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Proportional Image Resizing This is quite a simple one I guess (but my mind is dead currently and is getting cluttered by other numbers).
I have an square image that is currently 256 width x 256 height.
I am variably resizing this image to a bigger size proportionally, so lets say I resize the image to 350 width to 350 height.
I then take a seperate image that is 416 width by 416 height and want to scale this to match the translation that was previously done in the last sentance. So I'm mirroring whats going on...
How would I calculate the 416x416's final size?
|
The first image has changed from $256$ to $350$, a $350/256$ change.
The second image should change by the same factor: from $416$ to $416 \cdot 350/256 = 568.75$. So choose $568$ or $569$ as the final size.
|
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If $f$ is continuous, why is $f$ with the property $f\left(\frac{x+y}{2}\right)\le \frac{1}{2}f(x)+\frac{1}{2}f(y)$ is convex?
If $f$ is continuous, why is $f$ with the property
$$f\left(\frac{x+y}{2}\right)\le \frac{1}{2}f(x)+\frac{1}{2}f(y),$$ where $0\le x,y\le 1$
is convex?
|
By induction we can prove that: if $k,m, l\in\mathbb{N} , k+m=2^l , x,y\in \mbox{domain} f $ then $$f\left( \frac{k}{2^l} \cdot x +\frac{m}{2^l} \cdot y \right)\leq \frac{k}{2^l} \cdot f(x) +\frac{m}{2^l} \cdot f(y). $$
Indeed the asertion is true when $l=1 .$ Suppose that it is true for some $l\geq 1 .$ And let $k+m=2^{l+1} , k=2^l +s , m=2^l-s ,s\in\mathbb{N} . $ We have:
$$f\left( \frac{k}{2^{l+1}} \cdot x +\frac{m}{2^{l+1}} \cdot y \right) =f\left( \frac{1}{2} \cdot x +\frac{1}{2}\cdot \left(\frac{s}{2^l} \cdot x+\frac{2^l -s}{2^l} \cdot y\right) \right) \leq \frac{1}{2} \cdot f(x) +\frac{1}{2} \cdot f\left(\frac{s}{2^l} \cdot x+\frac{2^l -s}{2^l} \cdot y\right)\leq \frac{1}{2} \cdot f(x) +\frac{1}{2} \cdot \frac{s}{2^l} \cdot f(x)+\frac{1}{2}\cdot\frac{2^l -s}{2^l} \cdot f(y) = \frac{k}{2^{l+1}} \cdot f(x) +\frac{m}{2^{l+1}} \cdot f(y) .$$
Hence by Induction the asertion holds true for any $l\in\mathbb{N} .$
Now let $1>\alpha >0 , $ and let $\frac{k_l}{2^l} \rightarrow \alpha $ as $l\to \infty .$ Since $f$ continuous we have $$f(\alpha x +(1-\alpha )y ) =\lim_{l\to\infty } f\left(\frac{k_l}{2^l} \cdot x + \left(1-\frac{k_l}{2^l} \right) y\right) \leq \lim_{l\to\infty } \left(\frac{k_l}{2^l} \cdot f(x) + \left( 1-\frac{k_l}{2^l} \right) f(y)\right) =\alpha f(x) +(1-\alpha )f(y) .$$
|
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Different ways of proving a polynomial is irreducible in finite field. Is there a general characterization of irreducible polynomials over a finite field?
I was going through a problem in finding whether $p(x):=x^7+x^5+1$ is irreducible over $\mathbb F_2[x]$ or not.
If the polynomial is of degree less than or equal to $3$ then we can easily find out if its irreducible or not by finding whether it has a root or not.
In this case considering the polynomial $p(x)=f(x)\cdot g(x)$ we may be able to show the irreducibility but this doesn't seem to be a very great idea.
Can anyone suggest a better idea ?
|
The given polynomial is in fact not irreducible. There is at least one decomposition:
$$
(x^2+x+1)\cdot(x^5+x^4+x^3+x+1) = x^7+x^5+1
$$
This can be found by resolving the equality for coefficients:
$$
(x^2+ax+1)\cdot(x^5+bx^4+cx^3+dx^2+ex+1) = x^7+x^5+1
$$
which, equating term by term, and ignoring the terms of degree $0$ and $7$:
$$
(a+b)x^6 + (1+ab+c)x^5 + (b+ac+d)x^4 + (c+ad+e)x^3 + (d+ae+1)x^2 + (e+a)x = x^5
$$
so
$$
a+b=0 \\
ab+c=1 \\
b+ac+d = 0\\
c+ad+e=0\\
d+ae=1\\
e+a=0
$$
which are more than enough to find the solution, given that in $\mathbb{Z}_2$, $a^2=a$ and $a+a=0$ for any $a$.
|
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Most important Linear Algebra theorems? I was reading up on symmetric matrices and the textbook noted that the following is a remarkable theorem:
A matrix $A$ is orthogonally diagonalizable iff $A$ is a symmetric matrix.
This is because it is impossible to tell when a matrix is diagonalizable, or so it seems.
I haven't gotten to realize yet how important this is, but I will soon. What, in your opinion , is the most important linear algebra theorem and why?
|
The two main candidates are:
*
*The fundamental theorem of linear algebra, as popularised by Strang.
*The singular value decomposition.
From these, lots of important results follow.
|
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Question regarding 3 x 3 matrices If $A$ is a $3 \times 3$ matrix with real elements and $\det(A)=1$, then are these affirmations equivalent:
$$
\det(A^2-A+I_3)=0 \leftrightarrow \det(A+I_3)=6 \text{ and } \det(A-I_3)=0?
$$
|
$\Leftarrow)$ From $\det (A-I)=0$ we know that $1$ is an eigenvalue. Let $x,y$ be the two others (possibly complex, and counting multiplicities). From $\det A=1$ we know that $xy=1$. And $6=\det(A+I)=2(x+1)(y+1)$, so
$$
3=xy+x+y+1=2+x+y,
$$
so $x+y=1$. We get a system of two equations on $x,y$, namely
$$
x+y=1,\ \ xy=1.
$$
That is, $x(1-x)=1$, or $x^2-1x+1=0$. Note that $y$ satisfies the same equation. In any case $A^2-A+I$ has zero as eigenvalue, so $\det(A^2-A+I)=0$.
$\Rightarrow)$ We consider the three eigenvalues of $A$, $x,y,z$; they satisfy $xyz=1$ by hypothesis. Since $\det(A^2-A+I)=0$, one of them, say $y$, satisfies $y^2-y+1=0$. This implies that $y$ is non-real, and its conjugate is also an eigenvalue (since $A$ has real entries). So $y=\frac12+i\frac{\sqrt3}2$, $z=\frac12-i\frac{\sqrt3}2$, and
$$
x=\frac1{yz}=\frac1{\frac14+\frac34}=1.
$$
Now $\det(A-I)=0$ (since $1$ is an eigenvalue) and
$$
\det(A+I)=2(y+1)(z+1)=2\left(\frac32+i\frac{\sqrt3}2\right)\left(\frac32-i\frac{\sqrt3}2\right)=2\,\left(\frac94+\frac34\right)=6.
$$
|
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Context free grammar question i have two context free grammar questions and I don't know how to do them.
*
*$$\{(a^n)b(c^n) \mid n >0 \}$$ I'm having trouble with this one because I don't know how to account for $a$ or $b$ not having empty set.
*$$\{(a^n)(b^m)(c^n) \mid n,m \ge 0\}$$
|
You are close for the first:
$$S\to aAc,\qquad A\to S \mid b$$
and for the second
$$S\to aSc \mid A\qquad A\to bA\mid\epsilon $$
should work
|
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How can I calculate $\displaystyle\lim_{x \to \infty}x^{2}\ln\left(\cos \left(\pi/x\right)\right) $? Does anybody know how to solve this?
$$\lim_{x \to \infty}x^{2}\ln\left(\cos\left(\pi \over x\right)\right)$$
|
Let $x=1/t$. So we are interested in the limit as $t\to 0^+$ of $\ln(\cos(\pi t))/t^2$.
One round of L'Hospital's Rule brings us to
$$\lim_{t\to 0^+} -\frac{\pi}{2\cos(\pi t)} \frac{\sin(\pi t)}{t}.$$
The first part is nicely behaved near $0$. For $\lim_{t\to 0^+} \frac{\sin(\pi t)}{t}$, use L'Hospital's Rule, or simpler tools.
|
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Inverse image of a closed subscheme Let $f:X\to Y$ be a surjective morphism of schemes, and $Z\subset Y$ a closed subscheme with short exact sequence
$$ 0\to I_Z \to \mathcal{O}_Y \to \mathcal{O}_Z \to 0. $$
What are sufficient conditions on $X$, $Y$, $Z$ and $f$ such that the scheme-theoretic inverse image $W$ of $Z$ is a closed subscheme of $X$ with short exact sequence
$$ 0\to f^*(I_Z) \to \mathcal{O}_X \to \mathcal{O}_W \to 0 \quad? $$
|
Do you see that $f^*(I_Z) \to \mathcal{O}_X \to \mathcal{O}_W \to 0$ is always exact? So the question is basically only if $f^*(I_Z) \to \mathcal{O}_X$ is injective. This holds when $f$ is flat.
|
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Closed sets, boundary, topology. Let A be a closed subset of the real numbers. It is always possible to find a subset B of the real numbers such that A is equal to the boundary of B? Prove if true, find a counterexample if not.
I think it's false but have no idea how to find a counterexample. Any help will be much appreciated.
|
Given $A$, let $B$ be any countable dense subset of $A$. (Such a $B$ exists; just take one point from each nonempty set of the form $A\cap(p,q)$ where $p<q$ are rational numbers.) So the closure of $B$ is $A$. The interior of $B$ is empty, because $B$ is countable. So the boundary of $B$ is $A$.
|
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How to prove that either $2^{500} + 15$ or $2^{500} + 16$ isn't a perfect square? How would I prove that either $2^{500} + 15$ or $2^{500} + 16$ isn't a perfect square?
|
The question is not about that big numbers...
Question is about a result that two consecutive numbers can not be squares simultaneously..
Suppose $a=b^2\text { and }a+1=c^2\Rightarrow b^2+1=c^2\Rightarrow c^2-b^2=1\Rightarrow (c+b)(c-b)=1$
w.l.o.g. assume $c+b=1$ which implies
$a+1=c^2=(1-b)^2=1+b^2-2b\Rightarrow a=b^2-2b$ but then we have $a=b^2$
you should now be able to see some contradiction...
So...
|
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Number of involution in symmetric group Can any one show me on how to prove that the number of involutions in the symmetric group on $n$ letters, $$\sum_{k=0}^{\lfloor n/2 \rfloor} {n \choose 2k} (2k-1)!! = \sum_{k=0}^{\lfloor n/2 \rfloor} \frac{n!}{2^kk!(n-2k)!}$$
I've tried proving it via Young tableaux, knowing that the number of tableaux is the number of involutions, but I haven't got the right answer so far.
|
Let $\tau$ be an involution. Writing $\tau$ as a product of disjoint cycles, we see that no cycle can have length greater than $3$ since $\tau$ has order $2$. It follows that $\tau$ is a product of disjoint transpositions.
Suppose that $\tau$ is a product of $k$ disjoint transpositions. Then it permutes precisely $2k$ letters so we must choose those letters out of the $n$ available. Next we must group the $2k$ letters into $k$ pairs. There are precisely
$$a_k = \binom{n}{2k}\frac{(2k)!}{k!2^k} = \binom{n}{2k}(2k-1)!!$$
ways to do this so there are precisely $a_k$ involutions with $k$ disjoint transpositions.
Clearly $k$ can range from $0$ (the identity) to $\lfloor \frac{n}{2}\rfloor$ transpositions. Summing through the different values of $a_k$ gives the desired result.
|
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Solve PDE in 2D Problem How should I go about solving this PDE:
$$ \phi_x+\phi_y=x+y-3c $$
Where $\phi = \phi(x,y)$, $c$ is a constant, and $\phi$ is specified on the circle
$$ x^2+y^2=1 $$
My Attempt to solve it I would like to use the method of characteristics, but then I get stuck because of the given initial condition. In fact, so far I have the characteristics equations
$$ \dot{{z}}(s)=x+y-3c $$
$$ \dot{{x}}(s)= 1 $$
$$ \dot{{y}}(s)= 1 $$
The last two are easy to solve but then I am not sure how to use the initial condition.
If you know of a different/easier method to solve this PDE, feel free to let me know, thanks!
|
As you say, the characteristic equations are $\dot{z} = x + y - 3 c$, $\dot{x} = 1$, $\dot{y} = 1$. So the characteristic curves are $x = x_0 + s$, $y = y_0 + s$, i.e. $x - y = \text{constant}$. But there's a problem with specifying the initial conditions on the circle $x^2 + y^2 = 1$: the characteristic curves through most points either don't intersect the circle at all (so the initial condition doesn't determine $\phi$ there) or intersect it in two points (so
the initial conditions might not be consistent).
|
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Constrained optimisation question
Since $f$ has a local extremum at $x_1$, then surely the LHS of equation (3) always zero? If so, then isn't lambda always simply zero too? But this cannot be, otherwise the last sentence of the theorem wouldn't be phrased the way it is. What am I missing?
|
It's not $f$ that has a local extremum at $x_{1}$, but rather $f|_{S}$. Consider, for example, $f(x,y)=xy$ restricted to the line $y=1-x$. This restricted function has a local maximum at $(\frac{1}{2},\frac{1}{2})$, but the full function $f$ is not at an extremum at $(\frac{1}{2},\frac{1}{2})$.
|
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Why/How are there infinite points in a line segment? A line may have infinite points becauase it may be expanded.But in case of a line segment it has 2 distinct points which are not movable.The distance between the end points in finite and known.
But still why do people(in my school) say that there are infinite points in a line segment.When I ask the teacher,she says you will learn at higher levels(what a genius way to get rid of question).
|
The concept of infinity is used for many different things, and one should not confuse them. A line, be it closed, open, straight, curved, finite in length, or infinite in length always consists of infinitely many individual points (at least for any reasonable notion of 'line'). This can be proven rigorously, and it's not hard at all. Basically, between any two distinct points on a line there is a third point between these two points (between should not necessarily mean the mid-point on a straight path connecting the points, and this may get a bit tricky, but not too tricky). The length of the line is a different issue. It may be finite in length or infinite in length. The totality of the points comprising the line is in any case infinite. In fact it can be shown (quite easily) that the cardinality of points of any line is always the same, so in a sense all lines have the same "amount" of points in them, though the way these points are arranged may give totally different geometric qualities to the line.
|
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Simplify $2^{(n-1)} + 2^{(n-2)} + .... + 2 + 1$ Simplify $2^{(n-1)} + 2^{(n-2)} + .... + 2 + 1$
I know the answer is $2^n - 1$, but how to simplify it?
|
Another "proof" (this was actually the way I had "proved" it myself until I saw the GP proof):
The first summand is 1 followed by a $n-1$ in base 2. The later ones are with one less zero every time. Adding them up results in a number composed of 1 followed by n-1 zeros, which is $2^n-1$ (1 followed by n zeros minus one).
|
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Does De Moivre's Theorem hold for all real n? I have seen the proof by induction for all integers, and I have also seen in a textbook that we can use Euler's formula to prove it true for all rational n, but nowhere in the book does it say its true for irrational n.
I have also looked over the internet and there seems to be some problem with non-integer values for n (as I understand, a problem of uniqueness, but I'm not sure).
I would appreciate it if someone could just clarify this for me.
Thanks in advance!
|
The formula is actually true in a more general setting: if $z$ and $w$ are complex numbers, then $\left(\cos z + i\sin z\right)^w$ is a multi-valued function while $\cos (wz) + i \sin (wz)$ is not. However, it still holds that $\cos (wz) + i \sin (wz)$ is one value of $\left(\cos z + i\sin z\right)^w$.
|
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How to show that $A_k=(-1)^k\binom nk$?
In the identity $$\frac{n!}{x(x+1)(x+2)\cdots(x+n)}=\sum_{k=0}^n\frac{A_k}{x+k},$$prove that $$A_k=(-1)^k\binom nk.$$
My try: The given identity implies $$\frac{1\cdot2\cdots n}{x(x+1)(x+2)\cdots(x+n)}=\frac{A_0}{x}+\frac{A_1}{x+1}+\dots+\frac{A_n}{x+n}.$$
Now putting $A_k=(-1)^k\binom nk,$$$\frac{1\cdot2\cdots n}{x(x+1)(x+2)\cdots(x+n)}=\frac1x-\frac{n}{x+1}+\dots+\frac{(-1)^n}{x+n}.$$ How to proceed further?
|
HINT: Multiply both side by $x+k$ and then put $x=-k.$
|
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prove $f^{-1}(C \cap D) = f^{-1}(C) \cap f^{-1}(D)$ I need help with this proof:
$f: X\rightarrow Y$
$C,D\subseteq Y$
$f^{-1}(C \cap D) = f^{-1}(C) \cap f^{-1}(D)$
Thanks.
|
To solve these kinds of questions, you want to show that $$f^{-1}(C\cap D) \subseteq f^{-1}(C) \cap f^{-1}(D)$$ and $$f^{-1}(C\cap D) \supseteq f^{-1}(C) \cap f^{-1}(D).$$
I will show $\subseteq$ here; you should try the other direction.
Suppose $x \in f^{-1} (C \cap D)$. This means $f(x) \in (C\cap D)$, which further implies $f(x) \in C$ AND $f(x) \in D$. Thus, $x \in f^{-1}(C)$ AND $x \in f^{-1}(D)$.
|
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non degenerate bilinear map for modules Let $R$ be a commutative ring with 1. Suppose $A,B$ are $R$-modules, $P:A\times B\to R$ is
a bilinear map that satisfies the following property: if $P(a,b)=0$ for all $b\in B$, then $a=0$. Then is the $R$-linear map $P':A\to B^*$, $P':a\mapsto P(a,b)$ an isomorphism? I could prove the injective part, but got stuck on proving $P'$ is surjective.
Plus, if I from these assumptions surjection is not necessarily true, what else should be assume about $P$ to ensure that $P'$ is surjective?
|
Hint: What you're talking about are the induced isomorphisms of a non-degenerate (not necessarily perfect) pairing. The definition of $P'$ is, however, a bit different. Here is the right one:
$$P': A \to B^*, a \mapsto P(a,-)$$
So the image of an $a \in A$ is really a dual map $B \to R$.
Are those modules finitely generated over $R$?
|
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How to find the derivative of $\operatorname{arcsinh}(3x)$? I know
The derivative of $\operatorname{arcsinh}(x) = 1/(x^2+1)^{1/2}$
But if I derivative $\operatorname{arcsinh}(3x)$
Why it doesn't equal to
$(\operatorname{arcsinh}(3x))^{-1} (1/(9x^2+1)) (3)$ ??
Thanks all
|
Where did the factor $(\operatorname{arcsinh} (3x))^{-1}$ come from? Alpha agrees with $\frac 3{\sqrt{9x^2+1}}$
|
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Why is $K(\alpha) = \left\lbrace \frac{f(\alpha)}{g(\alpha)} : f,g\in K[X],\, g(\alpha)\neq 0\right\rbrace$? I know from definition that: $K(\alpha)$ denotes the smallest subfield of $L$ that contains both $K$ and $\alpha$.
I've read here that this is equivalent with:
$$K(\alpha) = \left\lbrace \frac{f(\alpha)}{g(\alpha)} : f,g\in K[X],\, g(\alpha)\neq 0\right\rbrace.$$
But I don't see how this makes sense.
|
Let $K(\alpha)$ be the smallest subfield containing $\alpha$ and let
$$K'(\alpha) = \left\lbrace \frac{f(\alpha)}{g(\alpha)} : f,g\in K[X],\, g(\alpha)\neq 0\right\rbrace.$$
You want $K'(\alpha) = K(\alpha)$.
First convince yourself that $K'(\alpha)$ is a subfield containing $\alpha$. Then by definition you have $K(\alpha) \subseteq K'(\alpha)$.
To get $K'(\alpha) \subseteq K(\alpha)$ take an element $f(\alpha)/g(\alpha) \in K'(\alpha)$. Note that the coefficients of $f(x)$ are contained in $K(\alpha)$ and $\alpha$ is also contained in $K(\alpha)$ so $f(\alpha) \in K(\alpha)$. Similarly $g(\alpha) \in K(\alpha)$. As $K(\alpha)$ is a field we then get $f(\alpha)/g(\alpha) \in K(\alpha)$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How I can calculate $\sum_{n=0}^\infty{\frac{n}{3^n}}$? For more than one series of any search function known not seem to find the sum of this series:
$$\sum_{n=0}^\infty{\frac{n}{3^n}}$$
I've found that converges with the quotient criterion.
Could you give me some suggestions for finding the sum of this series?
Thanks!
|
HINT:
If $|a|<1$,$$a\sum_{0\le n<\infty}r^n=\frac a{1-r}$$ (Proof)
Differentiate wrt $r$
|
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|
Expressing a factorial as difference of powers: $\sum_{r=0}^{n}\binom{n}{r}(-1)^r(l-r)^n=n!$? The successive difference of powers of integers leads to factorial of that power. Here's the formula:
$$\sum_{r=0}^{n}\binom{n}{r}(-1)^r(n-r)^n=n!$$
Can anyone give a proof of this result?
Note: The original question was to prove the more general
$$\sum_{r=0}^{n}\binom{n}{r}(-1)^r(l-r)^n=n!$$
for an integer $l$, which some of the answers address.
|
Let $A:=\{ x_1,..,x_n \}$ and $B=\{y_1,..,y_m \}$.
Lets count the number of onto functions $f:A \to B$. There are $m^n$ functions from $A$ to $B$. Lets count now the ones which are not onto:
Define
$$P_i= \{ f : A \to B |y_i \notin f(A) \}$$
Then we need to figure out the cardinality of $\cup_i P_i$.
By the inclusion exclusion principle
$$|P_1 \cup P_2 ..\cup P_m |=\sum |P_i|-\sum |P_i \cap P_j|+\sum |P_i \cap P_j \cap P_k| -... \\=\binom{m}{1}(m-1)^n-\binom{m}{2}(m-2)^n+\binom{m}{3}(m-3)^n-...
$$
Thus in total there are
$$m^n-\binom{m}{1}(m-1)^n+\binom{m}{2}(m-2)^n-\binom{m}{3}(m-3)^n-...=\sum_{k=0}^m (-1)^k\binom{m}{k}(m-k)^n$$
When $n=m$ the number of onto functions is
$$\sum_{k=0}^n (-1)^k\binom{n}{k}(n-k)^n$$
But any function $f: \{ x_1,..,x_n \} \to\{y_1,..,y_n \}$ is onto if and only if it is a bijection. Thus the number of onto functions is equal to the number of bijections, which is $n!$.
Hence
$$\sum_{k=0}^n (-1)^k\binom{n}{k}(n-k)^n=n!$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Subgroup of order $n-1$ of a group of order $n$ Here is question 2.1.5 from Dummit and Foote : Prove that $G$ cannot have a subgroup $H$ with $|H| = n-1$, where $n = |G| > 2$.
How can one show this without using Lagrange's theorem (which is in chapter 3 of Dummit).
Thank you
|
Let $g$ be an element of $G$ that is not in $H$. Let $h$ be any nonidentity element of $H$. Now show that $gh$ is not in $H$.
|
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|
How does one get better at real analysis proofs? How does one proceed through a math proof in real analysis? My instructor always says make a diagram, but I am not a visual learner. It seems that whenever I write out the definition of an assumption, then I cannot make the next logical step. Also, when I go to try to verify that my proof is correct, I ask myself questions like, "why must this be true?" but the proof does not end up not being air tight. For those that have had real analysis, what did you do to master proofs and do the exercises?
|
I think the best thing to do is to learn and understand the proofs of the theorem you do in class. The key thing about analysis (as opposed to algebra) is that all the proofs have a pattern to them.
For example: The proof of sequence of continuous functions converges uniformly to a continuous function uses the idea called $\frac{\epsilon}{3}$ argument. By knowing this technique, you can do a number of other proof that involves convergence and continuity.
|
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|
Brouwer Fixed Point Theorem $f(S^1)\subset B$ I have a question about the Brouwer Fixed Point Theorem:
Theorem 1.(Brouwer Fixed Point Theorem) Let $B=\{x\in \mathbb R^2 :∥x∥≤1\}$ be the closed unit ball in $\mathbb R^2$ . Any continuous function $f:B\rightarrow B$ has a fixed point.
Theorem 2. Let $B=\{x\in \mathbb R^2 :∥x∥≤1\}$ be the closed unit ball in $\mathbb R^2$ . Any continuous function $f:B\rightarrow \mathbb{R}^2$ such that $f(\partial B)\subset B$ has a fixed point.
Note that Theorem $2$ implies Theorem $1$.
We can prove Theorem $2$ using Theorem $1$(or a similar proof)?
$\partial B=S^{1}$
Any hints would be appreciated.
|
Since Theorem $1$ obviously generalises to arbitrary finite radii, we can deduce Theorem $2$ from Theorem $1$ in a simple way:
Since $B$ is compact, $f(B)$ is contained in a ball $B_R$ of some finite radius $R > 0$.
Extend $f$ to $B_R$ in the following way:
$$F(x) = \begin{cases}f(x) &, \lVert x\rVert \leqslant 1\\
f\left(\lVert x\rVert^{-1}\cdot x\right) &, 1 < \lVert x\rVert \leqslant R. \end{cases}$$
Then $F \colon B_R \to B_R$ is continuous, hence it has a fixed point by the generalisation of Theorem $1$. Since $\lVert F(x)\rVert \leqslant 1$ for $\lVert x\rVert > 1$, all fixed points of $F$ must lie in $B$, and therefore are fixed points of $f$.
|
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|
How to find $\lim_{n \rightarrow +\infty } \left(\sqrt[m]{\prod_{i=1}^{m}(n+{a}_{i})}-n\right)$? I think it is zero;
$$\lim_{n \rightarrow +\infty } \left(\sqrt[m]{\prod_{i=1}^{m}(n+{a}_{i})}-n\right)$$
we can make that steps:
$$\lim_{n \rightarrow +\infty } \left(\sqrt[m]{{n}^{m}\prod_{i=1}^{m}\left(1+\frac{a_i}n\right)}-n\right)$$
and then:
$$\lim_{n \rightarrow +\infty } \left(n \cdot \sqrt[m]{\prod_{i=1}^{m}\left(1+\frac{{a}_{i}}{n}\right)}-n\right)$$
each bracket aspires to unit
so we have:
$$\lim_{n\rightarrow \infty} (n-n)$$
and..
$$\lim_{n\rightarrow \infty} (0) = 0$$
|
The hard part is
showing that
$ P_n
=\sqrt[m]{\prod_{i=1}^{m}\left(1+\frac{{a}_{i}}{n}\right)}
\to 1$
as $n \to \infty$.
If $n > km\max(|a_i|)$,
$1+\frac{{a}_{i}}{n}
< 1+\frac1{km}
$
so $P_n
<\sqrt[m]{\prod_{i=1}^{m}\left(1+\frac1{km}\right)}
< 1+\frac1{km}
$.
By choosing $k$ large enough,
$P_n$ can be made as close to $1$
as wanted,
showing that
$\lim_{n \to \infty} P_n = 1$.
|
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|
Reduction Transitive Relation Problem I have this problem on my homework, it's my last one left but I'm having trouble with it. Any help would be appreciated.
|
Definition:
Let $P_1$ and $P_2$ be two problems and $\mathbb{A}$ be the space of all poly-time algorithms, then
$P_1 \leq_P P_2$ if $((\exists \ A_1,\ A_2 \in \mathbb{A}):
(A_1 \text{ maps any instance } I \text{ of } P_1 \text{ to an instance } A_1(I) \text{ of } P_2) \text{ and } (A_2 \text{ maps any solution } S \text{ to } A_1(I) \text{ to a solution } A_2(S) \text{ to } I))$
Proposition:
Show that
$$P_1 \leq_P P_2 \text{ and }P_2 \leq_P P_3 \Rightarrow P_1 \leq_P P_3$$
Proof:
Assume that $P_1 \leq_P P_2 \text{ and }P_2 \leq_P P_3$
By def. $(\exists\ A_1,\ A'_1,\ A_2,\ A'_2 \in \mathbb{A}): (A_1(p_1) = p_2 \text{ and } A'_1(s_{p_2})=s_{p_1}) \text{ and } (A_2(p_2) = p_3 \text{ and } A'_2(s_{p_3})=s_{p_2})$
Illustration:
The goal is to show the existence of two poly-time algorithms which satisfy the definition of reduction. Formally, we need to proof $(\exists\ A,\ A' \in \mathbb{A}): A(p_1) = p_3 \text{ and } A'(s_{p_3}) = s_{p_1}$
$A(p_1) = p_3 \Leftrightarrow A(p_1) = A_2(p_2) \Leftrightarrow A(p_1) = A_2(A_1(p_1)) $
As you can see $A$ maps $p_1$ to $p_3$
Since $A_2 \in \mathbb{A}$, then $A \in \mathbb{A}$, That is, $A$ is a poly-time algorithm.
Again we have $A'(s_{p_3}) = s_{p_1} \Leftrightarrow A'(s_{p_3}) = A'_1(s_{p_2}) \Leftrightarrow A'(s_{p_3}) = A'_1(A'_2(s_{p_3}))$
So $A'$ maps $s_{p_3}$ to $s_{p_1}$
Since $A'_1 \in \mathbb{A}$, then $A' \in \mathbb{A}$, That is, $A'$ is a poly-time algorithm.
Illustration:
|
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|
Relations $R^2, R^3, R^i and R^*$ Consider the relation on R on the reals where $xRy$ iff $xy=1$
I need to find $R^2, R^3, R^i $ and $R^*$
Ok, so I first started off with the following:
$$xR^2z \equiv \exists y: xRy\land yRz \\ \equiv\exists y: xy=1 \land yz=1 \\ \equiv xy + yz =y(x+z)=2$$
This to me doesn't seem right. Does the $\land$ represent addition in cases where you'
re trying to find compositions of $R$ ? How would I find $R^3$ in this case? If my assumption of $R^2 = y(x+z)=2$ is correct, then would $R^i = yi(x+z)=2$ for some $i\ge1$?
I'm getting confused with the definition of $R \circ R$ and have a hard time applying it for $i\gt2$ and subsequent compositions.
Another point of confusion is with $R^*$. Given:
$$R^*= R\cup R^2 \cup R^3 \cdots\cup R^n = \bigcup_{i=1}^n R^n $$
How would you define what it $R^*$ look like if the set is infinite? Suppose we had $n=3$ then you would have 3x3 matrix and in this case $R^*=R\cup R^2 \cup R^3$. At this point if you're final matrix $R^3$ differs from $R$ then it's said that it wouldn't be transitive.
|
The definition of $R \circ S$ is: $x(R \circ S)z$ if and only if there exists a $y$ such that $xRy$ and $ySz$.
This symbol $\wedge$ is not a plus! It means "and", and like the name suggests, it means both statements must be true.
So, to the actual problem. Say $xR^2z$. Then there exists a $y$ such that $xy = 1$ and $yz = 1$. This implies $y = \frac{1}{x}$ and $y = \frac{1}{z}$. Because both are true, $\frac{1}{x} = \frac{1}{z}$, which means $x = z$. So $xR^2z$ iff $x = z$.
For $R^3$: Say $xR^3z$. Then there exists some $y$ such that $xR^2y$ and $yRz$. But we know that $xR^2y$ implies $y = x$. So we substitute into the second and get $xRz$. So $xR^3z$ iff $xRz$ (which makes intuitive sense, composing with $R^2$, the $=$ relation, shouldn't do anything)
To define $R^\ast$ when $n$ is infinite, you have $R^\ast = \bigcup_{i = 1}^\infty R^i$. In a union, you keep all the elements, so this means: $xR^\ast y$ iff $xR^i y$ for some $i$. In this case, we know that $xR^iy \iff xy = 1$ or $x = y$, depending if $i$ is even or odd. But $R^\ast$ is still transitive. Can you verify that?
|
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|
Change of sign for Brownian motion For a fixed $\epsilon > 0$, I want to show that almost surely (i.e., with probability $1$), a standard Brownian motion $W_t$ would change sign over $[0,\epsilon]$.
I thought about defining a random variable $U_t = \mbox{sign} (W_t)$ which would satisfy $(1+U_t)/2$ follows symmetric Bernoulli distribution
But I can't conclude with this. Am I on a good track?
|
Another way to solve this problem is with the Blumenthal zero-one law. Let $\mathcal{F}_t = \sigma(B_s : 0 \le s \le t)$, and $\mathcal{F}_{0+} = \bigcap_{t > 0} \mathcal{F}_t$. The Blumenthal zero-one law asserts that $\mathcal{F}_{0+}$ is almost trivial; for each $A \in \mathcal{F}_{0+}$ we have $\mathbb{P}(A) = 0$ or $1$. Consider the event $A_+$ that $B_{1/n} > 0$ for infinitely many $n$. Show that $\mathbb{P}(A_+) \ge 1/2$, and that $A_+ \in \mathcal{F}_{0+}$. Conclude that $\mathbb{P}(A_+) = 1$. Show the same for $A_-$, the event that $B_{1/n} < 0$ for infinitely many $n$. On the event $A_+ \cap A_-$, $B_t$ changes sign in every interval $[0,\epsilon]$.
|
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How to solve this reccurence relation? Let a,b,c be real numbers. Find the explicit formula for $f_n=af_{n-1}+b$ for $n \ge 1$ and $f_0 = c$
So I rewrote it as $f_n-af_{n-1}-b=0$ which gives the characteristic equation as $x^2-ax-b=0$. The quadratic formula gives roots $x= \frac{a+\sqrt{a^2+4b}}{-2}, \frac{a-\sqrt{a^2+4b}}{-2}$
Then $f_n=P_1(\frac{a+\sqrt{a^2+4b}}{-2})^n+P_2(\frac{a-\sqrt{a^2+4b}}{-2})^n$ and using the initial condition $t_0=c$ gives $C=P_1+P_2 \Rightarrow P_1=C-P_2$
So $(C-P_2)(\frac{a+\sqrt{a^2+4b}}{-2})^n+P_2(\frac{a-\sqrt{a^2+4b}}{-2})^n$ what next? I tried expanding but that didn't help. I know the answer is something like $cd^n-\frac{b}{a-1}+\frac{bd^n}{a-1}$
|
Why not consider this?
$f_n + m = a(f_{n-1} + m) \Longrightarrow (a-1)m=b$
1) $a=1$, simple recurrence $f_n = f_{n-1} + b$, $f_n = bn+c$
2) $a\neq 1$, $m=\frac{b}{a-1}$, $f_n+m = a(f_{n-1}+m)$, geometric sequence $f_n+m=a^n(c+m)$
Hope it is helpful!
|
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|
Bijective check with matrix My book doesn't cover the criterion for bijective transformations very well. I just want to check my understanding: is this statement true?
Let F be a linear transformation. Let A be the matrix that represents that transformation (which means that that $F(v)=Av$ for any vector $v$). We now have that F is bijective iff $\det(A)\not=0$.
|
I think about this in the following manner. A bijective linear transformation should have an inverse. Hence the associated matrix should also be invertible. Therefore it's determinant is non-zero. Hope this helps.
|
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|
Prove intersection of all inductive sets is inductive How to prove that the intersection of all inductive sets is inductive?
Subset A of ordered field F is inductive, when:
1) $1 \in A$
2) if $a \in A$ then $a+1 \in A$
Prove that $$\mathbb{I} = \bigcap \left\{A\in F\ {\large|} \ A \text{ is inductive }\right\}$$ is inductive.
|
HINT: Recall that $x\in\bigcap\cal A$ if and only if for every $A\in\cal A$, $x\in A$.
|
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|
Construction of Homogenous Differential Equation We have to construct a homogenous differential equation of second order which has y(t) = e^t cos(t) as solution.
I know have some knowledge of how to solve (some) differential equations, but how do I construct one (systematically)?
|
You want $y=e^t cos(t)$ as a solution so lets take charecteristic equation to be the one with $i+1 $ as its root.
This will be your charecteristic equation $(\lambda-(i+1))(\lambda+(1-i)) \to \lambda ^2 - 2\lambda +2$.
And this is your ode $y''-2y'+2y=0$ .
|
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Integrals involving Hermite Polynomials Could you please tell me, How to evaluate this integral which involve hermite polynomials, $\int_{-\infty}^\infty e^{-ax^2}x^{2q}H_m(x)H_n(x)\,dx=?$ where $H_n$ is the $n$-th Hermite polynomial (Physicist's version) and $q,\,m$ and $n$ are positive integers.
If $x^{2q}$ term were absent, I am able to perform the integral by writing the product of the Hermite polynomials into a single Hermite polynomial with a higher degree.
Can anybody give me a hint to perform this integral?
|
As noted in the comments, the solution for arbitrary $q$
$$
I_{nm}(q)=\int_{-\infty}^\infty e^{-ax^2}x^{2q}H_n(x)H_m(x)dx
$$
are the derivatives of the integral
$$
I_{nm}(0)=\int_{-\infty}^\infty e^{-ax^2}H_n(x)H_m(x)dx
$$
with respect to $a$
$$
I_{nm}(q)=(-1)^q\frac{\partial^q}{\partial a^q}I_{nm}(0).
$$
The $q=0$ integral can be done by linearization and scaling. First, transform to the probabilists Hermite polynomials and use linearization to obtain
$$
I_{nm}(0) = 2^{\frac{n+m}{2}}\sum_{k=0}^{\operatorname{min}(n,m)}{n\choose k}{m\choose k}k!\int_{-\infty}^\infty e^{-ay^2}He_{n+m-2k}(\sqrt{2}y)dy.
$$
Now change the integration variable to get the probabilists weighting function $\frac{x}{\sqrt{2}}=\sqrt{a}y$ leading to
$$
I_{nm}(0) = 2^{\frac{n+m-1}{2}}\frac{1}{\sqrt{a}}\sum_{k=0}^{\operatorname{min}(n,m)}{n\choose k}{m\choose k}k!\int_{-\infty}^\infty e^{-\frac{x^2}{2}}He_{n+m-2k}\left(\frac{x}{\sqrt{a}}\right)dy.
$$
Now use the scaling formula
$$
\operatorname{He}_{n+m-2k}(\gamma x) =(n+m-2k)!\sum_{s=0}^{\frac{n+m}{2}-k}\frac{1}{2^ss!(n-m-2k-2s)!}\gamma^{n-m-2k-2s}\left(\gamma^2-1\right)^s \operatorname{He}_{n-m-2k-2s}(x),
$$
the floor function on the upper limit of the sum is not necessary since this integral is zero if $n$ and $m$ have opposite parity. Therefore the sum is always even. Also, since this sum is integrated, only the maximum $s$ term is retained by orthogonality. Therefore the integral becomes
$$
I_{nm}(0)=\sqrt{\frac{\pi}{a}}n!m!\sum_{k=0}^{\operatorname{min}(n,m)}\frac{2^k(n+m-2k)!}{k!(n-k)!(m-k)!\left(\frac{n+m}{2}-k\right)!}\left(\frac{1-a}{a}\right)^{\frac{n+m}{2}-k}
$$
This is actually the $b=0$ limit of the question Integration involving Hermite Polynomials through generating functions. To see this, the parity has to be even $p=0$ and $b^{2a}$ is only $1$ when $a=0$, in which case this integral reduces to this one.
|
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|
$R \setminus (S \cup T)$ . Where is $x$? I am sorry for the messy math symbols.
If I have the set: $R \setminus(S \cup T)$ , is it correct to assume that:
$$R \setminus(S \cup T) = \{x: x∈ \mathbb{R} \text{ and } ( x \notin S \text{ and } x \notin T) \}$$
I am confused because if I had the set $S \cup T$ I would assume that:
$$(S \cup T)= \{x: x \in S \text{ or } x \in T\}$$
Thanks!
|
Your statement is correct.
I think where you're getting confused is when you have to use De Morgan's Law.
This can be converted logically as follows:
$$
x \in R \setminus (S \cup T) \to x \in R \wedge \neg (x \in S \vee x \in T)\\
x \in R \wedge (\neg x \in S \wedge \neg x \in T) \\
x \in R \wedge x \notin S \wedge x \notin T
$$
De Morgan's Law is applied in the last step and it converts the or to an and.
|
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|
Integrating this indefinite integral $$\int\dfrac{t-2}{t+2-3\sqrt{2t-4}}dt$$
I'm not sure whether to use substitution or what.
|
Hint
With the change of variable $x=\sqrt{2t-4}$ the integral becomes
$$\int\frac{x^3}{x^2-6x+8}dx$$
then use the partial fraction decomposition (notice that $x^2-6x+8=(x-2)(x-4)$).
|
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|
Stats is not maths? How mainstream is the claim that stats is not maths? And if it's right, how many people don't agree?
Given that it's all numbers, taught by maths departments and you get maths credits for it, I wonder whether the claim is just half-jokingly meant, like saying it's a minor part of maths, or just applied maths.
|
Statistics can be thought of as an application of mathematics towards the rather specific goal of examining numerical data.
In order to understand statistics, you should probably know some mathematics. In a similar manner, in order to understand engineering, you should probably know some mathematics.
But just like in engineering (or any other applied field), there are domain-specific definitions, terms, and conventions that arise extrinsically from mathematics. For example, the $p$-value is a concept that arose to make sense of data. It can be mathematically studied, but it did not necessarily arise as a natural extension of some prior mathematical definition.
|
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|
Necessity of a hypothesis in the fundamental theorem of calculus Baby Rudin's Fundamental Theorem of Calculus (Theorem 6.21), in my professor's words,states:
Let $f: [a,b] \to \mathbb{R}$ be a Riemann integrable function. If $F: [a,b] \to \mathbb{R}$ is an antiderivative of $f$, then $\int_a^b \! f(x) \, \mathrm{d}x = F(b)-F(a)$.
During the proof, one of my peers asked if the hypothesis that $f$ is Riemann integrable was needed since we have right after that the derivative of $F$ is little $f$. That is, does the second hypothesis imply the first? $F$ is differentiable, so it's continuous on $[a,b]$, and furthermore bounded. Does this then imply that $f$ is also continuous and bounded? If it does, that mean we can exclude the first hypothesis, or is necessary?
|
The derivative of a bounded differentiable function isn't necessarily bounded or continuous. A standard example is to let $1 < \alpha < 2$ and define $f(x) = x^\alpha \sin \frac 1x$ if $x \not= 0$, and $f(0) = 0$. In this case $f'(0) = 0$ but $f'$ is unbounded in every neighborhood of $0$.
This doesn't provide a counterexample to the fundamental theorem, though. An example of a differentiable function whose derivative is not Riemann integrable is Volterra's function.
|
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|
Factorial, but with addition Is there a notation for addition form of factorial?
$$5! = 5\times4\times3\times2\times1$$
That's pretty obvious. But I'm wondering what I'd need to use to describe
$$5+4+3+2+1$$
like the factorial $5!$ way.
EDIT: I know about the formula. I want to know if there's a short notation.
|
We should also note that the factorial function has a similar look to it as the sigma summation notation; as
$$\frac{n(n+1)}{2}=1+2+3+...+n=\sum_{k=1}^nk$$
$$n!=1 \cdot 2 \cdot 3 \cdot ... \cdot n=\prod_{k=1}^nk$$
|
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|
Given $n=p_1^{k_1},...,p_r^{k_r}$. Show that $a^{lcm(\phi{(p_1^{k_1})},\phi{(p_2^{k_2})},...,\phi({p_r^{k_r})})}\equiv 1\pmod n$ I am doing revision for my final exam. This is a sample question given by TA.
I actually have some idea, but wondering it is true or not.
I guess $lcm(\phi{(p_1^{k_1})},\phi{(p_2^{k_2})},...,\phi({p_r^{k_r})})$ here is just equal to $\phi{(n)}$, right? Then, it is just another form of Euler's Theorem.
Can anyone answer this for me?
|
It is not the case that $\text{lcm}(\phi(p_1^{k_1}), \ldots, \phi(p_r^{k_r}))$ is necessarily equal to $\phi(n)$: Take $n=12$, for instance. In order to prove your statement, try showing that the congruence holds modulo each of the prime powers $p_i^{k_i}$ using Euler's theorem for those numbers.
|
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|
Understanding a statement: Uniform convergence of functions I'm trying to understand the following statement:
Let $f$ be a real function defined on an interval $I$ and let $\{f_n\}$ a sequence of functions that converges uniformly to $f$ on $I$. For each $k \in \Bbb N$, there is a subsequence $\{f_{n_k}\}$ such that $|f_{n_k}(x)-f(x)|<2^{-k}$ for all $x \in I$.
I can't understand it because they say that for every fixed $k \in \Bbb N$, there is a subsequence $\{f_{n_k}\}$, and here the subsequence's index is $k$, so I think it doesn't make sense, shouldn't they use an index other than $k$? Maybe I'm misinterpreting the statement, but I'm not sure.
Could someone please explain this to me?
Thank you so much in advace.
|
You're right, the way it is written is simply wrong. What they are saying is that we can find a subsequence $\langle f_{n_k}\rangle$ such that $|f_{n_k}-f|<2^{-k}$ over $I$.
Proof Since $f_n\to f$ uniformly on $I$, for each $\varepsilon >0$ there exists $n_\varepsilon$ such that $n\geqslant n_\varepsilon$ implies $$|f_n(t)-f(t)|<\varepsilon$$ over $I$. Now choose $\varepsilon =2^{-k}$. Then you get for each $k=1,2,3,\ldots$ an integer $n_k$ such that...? You have to take care in making $n_1<n_2<n_3<\cdots$, but that is not hard: if $n_1$ has been chosen and you get a $n_2'$, let $n_2=\max\{n_1,n_2' \}+1$. Continue in this fashion.
|
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|
Confusion in proof that primes $p = 4k + 1$ are uniquely the sums of two squares I'm reading a proof in my number theory textbook that all primes of the form $p = 4k+1$ are uniquely the sum of two squares. I'm stuck right at the beginning of the proof, where they say:
To establish the assertion, suppose that
$$
p = a^2 + b^2 = c^2 + d^2
$$
where $a,b,c,d$ are all positive integers. Then
$$
a^2 d^2 - b^2 c^2 = p(d^2 - b^2).
$$
Perhaps I'm just missing something obvious, but I can't figure out how they managed to conclude that $a^2 d^2 - b^2 c^2 = p(d^2 - b^2).$ Please advise.
|
Have you studied Gaussian integers at all in the years since you asked this question, or prior to that? (I'm here because of a duplicate).
It might help. For example, suppose $$29 = (2 - 5i)(2 + 5i) = 2^2 + 5^2 = c^2 + d^2,$$ where $c \neq 2$ and $d \neq 5$. Then $$2^2 d^2 - 5^2 c^2 = 29(d^2 - 5^2).$$ If we set $d = 0$, then we have $$-5^2 c^2 = -725$$ and so $c = \pm\sqrt{29}$, both of which are outside the domain of Gaussian integers. $d = 1$ also leads to a dead end. Now try $d = 2$, giving us $$16 - 5^2 c^2 = -609.$$ Subtracting 16 from both sides, we have $$-5^2 c^2 = -625.$$ Solving for $c$ we get $c = 5$, which does satisfy the stipulation that $c \neq 2$.
So, all we've done is switch $a$ and $b$ around. Similar results can be obtained by multiplying by $i$ or $-i$.
|
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|
Linear transformations and eigenvalues Let $T: \mathbb C^n \rightarrow \mathbb C^n$ be linear. Let $\beta$ and $\gamma$ be any two ordered bases. Prove that the eigenvalues of $[T]_\beta$ and $[T]_\gamma$ are the same.
Can anyone provide tips/hints in the right direction? I'm struggling as I try to understand this intuitively....thank you
|
Hint:
$[T]_\beta$ and $[T]_\gamma$ are similar matrices, i.e. there is a matrix $Q$ (change of coordinate matrix) so that $[T]_\beta = Q^{-1}[T]_\gamma Q$.
|
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|
Let $F$ be a finite field and $\tau$ an element of $F$. Prove that there exists $a,b\in F$ such that $\tau=a^2+b^2$. Let $F$ be a finite field and $\tau$ an element of $F$. Prove that there exists $a,b\in F$ such that $\tau=a^2+b^2$. It suffices to prove for the case $F=\mathbb{Z}_p$. How to prove?
|
Let $|F| = p^n$, where $p$ is a prime; and consider $\varphi : F\to F$ given by $x\mapsto x^2$.
*
*If $p=2$, $\varphi$ is an isomorphism, so we're done.
*If $p > 2$, check that $\varphi(x) = \varphi(y)$ iff $x = \pm y$, and hence (why?)
$$
|Im(\varphi)| \geq \frac{p^n+1}{2} := k
$$
For $z \in F$, consider $S:= \{z - \varphi(x) : x \in F\}$. Since $2k > p^n$, one has that $S\cap Im(\varphi) \neq \emptyset$, so we are done.
|
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|
How to prove that if $\forall x \in (a,b)$ Lebesgue integral $\int_{(a,x)}fd\lambda=0$, then $f(x)=0$ $\lambda$-almost everywhere? Let $f:(a,b)\rightarrow\mathbb{R}$.
The statement to prove is that if $\forall x \in (a,b)$ Lebesgue integral $\int_{(a,x)}fd\lambda=0$, then $f(x)=0$ $\lambda$-almost everywhere.
So if it wouldn't be true then we would have $\forall x \in (a,b):\int_{(a,x)}f_+d\lambda=\int_{(a,x)}f_-d\lambda\neq0$, so every interval contains positive and negative values of the function. It is possible to construct such an $f$ so my guess would be that this function is not Lebesgue measurable and we can't take this integral.
Thanks in advance!
|
Consider arbitrary $x \in (a,b)$. Then for every suitably small $\epsilon > 0$,
Let, $$ I_\epsilon : = \int_{((x-\epsilon),(x + \epsilon))} f d\lambda = \int_{(a,x+\epsilon)} f d\lambda - \int_{(a,x - \epsilon)} f d\lambda = 0 - 0 = 0. $$
Also, $$ \lim_{\epsilon \rightarrow 0} \frac{I_{\epsilon}}{2\epsilon} = \lim_{\epsilon \rightarrow 0} 0 = f(x) \: (\lambda) \mbox{ - a.e. by Lebesgue differentiation theorem }. $$
Hence we are done.
|
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|
Injective but not divisible module
I want to find an injective but not divisible $R$-module.
If $R$ is integral domain, every injective is divisible so it should be $R$ is not an integral domain. Is there any example?
|
Consider the ring $R=F_2[Z]/(Z^2)=M$ where $F_2$ is the field of two elements. This is a self-injective ring, so $M$ is an injective $R$-module.
But now consider $x=1$ and $r=Z$, where I abuse notation for the images of $1$ and $Z$ in this ring. Saying that there exists $y\in R$ such that $yZ=1$ implies that $Z$ is a unit, but it is clearly not since it is a nilpotent element. So this module is not divisible in your sense.
What user says is good information: while this definition of "divisible" seems like the simplest extension, it turns out to be undesirable. T.Y. Lam's definition cited by user is the nicest one I know.
Notice that in the example above, $ann(Z)=(Z)\not\subseteq ann(1)=\{0\}$, so this is no longer a counterexample for Lam's definition. Indeed, all injective modules are divisible by that definition.
For domains this is not an issue since $ann(r)=\{0\}$ for every nonzero $r$, and therefore the condition that $ann(r)\subseteq ann(x)$ is met automatically. That's how this defintion reduces to the domain definition.
|
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|
How to prove $\sum_{n=0}^{\infty} \frac{n^2}{2^n} = 6$? I'd like to find out why
\begin{align}
\sum_{n=0}^{\infty} \frac{n^2}{2^n} = 6
\end{align}
I tried to rewrite it into a geometric series
\begin{align}
\sum_{n=0}^{\infty} \frac{n^2}{2^n} = \sum_{n=0}^{\infty} \Big(\frac{1}{2}\Big)^nn^2
\end{align}
But I don't know what to do with the $n^2$.
|
Let me show you a slightly different approach; this approach is very powerful, and can be used to compute values for a large number of series.
Let's think of this in terms of power series. You noticed that you can write
$$
\sum_{n=0}^{\infty}\frac{n^2}{2^n}=\sum_{n=0}^{\infty}n^2\left(\frac{1}{2}\right)^n;
$$
so, let's consider the power series
$$
f(x)=\sum_{n=0}^{\infty}n^2x^n.
$$
If we can find a simpler expression for the function $f(x)$, and if $\frac{1}{2}$ lies within its interval of convergence, then your series is exactly $f(\frac{1}{2})$.
Now, we note that
$$
f(x)=\underbrace{0}_{n=0}+\sum_{n=1}^{\infty}n^2x^n=x\sum_{n=1}^{\infty}n^2x^{n-1}.
$$
But, notice that $\int n^2x^{n-1}\,dx=nx^n$; so,
$$\tag{1}
\sum_{n=1}^{\infty}n^2 x^{n-1}=\frac{d}{dx}\left[\sum_{n=0}^{\infty}nx^n\right].
$$
Now, we write
$$\tag{2}
\sum_{n=0}^{\infty}nx^n=\underbrace{0}_{n=0}+x\sum_{n=1}^{\infty}nx^{n-1}=x\frac{d}{dx}\left[\sum_{n=0}^{\infty}x^n\right].
$$
But, this last is a geometric series; so, as long as $\lvert x\rvert<1$,
$$
\sum_{n=0}^{\infty}x^n=\frac{1}{1-x}.
$$
Plugging this back in to (2), we find that for $\lvert x\rvert<1$,
$$
\sum_{n=0}^{\infty}nx^n=x\frac{d}{dx}\left[\frac{1}{1-x}\right]=x\cdot\frac{1}{(1-x)^2}=\frac{x}{(1-x)^2}.
$$
But, plugging this back in to (1), we find that
$$
\sum_{n=1}^{\infty}n^2x^{n-1}=\frac{d}{dx}\left[\frac{x}{(1-x)^2}\right]=\frac{x+1}{(1-x)^3}
$$
So, finally,
$$
f(x)=x\cdot\frac{x+1}{(1-x)^3}=\frac{x(x+1)}{(1-x)^3}.
$$
Plugging in $x=\frac{1}{2}$, we find
$$
\sum_{n=0}^{\infty}\frac{n^2}{2^n}=f\left(\frac{1}{2}\right)=6.
$$
|
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Matrix and eigenvectors
$\quad$The matrix $\mathbf A=\frac19\begin{bmatrix}
7 & -2 & 0 \\
-2 & 6 & 3 \\
0 & 2 & 5 \\
\end{bmatrix}$ has eigenvalues $1$, $\frac23$ and $\frac13$n with the corresponding eigenvectors $ \mathbf v_1=\begin{bmatrix}
-2 \\
2 \\
1 \\
\end{bmatrix}$, $\mathbf{v}_2=\begin{bmatrix}
2 \\
1 \\
2 \\
\end{bmatrix}$, $\mathbf v_3=\begin{bmatrix}
1 \\
2 \\
-2 \\
\end{bmatrix}$.
$\text{(i)}$ Find the general solution of the equation $\mathbf x_{k+1}=\mathbf{Ax}_k$ if $\mathbf x_0=\begin{bmatrix}
1 \\
11 \\
-2 \\
\end{bmatrix}$.
For this question, does it mean I should find out what $x_k$ is in terms of $A$?
Can anyone help me please?
|
Here is a HINT: $$x_k = A^k x_0$$
|
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|
Prove $a^\alpha b^{1-\alpha} \le \alpha a + (1 - \alpha)b, \; a,b > 0,\; 0 < \alpha < 1$ I have no idea how to do this. Any help would be appreciated. The chapter I'm on is about differentiation and the mean value theorem.
Prove $a^\alpha b^{1-\alpha} \le \alpha a + (1 - \alpha)b, \; a,b > 0,\; 0 < \alpha < 1$
|
Hint: Notice that if we divide by $b$ the inequality is equivalent to $$(a/b)^\alpha \leq \alpha \cdot (a/b) + (1-\alpha)$$
set $t = a/b > 0$...
|
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|
Proof that $n^3-n$ is a multiple of $3$. I'm struggling with this problem of proof by induction:
For any natural number $n$, prove that $n^3-n$ is a multiple of $3$.
I assumed that $k^3-k=3r$
I want to show that $(k+1)^3-(K+1)=3r$
The final statement is $K^3 +3K^2+2K$
Am I missing something ?
|
Use Fermat's little theorem.
Case1: if n is a multiple of 3 then trivially $n^3 -n$ is a multiple of $3$.
Case 2: If $n$ is not a multiple of $3$ you shall get $n^2 \equiv 1 \pmod 3$. Multiply $n$ and get $n^3 -n \equiv 0 \pmod 3$.
Now get your result.
|
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|
Why is ($\mathbb R$,usual) not homeomorphic to ($\mathbb R$,discrete)?
*
*Why is ($\mathbb R$,usual) not homeomorphic to ($\mathbb R$,discrete)?
($\mathbb R$,discrete) means $d(x,y) =1$ for any $x\neq y$ and $d(x,y) =0$ for all $x=y$, both $x$ and $y$ are in $\mathbb R$.
*Is ($\mathbb R$,discrete) homeomorphic to ($\mathbb R$,usual)? Why?
*$M$ is homeomorphic to $N$ means there exists a mapping $f$, both one to one and continuous while onto not necessary, from $M$ to $N$, is that right?
|
Because everything is clopen in $\mathbb{R}$ with the discrete topology.
$(0,1)$ is not closed in the usual.
2. The answer is no. Homeomorphisms are open maps.
3. You need a continuous bijection whose inverse is also continuous.
|
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|
How many four digit numbers are there? Assume that 0 can't be a first digit. I got 9,000. Is that right?
Follow up question: How many of those four digit numbers have no repeated digits?
|
Equivalent question: You have a bag with 10 marbles in it labeled 0 through 9. You pick out a marble 4 times and place it back in the bag each time. The first time, there is no marble with a 0. The total combinations of picking and replacing 4 marbles with the no-zero-condition on the first pick is equal to the product of how many marbles are in the bag for each pick:
$$9 \times 10 \times 10 \times 10 = 9000.$$
For no repeated digits, it's the same problem, but you don't replace the marble (but on the second pick, you place the 0 marble back in):
$$9 \times 9 \times 8 \times 7 = 4536$$
|
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|
$G\times H\cong G$ with $H$ non-trivial Aluffi II.3.4 asks the reader to find groups $G$ and $H$ such that $G\times H\cong G$ but $H$ is not trivial. I believe I have found a solution, but would like someone to check. Also, is there a more elementary example? I tried mucking around with $S^1\times (\Bbb Z/2\Bbb Z)$, which feels like it should be isomorphic to $S^1$ (the unit circle under complex multiplication), but I kept getting confused.
My solution
Let $H$ be a non-trivial group and let $G=H^\omega$. Then $\phi\colon H\times G\to G$ defined by
$$\phi(h,g)_n=\begin{cases}
h &n=0\\
g_{n-1} &n\ne 0
\end{cases}$$
looks like a group isomorphism to me. It's obviously a bijective function, and \begin{align*}
\phi((h_1,g_1)(h_2,g_2))_n
&=\phi(h_1h_2,g_1g_2)_n \\
&=\begin{cases}
h_1h_2&n=0\\
(g_1g_2)_{n-1}&n\ne 0
\end{cases}\\
&=\phi(h_1,g_1)_n\phi(h_2,g_2)_n \\
&=(\phi(h_1,g_1)\phi(h_2,g_2))_n.
\end{align*}
|
A simple example would be the natural numbers (including zero) under the operation "bitwise exclusive or." This is isomorphic to $$G=\oplus_{i=0}^{\infty} \mathbb Z/\langle 2\rangle$$
|
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|
Lines in a p-adic plane The geometry of lines in $\mathbb{R}^2$ is fundamental to mathematics and likewise for lines in $\mathbb{C}^2$ since $\mathbb{C}^2 \cong \mathbb{R}^4$. But is there a good treatment of lines in $\mathbb{Q}_p^2$, i.e. the $p$-adic plane?
For example, suppose $y= \zeta x$ is a $p$-adic line. How does it interact with the lattice $\mathbb{Z}^2 \subset \mathbb{Q}_p^2$? Certainly the line will pass through a well-determined set of lattice points if and only if $\zeta \in \mathbb{Q} \subset \mathbb{Q}_p$. So what if $\zeta \not\in \mathbb{Q}$? Can we make a statement (similar to irrational slopes in the reals) that the line passes arbitrarily close to lattice points?
Perhaps more importantly, is there any value added by thinking geometrically since most of these questions could be answered algebraically and the non-archimedean space is difficult to imagine?
|
Unfortunately, there are no lattices in $p$-adic vector spaces. The reason is that if $v$ is a nonzero vector, then $p^nv$ approaches $0$ in the $p$-adic topology, so $0$ is an accumulation point of every subgroup.
|
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|
How to put $2x^2 + 4xy + 6y^2 + 6x + 2y = 6$ in canonical form We are given the equation $2x^2 + 4xy + 6y^2 + 6x + 2y = 6$
We did an example of this in class but the equation had less terms.
I took a note in class that says : if there are linear terms, I have to rotate...
This is what I think I have to do.
*
*Put the coefficients of this equation in a matrix $A$, then $B =
det(A - I \lambda)$
*$det(B)$ must be $ = 0$
*Evaluating this determinant will give me a value for lambda that is
the root of the equation.
*I can then go on and find eigen lines or eigen vectors.
My questions :
*
*Am I right about the steps to solve it ? Shouldn't it give me a canonical form at the end ? because it my notes, it doesn't.
*How do I place the coefficients in the matrix A ?
this is my try :
$A = \begin{bmatrix}2 & 2\\2 & 6 \\\end{bmatrix}$
the first 2 is for the $2x^2$ term, the second 2 next are for the $4xy$ term which I divided by 2...( I dont really know why btw :\ ) and the 6 is for the $6y^2$ term. Now how do I place the 6x and 2y terms ?
$A = \begin{bmatrix}2 & 2\\2 & 6\\3 & 1 \\\end{bmatrix}$
I add another line with those 2 divided by 2 ?
edit :
Here's my current work
I rewrote A as :
$ \begin{bmatrix} x & y \\\end{bmatrix} $
$ A = \begin{bmatrix}2 & 2\\2 & 6 \\\end{bmatrix}$
$ \begin{bmatrix} x \\ y \\\end{bmatrix} $ +
$ \begin{bmatrix} 6 & 2 \\\end{bmatrix} $
$ \begin{bmatrix} x \\ y \\\end{bmatrix} $ - 6
$B = \begin{bmatrix}2 - \lambda & 2\\2 & 6 - \lambda \\\end{bmatrix}$
$det(B) = (2 - \lambda)(6 - \lambda) - 4$
$det(B) = 12 - 2 \lambda - 6 \lambda + \lambda^2 - 4 $
And I'm stuck here. How do I factor this ?
$\lambda^2 - 8 \lambda + 8$
edit : so with the quadration formula I found two roots.
$4 + \sqrt{\frac{32}{2}}$ and
$4 - \sqrt{\frac{32}{2}}$
|
Not knowing what exactly "canonical form" is, here is what I get.
Translating to get rid of the linear terms:
$$
2(x+2)^2+4(x+2)(y-1/2)+6(y-1/2)^2=\frac{23}{2}\tag{1}
$$
With $P=\dfrac{\sqrt{2+\sqrt2}}{2}\begin{bmatrix}1&1-\sqrt2\\-1+\sqrt2&1\end{bmatrix}=\begin{bmatrix}\cos(\pi/8)&-\sin(\pi/8)\\\sin(\pi/8)&\hphantom{+}\cos(\pi/8)\end{bmatrix}$ we have
$$
\begin{bmatrix}2&2\\2&6\end{bmatrix}
=P^T
\begin{bmatrix}4-\sqrt8&0\\0&4+\sqrt8\end{bmatrix}
P\tag{2}
$$
Therefore,
$$
\begin{bmatrix}x+2\\y-1/2\end{bmatrix}^TP^T
\begin{bmatrix}4-\sqrt8&0\\0&4+\sqrt8\end{bmatrix}
P\begin{bmatrix}x+2\\y-1/2\end{bmatrix}
=\frac{23}2
$$
Thus, the curve is an ellipse with its center at $(-2,1/2)$ and its major axis tilted $\pi/8$ clockwise from the $x$-axis. The major and minor axes are
$$
\sqrt{\frac{23}8(2\pm\sqrt2)}
$$
To get $(1)$, we translated to get rid of the linear terms. So
$$
2(x+h)^2+4(x+h)(y+k)+6(y+k)^2\\
=(2x^2+4hx+2h^2)+(4xy+4kx+4hy+4hk)+(6y^2+12ky+6k^2)\\
=(2x^2+4xy+6y^2)+(4h+4k)x+(4h+12k)y+(2h^2+4hk+6k^2)
$$
To match the linear terms, we need $4h+4k=6$ and $4h+12k=2$. Thus, $h=2$ and $k=-1/2$:
$$
2(x+2)^2+4(x+2)(y-1/2)+6(y-1/2)^2\\
=2x^2+4xy+6y^2+6x+2y+\frac{11}2=6+\frac{11}2=\frac{23}2
$$
To get $(2)$, we diagonalized $\begin{bmatrix}2&2\\2&6\end{bmatrix}$. The matrix $P$ rotates counterclockwise by $\pi/8$. After translating by $(2,-1/2)$ and rotating $\pi/8$ counterclockwise, we are left with the ellipse
$$
(4-\sqrt8)x^2+(4+\sqrt8)y^2=\frac{23}2
$$
|
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Number of subgroups and elements I have a question that I feel I am going about in a roundabout way, and would like some help on. I am preparing for an exam.
Problem: Let $G$ be a group with $|G|=150.$ Let $H$ be a non-normal subgroup in $G$ with $|H|=25$.
(a) How many elements of order 5 does $G$ have?
(b) How many elements of order 25 does $G$ have?
My attempt:
$G$ has 6 subgroups of order 25 because the number of subgroups of order 25 has to divide $150/25=6$ and $6 \cong 1 (\mod 5)$. So there are 6 such subgroups: $\{H_1,H_2,H_3,H_4,H_5,H_6\}$. Let $G$ act on this set by conjugation. The permutation representation of the group action of conjugation on this set is $\phi:G \rightarrow S_6$, and $|S_6|=720$. Also, $|\ker(\phi)||im(\phi)|=150$. Because $|im(\phi)|$ has at most one factor of $5$, $5$ divides $|\ker(\phi)|$. Suppose $25$ divides $|\ker(\phi)|$. Then $\ker(\phi)$ has a Sylow 5-subgroup $H_i$. So $H_i \subset \ker(\phi) \implies aH_ia^{-1} \subset a(\ker(\phi))a^{-1} = \ker(\phi)$. By the Second Sylow Theorem, $H_1 \cup \ldots \cup H_6 \subset \ker(\phi) \implies G = \ker(\phi) \implies \phi$ trivial. So $\ker(\phi)$ has a subgroup $K$ with $|K|=5$. So for some $H_i, \ker(\phi) \cap H_i=K \implies a(\ker(\phi))a^{-1} \cap aH_ia^{-1} = aKa^{-1}$
What next? Surely this shouldn't be so long-winded.
|
Your first part is correct $G$ has 6 Sylow 5-subgroups. $H$ is abelian and either cyclic or isomorphic to $C_5 \times C_5$. So ...
|
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|
Prove-If n is prime $\mathbb{Z}_n$ is a field. I need to prove that $\mathbb{Z}_n$ is a field if and only if $n$ is prime. And I proved the forward.
But I am not sure how to prove the backward, 'if n is prime $\mathbb{Z}_n$ is a field.
'
What can I assume from the assumption $n$ is prime?
and What do I need to show?
|
Other than showing trivial axioms, you need to show that every element in $Z/pZ$ has a multiplicative inverse. In other words, given $k$, you need to find $m$ such that $k \cdot m \equiv 1 \pmod p$
This, though, is equivalent to showing that there exists $n$ such that $km = pn + 1$. Because $p$ is a prime, we know that $\gcd(k, p) = 1$. By Euclid's Algorithm/Linear Diophantine Equations, there must exist $(m_0 , n_0)$ such that $km_0 +pn_0 = 1$. The result follows.
I can clarify any of this if you'd like - feel free to comment.
Cheers!
|
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|
Indented Path Integration The goal is to show that
$$\int_0^\infty \frac{x^{1/3}\log(x)}{x^2 + 1}dx = \frac{\pi^2}{6}$$
and that
$$\int_0^\infty \frac{x^{1/3}}{x^2 + 1}dx = \frac{\pi}{\sqrt{3}}.$$
So, we start with the function
$$f(z) = \frac{z^{1/3}\log(z)}{z^2 + 1}.$$
Let $c_r$ be the upper semicircle with radius r such that $r< 1$ and let $c_R$ be the upper semicircle with radius R such that $R>1$. Let $L_1$ be the interval $(r,R)$ on the reals and let $L_2$ be the interval $(-R,r)$ on the reals.
Then
$$\int_{c_r} f + \int_{c_R} f + \int_{L_1} f + \int_{L_2} f$$
taken counterclockwise is equal to
$$2\pi i\text{ res}(f,i) = -\frac{\pi^2 e^{\frac{\pi i}{6}}}{2}$$
The integrals along $c_r$ and $c_R$ should go to $0$ as $r$ goes to $0$ and $R$ goes to infinity.
So:
$$\int_{L_1} + \int_{L_2}= \int_0^{\infty}\frac{x^{1/3}\log(x)}{x^2 + 1}dx - \int_{0}^{\infty} \frac{x^{1/3}e^{i\pi/3}(ln(x) + i\pi) }{x^2 + 1}dx$$
where x is the polar radius.
Let $A = \int_0^\infty \frac{x^{1/3}\log(x)}{x^2 + 1}dx$
and let $B = \int_0^\infty \frac{x^{1/3}}{x^2 + 1}dx $.
Then we have $A*(1-e^{\frac{i\pi}{3}}) - \pi i e^{\frac{i\pi}{3}}B = $-$\frac{\pi^2 e^{\frac{\pi i}{6}}}{2}$.
This doesn't mesh with the given values for $A$ and $B$, so I'm wondering where I went wrong. Thanks in advance.
|
First, make the substitution $t=\displaystyle\frac1{1+x^2}$ , and rewrite $I_1=\displaystyle\frac14\int_0^1t^{^{-\frac16-\frac12}}\cdot(1-t)^{^{\frac16-\frac12}}\ln{1-t\over t}dt$ and $I_2=\displaystyle\frac12\int_0^1t^{^{-\frac16-\frac12}}\cdot(1-t)^{^{\frac16-\frac12}}dt$ . Then recognize the expression of Euler's Beta function in each, and use the fact that $\displaystyle\ln\frac ab=\ln a-\ln b$ , in order to break up the first integral into two parts, and lastly employ the Taylor series expansion for the natural logarithm, $\displaystyle\ln(1-t)=-\sum_{n=1}^\infty\frac{t^n}n$ to deliver the final blow. Also, for the second integral, you will have to make use of Euler's reflection formula for the Gamma function, $\displaystyle\Gamma(z)\Gamma(1-z) = \frac{\pi}{\sin{(\pi z)}}$ .
|
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|
Prove that at least one of the real numbers $a_1 , a_2 , … , a_n$ is greater than or equal to the average of these numbers Prove that at least one of the real numbers $\,a_1 , a_2 , … , a_n$ is
greater than or equal to the average of these numbers. What kind of
proof did you use?
I think I should use contradiction but I don't know how should I use that.
Thank you so much.
|
Let average $g$ and $a_i<g$ for $1\le i\le n$
$$\implies g\cdot n=\sum_{1\le i\le n}a_i<\sum_{1\le i\le n}g=g\cdot n$$
|
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|
Find the value of $\cos(2\pi /5)$ using radicals This is homework so if there is another example that can illustrate the technique I would happily accept that as guidance. The only thing I have been able to find is a question asking about $\cos(2\pi/7)$, which I think is a much harder problem.
I dont have the faintest idea how to solve this and the textbook (Hungerford) doesn't have any examples at all. Ive tried looking for resources online but havent found any that I was able to understand.
So can anyone show me how to solve these types of problems? Thanks a bunch.
|
Note that
$2\cdot \dfrac{2\pi}{5} + 3\cdot \dfrac{2\pi}{5} = 2\pi,$
therefore
$\cos(2⋅\dfrac{2\pi}{5})=\cos(3⋅\dfrac{2\pi}{5})$.
Put $\cos(\dfrac{2\pi}{5})=x$. Using the formulas
$\cos2x=2\cos2x−1,\cos 3x=4\cos 3x−3\cos x$,
we have
$4x^3−2x^2−3x+1=0⇔(x−1)(4x^2+2x−1)=0$.
Because $\cos(\dfrac{2\pi}{5})≠1$, we get
$4x^2+2x−1=0.$
Another way, $\cos(\dfrac{2\pi}{5})>0$, then $\cos \dfrac{2\pi}{5} = \dfrac{-1 + \sqrt{5}}{2}$.
|
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|
Question about finite ring with more than one element; division ring
A finite ring with more than one element and no zero divisors is a division ring. (Hungerford, Algebra, Exercise 6, Chapter 3, Section 1.)
This a problem taken from Hungerford's graduate algebra text. Hungerford defines left and right zero divisors and then says that an element is a zero divisor if it is both a left and right zero divisor. I took a peek at a solution to the exercise and the author of the solution says that for $r \in R-\{0\}=R^*$, $rR^*=R^*=R^*r$. I might be missing something.
Say I want to show $rR^*=R^*$. Pick $ry \in rR^*$. If $ry=0$ then this implies that $r$ is a left zero divisor and $y$ is a right zero divisor.
$R$ has no zero divisors, but I am assuming this doesn't exclude the possibility of elements being a right zero divisor or a left zero divisor. I am thinking correctly here?
Just answering this step is fine, I think I can try to continue working on the problem from this point. Thanks.
|
Lemma: A ring with no nonzero two-sided zero divisors has no nonzero one-sided zero divisors.
Suppose $ab=0$ for nonzero $a,b$. Then $ba\neq 0$. But also $(ba)^2=0$, contradicting the hypothesis. Therefore such $a,b$ do not exist.
|
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Show that if $a$ and $b$ are positive integers then $(a, b) = (a + b, [a, b])$. Show that if $a$ and $b$ are positive integers then $(a, b)=(a + b, [a, b])$.
I was thinking that since $[a, b]=LCM(a, b)=\frac{ab}{(a, b)}$ that if $d= (a + b, [a, b])$, then $d|[a,b]$ and thus $d|(a, b)$ since $(a, b)|[a, b]$
Then I would just have to prove that $(a+b, (a, b)) = (a, b)$.
Is that the right way to go?
|
Another way : Let $(a,b)=d$ and $\displaystyle \frac aA=\frac bB=d\implies (A,B)=1$
So, $\displaystyle(a+b, [a,b])=(d(A+B), dAB)=d(A+B,AB)$
Now, if $D$ divides both $ A+B, AB; D$ will divide $A(A+B)-AB=A^2$ and $D$ will divide $B(A+B)-AB=B^2$
$\displaystyle\implies D$ will divide $(A^2,B^2)=(A,B)^2=1\implies D=1$
|
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Find an estimation (using polar coordinates) Consider the function
$$
f(x,y):=\lVert x\rVert^{1-n}\ln(\lVert x\rVert)(\arctan(\lVert x-y\rVert))^{-\alpha},~~0<\alpha<n,~~n>1,~~(x,y)\in\Omega\times\Omega,~~~\Omega\subset\mathbb{R}^n
$$
with $x\neq y$.
I am searching for an estimation
$$
\lvert f(x,y)\rvert\leq\frac{\lvert a(x,y)\rvert}{\lVert x-y\rVert^{\alpha}}
$$
with $a\in L^{\infty}(\Omega\times\Omega)$. The hint is, to use polar coordinates.
Here (Find a weakly singular kernel function for an estimation of a kernel) the same task was asked for the function $g(x,y):= (\arctan(\lVert x-y\rVert))^{-\alpha}$ and I found, using the main value theorem, that
$$
\lvert g(x,y)\rvert\leq\frac{(\lVert x-y\rVert^2+1)^{\alpha}}{\lVert x-y\rVert^{\alpha}}.
$$
So I think to estimate now f, I have to use this result, getting for now
$$
\lvert f(x,y)\rvert\leq\lvert\lVert x\rVert^{1-n}\ln(\lVert x\rVert)\rvert\frac{(\lVert x-y\rVert^2+1)^{\alpha}}{\lVert x-y\rVert^{\alpha}}
$$
But I do not know how I can continue now, especially using polar coordinates.
Hope you can help me.
Edit: Correction of the task!
There was a mistake in the function f!
It has to be
$$
f(x,y):=\lVert x\rVert\ln(\lVert x\rVert)(\arctan(\lVert x-y\rVert))^{-\alpha}.
$$
|
Easy Peasy. The new function ||x|| ln ||x|| is continuous (although negative near he orign), so it is bounded on $\Omega$ if $\Omega$ is bounded, so you can just use your old estimate times the bound on xlnx.
|
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What are the techniques to find the sum to infinity of a series? Given a series, what are the techniques to find a formula that sums the series to infinity?
I only know the method of multiplying the series with a factor and then taking their difference (like here). But today, I found out that we can also try to find such a formula by differentiation (see answer to this question).
Therefore, I was wondering what other techniques exist and if there is a book/online resource that I can read to understand the logic behind them?
|
There are many techniques, but unfortunately there is no general method which is guaranteed to succeed, hence the difficulty (perhaps impossibility) of finding things like a closed formula for $\zeta(3)$.
The techniques that I can recall using in the past include:
*
*Using known power series expansions (including things like geometric series)
*Differentiating or integrating power series
*Using complex analysis (look for summation of series by using residues) as in this question
*Fourier expansions, including Parseval's theorem - as in this question
However, there is no method which is always going to work, so it means lots of practice to gain experience!
I am sure others will be able to suggest more modern books, but one I have is:
Konrad Knopp - "Theory and Application of Infinite Series" (Dover edition) which has a whole chapter on closed and numerical expressions for sums of series.
|
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|
Real analysis question about limits? If $0 < x < 1$, show that $x^n → 0$ as $n → ∞$ .
I'm thinking it has something to do with rational numbers (declaring $x = p/q$) and then using exponent laws to show that $x^n = p^n/q^n$, and since $q > p$ ... but I'm not sure where to go with this. Any help is appreciated. Thanks
|
Hints:
$$0<x<1\implies x>x^2>x^3>\ldots >x^n\;,\;\;\forall\,n\in\Bbb N$$
$$a_n>0\;\;\forall\,n\in\Bbb N\implies \lim_{n\to\infty}a_n=\infty\iff \lim_{n\to\infty}\frac1{a_n}=0$$
|
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Evaluation of a line integral using Green's Theorem where P, Q, and partial derivatives of P & Q are not continuous
How can the author evaluate the below?
$$\oint_{C'}\frac{-y}{x^2+y^2}dx+\frac{x}{x^2+y^2}dy$$
Doesn't this contradict Theorem 9.12.1? P(0, 0) is undefined on region $R_2$ (corresponding to $C'$).
|
The author is correct that he is able to do this, but he does not justify himself very well. It is actually true that if two paths are homotopic (see wikipedia) through a region where the vector field is curl free, then the integrals over both paths will be equal.
I will draw some pictures to show you this is true only using version of greens theorem which you know. (sorry the picture is a bit dinky)
Green's theorem does apply to each of the regions $R_1,R_2,R_3,R_4$. Since $Pdx+Qdy$ is curl free in each region, we know that the integral
$$\int_{bR_i} Pdx + Qdy = 0$$ holds for the oriented boundary of each $R_i$. So
$$\int_{bR_1} Pdx + Qdy + \int_{bR_2} Pdx + Qdy + \int_{bR_3} Pdx + Qdy + \int_{bR_4} Pdx + Qdy = 0$$.
Now notice that there is some cancelation that occurs on the region of integrations! For example, $R_1$ and $R_2$ both share the upper vertical line segment, but with different orientations. In fact, looking at the second picture, you can see that all 4 line segments are present twice in the above sum of integrals, each time with opposite orientations. So those integrals cancel out. The remaining pieces are just those belonging to $C$ and $C'$, but with the opposite orientation!
$$\int_C Pdx +Qdy - \int_{C'} Pdx +Qdy = 0$$
$$\int_C Pdx +Qdy = \int_{C'} Pdx +Qdy$$
which is what your textbook claimed.
This is a powerful idea. If you have some region of integration, you can often cut it up like this, and because the boundaries are oriented, you have cancellation on the interior. In some sense this is the real reason for Green's theorem to begin with (It is true for rectangles by using the fundamental theorem twice, then decompose a region into rectangles and do this kind of thing to see the general result)
|
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Does the orientation you evaluate line integrals matter?
If instead of evaluating the above line integral in counter-clockwise direction, I evaluate it via the clockwise direction, would that change the answer? What if I evaluate $C_1$ and $C_3$ in the counter-clockwise direction, but I evaluate $C_2$ in the clockwise direction?
If the direction does matter, in which direction would I evaluate the below line integral?
|
Direction does not matter for the line integral of a function, but here you are dealing with a work integral (i.e. the integral of a vector field along the curve). In the latter case, orientation does matter.
The statement of Green's Theorem includes (or it should, to make sense) the orientation required for the equality to hold. The orientation for the curve is the one that leaves to region $R$ to your left as you traverse the curve.
|
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Evaluating $I(n) = \int^{\infty}_{0} \frac{\ln(x)}{x^n(1+x)}\, dx$ for real $n$ I am not sure how to handle the additional parameter $n$. I first need to find out for which real values of $n$ will the integral converge. Based on intuition and checking with mathematica, I believe it will converge only for $0 < n < 1$, although I am not sure and I would like to see exactly how to determine those values. For this integral, I would normally try using a branch cut, but again I am not sure how to apply it given my additional parameter $n$.
$$I(n) = \int^{\infty}_{0} \dfrac{\ln(x)}{x^n(1+x)}\, dx$$
|
Firstly, to determine the range of $n$ for which the integral converges, just look at the singularity at $0.$ To integrate, and put the pesky $n$ where it is easier to deal with, make the substitution $u=\log x.$ This will transform your integral to $\int_{-\infty}^{\infty} u \exp(-(n-1) u)/(1+\exp u) du,$ which should be easy by standard contour integration methods. To check your computation, the answer is $\pi^2 \cot(n \pi) \csc(n \pi),$ provided $0<n<1.$
|
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How to define a finite objects with parametric equations I never had seen parametric equations, but while trying to learn line integrals through Wikipedia, quickly found these equations are remarkable. Some can represent things for which more normal equations or functions are needed, if at all possible.
However, the page about them in Wikipedia is not very long, and I didn't learn as much about them as I would like. Being so, I didn't find described any limitation for them. They can make many more finite shapes than normal equations.
But can define stuff with clear ends, such as a curve with two separate end-points, or just a line segment? I have no idea how to find these hypothetical parametric equations, as I said, I didn't find limitations, nor examples to this. Is this possible, or does it require "parametric inequations"?
|
You do this by defining an interval for the parameter.
Take a simple line: $x = y = z = t.$ The (infinite) line implies that the parameter $t$ can take on any value.
The line segment from $(-1, -1, -1)$ to $(2, 2, 2)$ means restricting $t$ to the interval $[-1, 2]$.
|
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Why do we have to do the same things to both sides of an equation? Forgive me in advanced if this is a trivial question.
This convention makes perfect sense to me intuitively, but is there any rigorous underpinning to it?
I'm beginning to read through an abstract algebra textbook, and soon after establishing what a ring is, it all of the sudden added to both sides of the equation to prove a(0)=(0)a=0. It seemed a little premature to me.
|
Here's a short example of doing the same thing to both sides of an equation, and a justification for each step, using only substitution and the field axioms. The magic happens in the first two steps.
Statement:
\begin{align*}
a+c=b+c \Rightarrow a=b
\end{align*}
Proof:
\begin{align*}
(a+c)+(-c)&=(a+c)+(-c) \tag{property of =} \\
(a+c)+(-c)&=(b+c)+(-c) \tag{substitution}\\
a+(c+(-c))&=b+(c+(-c)) \tag{associativity}\\
a+0&=b+0 \tag{inverses}\\
a&=b \tag{identity}
\end{align*}
$\square$
|
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How to show that $2730\mid n^{13}-n\;\;\forall n\in\mathbb{N}$
Show that $2730\mid n^{13}-n,\;\;\forall n\in\mathbb{N}$
I tried, $2730=13\cdot5\cdot7\cdot3\cdot2$
We have $13\mid n^{13}-n$, by Fermat's Little Theorem.
We have $2\mid n^{13}-n$, by if $n$ even then $n^{13}-n$ too is even; if $n$ is odd $n^{13}-n$ is odd.
And the numbers $5$ and $7$, how to proceed?
|
HINT:
$$n^{13} \equiv n^5 \cdot n^5 \cdot n^3 \equiv n \cdot n \cdot n^3 \equiv n^5 \equiv n \pmod 5$$
$$n^{13} \equiv n^6 \cdot n^7 \equiv n \pmod 7$$
Also you've missed $3$ as prime factor. But that should be easy.
|
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|
Determinant without computing How can I show without computing the determinant that the equation is true?
$$\det \begin{pmatrix}
b1 + c1 & c1 + a1 & a1 + b1\\
b2 + c2 & c2 + a2 & a2 + b2\\
b3 + c3 & c3 + a3 & a3 + b3
\end{pmatrix} = 2 \det \begin{pmatrix}
a1 & b1 & c1\\
a2 & b2 & c2\\
a3 & b3 & c3
\end{pmatrix}$$
|
Hints:
(1) Determinant is a multilinear function. For example
$$\det\begin{pmatrix}a+b&c&d\\a'+b'&c'&d'\\a''+b''&c''&d''\end{pmatrix}=\det\begin{pmatrix}a&c&d\\a'&c'&d'\\a''&c''&d''\end{pmatrix}+\det\begin{pmatrix}b&c&d\\b'&c'&d'\\b''&c''&d''\end{pmatrix}$$
and likewise for each column/row.
(2) The determinant is an alternating function, i.e.: if a matrix has two equal columns/rows then its determinant is zero and if we interchange two rows/columns the determinant is multiplied by $\;-1\;$ .
|
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|
what are the eigenvalues in orthgonal matrix, How to explain? what are the possible eigenvalues of an orthogonal matrix?
I got the answer key which says its 1 and -1 but it doesn't explain well
|
Here is an approach. Recalling the fact
if $\lambda$ is the eigenvalue of $A$, then $\lambda^{-1}$ is the eigenvalue of $A^{-1}$,
we have
$$ \langle Ax,x \rangle = \langle x, A^T x \rangle = \langle x, A^{-1} x \rangle $$
$$ \implies \langle \lambda x,x \rangle = \langle x, \frac{1}{\lambda} x \rangle $$
$$ \implies \lambda \langle x,x \rangle =\frac{1}{\bar \lambda} \langle x, x \rangle $$
$$ |\lambda|^2 = 1 .$$
|
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|
Easy GRE question: Statistics I'm not sure how to set this statistics problem when they give me a group of arbitrary values. Can someone help?
A group of 20 values has a mean of 85 and a median of 80. A different group of 30 values has a mean of 75 and a median of 72.
a) What is the mean of the 50 values?
b) What is the median of the 50 values?
Thank you
|
The formula to find mean is
Mean=
(total number of values) ÷ (quantity of values)
So, as we know the mean and the quantity values of the first group is, 85 and 20. (It is given)
So we can substitute the values into the formula and get the (total number of values)
Same goes to the other group that contains 30 values...
The trick of the question is... They gave you the values, mean and median, of (group of 20 values) and (group of 30 values). And asked the mean and median of the combined group. (Group of 20 values and group of 30 values)
So workouts are...
85=(sum of x values)÷20
20(85)= (sum of x values)
X= 1700
75=(sum of x values)÷30
30(75)=(sum of x values)
X=2250
So... 30+20=50
So... Mean of the group of 50values is..
Mean= (1700+2250)÷50
= 3950÷50
= 79
And.. For question b)..
The median cannot be identified with the given information in the question.
|
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|
Visualizing Non-Zero Closed-Loop Line Integrals Via Sheets? How do I visualize $\dfrac{xdy-ydx}{x^2+y^2}$?
In other words, if I visualize a differential forms in terms of sheets:
and am aware of the subtleties of this geometric interpretation as regards integrability (i.e. contact structures and the like):
then since we can interpret a line integral as counting the number of sheets you cross through:
we see we can interpret the notion of a closed loop line integral as not being zero in terms of this contact structure idea, i.e. as you do the closed line integral you do something like enter a new field of sheets, what exactly are you doing & how does this explain the non-zero line integral around a closed loop at the origin for the differential form I've given above? Thanks!
|
First, it is important to note that those are oriented sheets: they have two sides, one painted red and another blue. When a curve crosses a sheet, the red-blue crossing counts as $1$ while the blue-red crossing counts as $-1$.
Picture a stack of vertical planes passing through the $z$-axis, like a book that opens 360 degrees wide. Here is an example. Odd-numbered pages are red, even-numbered are blue. A loop around the spine of the book crosses all pages in the same direction (either red-blue or blue-red). There is no cancellation, hence the integral is nonzero.
|
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|
Finding no-self-intersecting path in geometric graphs Is there a polynomial algorithm to determine whether there exists no-self-intersecting path between given vertices $s$ and $t$ in a geometric graph $G$?
Geometric graph is an image of a graph on a plane where vertices are represented as points and edges are drawn as straight line segments (possibly intersecting with each other).
A path is called no-self-intersecting if every two edges from the path do not intersect.
For example, there exists no-self-intersecting path between highlighted vertices on the following picture,
but there is no such path on the second picture.
Here is a stronger version of the problem: given a conventional graph $G$, a list of "crossings", where a "crossing" is a pair of edges, and vertices $s$ and $t$, determine whether there is a path between $s$ and $t$ in which no edges form a "crossing". I have a proof it's NP-complete, but I don't know a way to convert an arbitrary "graph with a list of crossings" to corresponding geometric graph, in which edges intersect iff they formed a "crossing" in original graph.
|
The second problem is a variant of problem called Path avoiding forbidden pairs (PAFP) (see Computers and intractability by Garey and Johnson, 1979, page 203).
I could not find any mention of the main problem, though, so I prooved it is NP-complete. I uploaded the paper to my Github (sorry, only in Russian).
|
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Finding the limit of $\left(\frac{n}{n+1}\right)^n$
Find the limit of: $$\lim_{n\to\infty}\left(\frac{n}{n+1}\right)^n$$
I'm pretty sure it goes to zero since $(n+1)^n > n^n$ but when I input large numbers it goes to $0.36$.
Also, when factoring: $$n^{1/n}\left(\frac{1}{1+\frac1n}\right)^n$$ it looks like it goes to $1$.
So how can I find this limit?
|
$$ \left(\dfrac{n}{n+1}\right)^n = \frac {1}{\left(\dfrac{n+1}{n}\right)^n}=\frac {1}{\left(1+\dfrac{1}{n}\right)^n}$$
|
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|
If $A$ has two eigenvalues $\lambda _1, \lambda_2$ and $\dim (E_{\lambda_1})=n-1$, then $A$ is diagonalizable Suppose that $A \in M_{n\times n}(\Bbb F)$ has two distinct eigenvalues $\lambda_{1}$ and $\lambda_{2}$ and that $\dim (E_{\lambda_1})=n-1$ show that $A$ is diagnolizable.
|
Hint: Since $\dim(E_{\lambda _1})=n-1$, there exist $v_1, \ldots , v_{n-1}$ linearly independent eigenvectors of $\lambda _1$. Let $v_n$ be an eigenvector of $\lambda _2$. Now consider the $n\times n$ matrix $P$ whose $i^{\text{th}}$column is $v_i$. The invertibility of $P$ follows from this.
Can you take it from here?
|
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|
Proving a relation with induction I have a problem:
Let $p_n$ be the $n:th$ prime number ($p_1=2, p_2=3, p_3=5$ and so on). With induction, show that $p_{n+2}>3n$ for each integer $n\geq1$.
I can't figure this out because the primes are confusing me, making me unable to show the inductive step.
|
Proof by induction.
Let's start with the case $n=1$:
$p_{1+2}=p_3=5>3*1=3$
So the equality is true for $n=1$.
Now, we assume that the equality is true for the case $n-1$, that is,
$p_{n-1+2}=p_{n+1}>3(n-1)=3n-3$
and let's see that it is also true for n:
$p_{n+2}\geq p_{n+1}+2>3(n-1)+2=3n-1$; so $p_{n+2}\geq3n$
Now,you have to see that $p_{n+2}>3n$
|
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|
Finding new probability density function with change of variable Y=sqrt(X) Say we have a given distribution, such as X~No(a, b). I am trying to find the pdf and mean for $Y=\sqrt{X}$.
I know the steps for finding the PDF, but since Y can only take on positive values, then the new PDF is only valid for Y>0.
Then how does this affect the mean, and how would I go about finding the mean?
|
If $Y=\sqrt{|X|}$, then, for every $y\gt0$,
$$
f_Y(y)=2y\cdot(f_X(y^2)+f(-y^2)).$$
|
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|
Area of a square in polar coordinates? I was attempting, for the exercise of it, to find the area of the a simple square with an infinite number of infinitesimal circle sectors. Let us say this square is $[5 x 5]$.
Alas, it's been proving as awkward to compute this as it sounds. First, I did the integral in rectangular coordinates, (much easier) $x=5$, so the area becomes $$\int_0^5 5 \space \mathrm{d}x = 25$$.
I then attempted to describe a rectangular area in polar coordinates. I did this using $r\cos(\theta)=5$, which simplifies in terms of $\theta$ to $r=\frac{5}{\sin{\theta}}$. Because this is a square, the angle the hypotenuse will make from the bottom left corner to the top right corner is going to be, surely, $45^{\circ}$. This produces the integral:
$$\frac{25}{2}\int_0^\frac{\pi}{4} \frac{1}{\sin^2 (\theta)} \mathrm{d}\theta$$.
This integral does not converge according to wolfram alpha. Would anyone be so kind as to tell me what I've done wrong?
|
The area of a circular sector of radius $r$ and angle $d\theta$ is $\pi r^2 \frac{d\theta}{2\pi} = \frac{1}{2} r^2 d\theta$.
The right side of the square ($0 < \theta < \pi/4$) is the line $x = 5$, which in polar coordinates is $r = \frac{5}{\cos \theta}$ (not $\sin$).
Putting this together, the integrand should be $\frac{1}{2}(\frac{5}{\cos \theta})^2 d\theta = \frac{25}{2 \cos^2 \theta} d\theta$.
This works out to $12.5$. Since it only covers half of the square, double it to get the $25$ you expect.
|
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|
Compute the minimum distance between the centre to the curve $xy=4$. I wish to solve the following problem:
Compute the minimum distance between the center to the curve $xy=4$.
But I don't know where to start from?
|
You need to minimize the function
$$d= \sqrt{x^2 +y^2} = \sqrt{x^2+ 16/x^2}.$$
Which we got by considering the distance from a point on the curve to the origin. Now, use the derivative test. See related problem.
|
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|
Least prime of the form $38^n+31$ I search the least n such that
$$38^n+31$$
is prime.
I checked the $n$ upto $3000$ and found none, so the least prime of that form must have more than $4000$ digits. I am content with a probable prime, it need not be a proven prime.
|
This is not a proof, but does not conveniently fit into a comment.
I'll take into account that $n=4k$ is required, otherwise $38^n+31$ will be divisible by $3$ or $5$ as pointed in the comments.
Now, if we treat the primes as "pseudorandom" in the sense that any large number $n$ has a likelihood $1/\ln(n)$ of being prime (which is the prime number density for large $n$), the expected number of primes for $n=4,8,\ldots,4N$ will increase with $N$ as
$$
\sum_{k=1}^N\frac{1}{\ln(38^{4k}+31)}
\approx\frac{\ln N+\gamma}{4\ln 38}
\text{ where }\gamma=0.57721566\ldots
$$
and for the expected number of primes to exceed 1, you'll need $N$ in the order of 1,200,000.
Of course, you could get lucky and find it at much lower $n$, but a priori I don't see any particular reason why it should be...or shouldn't.
Basically, in general for numbers $a^n+b$, the first prime will usually come fairly early, otherwise often very late (or not at all if $a$ and $b$ have a common factor).
Of course, this argument depends on assuming "pseudorandom" behaviour of the primes, and so cannot be turned into a formal proof. However, it might perhaps be possible to say something about the distribution of $n$ values giving the first prime over different pairs $(a,b)$.
|
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Inner product doubt I have a doubt about a problem involving inner product spaces. The exercise is:
Given the subspace generated by the vectors $ (1,1,1) $ and $ (1,-1,0) $, find the orthogonal subspace and give a basis.
Now, what I understood of this problem is that I have to find the orthogonal complement. But they don´t give me any inner product to work with. Does the orthogonal complement vary with the product? Which one should I use?
I haven´t seen much of inner product, just the basic, but I should be able to solve this. Thank you! If you do not understand something, please ask me.
|
The orthogonal component depends on the inner product. However, typically the inner product that is chosen (assuming you are dealing with a vector space over real numbers) is the following one:
$$\langle(x_1,x_2,x_3), (y_1,y_2,y_3) \rangle = x_1y_1 + x_2y_2 + x_3y_3$$
|
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|
Finding a 2x2 Matrix raised to the power of 1000 Let $A= \pmatrix{1&4\\ 3&2}$. Find $A^{1000}$.
Does this problem have to do with eigenvalues or is there another formula that is specific to 2x2 matrices?
|
Perform an eigenvalue decomposition of $A$, we then get
$$A =
\begin{bmatrix}
-4/5 & -1/\sqrt2\\
3/5 & -1/\sqrt2
\end{bmatrix}
\begin{bmatrix}
-2 & 0\\
0 & 5
\end{bmatrix}
\begin{bmatrix}
-4/5 & -1/\sqrt2\\
3/5 & -1/\sqrt2
\end{bmatrix}^{-1}
=VDV^{-1}
$$
where $V = \begin{bmatrix}
-4/5 & -1/\sqrt2\\
3/5 & -1/\sqrt2
\end{bmatrix}$ and $D = \begin{bmatrix}
-2 & 0\\
0 & 5
\end{bmatrix}$.
Hence,
$$A^n = \underbrace{\left(VDV^{-1} \right)\left(VDV^{-1} \right)\cdots \left(VDV^{-1} \right)}_{n \text{ times}} = VD^n V^{-1}$$
|
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Let $n=2m$, where $m$ is odd. How many elements of order $2$ does the group $D_n/Z(D_n)$ have?
Let $n=2m$, where $m$ is odd. How many elements of order $2$ does the group $D_n/Z(D_n)$ have?
I don't how to begin this proof. All I have so far is that Dn/Z(Dn) should have one element of order 2.
|
SO I have that the Dn/Z(Dn)={R0,R180}. I also figured out that the order of Dn=(m+1)(n+1)=(mn)+2n+m+1. But I don't know where to go from here.
|
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Exponential distribution function. I am having trouble with this question as well. Any help or hints will be really appreciated.
On a TV game show contestants have to complete a particular task. Suppose that the time take for a typical contestant to complete the task is a random variable with an exponential distribution sigma=2, measured in minutes. This means that the probability density is given by.
$$f(x)=\left\{
\begin{array}{lr}
\frac{1}{2}e^{-\frac{x}{2}} & x ≥0\\
0 & \text{otherwise}
\end{array}
\right.
$$
a) Calculate the probability that a typical contestant fails to complete the task in 3 minutes.
b) Calculate the probability that a contestant takes over 2 minutes to complete the task given that they do not fail
I keep getting the wrong answers for both of a and b .Any sort of hint for me to understand this better would be really appreciated .
thanks
|
For the first question, try integrating that distribution from 3 to $\infty$. This is the probability that the contestant takes 3 minutes or more to complete the task.
For the second question, I'm going to assume that it is considered a failure if the contestant takes more than 3 minutes to complete a task. In this case, integrate the same distribution from 2 to 3 this time.
Does this make sense?
|
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On Pr(X>Y) when X and Y are independent normal Let X∼N(6,1) and Y∼N(7,1) be two independent normal variables. Find Pr(X>Y). the answer is 0.2389 but I do not know how to do it.I have tried adding them and subtracting but i am still clueless.
|
We have $Pr(X>Y)=Pr(X-Y>0)$. Because $X$ and $-Y$ are independent Gaussians, their sum is normal (mean the sum of the means and variance the sum of the variances). Therefore $X+(-Y)\sim N(6-7,1+1)=N(-1,2)$.
Depending on what calculational tools you have available to you, you may be able to evaluate $P(N(-1,2)>0)$ directly. However, it is instructional to convert this quantity into something you can look up in a table for the standard normal. Therefore, we can compute
$$P(N(-1,2)>0)=P(N(0,2)>1) = P\left(\frac{N(0,2)}{\sqrt{2}}>\frac{\sqrt{2}}{2}\right)=P\left(N(0,1)>\frac{\sqrt{2}}{2}\right). $$
|
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|
system of linear congruences when moduli are not coprime $\begin{cases}x\equiv 1 \pmod{3}\\
x\equiv 2 \pmod{5}\\
x\equiv 3 \pmod{7}\\
x\equiv 4 \pmod{9}\\
x\equiv 5 \pmod{11}\end{cases}$
I am supposed to solve the system using the Chinese remainder theorem but $(3,5,7,9,11)\neq 1$
How can I transform the system so that I will be able to use the theorem?
|
Solving the first two equations simultaneously you get X = 7(mod 15).
Solving the third and fourth simultaneously you get X = 31(mod 63).
Solving these two results simultaneously you get X = 157(mod 315).
Solving this result with the fifth equation simultaneously,
you get the final answer X = 1732(mod 3465).
|
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Primes and perfect square on numbers I have two equations, which may play an important role in my further studies on theory of numbers.
1) How many pairs of (A, B, x) we can make in $A^x + B ^x = prime$? Here $x$ is $> 2$ and A, B are positive integers.
2) can we find a number(s) with one hundred 0′s, one hundred 1′s and one hundred 2′s be a perfect square. If yes, what is that number or otherwise how to disprove it about such number does not existence?
Please...
|
For the second question the answer is no.
The number you want is $222\cdots 111\cdots 000\cdots$ which is divisible by $3$ (because the sum of its digits is $2\cdot 100+1\cdot 100+0\cdot 100=300$ divisible by $3$)
But,it is not divisible by $9$ (because also the sum of its digits is not)
There for it must not be a square.
|
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Equations and inequalities as parameters: proving that an equation holds. I have $x-y=3$ and $y\le1$ and $x\ge\frac12$.
I proved that $\sqrt{(2x-1)^2}+\sqrt{(2y-2)^2}=7$ and that $-\frac52\le y\le 1$ and $\frac12\le x\le4$.
How can I prove that $|x+y-5|+|x+y+2|=7$?
|
ok let us consider following cases :
first side is just $-x-y+5$,second case is $x+y+2$,so sum is $5+2=7$
to be more deeply,let us take such situations
*
*$y=1$
2.$x=4$
we have
$|x+y-5]+|x+y+2|=x+y-5+x+y+2=2*x+2*y-3=2*(x+y)-3
=2*(5)-3=7$
can you continue from this?
just consider this situation when $x<0$ then $|x|=-x$,else if $x>=0$ then $|x|=x$.ok let us consider following case,when
$y=-5/2$ and $x=1$ ,we have that $-5/2+1-5<0$,that why we will have
$-x-y+5$,
for second we have $-5/2+1+2>0$ ,
that why we have second as $x+y+2$,together is is still $7$
|
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Assume $X$ and $Y$ be nonempty subsets of $R$ such that $xWe have that both sets are nonempty, by the completeness axiom, both sup and inf exist for both sets. Since $x < y$, $y$ is an upper bound for $X$ and hence by def. of sup, sup$X \le y$. So $\sup X$ is a lower bound for $Y$ and thus by def. of inf, $\sup X \leq \inf Y$.
|
That is basically a good proof, but could use some clarifications. In particular you should try to distinguish more clearly between numbers and sets.
*
*sup and inf do not necessarily exist for a nonempty set of real numbers. However, it is true that a nonempty set which has an upper bound has a sup, and similarly a nonempty set which has a lower bound has an inf, and that's clearly true here.
*When you say $x < y$, you're considering a single $y$ and all $x \in X$ to conclude that sup $X$ $\le y$.
*Because sup $X$ $\le y$ is true for all $y \in Y$, we can also conclude that sup $X$ is a lower bound for $Y$. It is not a lower bound for $y$: we say that sets have bounds, but numbers do not.
|
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|
Compact hypersurface of $\mathbb{R}^{n+1}$ with positive curvature is diffeomorphic to $S^n$. I have a compact hypersurface $M$ of $\mathbb{R}^{n+1}$ with positive curvature. I need to show that it is diffeomorphic to $S^n$.
The hint is to consider the shape operator $A_{\nu_p} x$, where $\nu$ is a smooth unit normal vector field regarded as a map $\nu: M \rightarrow S^n$, and then show that it is a covering map.
Unfortunately, I don't think the hint really made the approach any clearer. Can anyone help to shed some light on this exercise for me?
|
The shape operator is selfadjoint as an endomorphism of the tangent space. Therefore it can be diagonalized over $\mathbb{R}$. The hypothesis of positive (sectional) curvature then implies that all the eigenvalues are positive. Therefore the determinant is positive and in particular nonzero. It follows that the Gauss map (assigning to each point, the unit normal vector at that point) is a regular map. This is because the shape operator is the tangent map of the Gauss map. But any regular cover of the sphere is the identity because the sphere is simply connected.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Grammar outside the Chomsky Hierarchy This grammar describes a language that may fall outside the Chomsky Hierarchy (CH):
\begin{array}{l}
S \to abAbba \\
A \to abA \mid bbaB \\
B \to aab \\
\lambda \to Aab \mid aB \\
\end{array}
Going down the list, it's not context-free, it's not regular, etc. Getting all the way to RE languages, the production starting with $\lambda$ seems outside any possible language within the CH. Does this grammar not fall within the CH?
|
The last rule of your grammar does not satisfy the requirements of a formal grammar (every rule in formal grammar must have at least one non-terminal symbol on the left-hand side), so it clearly cannot belong to any hierarchy of formal grammars.
That being said, the language produced by your grammar (under the intuitive semantics) falls into the class of recursively enumerable languages. At a glance, it should be sufficient to add a non-terminal symbol $L$, along with rules $L\rightarrow \lambda\ | LL$ and add $L$ before, after and between each pair of characters on the right-hand sides of the rules (e.g. the third rule would become $L\rightarrow LaLaLbL$).
As coffeemath correctly observed, the resulting grammar will be context-free, so the language itself is context-free, instead of being just recursively enumerable.
|
{
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"url": "https://math.stackexchange.com/questions/598427",
"timestamp": "2023-03-29T00:00:00",
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|
Concentric and Tangent Ellipse from 2 Hyperbolas Find the equation of the ellipse that is concentric and tangent to the following hyperbolas:
$$\begin{align}
-2x^2 + 9y^2 - 20x - 108y + 256 &= 0 \\
x^2 - 4y^2 + 10x + 48y - 219 &= 0
\end{align}$$
I did the math for both equations and the center is the same: $(-5,6)$.
I have the equations of each:
For the first:
$$-2x^2 + 9y^2 - 20x - 108y + 256 = 0$$
$$-\frac{(x+5)^2}{9} + \frac{(y-6)^2}{2} = 1$$
For the 2nd:
$$x^2 - 4y^2 + 10x + 48y - 219 = 0$$
$$\frac{(x+5)^2}{100} - \frac{(y-6)^2}{25} = 1$$
I know concentric means the new Ellipse has to have the same center, but I don't know how to make it tangent. Please help. Thanks
|
Not a complete solution, but this approach will work:
An ellipse with centre at the point $(-5,6)$ would be
$$\frac{(x+5)^2}{a^2}+\frac{(y-6)^2}{b^2} = 1$$
Now change to a new set of axes ($u, v$) parallel to the $x,y$ axes, but with origin at the point $(-5,6)$. In other words, put $u = x+5$ and $v = y-6$.
Referred to the new axes, the equations of the three conics become:
$$-\frac{u^2}{9} + \frac{v^2}{2} = 1,$$
$$\frac{u^2}{100} - \frac{v^2}{25} = 1,$$
$$\frac{u^2}{a^2}+\frac{v^2}{b^2} = 1$$
(Note that the values $a$ and $b$ are unchanged by the change in coordinates.)
Now, looking at a hyperbola $u^2/a^2-v^2/b^2 = \pm 1$ and an ellipse $u^2/c^2+v^2/d^2 = 1$, you can see (from a sketch) that the ellipse will be a tangent to the hyperbola only when they meet at a point on the $u$ or $v$ axis.
Can you finish it from there?
|
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|
A curious problem about Lebesgue measure. The Problem:
Let $(B(x_{m},0.5))_{m}$ be a sequence of disjoint open discs in $\mathbb{R}^{2}$ centered in $x_{m}$ and with radius 0.5. Let $\psi(n)$ be the number of these discs contained in the open disc $B(0,n)$ (that is, the disc centered in (0,0) and with radius $n$).
Prove that if $\lim \inf \frac{\psi(n)}{n^{2}} = k > 0$, then there exists a ray starting from (0,0) that crosses an infinite number of the discs $(B(x_{m},0.5))_{m}$.
My Thoughts:
I find this problem particularly curious. There are several hints below the problem:
*
*Use that if $A \subset \mathbb{R}^{2}$ is Lebesgue-measurable and $k \geq 0$ then $kA= \{ kx:x \in A \} $ is Lebesgue-measurable too and $\lambda(kA)=k^{2}\lambda(A)$.
*Use that $\mu( \cup _{n} A_{n}) < +\infty$ implies $\mu( \lim \sup A_{n}) \geq \lim \inf _{n} \mu (A_{n})$ for any measure $\mu$.
I have thought about calling $R_{\alpha}$ to the ray with angle $\alpha$ and $A_{n} = \{ \alpha : R_{\alpha}$ crosses $B(x_{n},0.5) \}$. Then it would be enough to prove that $\lim \sup A_{n} \neq \emptyset$. Using the second hint, it is enought to prove that $\lim \inf \mu(A_{n}) >0$ for certain measure $\mu$.
It would be done if I could find a measure such that $\mu(A_{n})=\frac{\psi(n)}{n^{2}}$. I feel it is almost done but I'm stuck for nearly a week. Thanks in advance!
|
The idea is to rescale the disks contained in $B(0,n)$ by $n^{-1}$, thus obtaining a set contained in $B(0,1)$ with measure bounded from below. Some care must be taken to produce a sequence of such sets without double-counting disks.
By the way, it suffices to assume that $$\limsup \frac{\psi(n)}{n^{2}} = k > 0$$ instead of $\liminf$.
Proof. Choose the sequence $n_j$ inductively, so that $\psi(n_1)>\frac{k}{2}n_j^2$ and for $j\ge 2$, $\psi(n_j)-\psi(n_{j-1})>\frac{k}{2}n_j^2$. Let $U_j$ be the union of disks that are contained in $B(0,n_j)$ but are not contained in $B(0,n_{j-1})$. The set $A_j:=n_j^{-1}U_j$ has Lebesgue measure at least $\frac{k}{2}$, and is contained in $B(0,1)$. Use the hint to conclude that the Lebesgue measure of $\limsup A_j$ is at least $k/2$. Any ray crossing $\limsup A_j$ crosses infinitely many disks.
|
{
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"url": "https://math.stackexchange.com/questions/598618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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|
isomorphism, integers of mod $n$. Hello I think this is true, but I'm not sure.
Setup:
If $n = p_{1}\cdot p_{2} \cdots p_{n}$ where $p_{i}$ prime for all $i\in\lbrace 1,\dots,n\rbrace$.
Define the ring $A = p_{j}\mathbb{Z}/n\mathbb{Z}$.
Question:
Is $A$ isomorphic to $\mathbb{Z}/(n/p_{j})\mathbb{Z}$? And why?
|
Hint: Does $\mathbb Z/(n/p_i)\mathbb Z$ have a multiplicative unit? Does $p_i\mathbb Z/n\mathbb Z$?
|
{
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Proving $f(x)=\pm x + c$ Using Differentiation Suppose $|f(x)-f(y)|=|x-y|$ for all $x,y \in R$. Prove there is a constant $c$ such that $f(x)=\pm x+c$. Suppose $|f(x)-f(y)|=|x-y|$ for all $x,y \in R$. Prove there is a constant $c$ such that $f(x)=\pm x+c$. What I did was make $|\frac{f(x)-f(y)}{x-y}|=1$. Thus, $f$ is differentiable for all $x,y \in R$, and that $\forall x \in R, f'(x)=1$. Thus, $f(x)=x+c$. But where would the $-x$ be? I was considering a case where the denominator is a different sign than the numerator, but the absolute value bars stop me dead in my tracks. Another thing I considered doing was distributing the absolute value. I would get $f(x)=x+(f(y)-y)$ and $f(x)=-x+(f(y)+y)$. Then could I make y a constant, differentiate $f(x)$, and get my results?
|
You cannot conclude from $\left|\frac{f(x)-f(y)}{x-y}\right|=1$ that $\lim_{y\to x}\frac{f(x)-f(y)}{x-y}=1$. In fact, even the existence of the limit is not that obvious.
You can solve the problem without differentiation for example by considering $f(0),f(1), f(x)$.
|
{
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"url": "https://math.stackexchange.com/questions/598767",
"timestamp": "2023-03-29T00:00:00",
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|
Complex Integration of DTFT Question
A discrete-time signal $u \in \mathcal{l}^2(\mathcal{Z})$ has DTFT
\begin{equation}
\hat{u}(\omega) = \frac{5+3\cos(\omega)}{17+8\cos(\omega)}
\end{equation}
Use complex integration to find $u(k)$ for $k\in\mathcal{Z}$
My Attempt
If I'm not mistaken, to find $u[k]$, I should find the inverse Fourier Transform
\begin{equation}
u(k) = \frac{1}{2\pi} \int_{-\pi}^{\pi} \frac{5+3\cos(\omega)}{17+8\cos(\omega)} e^{j\omega k} d\omega
\end{equation}
Is this where I use complex integration? In order to do so, doesn't my equation need poles? I can't think of when $17+8\cos(\omega)$ would equal zero for this to not be analytic.
|
Using the residue theorem (complex integration) on integrals of rational functions of sines and cosines is commonplace. The trick is to let $z=e^{i \omega}$. Then $d\omega = -i dz/z$ and the integral may be expressed as a complex integral over the unit circle:
$$\begin{align}u(k) &= -\frac{i}{2 \pi} \oint_{|z|=1} \frac{dz}{z} \frac{5+\frac{3}{2} (z+z^{-1})}{17+4 (z+z^{-1})} z^k\\ &= -\frac{i}{4 \pi} \oint_{|z|=1} dz \, z^{k-1} \frac{3 z^2+10 z+3}{4 z^2+17 z+4}\end{align}$$
Because $k \in \mathbb{Z}$, we have no branch points. Consider $k \ge 1$. Then the only poles in the integrand are where
$$4 z^2+17 z+4=0 \implies z_{\pm} = \frac{-17 \pm 15}{8} $$
i.e., $-1/4$ or $-4$, respectively. As only $z_+=-1/4$ is the only pole within the unit circle, this is the only pole contributing to the integral.
By the residue theorem, the integral is $i 2 \pi$ times the residue at the pole $z=-1/4$. Thus
$$u(k) = \frac12 \left (-\frac14 \right)^{k-1} \frac{11}{16 \cdot 15} = -\frac{11}{120} \left (-\frac14 \right)^{k} $$
when $k \ge 1$. When $k=0$, there is an additional pole at $z=0$, which contributes $3/8$ to add to the above piece, or
$$u(0) = \frac{3}{8}-\frac{11}{120} = \frac{17}{60} $$
For $k \lt 0$, we can use the fact that
$$u(-k) = u(k)^*$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/598854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
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