily closer to $L$}$
$\G{(generally notated }$
$\G{as $\lvert f(x)-L\rvert<“\epsilon”$)}$
$\G{when $x$ gets}$
$\G{arbitrarily closer}$
$\G{to $c$ (generally}$
$\G{notated as}$
$\G{$\lvert x-c\rvert<“\delta”$).}$ 5. Augment Your Mathematical Vocabulary You need obviously to know the terms of whatever unit you’re studying, but in addition to that, speaking like a mathematician will both prove to the grader that you are knowledgeable clarify your explanations. Some terms that come to mind include: if and only if (iff / $\iff$) for some ($\exists$) for all ($\forall$) implies ($\implies$) arbitrary I invite other users to expand this list with what comes to their minds. I hope these tips help, and I will expand this answer some more as more ideas come to mind.
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|
2,806,513 |
I am using function gsl_linalg_SV_decomp provided by the GNU Scientific Library to solve a least-squares problem $$\min \|Ax-b\|_2$$ where $A \in \mathbb R^{m \times n}$. The procedure is to first find the SVD of $A$ $$A = U\Sigma V^T$$ where $U \in \mathbb R^{m \times m}$ and $V \in \mathbb R^{n \times n}$ are orthogonal, i.e., $U^TU = UU^T = I_m$ and $V^TV = VV^T = I_n$. However, gsl_linalg_SV_decomp returns a thin SVD, i.e., it provides a $U \in \mathbb R^{m \times n}$. Therefore, I need to extend the thin matrix $U$ to a full orthogonal matrix. I've been thinking about randomly appending some orthonormal columns to $U$, but I'm not sure if it is appropriate. Update: After some derivations, I found that a thin SVD is actually enough for solving the least-squares problem if the residual will be ignored. But In my problem, there is a constraint $\|Ax-d\| < \gamma$ and a full SVD is required.
|
You mention mathematical writing and symbols/notation. Though I’ve never taken a linear algebra course, here are some relevant tips: Make sure every line of work has a meaningful relation (symbol) in it. This helps you label your answers. If you define your own notations, explain that. Aesthetically organize your work. Do not be afraid to blend English (or whatever language you write in) with mathematics. Use an ample and appropriate vocabulary of mathematical jargon. I will stick to calculus examples, because it is something I know and that most readers can probably follow. 1. Meaningful Relations & Labeled Answers This practice is the easiest to implement and comes directly from an instructor of my own, so I’ve put it first and foremost. I don’t have the clearest explanation, but connecting mathematical phrases without a relation symbol is illogical. Differentiate $\newcommand{\b}{\boldsymbol} \b{x^2+e}$ with respect to $\b x$. $\newcommand{\r}{\color{red}}\qquad\qquad\quad\r{ ↳ 2x}$ This is a bad answer because there no is meaningful relation specified—namely equality. You need that equals sign! Differentiate $\b{x^2+e}$ with respect to $\b x$. $$\newcommand{\g}{\color{green}} \g{\frac{d}{dx}\left(x^2+e\right)=2x}$$ 2. Explain Notations You can’t just assume your grader will intuitively follow your approach to problem solving. This is especially true with notation, where no absolutes exist! As a result you need to explain the meaning of any notation you make up (that wasn’t given in the problem). Prove that $\b{\displaystyle\frac{d}{dx}\int_a^x f(t) \, dt = f(x)}$ for constant $\b a$. $$\r{\frac{d}{dx}\bigl(F(x) - F(a)\bigr) = F’(x)-(F(a))’=f(x)}$$ The problem here is that the meaning of $F$ has not been explained explicitly. (Yes, you can intuit the meaning here , but that’s beside the point.) Also, the final value $f(x)$ has not been attached to the initial integral, so since there is no appropriate labeling, the whole thing is irrelevant! A better response might go something like: Prove that $\b{\displaystyle\frac{d}{dx}\int_a^x f(t) \, dt = f(x)}$ for constant $\b a$. $\g{\text{Taking $F’(t)=f(t)$},}$ $$\begin{align}
\g{\frac{d}{dx}\int_a^x f(t)\,dt} &\g= \g{\frac{d}{dx}\left(F(t)\bigr|_a^x\right)} \\[2ex]
&\g= \g{\frac{d}{dx}\bigl(F(x)-F(a)\bigr)} \\[2ex]
&\g= \g{F’(x) - \frac{d\,F(a)}{dx}} \\[2ex]
&\g= \g{F’(x)-0} \\[2ex]
&\g= \g{f(x)}
\end{align}$$ $\g{\text{since $F(a)$ evaluates as a constant.}}$ The above example also demonstrates point 3 by spacing lines of work appropriately and point 4 with its last corollary. 3. Aesthetically Organize Your Work Do not be afraid to use a bunch of paper! (I promise that deforestation of luscious ecosystems is not caused by mathematicians but rather greedy capitalists purchasing plots of the Amazon on which they implement slash-and-burn agriculture and which they abandon after one season of soy.) This is how I record my answers on math tests: This setup allows me to use ink (which is more visible), switch back and forth between questions, et cetera, because I merely cross out erroneous work with a single line and get more paper when I need it. If you’re doing stuff with matrices, you might invest in graph paper. DON’T try to solve an entire problem in the blank space under the prompt. That’s a mess! Also, be sure to align your writing appropriately. 4. Blend In Spoken Language Sometimes, a blend between spoken language and mathematical notation is advantageous as a more natural way to communicate yourself. You want your math and your language to supplement one another. What is the definition of a limit? $$\r{\lim_{x\to c}f(x)=L \iff \forall\epsilon>0:\exists\delta>0:\lvert x-c\rvert<\delta\Rightarrow\lvert f(x)-L\rvert<\epsilon}$$ If you were a grader, would you (a) spend time sorting through that mess and (b) be satisfied that you the student have an understanding of whatever that mess is? What is the definition of a limit? $\newcommand{\G}[1]{\color{green}{\text{#1}}} \G{Let $f(x)$ be a function,}$
$\G{and let $c$}$
$\G{be an $x$-value}$
$\G{to approach.}$
$\G{It is}$
$\G{defined that}$
$\G{$\lim_{x\to c}f(x)=L$}$
$\G{if the values}$
$\G{of $f(x)$ get}$
$\G{arbitrarily closer to $L$}$
$\G{(generally notated }$
$\G{as $\lvert f(x)-L\rvert<“\epsilon”$)}$
$\G{when $x$ gets}$
$\G{arbitrarily closer}$
$\G{to $c$ (generally}$
$\G{notated as}$
$\G{$\lvert x-c\rvert<“\delta”$).}$ 5. Augment Your Mathematical Vocabulary You need obviously to know the terms of whatever unit you’re studying, but in addition to that, speaking like a mathematician will both prove to the grader that you are knowledgeable clarify your explanations. Some terms that come to mind include: if and only if (iff / $\iff$) for some ($\exists$) for all ($\forall$) implies ($\implies$) arbitrary I invite other users to expand this list with what comes to their minds. I hope these tips help, and I will expand this answer some more as more ideas come to mind.
|
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|
2,806,957 |
Is there an easy geometric way to prove that the circle is tangent to the line $\overleftrightarrow {CD}$, where $C=(3,-6)$, $D=(6,-2)$, and $B=(6,0)$? I can do this by using calculus but I think there has to be a nicer/shorter solution.
Thanks in advance. Edit: I found that $CE$ has distance $3$, so maybe one could use the theorem of Thales to obtain a right angle in the triangle $(0,-6), E,(6,-6)$ and use this?
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Extrapolate the line until it it intersects both coordinate axes. The points of intersection are then $(15/2,0)$ and $(0,-10)$. The line and the exes now make a right triangle whose legs are $15/2$ and $10$. Now, apply the Pythagorean Theorem, find that the hypoteneuse is $25/2$ (proportional to a 3-4-5 right triangle). In a right triangle, the area is half the product of the legs and also half the product of the hypoteneuse and the altitude to that hypoteneuse. So the products must be equal and we must have $(10)(15/2)=(25/2)x$ where $x$ is the altitude to the hypoteneuse. Then $x=6$ matching the radius of the given circle; the line is tangent to the circle.
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2,808,744 |
Recently I have started reviewing mathematical notions, that I have always just accepted. Today it is one of the fundamental ones used in equations: If we have an equation, then the equation holds if we do the same to both sides. This seems perfectly obvious, but it must be stated as an axiom somewhere, presumably in formal logic(?). Only, I don't know what it would be called, or indeed how to search for it - does anybody knw?
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This axiom is known as the substitution property of equality . It states that if $f$ is a function, and $x = y$, then $f(x) = f(y)$. See, for example, Wikipedia . For example, if your equation is $4x = 2$, then you can apply the function $f(x) = x/2$ to both sides, and the axiom tells you that $f(4x) = f(2)$, or in other words, that $2x = 1$. You could then apply the axiom again (with the same function, even) to conclude that $x = 1/2$.
|
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2,808,804 |
In mathematics the existence of a mathematical object is often proved by contradiction without showing how to construct the object. Does the existence of the object imply that it is at least possible to construct the object? Or are there mathematical objects that do exist but are impossible to construct?
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Really the answer to this question will come down to the way we define the terms "existence" (and "construct"). Going philosophical for a moment, one may argue that constructibility is a priori required for existence, and so ; this, broadly speaking, is part of the impetus for intuitionism and constructivism , and related to the impetus for (ultra)finitism .$^1$ Incidentally, at least to some degree we can produce formal systems which capture this point of view (although the philosophical stance should really be understood as preceding the formal systems which try to reflect them; I believe this was a point Brouwer and others made strenuously in the early history of intuitionism) . A less philosophical take would be to interpret "existence" as simply "provable existence relative to some fixed theory" (say, ZFC, or ZFC + large cardinals). In this case it's clear what "exists" means, and the remaining weasel word is "construct." Computability theory can give us some results which may be relevant, depending on how we interpret this word: there are lots of objects we can define in a complicated way but provably have no "concrete" definitions: The halting problem is not computable. Kleene's $\mathcal{O}$ - or, the set of indices for computable well-orderings - is not hyperarithmetic. A much deeper example: while we know that for all Turing degrees ${\bf a}$ there is a degree strictly between ${\bf a}$ and ${\bf a'}$ which is c.e. in $\bf a$, we can also show that there is no "uniform" way to produce such a degree in a precise sense. Going further up the ladder, ideas from inner model theory and descriptive set theory become relevant. For example: We can show in ZFC that there is a (Hamel) basis for $\mathbb{R}$ as a vector space over $\mathbb{Q}$; however, we can also show that no such basis is "nicely definable," in various precise senses (and we get stronger results along these lines as we add large cardinal axioms to ZFC). For example, no such basis can be Borel. Other examples of the same flavor: a nontrivial ultrafilter on $\mathbb{N}$; a well-ordering of $\mathbb{R}$; a Vitali (or Bernstein or Luzin) set, or indeed any non-measurable set (or set without the property of Baire, or without the perfect set property); ... On the other side of the aisle, the theory ZFC + a measurable cardinal proves that there is a set of natural numbers which is not "constructible" in a precise set-theoretic sense (basically, can be built just from "definable transfinite recursion" starting with the emptyset). Now the connection between $L$-flavored constructibility and the informal notion of a mathematical construction is tenuous at best, but this does in my opinion say that a measurable cardinal yields a hard-to-construct set of naturals in a precise sense. $^1$I don't actually hold these stances except very rarely , and so I'm really not the best person to comment on their motivations; please take this sentence with a grain of salt.
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2,809,846 |
If I don't have a computer at hand (or an app), how can I know that $1637$ is a prime number? I factored the number $99857$ as $1637\times 61$ and the computer told me that $1637$ is a prime. So would it be easy at all to know this without it?
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If it was not prime, it would have a prime factor smaller than or equal to its square root. Since$$40^2=1\,600<1\,637$$and$$41^2-40^2=(41-40)\times(41+40)=81,$$it is clear that $40<\sqrt{1\,637}<41$. So, all you have to do is to check whether $1\,637$ is a multiple of one of the twelve numbers $2,3,5,\ldots,37$.
|
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|
2,811,197 |
Does the sum of a convex function and monotonically increasing function (not necessarily convex) yield a convex function?
|
$g(x)=2x+\sin x$ is strictly increasing, $f(x)=\frac15x^2$ is strictly convex. Yet, $f''(x)+g''(x)=\frac25-\sin x$, so $f+g$ is not convex.
|
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|
2,811,211 |
Consider the blowup of the affine plane at the origin: $f:\hat{X}\to X=\mathbb{A}^2_k$. I want to show that $\mathcal{L}:=R^1f_*\mathcal{O}_\hat{X}(\pm E)=0$ in the most elementary way possible. It's easily seen that $\mathcal{L}|_{X-0}=0$, since the morphism is affine when restricted to this open set, so the problem is to show that its stalk at the origin is zero as well. I've thought about two possible strategies, both which end up having to deal with some cumbersome commutative algebra I'm not comfortable with: 1) Either showing that the sheaf $\mathcal{L}$ is S2 (meaning that $\operatorname{depth}_x(\mathcal{L}_x)=\min(\dim \mathcal{L}_x,2)$ for all $x\in X$, and the depth of a module is the supremum of the lenghts of regular sequences) so that, by this answer in MO , we can conclude that the sheaf $\mathcal{L}|_{X-0}=0$ extends uniquely to the zero sheaf over the whole space. 2) Or on the other hand showing that $\mathcal{O}_\hat{X}(\pm E)$ is flat over $X$. Then all the hypothesis are met so that the stalk of the sheaf at the origin is:
$$
\mathcal{L}_0\cong H^1(E,\mathcal{O}_\hat{X}(\pm E)|_E)\cong H^1(\mathbb{P}^1,\mathcal{O}(\mp 1))=0.
$$ I guess my question is what is the best approach? Is it one of these two or am I missing something more straighforward? If it's either of these some additional pointers would be appreciated.
|
$g(x)=2x+\sin x$ is strictly increasing, $f(x)=\frac15x^2$ is strictly convex. Yet, $f''(x)+g''(x)=\frac25-\sin x$, so $f+g$ is not convex.
|
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|
2,811,213 |
A professor knows $9$ jokes and tells $3$ jokes per lecture.
Prove that in a course of $13$ lectures there is going to be a pair of jokes that will be told together in at least $2$ lectures. I've started with counting how many possibilities there are to tell jokes in a lecture. Let $$J := \{1,2,\dots,9\}$$ The amount of all different possible combinations for jokes is $9 \choose 3$ and for each lecture there are going to be $3$ unique pairs of jokes $\left(\frac{3!}{2!}=3\right)$. I'm not sure how to continue from here to get to the PHP, I think I might be doing something wrong here, any advice how to abstract it properly? This is an exercise from the Tel-Aviv University entry test preparation and I'm not a student yet so elementry combinatorics should do here.
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I will assume that he doesn't tell the same joke twice (or three times) in the same lecture. Else, here is a counterexample: Let $\{a_1, \ldots, a_9\}$ be the set of jokes. On the $i$-th day for $1 \leq i \leq 9$, tell jokes $(a_i, a_i, a_i)$. Then tell $(a_1, a_2, a_3)$, $(a_4, a_5, a_6)$, $(a_7, a_8, a_9)$ and $(a_1, a_4, a_7)$. So, now towards the exercise. Every day the professor tells $3$ distinct pairs of jokes. Which means in total he tells $39$ pairs of jokes over the $13$ days. There are $\frac{9!}{7!2!}= 36$ pairs of jokes he can tell. So he must tell at least one pair of jokes twice.
|
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|
2,811,951 |
I am currently trying to self-study linear algebra. I've noticed that a lot of the definitions for terms (like eigenvectors, characteristic polynomials, determinants, and so on) require a square matrix instead of just any real-valued matrix. For example, Wolfram has this in its definition of the characteristic polynomial: The characteristic polynomial is the polynomial left-hand side of the characteristic equation $\det(A - I\lambda) = 0$, where $A$ is a square matrix. Why must the matrix be square? What happens if the matrix is not square? And why do square matrices come up so frequently in these definitions? Sorry if this is a really simple question, but I feel like I'm missing something fundamental.
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Remember that an $n$-by-$m$ matrix with real-number entries represents a linear map from $\mathbb{R}^m$ to $\mathbb{R}^n$ (or more generally, an $n$-by-$m$ matrix with entries from some field $k$ represents a linear map from $k^m$ to $k^n$). When $m=n$ - that is, when the matrix is square - we're talking about a map from a space to itself. So really your question amounts to: Why are maps from a space to itself - as opposed to maps from a space to something else - particularly interesting? Well, the point is that when I'm looking at a map from a space to itself inputs to and outputs from that map are the same "type" of thing, and so I can meaningfully compare them . So, for example, if $f:\mathbb{R}^4\rightarrow\mathbb{R}^4$ it makes sense to ask when $f(v)$ is parallel to $v$, since $f(v)$ and $v$ lie in the same space; but asking when $g(v)$ is parallel to $v$ for $g:\mathbb{R}^4\rightarrow\mathbb{R}^3$ doesn't make any sense, since $g(v)$ and $v$ are just different types of objects. (This example, by the way, is just saying that eigenvectors/values make sense when the matrix is square, but not when it's not square.) As another example, let's consider the determinant. The geometric meaning of the determinant is that it measures how much a linear map "expands/shrinks" a unit of (signed) volume - e.g. the map $(x,y,z)\mapsto(-2x,2y,2z)$ takes a unit of volume to $-8$ units of volume, so has determinant $-8$. What's interesting is that this applies to every blob of volume: it doesn't matter whether we look at how the map distorts the usual 1-1-1 cube, or some other random cube. But what if we try to go from $3$D to $2$D (so we're considering a $2$-by-$3$ matrix) or vice versa? Well, we can try to use the same idea: (proportionally) how much area does a given volume wind up producing? However, we now run into problems: If we go from $3$ to $2$, the "stretching factor" is no longer invariant. Consider the projection map $(x,y,z)\mapsto (x,y)$, and think about what happens when I stretch a bit of volume vertically ... If we go from $2$ to $3$, we're never going to get any volume at all - the starting dimension is just too small! So regardless of what map we're looking at, our "stretching factor" seems to be $0$. The point is, in the non-square case the "determinant" as naively construed either is ill-defined or is $0$ for stupid reasons.
|
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|
2,811,957 |
I'm quite close to proving a particular function has infinitely many zeros of the form $2^{2^k}-5$, $k\ge 3$. The exceptional cases happen when $2^{2^k}-5$ can be written as $2^m+3^n$. Working mod 3, it's easy to show that $m$ must be odd. Are there solutions to $2^{2^k}-5=2^m+3^n$ over the positive integers with $k\ge 3$? How would one go about finding them, or proving them impossible? I know the trio $(2, 3, 1)$ is a solution, hence the $k\ge 3$ restriction. It would be sufficient for my purpose to prove there are infinitely many fixed $k$ where there is no solution, but seeing as how I have found no solutions at all with higher $k$, I am curious about ways to prove anything conclusive. Thanks in advance.
|
Remember that an $n$-by-$m$ matrix with real-number entries represents a linear map from $\mathbb{R}^m$ to $\mathbb{R}^n$ (or more generally, an $n$-by-$m$ matrix with entries from some field $k$ represents a linear map from $k^m$ to $k^n$). When $m=n$ - that is, when the matrix is square - we're talking about a map from a space to itself. So really your question amounts to: Why are maps from a space to itself - as opposed to maps from a space to something else - particularly interesting? Well, the point is that when I'm looking at a map from a space to itself inputs to and outputs from that map are the same "type" of thing, and so I can meaningfully compare them . So, for example, if $f:\mathbb{R}^4\rightarrow\mathbb{R}^4$ it makes sense to ask when $f(v)$ is parallel to $v$, since $f(v)$ and $v$ lie in the same space; but asking when $g(v)$ is parallel to $v$ for $g:\mathbb{R}^4\rightarrow\mathbb{R}^3$ doesn't make any sense, since $g(v)$ and $v$ are just different types of objects. (This example, by the way, is just saying that eigenvectors/values make sense when the matrix is square, but not when it's not square.) As another example, let's consider the determinant. The geometric meaning of the determinant is that it measures how much a linear map "expands/shrinks" a unit of (signed) volume - e.g. the map $(x,y,z)\mapsto(-2x,2y,2z)$ takes a unit of volume to $-8$ units of volume, so has determinant $-8$. What's interesting is that this applies to every blob of volume: it doesn't matter whether we look at how the map distorts the usual 1-1-1 cube, or some other random cube. But what if we try to go from $3$D to $2$D (so we're considering a $2$-by-$3$ matrix) or vice versa? Well, we can try to use the same idea: (proportionally) how much area does a given volume wind up producing? However, we now run into problems: If we go from $3$ to $2$, the "stretching factor" is no longer invariant. Consider the projection map $(x,y,z)\mapsto (x,y)$, and think about what happens when I stretch a bit of volume vertically ... If we go from $2$ to $3$, we're never going to get any volume at all - the starting dimension is just too small! So regardless of what map we're looking at, our "stretching factor" seems to be $0$. The point is, in the non-square case the "determinant" as naively construed either is ill-defined or is $0$ for stupid reasons.
|
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2,812,075 |
In a theorem such as this, found in Rudin: Let $u$ and $v$ be real measurable functions on a measurable space
$X$, let $\phi$ be a continuous mapping of the plane into a topological space $Y$, and define $h(x) = \phi (u(x), v(x))$ for all $x \in X$. Then $h: X \to Y$ is measurable. Isn't it redundant/unnecessary to refer to $Y$ as a topological space? Every set can be a topological space, because you can generate a topology, no matter how trivial, from every set, right? Since the theorem makes no explicit use of any topology on $Y$, why does $Y$ need to be labeled as a topological space? In the same way, why are any spaces labeled as "measurable spaces" when every set, at least trivially, is a measure space? Note:
I am aware that both a topological space and a measurable space refer more technically to the ordered pairs of a set and a collection of subsets, known as the topology or measure set respectively. However, when no explicit reference is made to these collections of subsets I don't understand how sets can be meaningfully classified as "topological" or "measurable."
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When we say "topological space $Y$", that is an abbreviation for "topological space $(Y,T)$" for some unspecified (but fixed ) topology $T$. In other words, we have some specific topology we are using on $Y$ (and will use this topology to make sense of words like "open set", "continuous", etc), but we are not bothering to explicitly name this topology. It is very important that we are not just saying there exists a topology on $Y$ (as you say, that is true of any set), but that we are actually picking some particular topology, because we are using that specific topology when we say $\phi$ is continuous, for instance. This sort of abbreviation is extremely common throughout mathematics and is done with pretty much any kind of mathematical structure. For instance, when we say "Let $G$ be a group" we actually mean that we are letting $(G,\cdot)$ be a group for some binary operation $\cdot$ that we will not explicitly name. Basically, it would get extremely tiresome to constantly be writing long tuples for every single mathematical structure we talk about, so we abbreviate and refer to them just by their underlying set.
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2,812,627 |
Is there a notation in English written mathematics for $$\textit{the interval of all points lying between two real numbers $a$ and $b$}$$ when you don't know which of $a$ and $b$ is greater? Which one is greater is completely irrelevant for what I am writing, and I would like to avoid making the text heavier as much as possible. Suggestions that have been made so far that rely on external notions:
$$[\min\{a,b\}, \max\{a,b\}]\qquad \operatorname{Conv}(a,b)$$ Suggestions for a brand new notation: $$(a,b]^*\qquad (\{a,b\}]\qquad (a\nearrow b]\qquad /a,b/\qquad \left(\begin{matrix}a\\b\end{matrix}\right]^\star$$ $^\star$ intervals open at the lower bound and closed at the higher bound, whichever of $a$ and $b$ they are. Some other options: Assume wlog that $a<b$ Make explicit that the notation $[a,b]$ doesn't imply $a<b$.
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Assuming you're meaning the closed interval for the notation I'm going to write, something that will always work is $$[\min\{a,b\}, \max\{a,b\}]$$ Another possibility is $$[a,b] \cup [b,a]$$ But I think that there is no standard notation, so you could create yours explaining it.
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2,814,041 |
Let $(a_n)_{n\in\mathbb{N}}$ be a positive sequence. Assume, that $\lim_{n \to \infty} a_n = \infty$. $\forall\ \zeta > 0, \exists\ n=n(\zeta) \geq 1: a_n < \zeta.$ From 1. we know, that $\forall\ C > 0, \exists\ N = N(C) \geq 1: a_n \geq C, \forall\ n \geq N.$ The difference $N(C) - n(\zeta)$ can be interpreted as the 'recovery time', i.e. the time the sequence needs from being arbitrarily small to become larger than an arbitrary positive constant $C$ for the rest of its existence. Question: Can somebody think of an example where $\forall\ C > 0: N(C) - n(\zeta)$ is arbitrarily large? (Is this even possible?)
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There is no sequence satisfying both of your conditions 1 and 2. Proof: given that $a_n \to \infty$, there is a constant $N = N(1)$ such that $a_n \geq 1$ for all $n \geq N$. Then there are only finitely many terms smaller than 1, and you can choose $\zeta$ to be smaller than any of these.
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2,815,845 |
Which is Greater $31^{11}$ or $17^{14}$ My try: Consider $$x=11 \ln(31)=11 \ln\left(32\left(1-\frac{1}{32}\right)\right)=55 \ln 2+11\ln\left(1-\frac{1}{32}\right)$$ hence $$x=55\ln2 -11\left(\frac{1}{32}+\frac{1}{2048}+\cdots \infty\right) \lt 55 \ln 2$$ Now consider $$y=14 \ln (17)=14 \ln\left(16 \left(1+\frac{1}{16}\right)\right)$$ hence $$y=56 \ln 2+14\ln \left(1+\frac{1}{16}\right) \gt 56 \ln 2$$ Now we have $$55 \ln 2 \lt 56 \ln 2$$ which implies $$11 \ln(31) \lt 14 \ln (17)$$ $\implies$ $$31^{11} \lt 17^{14}$$ is there any Elementary way to solve this?
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Yes:$$31^{11}<(2^5)^{11}=2^{55}<2^{56}=(2^4)^{14}<17^{14}.$$
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2,817,147 |
I noticed that there is an axiom that says that if $S(n)\implies S(n+1)$, and $S(1)$ is true, then $\forall n \in \Bbb N, S(n).$ My question is why is this an axiom? why can't we derive this from the other axioms?
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If you're looking at the Peano axioms for Arithmetic, and by 'the other axioms' you mean: $1. \bf{\forall x \ \neg s(x) = 0}$ $2. \bf{\forall x \forall y (s(x) = s(y) \rightarrow x = y)}$ $3. \bf{\forall x \ x+0=x}$ $4. \bf{\forall x \forall y \ x+s(y)=s(x+y)}$ $5. \bf{\forall x \ x\cdot0=0}$ $6. \bf{\forall x \forall y \ x\cdot s(y)=(x \cdot y) + x}$ then no, we can't derive the axiom of induction, which I'll formalize as: $\bf{(S(0) \land \forall x (S(x) \rightarrow S(s(x))))\rightarrow \forall x \ S(x)}$ (I include $0$ in $\mathbb{N}$, so base is $\bf{S(0)}$) Here is a counterexample: Take $\bf{S(x): s(x) \not = x}$ By axiom $1$, we have $\bf{\neg s(0)=0}$, and hence we have $\bf{S(0)}$ Also, if we have $\bf{S(k)}$, i.e. if we have $\bf{s(k) \not = k}$, then if it would be true $\bf{s(s(k)) = s(k)}$, then by Axiom 2 we have $\bf{s(k)=k}$, which contradicts $\bf{S(k)}$, and so we have $\bf{s(s(k)) \not = s(k)}$, i.e. $\bf{S(s(k))}$ OK, so with this $S(x)$, we have $\bf{S(0) \land \forall x (S(x) \rightarrow S(s(x)))}$ simply by virtue of Axioms 1 and 2 alone. But, we do not necessarily have $\bf{\forall x \ S(x)}$ Consider a model with domain $\mathbb{N} \cup \{ q \}$, i.e. the natural numbers together with some 'extra' element $q$. Interpret the $\bf{0}$ constant symbol as $0$ Interpret the $\bf{s}$ function symbol as a function $f$ for which $f(n)=n+1$ for any $n \in \mathbb{N}$, and for which $f(q)=q$ Interpret the $\bf{+}$ function symbol as a function $g$ for which $g(m,n)=m+n$ and $g(m,q)=g(q,n)=g(q,q)=q$ for any $m,n \in \mathbb{N}$ Interpret the $\bf{\cdot}$ function symbol as a function $h$ for which $h(m,n)=m\cdot n$ for any $m,n \in \mathbb{N}$, for which $h(0,q)=h(q,0)=0$, and for which $h(m,q)=h(q,n)=h(q,q)=q$ for any $m, n \in \mathbb{N}\setminus \{ 0 \}$ Then it is easily verified that all $6$ axioms are satisfied, and hence under this interpretation it is true that $\bf{S(0) \land \forall x (S(x) \rightarrow S(s(x)))}$. But, since $f(q)=q$, it is false that $\bf{\forall x \ s(x) \not = x}$. Hence, it is false that $\bf{S(0) \land \forall x (S(x) \rightarrow S(s(x))))\rightarrow \forall x \ S(x)}$, i.e. the axiom of induction would not hold under this interpretation. Therefore, the axiom of Induction does not follow from axioms $1$ through $6$.
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2,817,635 |
Derivatives, both ordinary and partial, appear often in my mathematics courses. However, my teachers have never really given a good example of why the derivative is useful. My questions: Other than the usual instantaneous rate of change, what are some common uses of the derivative? What does the partial derivative tell us? And what does the total derivative tell us? I find that often times, the derivative is simply explained as "the instantaneous rate of change". I am thinking about switching my major, because the applications of math at such an elementary level seem trivial when professors just push symbols and don't have any real world motivation included in their lectures. P.S. This question is not a duplicate of Why do we differentiate? I do not want to know why we differentiate. I want to know why it is important past our undergraduate learning. What are the applications beyond Calculus 3? Beyond academia, what makes the derivative important in complex situations?
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The derivative has many important applications both from elementary calculus, to multivariate calculus, and far beyond. The derivative does explain the instantaneous rate of change, but further derivatives can tell the acceleration amongst other things. With optimization, the derivative can tell us where the best place to sit in a room is, if the room is filling up with smoke, and at what time it is the best to sit there. The derivative can help with many optimization problems. The partial derivative tells us the direction of variables at a given time and the total derivative tells us where the slope increases the most and where. This is one way we can optimize in $\mathbb{R}^3$. The derivative can be applied to water flow and generally tells us much about how things change with respect to another variable. The derivative further can help in industry with economics, healthcare, engineering (especially), and many other things. Business has many applications as well. Your professor might not have time to delve into these applications as much as you would like because it is a calculus class, not an "application of the derivative" class. Although, he should definitely discuss these issues at some point. I have had some professors in my time who glossed over such subjects, but in multivariable calculus, they go way more in depth with them. I don't suggest switching your major without speaking directly with your professor about your difficulties. If you have further questions, I encourage you to ask your professor in office hours the same exact question and voice your concerns there. A good professor will encourage and motivate your learning outside of the classroom if you show initiative and ask.
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|
2,817,739 |
We know that there exist some functions $f(x)$ such that their derivative $f'(x)$ is strictly greater than the function itself. for example the function $5^x$ has a derivative $5^x\ln(5)$ which is greater than $5^x$. Exponential functions in general are known to be proportional to their derivatives.
The question I have is whether it is possible for a function to grow "even faster" than this. To be more precise let's take the ratio $f'(x)/f(x)$ for exponential functions this ratio is a constant. For most elementary functions we care about, this ratio usually tends to 0. But are there functions for which this ratio grows arbitrarily large? If so, is there an upper limit for how large the ratio $f'(x)/f(x)$ can grow? I also ask a similar question for integrals.
|
Consider the differential equation
$$
\frac{f'}{f} = g
$$
where $g$ is the fast-growing function you want. For instance, for $g(x) = e^x$ (and say the initial condition $f(0) =1$) you get
$$f(x) = e^{e^x-1}
$$
The ratio $f'/f$ grows arbitrarily large.
|
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|
2,819,107 |
All I have so far is $xy=168$, and I know I need a second equation to make a quadratic formula. So how do you write "$2$ consecutive even integers" as a formula? Answer: 12 and 14
|
Call the odd integer between the two even integers $n$. The even integers
are then $n-1$ and $n+1$, so that
$$168=(n-1)(n+1)=n^2-1$$
so that $n^2=169$ etc.
|
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|
2,819,108 |
What day will it be in 48 hours?
$$\frac{48}{24}=\frac{6}{3}(mod7)≡6∗3^{−1}(mod7)≡6∗5(mod7)≡2(mod7)$$
is the right answer. But the same technique does not work for 25 hours:
$$\frac{25}{24}≡\frac{25}{24}(mod7)≡{25}*{24^{-1}}(mod7)≡25∗5(mod7)≡6(mod7)$$ Why am I unable to get $1(mod7)$?
|
Call the odd integer between the two even integers $n$. The even integers
are then $n-1$ and $n+1$, so that
$$168=(n-1)(n+1)=n^2-1$$
so that $n^2=169$ etc.
|
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|
2,821,112 |
I begin this post with a plea: please don't be too harsh with this post for being off topic or vague. It's a question about something I find myself doing as a mathematician, and wonder whether others do it as well. It is a soft question about recreational mathematics - in reality, I'm shooting for more of a conversation. I know that a lot of users on this site (e.g. Cleo, Jack D'Aurizio, and so on) are really good at figuring out crafty ways of solving recreational definite integrals, like
$$\int_{\pi/2}^{\pi} \frac{x\sin(x)}{5-4\cos(x)} \, dx$$
or
$$\int_0^\infty \bigg(\frac{x-1}{\ln^2(x)}-\frac{1}{\ln(x)}\bigg)\frac{dx}{x^2+1}$$
When questions like this pop up on MSE, the OP provides an integral to evaluate, and the answerers can evaluate it using awesome tricks including (but certainly not limited to): Clever substitution Exploitation of symmetry in the integrand Integration by parts Expanding the integrand as a series Differentiating a well-know integral-defined function, like the Gamma or Beta functions Taking Laplace and Inverse Laplace transforms But when I play around with integrals on my own, I don't always have a particular problem to work on. Instead, I start with a known integral, like
$$\int_0^\pi \cos(mx)\cos(nx) \, dx=\frac{\pi}{2}\delta_{mn},\space\space \forall m,n\in
\mathbb Z^+$$
and "milk" it, for lack of a better word, to see how many other obscure, rare, or aesthetically pleasing integrals I can derive from it using some of the above techniques. For example, using the above integral, one might divide both sides by $m$, getting
$$\int_0^\pi \frac{\cos(mx)}{m}\cos(nx) \, dx=\frac{\pi}{2m}\delta_{mn},\space\space \forall m,n,k\in
\mathbb Z^+$$
Then, summing both sides from $m=1$ to $\infty$, and exploiting a well-known Fourier Series, obtain
$$\int_0^\pi \cos(nx)\ln(2-2\cos(x)) \, dx=-\frac{\pi}{n},\space\space \forall n\in
\mathbb Z^+$$
or, after a bit of algebra, the aesthetically pleasing result
$$\int_0^{\pi/2} \cos(2nx)\ln(\sin(x)) \, dx=-\frac{\pi}{4n},\space\space \forall n\in
\mathbb Z^+$$
After pulling a trick like this, I look through all of my notebooks and integral tables for other known integrals on which I can get away with the same trick, just to see what integrals I can "milk" out of them in the same way. This is just an example - even using the same starting integral, countless others can be obtained by using other Fourier Series, Power Series, integral identities, etc. For example, some integrals derived from the very same starting integral include
$$\int_0^\pi \frac{\cos(nx)}{q-\cos(x)} \, dx=\frac{\pi(q-\sqrt{q^2-1})^{n+1}}{1-q^2+q\sqrt{q^2-1}}$$
$$\int_0^\pi \frac{dx}{(1+a^2-2a\cos(x))(1+b^2-2b\cos(mx))}=\frac{\pi(1+a^m b)}{(1-a^2)(1-b^2)(1-a^m b)}$$
and the astounding identity
$$\int_0^{\pi/2}\ln{\lvert\sin(mx)\rvert}\cdot \ln{\lvert\sin(nx)\rvert}\, dx=\frac{\pi^3}{24}\frac{\gcd^2(m,n)}{mn}+\frac{\pi\ln^2(2)}{2}$$
Everyone seems to be curious about the proof of this last identity. A proof can be found in my answer here . I just pick a starting integral, and using every technique I know as many times as possible, try to come up with the most exotic integrals as I can, rather than picking a specific integral and trying to solve it. Of course, integrals generated this way would be poor (or at least extremely difficult) candidates for contest problems or puzzles to evaluate given the integral, since they are derived "backwards," and determining the derivation given the integral is likely much harder than pursuing the vague goal of a "nice-looking integral" with no objective objective (ha ha). QUESTION: Do you (residents of MSE who regularly answer/pose recreational definite integral questions) do this same activity, in which you try to generate , rather than solve, cool integrals? If so, what are some integrals you have come up with in this way? What strategies do you use? Does anyone care to opine on the value (or perhaps lack of value) of seeking integrals in this way? Cheers!
|
Yes, definitely. For example, I found that
$$ m\int_0^{\infty} y^{\alpha} e^{-y}(1-e^{-y})^{m-1} \, dy = \Gamma(\alpha+1) \sum_{k \geq 1} (-1)^{k-1} \binom{m}{k} \frac{1}{k^{\alpha}} $$
(and related results for particular values of $\alpha$) while mucking about with some integrals. Months later, I was reading a paper about a particular regularisation scheme (loop regularisation) useful in particle physics, and was rather surprised to recognise the sum on the right! I was then able to use the integral to prove that such sums have a particular asymptotic that was required for the theory to actually work as intended, which the original author had verified numerically but not proved. The resulting paper's on arXiv here . Never let it be said that mucking about with integrals is a pointless pursuit!
|
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|
2,823,128 |
I have a statement that says: If $a^2 + b^2 + c^2 = 2$ and $(a + b + c)(1 + ab + bc + ac) = 3^2$ What is the value of $( a + b + c )$ ? My reasoning was: $a^2 + b^2 + c^2 = 2$, rewritten as: $(a + b + c)^2 = 2 + 2ab + 2ac + 2bc$ Since, $(a + b + c)(1 + ab + bc + ac) = 3^2$ $(1 + ab + bc + ac) = \frac{9}{(a + b + c)}$ Now, replacing in the 1. Factorize $(a + b + c)^2 = 2(1 + ab + ac + bc)$ Replacing $(a + b + c)^2 = \frac{18}{a + b + c}$ Multiplying by $(a + b + c)$ $(a + b + c)^3 = 18$. Thus $a + b + c = \sqrt[3]{18}$. That is my result. But, the correct result should be $4$, then where is my mistake ?
|
Your work is perfectly correct. But I'm pretty sure you misinterpreted the original question: there should be $\color{green}{32}$, not $\color{red}{3^2}$, in the given data. With that value, the same work as you did will indeed yield $4$ as the final answer.
|
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|
2,823,174 |
I have been working hard on this limit:
$$(n^6 + n^5 + n^4)^{1/6} - n$$
as n goes to infinity. I have no clue of what i need to do. I always get indeterminante forms.
|
Your work is perfectly correct. But I'm pretty sure you misinterpreted the original question: there should be $\color{green}{32}$, not $\color{red}{3^2}$, in the given data. With that value, the same work as you did will indeed yield $4$ as the final answer.
|
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|
2,825,002 |
I sat an Algebra test yesterday, which consisted of 30 questions and a total time of 45 minutes (an average of 1 min 30 secs per question). One question of the test was this: Given the matrix: $$A=\begin{bmatrix}
-2 & 4 & 2 & 1 \\
4 & 2 & 1 & -2 \\
2 & 1 & -2 & 4 \\
1 & -2 & 4 & 2 \end{bmatrix}$$
Which of the following options is correct? (A) $A^{-1}=\dfrac{A}{25}$ (B) $A^{-1}=\dfrac{A}{5}$ (C) $A^{-1}=\dfrac{A}{15}$ (D) It has no inverse. I do know how to compute the inverse of a matrix. For example, using the Gauss-Jordan elimination method . Or for example, using this formula: $$A^{-1}=\dfrac{\text{Adj}(A^T)}{\text{det}(A)}$$ I calculated the determinant and it is $625$. However, this won't help me pick the correct option (of course I can eliminate option D, which is false). How in the world am I supposed to solve this problem in around 90-100 seconds, without using a calculator? Is there any shortcut or trick or something I missed? 90 seconds was the average time per question in the test. Given how little time I was given to solve the problem, this leads me to think that the structure of A could simplify the answer.
|
If $A^{-1}=\dfrac{A}{25}$ then $A^2=25I$. Similar for other options. Now you need to calculate only one diagonal element of $A^2$ to find the correct option.
|
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|
2,825,005 |
Given a covering $c:\tilde{X}\to X$, is there any set of conditions in order to have $T:\tilde{X}\to \tilde {X}$ to be a covering transformation provided $p=p\circ T$? I know that if T is the identity, the problem is trivial. But is it true for any other T that is NOT the identity?
|
If $A^{-1}=\dfrac{A}{25}$ then $A^2=25I$. Similar for other options. Now you need to calculate only one diagonal element of $A^2$ to find the correct option.
|
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|
2,825,505 |
Theorem: Suppose that $f : A \to \mathbb{R}$ where $A \subseteq \mathbb{R}$. If $f$ is differentiable at $x \in A$, then $f$ is continuous at $x$. This theorem is equivalent (by the contrapositive) to the result that if $f$ is not continuous at $x \in A$ then $f$ is not differentiable at $x$. Why then do authors in almost every analysis book, not take continuity of $f$ as a requirement in the definition of the derivative of $f$ when we (seemingly) end up with equivalent results? For example I don't see why this wouldn't be a good definition of the derivative of a function Definition: Let $A \subseteq \mathbb{R}$ and let $f : A \to \mathbb{R}$ be a function continuous at $a$. Let $a \in \operatorname{Int}(A)$. We define the derivative of $f$ at $a$ to be $$f'(a) = \lim_{t \to 0}\frac{f(a+t)-f(a)}{t}$$
provided the limit exists. I know this is probably a pedagogical issue, but why not take this instead as the definition of the derivative of a function?
|
Definitions tend to be minimalistic, in the sense that they don't include unnecessary/redundant information that can be derived as a consequence. Same reason why, for example, an equilateral triangle is defined as having all sides equal, rather than having all sides and all angles equal.
|
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|
2,825,625 |
I am trying to describe how irrational numbers, which are all modeled as a series of fractions , can themselves not be fractions, and are instead part of a unique group of "decimal numbers" outside of fractions, called the irrational numbers. I am confused atm. From Wikipedia , some example irrational numbers include: $\sqrt 2$ the golden ratio The sqrt of all natural numbers which are not perfect squares Logarithms Then they say: Almost all irrational numbers are transcendental and all real transcendental numbers are irrational. Examples include $e^\pi$. Rational numbers are fractions, which are included in the set of irrational numbers. Irrational numbers, however, are decimals and include things that "can't be represented as fractions" it seems. But where I'm confused is, sqrt 2 can be represented by a series of fractions: $${\displaystyle {\sqrt {2}}=\prod _{k=0}^{\infty }{\frac {(4k+2)^{2}}{(4k+1)(4k+3)}}=\left({\frac {2\cdot 2}{1\cdot 3}}\right)\left({\frac {6\cdot 6}{5\cdot 7}}\right)\left({\frac {10\cdot 10}{9\cdot 11}}\right)\left({\frac {14\cdot 14}{13\cdot 15}}\right)\cdots }$$ Similarly, $\pi$ can be represented by a series of fractions: $${\displaystyle 1\,-\,{\frac {1}{3}}\,+\,{\frac {1}{5}}\,-\,{\frac {1}{7}}\,+\,{\frac {1}{9}}\,-\,\cdots \,=\,{\frac {\pi }{4}}.}$$ Finally, the natural logarithm can be written as a series of fractions: $${\displaystyle \ln(1+x)=\sum _{k=1}^{\infty }{\frac {(-1)^{k-1}}{k}}x^{k}=x-{\frac {x^{2}}{2}}+{\frac {x^{3}}{3}}-\cdots }$$ It has been a while since I have added/divided/subtracted/multiplied fractions, but from what I remember doing any of those operations results in a new fraction. So I'm wondering what I'm missing when it comes to understanding irrational numbers. If irrational numbers can represent non-fraction numbers, yet they are themselves represented by a series of fractions, it seems the result of the series would itself be a fraction, and so the irrational numbers are all rational numbers. Looking for an understanding of how to explain the difference between rational and irrational numbers. I tried saying "irrational are decimal numbers you can't represent with a fraction", but then when getting into the definition of a rational numbers (fraction numbers), I was unable to explain how if all irrational numbers are themselves definable as a series of fractions , how they themselves aren't representable as fractions. Thank you for your help.
|
Rational numbers are values the can be written as a ratio (fraction) of two whole numbers. Irrationals are those values that can not. End of story. The first trick is to realize that there are values that can't be (Pythagoras didn't want to believe it). But if you toss a dart at something a mile wide what is the likelihood that it will hit a value that is exactly a ratio of two whole numbers. If one puts it that way it seems slim. But the second trick is to wonder, if the value isn't a ratio of whole numbers, then what is it and how can we express it? The hardest thing for novices to get is that question doesn't have an answer. We can't express them. But fractions can be made to be at least as small and smaller than anything we like. That means although we can not express all values, we can express a rational fraction that can be at least and closer to the value as we'd like. Now decimals are just fractions. $0.1=\frac 1 {10} $ and $0.3769=\frac {3769}{10000} $. We can get as close to expressing all values by taking longer and longer decimals. But we can never actually express an irrational with decimals. Instead ever Irrational has an infinite string of decimals that get close to it to any possible degree. And that is what we mean be an irrational number being a series of rational fractions. For every irrational number we can find a series (not just one; many series- the decimal expression is one of them) of rational numbers getting closer and closer together and honing in on the irrational and not honing in on anything else. The series is infinite and if we were immortal gods existing out of time and space we could see the entire infinite series at once and see it converges precisely to the irrational numbers. But we are not immortal gods outside of space so we can not. But we know it does.
|
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|
2,825,726 |
Background: I am currently studying real analysis using Tao's Analysis Volume One and so far I am really enjoying myself though I seem to have run into some confusion regarding professor Tao's construction of the reals using rationals. The following is the definition of reals that he provides in the text: $\DeclareMathOperator*{\LIM}{LIM}$ Definition 5.3.1 (Real numbers). A real number is defined to be an object of the form $\LIM\limits_{n → ∞} a_n$, where $(a_n)_{n = 1}^∞$ is a Cauchy sequence of rational numbers. Two real numbers $\LIM\limits_{n → ∞} a_n$ and $\LIM\limits_{n→∞} b_n$ are said to be equal iff $(a_n)_{n = 1}^∞$ and $(b_n)_{n = 1}^∞$ are equivalent Cauchy sequences. The set of all real numbers is denoted $\mathbb{R}$. Problem: While snooping around the internet I have found that a real number is in fact an equivalence class of sequences of rationals whose corresponding terms can be arbitrary close to each other i.e $(a_n)_{n=0}^{\infty}$ and $(b_n)_{n=0}^{\infty}$ are equivalent if and only if
$$\forall ε>0, \ \exists N \in \mathbb{N} \ \text{such that} \ \forall n \ge N, \ |a_n-b_n|\leq ε.$$
But Tao's definition seems to suggest that real numbers are limits of said sequences so what are they ?
|
This is a very formal definition of the real numbers (BTW there are others, look up "Dedekind cuts"). As to "what are they"? - well, they are exactly what he said: objects of the form ${\rm LIM}_{n\to\infty}a_n$. That is, they are nothing more nor less than a capital L, followed by a capital I, followed by a capital M, followed by... you get the point. And as this is the definition of real numbers, there is (at this point, and within the context of Tao's book) nothing else that we know about them. Of course, Tao did not choose the letters L,I,M at random: he wants to help you make the connection between
$${\rm LIM}_{n\to\infty}a_n$$
for rational $a_n$, which is the definition of a real number, and
$$\lim\nolimits_{n\to\infty}a_n$$
for possibly real $a_n$, which is the definition of a limit (Tao 6.1.8). Note that here we have lowercase l,i,m because it's a different concept. In other words, it is just as you stated in your question: But Tao's definition seems to suggest that real numbers are limits of said sequences... ...he wants to suggest this before he has actually defined the concept of a limit. (So, whether deliberately or not, you used exactly the right word!!!) You probably know lots about limits from previous courses: you should keep in mind all that you know and see how it fits in with what Tao is doing, but remember that "officially" you don't know what limits are because Tao hasn't defined them yet.
|
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