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H: Prove that $R(P) = N (I − P) = X$ and $R(I − P) = N (P) = Y$.
Suppose that $V = X ⊕Y$, and let $P$ be the projector onto $X$ along
$Y$. Prove that
$R(P) = N (I − P) = X$ and $R(I − P) = N (P) = Y$.
I know that from $V = X ⊕Y$ I got $v=x+y$ for $v,x,y$ are element of $V,X,Y$ and the intersect of $X$ and $Y$ is zero. But I don't know what to do next.
AI: Hint: For any $v\in V$, we have $v=P(v)+y,$ where $y=v-P(v)=(I-P)(v).$
Recall that for any linear transformation $T:V\to W,$ we have $$N(T)=\{v\in V:T(v)=0_W\}$$ and $$R(T)=\{w\in W:T(v)=w\text{ for some }v\in V\}.$$
In particular, then, recalling that all $v\in V$ can be uniquely written in the form $x+y$ where $x\in X,y\in Y,$ and so $P(v)=x,$ then $$N(P)=\{v\in V:x=0\}\\R(I-P)=\{w\in V:w=(I-P)(v)=v-P(v)=v-x=y\text{ for some }v\in V\}.$$ It shouldn't be too difficult to show that these two sets are equal, and in particular are equal to $Y$. Similarly, we can show that $N(I-P)=R(P)=X$. |
H: Show that $\{1, \sqrt{2}, \sqrt{3}\}$ is linearly independent over $\mathbb{Q}$.
My apologies if this question has been asked before, but a quick search gave no results. This is not homework, but I would just like a hint please. The question asks
Show that $\{1, \sqrt{2}, \sqrt{3}\}$ is linearly independent over $\mathbb{Q}$.
To begin, I consider some linear relation $a + b \sqrt{2} + c \sqrt{3} = 0$, where $a, b, c \in \mathbb{Q}$. There are several cases to consider.
If $c = 0$, then it must be the case that $a = b = 0$, because $\sqrt{2}$ is not rational. Similarly, if $b = 0$, then $a = c = 0$ because $\sqrt{3}$ is not rational. If $a = 0$, then we must have that $b = c = 0$ because $\frac{\sqrt{2}}{\sqrt{3}}$ is not rational (because $\sqrt{6}$ is not rational).
My issue is drawing a contradiction when $a$, $b$, and $c$ are all nonzero. It is not always the case that the sum of two irrational numbers is irrational (e.g. $1 - \sqrt{2}$ and $\sqrt{2}$). Hints or suggestions would be greatly appreciated!
AI: Hint:
$$a+b\sqrt2+c\sqrt3=0\;,\;\;a,b,c\in\Bbb Q\implies 2b^2+3c^2-a^2=-2bc\sqrt6$$
so if $\,bc\neq 0\;$ we get that $\;\sqrt6\in\Bbb Q\;$ , contradiction. So either
$$a+c\sqrt3=0\;\;or\;\;a+b\sqrt2=0$$
and then if $\;b\neq0\;$ or $\;c\neq0\;$ we get a straightforward contradiction again, so $\;b=c=0\;$ and etc. (there still lacks half a line to end the proof.) |
H: Existence of isomorphism between tensor products.
In multilinear algebra many maps are usually proven to exist rather than simply defined. For example, commutativity is one such example. In the book I'm studying the author says: let $V_1,\dots,V_k$ be a collection of vector spaces over $K$, then if $\sigma \in S_k$ there is a linear isomorphism
$$f_\sigma : V_1\otimes\cdots\otimes V_k\to V_{\sigma(1)}\otimes\cdots\otimes V_{\sigma(k)}$$
such that $f_\sigma(v_1\otimes\cdots\otimes v_k)=v_{\sigma(1)}\otimes\cdots\otimes v_{\sigma(k)}$. Then to show this, he defines
$$g : V_1\times\cdots\times V_k\to V_{\sigma(1)}\otimes\cdots\otimes V_{\sigma(k)}$$
by $g(v_1,\dots,v_k)=v_{\sigma(1)}\otimes\cdots\otimes v_{\sigma(k)}$. He argues then that $g$ is multilinear and by the universal property there is a linear $f_\sigma$ such that
$$f_\sigma(v_1\otimes\cdots\otimes v_k)=v_{\sigma(1)}\otimes \cdots\otimes v_{\sigma(k)}$$
and since this turns basis into basis is isomorphism.
That's pretty fine, I understand this proof, I just don't understand the need for it. Why do we need to show that $f_\sigma$ exists this way? Couldn't we simply say: pick $f_\sigma$ defined on the factorizable tensors by
$$f_\sigma(v_1\otimes\cdots \otimes v_k)=v_{\sigma(1)}\otimes\cdots\otimes v_{\sigma(k)}$$
and extend by linearity. Then $f_\sigma$ is linear, turn basis into basis and is isomorphism.
Why do we need to first define a multilinear map, then show that $f_\sigma$ exists using the universal property? This just one example, there are many others where the same thing happens (associativity, existance of tensor product of linear maps and so on).
Thanks very much in advance!
AI: It's because not everything you can formally write down is an actual linear map, or even well defined. Consider the following example:
Define $\newcommand{\RR}{\mathbb{R}}f : \RR \otimes \RR \to \RR$ by $$f(x \otimes y) = x + y$$ Sounds reasonable, right? Think again... Because this map is not well defined! Indeed, $f((1+1) \otimes 1) = (1+1)+1 = 3$. But $(1+1) \otimes 1 = 1 \otimes 1 + 1 \otimes 1$, and if you wanted to extend by linearity, you would find $f(1 \otimes 1 + 1 \otimes 1) = f(1 \otimes 1) + f(1 \otimes 1) = 2+2 = 4$.
So to define a map on a tensor product, you really have to check multilinearity. Once you're sufficiently accustomed to tensor products, you can skip this step, but until then, I advise you to always check multilinearity. |
H: Zeros of complex function sequence (Application of Rouche's Theorem).
For a given sequence of complex functions: $\phi_n(z)= 1+\frac1n-z-e^{-z}$; here $z\in${$z| Rez>0$}.
I want to prove that :
(1).
$\phi_n $ has a unique zero $z_n$ in the half plane. (i.e. there exists a unique $z_n$ in half plane {$z| Rez>0$} s.t $\phi_n(z_n)=0)$.
[Some how by Rouche's theorem, I almost proved that the function $1+\frac1n-z-e^{-z}$
has only one zero in half plane {$z| Rez>0$}. But I don't know how to show that $z_n$ is unique s.t. $f_n(z_n)=0$ ]
(2).
And further more, I have to show that $z_n\in \Bbb R$.
(3).
Does $lim_{x\to \infty} z_n$ exist? And If limit exists then what is it?
(I don't have any idea about last two parts.)
Thanks in advance.
AI: I would suggest that you start by showing that each $\phi_n$ has a positive real zero. That is an easy consequence of the intermediate value theorem, since all $\phi_n$ are real-valued on $\mathbb{R}$. That way, you are not distracted by thinking about the uniqueness.
Then show that each $\phi_n$ has only one zero in the right half plane. Rouché's theorem is a good option for that, one just has to find the right bounded subregion for it to be applicable. For that, note that no $\phi_n$ can have a zero $z_0$ with $\operatorname{Re} z_0 \geqslant 3$ (arbitrarily chosen value, others work too, just not too small) or $\operatorname{Im} z_0 \geqslant 3$ by the triangle inequality.
For the application of Rouché's theorem, let us choose the rectangle $R$ with vertices $-i\pi,\, \pi - i\pi,\, \pi +i\pi,\, i\pi$. All zeros of $\phi_n$ in the right half plane must lie in the interior of $R$, because
$$\lvert \phi_n(z)\rvert = \left\lvert 1 + \frac1n - z - e^{-z}\right\rvert \geqslant \lvert z\rvert - 1 - \frac1n - \lvert e^{-z}\rvert \geqslant \lvert z\rvert - 3 \geqslant \pi-3 > 0$$
for $\lvert z\rvert \geqslant \pi$ and $\operatorname{Re} z \geqslant 0$.
We apply Rouché's theorem to $f_n(z) = \frac1n - z$ and $g(z) = 1 - e^{-z}$. On the three sides of $\partial R$ that lie in the right half plane, we have
$$\lvert g(z)\rvert \leqslant 1 + \lvert e^{-z}\rvert \leqslant 2 < \pi - 1 \leqslant \lvert z\rvert - \frac1n \leqslant \lvert f_n(z)\rvert.$$
On the side of the rectangle that lies on the imaginary axis, we compute in a different way, we have
$$\left\lvert 1 - e^{-it}\right\rvert^2 = \left\lvert (1-\cos t) + i\sin t \right\rvert^2 = (1-\cos t)^2 + \sin^2 t = 2(1-\cos t) = 4 \sin^2 (t/2),$$
and
$$\left\lvert\frac1n - it \right\rvert^2 = \frac{1}{n^2} + t^2 > t^2 = 4(t/2)^2 \geqslant 4\sin^2(t/2).$$
So we have established that $\lvert g(z)\rvert < \lvert f_n(z)\rvert$ on $\partial R$, and can apply Rouché's theorem to the functions $f_n$ and $\phi_n = f_n+g$ on the rectangle, and conclude that both have the same number of zeros inside $R$. Evidently, $f_n$ has the single zero $\frac1n$ in $\mathbb{C}$, which lies inside $R$, so $\phi_n$ also has a unique zero in $R$.
Finally, for the third point, knowing that the zeros $z_n$ are real allows to approximately locate them with ease, finding $a_n,b_n > 0$ such that $a_n < z_n < b_n$ and such that $\lim (b_n-a_n) = 0$. Then you can see whether $\lim a_n$ exists. $\lim z_n$ exists if and only if $\lim a_n$ exists, and then the two are equal. |
H: Probability - $a$ white balls, $b$ black balls
From a urn containing $a$ - white balls and $b$ - black balls are pulled out $k$ balls ($k < a+b$) which are putted aside without knowing their color. Then, it is pulled out another ball. Which is the probability this last ball to be white?
I don't know very much about probability. I would be very happy if you can help me with this exercise and some materials.
thanks:)
AI: The easy argument is to imagine drawing all $a+b$ balls one at a time and choosing the $(k+1)$-st ball drawn. This is clearly the same as the probability that the first ball drawn is white: just interchange the first and $(k+1)$-st balls in each possible permutation of the $a+b$ balls. And that probability is simply
$$\frac{a}{a+b}\;.$$
Here’s a computational argument, if you prefer that:
Let $w$ be the number of white balls among the $k$ balls that are removed in the first step. Then $p_w$, the probability that the next ball is white, is
$$p_w=\frac{a-w}{a+b-k}\;.$$
There are $\dbinom{a}w\dbinom{b}{k-w}$ ways to draw $w$ white and $k-w$ black balls from the original collection of $a+b$ balls, so the probability of drawing $w$ white balls in the first step is
$$\frac{\dbinom{a}w\dbinom{b}{k-w}}{\dbinom{a+b}k}\;,$$
and the overall probability of drawing a white ball at the second step is
$$p=\sum_{w=0}^k\left(\frac{\dbinom{a}w\dbinom{b}{k-w}}{\dbinom{a+b}k}\cdot\frac{a-w}{a+b-k}\right)\;.$$
Now $$\binom{a}w(a-w)=a\binom{a-1}w$$ and $$\binom{a+b}k(a+b-k)=(a+b)\binom{a+b-1}k\;,$$
so
$$\begin{align*}
p&=\sum_{w=0}^k\frac{a\dbinom{a-1}w\dbinom{b}{k-w}}{(a+b)\dbinom{a+b-1}k}\\\\
&=\frac{a}{a+b}\binom{a+b-1}k^{-1}\sum_{w=0}^k\binom{a-1}w\binom{b}{k-w}\\\\
&=\frac{a}{a+b}\binom{a+b-1}k^{-1}\binom{a+b-1}k\\\\
&=\frac{a}{a+b}\;.
\end{align*}$$ |
H: Continuity and openness proof
I need some help proving the following theorem. My professor said that it was a "local" version of an important theorem:
Suppose f:X→Y and a ∈X. The function f is continuous at a if
and only if for every open set U containing b = f(a), there is an open set V containing a so that V ⊂ f−1(U).
What I know:
(i) Definition of continuity:Suppose X,Y are metric spaces, a ∈ X and f : X → Y. The function f is continuous at a if for every ε>0 there is a δ>0 such that if dX(a,x) < δ, then dY (f(a),f(x)) < ε.
If f is continuous at every point a ∈ X, then f is said to be continuous.
Thanks in advance for any help.
AI: For metric spaces $f$ is continuous at $a\in X$<=>for every $ε>0$ there is a $δ>0:$ for $x\in X $ that $d_{X}(x,a)<δ=>d_{Y}(f(x),f(a))<ε$.
=>)Let $U$ be an open neightbourhood of $f(a)$. Then for an $ε>0$ $(f(a)-ε,f(a)+ε)\subset U$ ($U$ is the union of such sets)we have an open neightbourhood of $f(a)$. Then for this $ε$ there is a $δ>0:d_{X}(x,a)<δ=>x\in B(a,δ)$ which is an open ball and thus we have $f(x)\in B(f(a),ε)$ for every $x\in B(a,δ)$.
<=)Let $ε>0$. Τhen $(f(a)-ε,f(a)+ε)$ is an open neightbourhood of $f(a)$ and there is an open set $V:f(V)\subset (f(a)-ε,f(a)+ε)$ that means that there is a $δ>0$ to do the job. |
H: Homework - combinatorics
How many solutions does this inequality have in non negative integers $x_i$:
$n \leq x_1+x_2+x_3+ \dots +x_n \le 2n$ ? Im stumped
I know I should add another variable but...I don't know.
AI: Suppose $s$ is positive integer. Now create a row of $n + s - 1$ blanks, and imagine you have $s-1$ $o$s and $n-1$ $x$s . There are ${n+s-1 \choose n-1}$ ways to position the $x$s in the blanks. There is one way to position the $o$s. Each of these permutations represents a solution to the equation
$$\sum_{k=1}^n x_k = s, \qquad x_k \ge 0, 1\le k \le n.$$
This correspondence is achieved by letting $x_1$ be the number of $o$s as the beginning, $x_2$, be the number of $o$s between $x_1$ and $x_2$, etc.
So this equation has ${n+s-1\choose n-1}$ solutions.
Now do some summation to get the solution to your problem. |
H: How to show that $\frac{\partial}{\partial y}\left(\int_{0}^{y}\frac{1}{x+it-2}dt\right)=\frac{1}{x+iy-2}$
I'm trying to show that $$\frac{\partial}{\partial y}\left(\int_{0}^{y}\frac{1}{x+it-2}dt\right)=\frac{1}{x+iy-2}$$ In an area that doesn't contain the point $2+0i$. If the function under the integral was a real function then it would have been immediate from the Fundamental Theorem but I haven't studied of an equivalent result for a complex function. I need to somehow show this result through a direct computation of some sort.
Edit: I would prefer not to use complex logarithms since I'm not really familiar with the subject but I do know it differs quite a bit from the real logarithm.
Help would be appreciated.
AI: Following Daniel's comment I realized that it's completely unnecessary to try and explicitly calculate the real/imaginary parts of $\frac{1}{x+it-2}$ and the result follows immediately from the following equalities: $$\frac{\partial}{\partial y}\left(\int_{0}^{y}\frac{1}{x+it-2}dt\right)=\frac{\partial}{\partial y}\left(\int_{0}^{y}\mathfrak{R}\left(\frac{1}{x+it-2}\right)dt+i\int_{0}^{y}\mathfrak{I}\left(\frac{1}{x+it-2}\right)dt\right)$$
$$=\frac{\partial}{\partial y}\int_{0}^{y}\mathfrak{R}\left(\frac{1}{x+it-2}\right)dt+i\frac{\partial}{\partial y}\int_{0}^{y}\mathfrak{I}\left(\frac{1}{x+it-2}\right)dt$$
$$\mathfrak{R}\left(\frac{1}{x+iy-2}\right)+i\mathfrak{I}\left(\frac{1}{x+iy-2}\right)=\frac{1}{x+iy-2}$$
Thanks for the help Daniel! |
H: Prove by induction that a^n+a^-n is an integer.
I am to prove by induction that given that $a+1/a$ is an integer (i.e. belongs to Z ) then $a^n+1/a^n$ is an integer too.
I'm pretty much clueless here. Thanks in advance.
AI: HINT:
We need Strong induction here
Use
$$\left(a^n+\frac1{a^n}\right)\left(a+\frac1a\right)=\left(a^{n+1}+\frac1{a^{n+1}}\right)+\left(a^{n-1}+\frac1{a^{n-1}}\right)$$
$$\implies \left(a^{n+1}+\frac1{a^{n+1}}\right)=\left(a^n+\frac1{a^n}\right)\left(a+\frac1a\right)-\left(a^{n-1}+\frac1{a^{n-1}}\right)$$
Assume $\displaystyle a^{n-1}+\frac1{a^{n-1}},a^n+\frac1{a^n}$ are integers
Base cases $$a^2+\frac1{a^2}=\left(a+\frac1a\right)^2-2$$
and $$a^3+\frac1{a^3}=\left(a+\frac1a\right)^3-3\left(a+\frac1a\right)$$ |
H: Simple Divisor Summation Inequality (with Moebius function)
Show that
$$\left| \sum_{k=1}^{n} \frac {\mu(k)}{k} \right| \le 1 $$
where $\mu$ is Moebius function and n is a positive integer.
The hard thing here is that the sum is not directly divisor sum; it's just a normal summation. What I know is that when $F(n)=\sum_{d|n}f(d)$,
$$\sum_{k=1}^N F(k)=\sum_{k=1}^N f(k)\left[\frac{N}{k}\right]$$
But it was hardly useful since the given inequality doesn't contain the Guass function, so it became really complicated(and I'm not very confident in using the doulbe-summation). One thing I also know is that
$$\frac{\phi(n)}{n}=\sum_{d|n}\frac{\mu(d)}{d}$$
I also tried to use this using Moebius Inversion Formula, but it also became quite complicated. For example, we can derive that
$$\frac{\mu(k)}{k}=\sum_{d|k}\frac{\phi(d)}{d} \mu \left(\frac{k}{d}\right)$$
I couldn't make it more compact. So by substitution, we get
$$\sum_{k=1}^{n} \frac {\mu(k)}{k}=\sum_{k=1}^{n}\sum_{d\mid k}\frac{\phi(d)}{d} \mu \left(\frac{k}{d}\right)$$
If it were not for the $k$ in the right term of Moebius function, we could use the first lemma I'd written down, but we can't. So I'm just stuck.
Could anybody help me solving this problem? It would be really grateful if someone could also give me general methods to tackle divisor-sum problems since I'm having a difficult time in changing the order of doulbe-summations.
Thanks!
AI: Let $e(k)=\sum_{d|k}\mu(d)$ and $s(n)=\sum_{k=1}^{n}e(k)$ , Then we have
$$s(n)=\sum_{k=1}^n\mu(k)\left\lfloor\frac{n}{k}\right\rfloor$$
and
$$\left|\sum_{k=1}^n\mu(k)\left\lfloor\frac{n}{k}\right\rfloor-n\sum_{k=1}^n\frac{\mu(k)}{k}\right|\leq\sum_{k=1}^n|\mu(k)|\left|\left\lfloor\frac{n}{k}\right\rfloor-\frac{n}{k}\right|\leq\sum_{k=1}^{n-1}|\mu(k)|\leq n-1$$
Hence,
$$\left|n\sum_{k=1}^n\frac{\mu(k)}{k}\right|\leq(n-1)+|s(n)|=(n-1)+1=n$$
and
$$\left|\sum_{k=1}^n\frac{\mu(k)}{k}\right|\leq1.$$ |
H: convergence of a function serie to norm 2
need to show if the following function serie converge on $||.||_2$ on [1,5]:
$$\sum _1^\infty {sin^2(nx) \over n^2} $$
I have no idea how to approach that one, would like
for some directions...
AI: In fact it converges uniformly. Let $f_n(x)=a_nsin^2(nx)$ with $a_n=1/n^2$. Then $\|f_n\|_{\infty}=max\|f_n(x)|=|a_n|$.
We have that $\sum_{n=1}^\infty |a_n|$ converges (exactly to $π^2/6$) and by Weierstrass Criterion of series convergance wwe have that $\sum_{n=1}^\infty \|f_n\|_{\infty}$ converges and thus $\sum_{n=1}^\infty f_n$ converges uniformly on $\Bbb R$. |
H: Question related to calculation of probality?
question given in my text book
There are three events A, B and C out of which only one and one can happen. The odds are 8 to 3 against A and 5 to 2 against B. Find odd against C.
Solution in my textbook
Let the total no. of cases = $m + n + p$
$m$ are in favor of $A$, $n$ are in favor of $B$ and $p$ in favor of $C$
we have $\frac{n+p}{m} = \frac{8}{3}$, $\frac{m+p}{n} = \frac{5}{2}$, so that
$$P(A') = \frac{n + p}{m+n+p}=\frac{8}{11}, P(B') =\frac{m +p}{m+n+p}=\frac{5}{7}$$
i am really not able to understand what they did here right from the beginning can anyone explain it to please
AI: Their use of numbers ‘in favor of’ an event seems to me a very odd way of expressing what’s going on, so I’ll use rather different language.
The odds against $A$ are $8$ to $3$; you can think of that as meaning that for every $3$ times that $A$ does happen, on average there are $8$ times when $A$ does not happen. That is, in $8$ out of $11$ tries, on average, $A$ does not happen, so $\frac8{11}$ is the probability that $A$ does not happen. $\Bbb P(A)$, the probability that $A$ does happen, is therefore $1-\frac8{11}=\frac3{11}$. Similarly, the odds against $B$ are $5$ to $2$, so on average $B$ fails to occur $5$ times out of $7$, and $\frac57$ is therefore the probability that $B$ does not occur. $\Bbb P(B)$, the probability that $B$ does occur, is therefore $1-\frac57=\frac27$.
Since $A$ and $B$ cannot occur simultaneously, the probability that one or the other occurs is simply $$\Bbb P(A)+\Bbb P(B)=\frac3{11}+\frac27=\frac{21+22}{77}=\frac{43}{77}\;.$$
The problem doesn’t actually say so, but apparently we are to assume that exactly one of $A,B$, and $C$ must occur. This means that $\frac{43}{77}$ is simply the probability that $C$ does not occur, and therefore the probability that $C$ does occur is
$$\Bbb P(C)=1-\frac{43}{77}=\frac{34}{77}\;.$$ |
H: Proving that there exists a local minimum between two local maximums of a continuous function
Suppose f is a continuous function that has local maximums at points $x_{1}$ and $x_{2}$. How do I prove that there is a third point between these two points which is a local minimum of f? Need some help thanks.
AI: Hint: Assume that there does not exist a local minimum in $(x_{1},x_{2})$ (assuming that $x_{1}<x_{2}$). Then inf{$f(x)|x∊[x_{1},x_{2}$]}=$f(x_{1})$ or $f(x_{2})$.
Now assume inf{$f(x)|x∊[x_{1},x_{2}$]}=$f(x_{1})$. Then $f(x)$≥$f(x_{1})$ when x∊[$x_{1},x_{2}$]. Also there exists $δ>0$ such that $f(x_{1})$∊$[x_{1},x_{1}+δ]$.
Then we get that $f(x)$=$f(x_{1})$. This is a contradiction. DO the same for $f(x_{2})$. |
H: How to solve the equation $x^2 + 4 |x| - 4 = 0$
How do I get the value of x from the equation that is provided
AI: Putting $x=a+ib$ where $a,b$ are real
we get $\sqrt{a^2+b^2}=4-(a+ib)^2=4+b^2-a^2-2abi$
Equating the imaginary parts $ b=0$
Now for real $x,$
$$|x|= \begin{cases} +x &\mbox{if } x\ge0 \\
-x & \mbox{if } x<0 \end{cases}$$ |
H: Can $\|f\| = \|a\|_{q}$ to arbitrary values of $p$ and $q$ satisfying ${1 \over p} + {1 \over q} = 1$
We all know that:
Suppose $a = (a_{1}, a_{2}, ..., a_{n})$ is a point in Euclide space $R^{n}$. Consider the mapping $f: R^{n} \rightarrow R$, $f(x) = \sum_{i=1}^{n}a_{i}x_{i}$. Then $\|f\| = \|a\|$.
So I wonder if we can make this to the general case like this:
Let ${1 \over p} + {1 \over q} = 1$, $p > 1$, $l_{p} = \{x = (\lambda_{k})_{k}: \sum_{k=1}^\infty|\lambda_{k}|^{p} < \infty\}$, $\|x\|_{p} = (\sum_{k=1}^{\infty}|\lambda_{k}|^{p})^{1 \over p}$. We define $a \in l_{p}$ and $f:l_{p} \rightarrow R$ like above (except the sum is countably infinite). Then is it right that $\|f\| = \|a\|_{q}$?
We can't use the way we prove for the case $p = q = 2$ in this general case. So I'm wondering whether this problem is right. If not, can we give a counter-example for this? Please help me clarify this. Thanks so much. I really appreciate.
AI: Yes, this is true. Using Holder inequlity you can show that $\Vert f\Vert\leq\Vert a\Vert_q$. Now consider
$$
x=(|a_1|^{q-1}\operatorname{sign}(a_1),\ldots,|a_n|^{q-1}\operatorname{sign}(a_n))
$$
then $f(x)=\Vert a\Vert_q^q$ and $\Vert x\Vert_p=\Vert a\Vert_{q-1}^q$. So $\Vert f\Vert\geq |f(x)|/\Vert x\Vert_p=\Vert a\Vert_q$. From these inequalities it follows that $\Vert f\Vert=\Vert a\Vert_q$ |
H: Maximal graph that does not contain Hamiltonian cycle
My lecture notes in Graph Theory states that a graph of order $n$ and with size (= number of edges) $\binom n 2-(n-2)$ is the maximal graph that does not contain a Hamiltonian cycle.
My question now is how to find such a graph? How can I construct a graph of order $n$ such that this is true?
For the proof that no greater size can be achieved there was a similar question on this site: Proving that a graph of a certain size is Hamiltonian
This can be used to prove that no greater size can be achieved, so I am not interested in this part, only in the constructon part.
AI: Hint: If a vertex has degree 1, it can't be part of a cycle. |
H: Question about an element at $\mathbb{Z}_{n}^{*}$
Assume that $n,q>1$ and $n=\frac{q^r-1}{q-1}$.
How do I prove that $q\in \mathbb{Z}_{n}^{*}$?
$$\mathbb{Z}_{n}^{*}= \left\{a\mid1\le a<n;\ \gcd(a,n)=1 \right\}$$
Thank you!
AI: You have $\frac{q^r-1}{q-1}=1+q+q^2+\ldots +q^{r-1} = n$.
So $q < n$ and $n + q(-1-q-\ldots -q^{r-2}) = 1$.
So $\gcd(n,q) = 1$ and $q\in \mathbb Z^{*}_n$ |
H: Limit at infinity -
I don't even know where to start on this one. It's obvious that it's infinity, but how to prove it? $$\lim_{n\to\infty}\frac{3^n+2n^n+n!}{(n+1)^4+\sin n+(3n)!}$$
AI: Hint: start rewriting it as
$$
\lim_{n\to\infty}
\frac{n^n
\left(
\dfrac{3^n}{n^n}+2+\dfrac{n!}{n^n}
\right)
}
{(3n)!
\left(
\dfrac{(n+1)^4)}{(3n)!}+\dfrac{\sin n}{(3n)!}+1
\right)
}
$$ |
H: Show that $\frac{(2n)!}{(n)!}=2^n(2n-1)!!$
Show that $\frac{(2n)!}{(n)!}=2^n(2n-1)!!$ is the question I am struggling with.
I started by saying: $(2n)!=2n(2n-1)(2n-2)(2n-3)...3*2*1$
But then I'm stuck.
AI: $$(2n)!=\prod_{k=1}^{2n}k=\prod_{k=1}^{n}(2k)\cdot\prod_{k=1}^{n}(2k-1)=2^n\cdot\prod_{k=1}^{n}k\cdot\prod_{k=1}^{n}(2k-1)=2^n\cdot n!\cdot(2n-1)!!$$ |
H: Prove that $n(n^2 - 1) = \frac{(n+1)!}{(n-2)!}$
Prove that for all $n \in \mathbb{N}$,
$$n(n^2 - 1) = \frac{(n+1)!}{(n-2)!}.$$
Thanks in advance.
AI: $\require{cancel}$
If you really want to express $n(n^2 - 1)$ using factorials, note that $$n(n^2 - 1) = n(n+1)(n-1) = (n+1)(n)(n-1) = \frac{(n+1)n(n-1)\cancel{(n-2)}\cdots \cancel{2}\cdot 1}{\cancel{(n - 2)}\cancel{ (n-3)}\cdots \cancel{2}\cdot 1}$$
So we have that $$n(n^2 - 1) = \dfrac{(n+1)!}{(n-2)!}$$
So the given identity you are asked to prove is incorrect: the numerator needs to be $(n + 1)!$. |
H: 0-1 Law in a sigma algebra Conditional Expectation
I get stuck on a simple question.
Let T be a sigma algebra with for all its elements A in T P(A)=0 or P(A)=1.
Assuming X is L1 find E[X/T]
I was thinking that T had to be a partition or not necessary?
Thx in advance
AI: $\mathcal T$ is not necessarily generated by a partition (for exemple, take the unit interval with the $\sigma$-algebra of countable and co-countable sets).
The random variable $\mathbb E[X\mid\mathcal T]$ has to satisfy for each $T\in\cal T$,
$$\int_TX\mathrm d\mathbb P=\int_T\mathbb E[X\mid\mathcal T]\mathrm d\mathbb P.$$
What is the integral of an integrable function over a set of measure $0$? Is the integral over a set of measure $1$ different from the integral over $\Omega$? |
H: function as a net?
One can see net as a generalization of a sequence.
As done in http://en.wikipedia.org/wiki/Net_(mathematics), in the special case where $f: M\backslash \{a\} \rightarrow X$ where M is a metric space and X is a topological space, one can see $f$ as a net from the directed set $M\backslash \{a\}$ with partial order given by the distance to $a\in M$. The closer a point is to $a$, the "greater" it is.
Question: It is possible to see an arbitrary continuous function from a topological space to another as a net?
(it's a question for free, i was not looking for any application of it)
Edit: actually one can define $f:M \rightarrow X$. it's just when we want to have a directed set that we take a point away.
AI: If you assume the Well-ordering theorem then yes! In that case, if we have $f:X\rightarrow Y$ then the Well-ordering theorem gives us a well-ordering of $X$ hence it becomes a directed set.
It is still interesting to see if one can use the topological structure on $X$ to turn it into a directed set. |
H: Prove that $\binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1} $
Prove that $\binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1} $
Thanks in advance, my professor asked us to this a couple weeks ago, but I was enable to get to the right answer.
Good luck!
Here is what I got up to;
$\frac{(n+1)!}{(n-r)!(r+1)!} = \frac{(n)!}{(r)!(n-r)!} + \frac{(n)!}{(r+1)!(n-r-1)!} $
AI: $$
\binom{n}{r} + \binom{n}{r+1} \\
\frac{n!}{(n-r)!r!} + \frac{n!}{(n-r-1)!(r+1)!} \\
\frac{n!}{(n-r)(n-r-1)!r!} + \frac{n!}{(n-r-1)!r!(r+1)} \\
\frac{n!}{(n-r-1)!r!}\left(\frac{1}{n-r} + \frac{1}{r+1}\right) \\
\frac{n!}{(n-r-1)!r!}\left(\frac{n+1}{(n-r)(r+1)}\right) \\
\frac{(n+1)!}{(n-r)!(r+1)!}\\
\binom{n+1}{r+1}
$$ |
H: System of linear equations with parameters, using a matrix
Let there be the following system of linear equations:
$$x+z+bw=a \\
ax+y+az+(a+ab)w=1+a^2 \\
bx+(a+b)z+(1+b^2)w=4+a\\
bx+bz+(a-ab+b^2)w=a+1+ab$$
a,b parameters. The question is, for which a,b there is no solution to the system, for which there are infinite and for which there is one. I put it into a matrix and with some row operations I got to:
$$\begin{array}{cccc|c} 1 & 0 & 1 & b&a \\
0&1&0&a&1\\
0&0&a&1&4\\
0&0&0&a-2b&a+1 \end{array}$$
How do I continue from here? I'm quite confused. For instance I thought that saying that for all b, if a=2b then there is no solution. But that doesn't sound quite right...
Thanks in advance for any help!
AI: You've done good until that row reduced form (I assume the calculations are correct).
Now, if $a=0$, you can do another row operation:
$$
\left[\begin{array}{cccc|c}
1 & 0 & 1 & b&0 \\
0&1&0&0&1\\
0&0&0&1&4\\
0&0&0&-2b&1 \end{array}\right]
\to
\left[\begin{array}{cccc|c}
1 & 0 & 1 & b&0 \\
0&1&0&0&1\\
0&0&0&1&4\\
0&0&0&0&1+8b \end{array}\right]
$$
What can you say about this case?
If, instead, $a\ne0$, you have to distinguish when $a-2b=0$ or not.
If $a-2b=0$ you have $a=2b$ and you have a solution if and only if $a+1=0$, that is $a=-1$ and $b=-2$.
If $a-2b\ne0$, then … |
H: is the set of all sequences that converge to zero first category
the metric space is: the set of all convergent sequences with metric supremum.
the subset is: all sequences that converge to zero.
is the subset first category in this metric space?
we have figured out that the subset is closed and it is rare (nowhere dense), but does this suffice that it is first category?
AI: A set is first category (meager) if it can be written as a countable union of nowhere dense sets. So if your set is nowhere dense, then it is automatically first category ($A = A \cup A \cup A \cup \cdots$ or $A = A \cup \emptyset \cup \emptyset \cup \cdots$). |
H: Integral from $0$ to $\infty$ of $\frac{1}{3}\ln\left(\frac{x+1}{\sqrt{x^2-x+1}}\right)+\frac{1}{\sqrt 3}\arctan \left(\frac{2x-1}{\sqrt 3} \right)$
Evaluate the integral $$ \int_0^\infty \left( \frac{1}{3}\ln\left(\frac{x+1}{\sqrt{x^2-x+1}}\right)+\frac{1}{\sqrt 3}\arctan \left(\frac{2x-1}{\sqrt 3} \right) \right) dx $$
I have read about improper integrals, and seen that I should do just like I would do with a and b, only that now, when I have $\infty$ I should take the limit. But I do not know hot to evaluate the limit of $$\ln\left(\frac{x+1}{\sqrt{x^2-x+1}}\right)$$ and $$\frac{1}{\sqrt 3}\arctan \left(\frac{2x-1}{\sqrt 3} \right)$$
AI: As $x\to\infty$, $\log\left(\frac{x+1}{\sqrt{x^2-x+1}}\right)\to0$. However, $\arctan\left(\frac{2x-1}{\sqrt3}\right)\to\frac\pi2$. Thus, the integrand tends to $\frac\pi{2\sqrt3}$, so the integral cannot converge.
To compute the limits:
$$
\lim_{x\to\infty}\log\left(\frac{x+1}{\sqrt{x^2-x+1}}\right)=\log\left(\lim_{x\to\infty}\frac{1+1/x}{\sqrt{1-1/x+1/x^2}}\right)
$$
and substitute $u=\frac{2x-1}{\sqrt3}$ to get
$$
\lim_{x\to\infty}\frac1{\sqrt3}\arctan\left(\frac{2x-1}{\sqrt3}\right)=\frac1{\sqrt3}\lim_{u\to\infty}\arctan(u)
$$ |
H: How do you prove that $\lim f(x) = 0$, when $f$ is rapidly decreasing?
Let $f: \Bbb{R} \to \Bbb{R}$ be rapidly decreasing in the sense than $\sup_{x \in \Bbb{R}} |x|^k |f^{(\ell)}(x)| \lt \infty$ for all $k, \ell \geq 0$, where $f^{(\ell)}$ is the $\ell$th derivative. It seems like $\lim_{x \to \infty} f(x) = 0$, but how do I prove it?
What's important is that $f$ is rapidly decreasing, not necessarily that its derivatives are too. But the Schwartz space just happens to be what I'm working with and it's closed under indefinite derivative taking.
AI: Hint:
$$f(x)=\frac1x\cdot x\,f(x)\xrightarrow[x\to\infty]{}\ldots$$
and remember:
$$g(x)\;\text{bounded in some neighborhood of $\,x_0\,$ and}\;\lim_{x\to x_0}h(x)=0\implies \lim_{x\to x_0}g(x)h(x)=0$$
and the above is easily generalized when $\;x_0=\infty\;$ ... |
H: How to solve this system of equations.
Solve the system of equations: $$\begin{cases}\dfrac{x^2+xy+y^2}{x^2+y^2+1}=\dfrac{1}{xy} \\\left(\sqrt{3}+xy\right)^{\log_2x}+\dfrac{x}{\left(\sqrt{3}-xy\right)^{\log_2y}}= 1+\dfrac{x}{y}\end{cases}$$
My try:
$\dfrac{x^2+xy+y^2}{x^2+y^2+1}=\dfrac{1}{xy}\\\Leftrightarrow xy\left(x^2+y^2\right)+\left(xy\right)^2= x^2+y^2+1\\\Leftrightarrow\left(xy-1\right)\left(x^2+y^2+xy+1\right)=0\\\Leftrightarrow xy=1\,\,\,\mbox{(because:}\,\, x^2+y^2+xy+1>0)$
And here I'm stuck!
AI: Subsitute $xy=1 , y=1/x,and \, log y=-logx$ :-
$log(3^{1/2}+1)^{logx}+ x*log(3^{1/2}-1)^{logx}=1+x^2$
then take Log at both sides of your equation then your 2nd equation becomes
$logx*log(3^{1/2}+1)+logx+ logx*log(3^{1/2}-1)=2logx$
$log(3^{1/2}+1)+ log(3^{1/2}-1)=1$
$log((3^{1/2}+1)(3^{1/2}-1))=1$
which is always true! |
H: combinatorics implementation in real life problems
How many ways there are to organize $7$ men in a row, if two insist on not standing next to each other? How do I approach this?
AI: There are a total of $7!$ ways to organize the $7$ men in a row with no restrictions. Now we must count all the ways in which $2$ specific men stand next to each other, say person $x$ and person $y$. Treat these $2$ men as a single person. We can place these $2$ men in $2\cdot{6\choose 1}$ ways. We multiply by $2$ because we can have their order switched ($xy$ or $yx$). Now we will fill in the remaining $5$ places with the remaining $5$ men in $5!$ ways. Thus we can organize $7$ men in a row where $2$ men refuse to stand next to each other in $7!-2\cdot{6\choose 1}\cdot5!$ ways. |
H: show that 210 is a triangular number
Show that 210 is a triangular number. Would it suffice to solve the equation 210=((n)(n+1))/2 ? Then n is equal to 20 and -21but n in this case must be positive, so 210 would be the 20th triangular number.
AI: Yes, indeed, your solution suffices. Well done! |
H: "closure preserves homeomorphism"
Let me explain the title of the problem and the problem very clearly :
If $X$ and $Y$ are subsets of a topological spaces $A$ and $B$ respectively, which are homeomorphic in the respective subspace topology, does it imply that their closure $\bar{X}$ and $\bar{Y}$ are homeomorphic ?
AI: Hint. Consider $\mathbb{R}$ and $(0, 1)$. |
H: How many different 8-letter words can be made with three $a$s, two $b$s, two $c$s and a $d$?
How many words, without making any reference to their meaning can be written from the letters: $ a,a,a,b,b,c,c,d$ ?
what is the best approach to solve this kind of problem ?
AI: Assuming you need to find the number of distinct permutations of the given eight letters:
There are $8$ letters to choose from: $a \times 3, \;b\times 2,\; c\times 2,\;d\times 1$.
That gives us the number of distinct 8-character strings: $$\binom{8}{3, 2, 2, 1} = \dfrac{8!}{3!2!2!1!} =\dfrac{8!}{24}= 1680$$ |
H: Why this two spaces do not homeomorphic?
Consider $\Bbb Q$ with subspace topology and $\Bbb Q\times \Bbb Q$ with product topology. Why this two spaces are not homeomorphic?($\Bbb Q$ is the rational numbers)
AI: It is a theorem of Sierpinski that every countable metric space without isolated points is homeomorphic to $\mathbb{Q}$. See eg here. $\mathbb{Q}^2$ is a countable metric space without isolated points. Therefore these two spaces are homeomorphic. |
H: Find the sum of the series
For any integer $n$ define $k(n) = \frac{n^7}{7} + \frac{n^3}{3} + \frac{11n}{21} + 1$ and $$f(n) = 0 \text{if $k(n)$ is an integer ; $\frac{1}{n^2}$ if $k(n)$ is not an integer } $$
Find $\sum_{n = - \infty}^{\infty} f(n)$.
I do not know how to solve such problem of series. So I could not try anything.
Thank you for your help.
AI: HINT:
Using Fermat's Little theorem, $$n^p-n\equiv0\pmod p$$ where $p$ is any prime and $n$ is any integer
$\displaystyle\implies n^7-n\equiv0\pmod 7\implies \frac{n^7-n}7$ is an integer
Show that $k(n)$ is integer for all integer $n$ |
H: Solving the differential equation $y'' + 2y' + 2y = 0$ given constraints
How can I solve this initial value problem?
$$ y'' + 2y' + 2y = 0,$$ given $y\,(\pi/4)=2$ and $y'(\pi/4)=0$.
I've found $y(t)=e^{-t} \left(C_1\cos t + C_2\sin t \right)$ but I wasn't able to find $C_1$ and $C_2$. How can I find them?
AI: Using the $y(t)$ you found (which is correct), substitute $\dfrac{\pi}{4}$ into $y(t)$, yielding:
$$\dfrac{1}{\sqrt{2}}e^{-\pi/4}(c_1 + c_2) = 2$$
Substitute $\dfrac{\pi}{4}$ into $y'(t)$, yielding:
$$-\sqrt{2}e^{-\pi/4} c_1 = 0$$
This gives us $c_1 = 0$.
Can you now find $c_2$? |
H: What is the meaning of "integral point"?
While reading this paper (http://cowles.econ.yale.edu/P/cd/d04b/d0473.pdf) I encountered the concept of "integral point", used first in definition 5.1, on page 34. Does anybody know more details about this?
AI: In this context it simply means a point in $\mathbb{R}^n$ with integer coordinates, i.e., a lattice point of the lattice defined by his unit vectors $e^i$. |
H: Finding $\lim_{x \to 0} x^x$ without l'Hôpital
I have to find the limit of $x^x$ as $x$ approaches $0$ without derivatives.
AI: We wish to find $\lim_{x\to0^{+}}x^{x}$. Notice
$$x^{x}=e^{x\ln(x)}=e^{(-1)\frac{\ln(\frac{1}{x})}{(\frac{1}{x})}}$$
so it suffices to find $\lim_{y\to\infty}\frac{\ln(y)}{y}$.
$$\frac{\ln(y)}{y}=\frac{1}{y}\int_{1}^{y}\frac{1}{t}dt.$$
For $y\ge1$ we have that $\sqrt{y}\le y$ so $\frac{1}{y}\le\frac{1}{\sqrt{y}}$ so:
$$\frac{\ln(y)}{y}\le\frac{1}{y}\int_{1}^{y}\frac{1}{\sqrt{t}}dt=\frac{1}{y}(2\sqrt{y}-2)=\frac{2}{\sqrt{y}}-\frac{2}{y}.$$
But also $\frac{\ln(y)}{y}\ge0$ for $y\ge1$. By squeeze theorem the limit is $0$. Hence,
$$\lim_{x\to0^{+}}x^{x}=1.$$ |
H: Test for convergence/divergence of $\sum_{n=1}^{\infty}(-1)^n\sin\left(\frac{n}{\pi}\right)$
Given the series
$$\sum_{n=1}^{\infty}(-1)^n\sin\left(\frac{n}{\pi}\right)$$
I need to test for convergence/divergence. I think the divergent test might work here. I could see that the $\lim_{n\rightarrow\infty}(-1)^n\sin(\frac{n}{\pi})$ might not exist, so the series is divergent. But I still need a solid proof here.
Any help is appreciated. Thanks.
AI: I guess the standard argument should work. If $S_n=\sum_{k=0}^na_k$ converges then $a_k\rightarrow 0$. The necessary condition is not satisfied. |
H: Poles of abelian differentials
Let $X$ be a smooth projective curve of genus $g$ over an algebraically closed field $k$. As a corollary of the Riemann-Roch theorem we know that for every abelian differential $\omega$ on $X$ we have
$$ \deg(\omega) = 2g - 2. $$
Now assume that $g\geq 2$, then $\deg(\omega)\geq 0$. Does this imply that an arbitrary differential has no poles on the whole $X$?
I guess no: a priori this just means that the number of zeros (with multiplicity) is greater than the number of poles. But anyway, is it always possible to find a function $f\in k(X)$ such that the form $f \omega$ has no poles?
Any idea or clarification is welcome!
AI: Yes, you can always find such a rational function $f\in k(X)$. Here is why:
Since the cotangent bundle $\Omega_X$ satisfies $\operatorname {dim} H^0(X,\Omega_X)=g\geq 2$, there certainly exists a regular differential form $ \eta\neq 0\in H^0(X,\Omega_X)$.
As is the case for all line bundles, the quotient $\frac \eta\omega$ of two rational sections (with $ \omega\neq0$) of $\Omega_X$ is a rational function $f\in k(X)$ and we have found our desired functoion $f$ with $\eta=f\omega \in H^0(X,\Omega_X)$, a differential form with no poles.
Notice that the result is true also for $g=1$ but of course false for $g=0$.
[As to your first question, you can of course force any differential form to acquire a pole at any designated point by multiplying it by a rational function having a sufficiently bad pole at that point] |
H: Proof of the properties of limits of CDFs
The cumulative distribution function is defined as $F(a) = \mu((-\infty,a])$ where $\mu$ is a probability measure on $(\mathbb{R},\mathcal{B}(\mathbb{R}))$. Given this definition, it is easy to prove right-continuity (I think).
We have also: $$\lim_{a\to -\infty} F(a) = 0$$
$$\lim_{a\to\infty} F(a) = 1$$
By using the above definition, I want to prove these properties. Some people on the web state things like following: Since $(-\infty,a] \to \emptyset$ as $a \to -\infty$, and since $\mu(\emptyset) = 0$, then we are done. Same thing can be written for $a \to \infty$. However, this type of proof makes me feel that something is problematic, i.e., not rigorous albeit intuitively makes sense. What is the rigorous proof of these properties?
AI: You will have to use two fundamental properties of measures. Namely, if $(X,\mathcal{E},\mu)$ is a measure space, then we have the following:
If $(A_n)$ is an increasing sequence of sets from $\mathcal{E}$, i.e. $A_1\subseteq A_2\subseteq\cdots$, then
$$
\mu\left(\bigcup_{n=1}^\infty A_n\right)=\lim_{n\to\infty}\mu(A_n).
$$
If $(A_n)$ is a decreasing sequence of sets from $\mathcal{E}$, i.e. $A_1\supseteq A_2\supseteq\cdots$ and $\mu(A_1)<\infty$, then
$$
\mu\left(\bigcap_{n=1}^\infty A_n\right)=\lim_{n\to\infty}\mu(A_n).
$$
Use this with $A_n$ being $(-\infty,-n]$ and $(-\infty,n]$ respectively. |
H: Validity of a Limit Proof
I am trying to show that if $\displaystyle{\lim_{s\to\infty}} s_n = s$, $\displaystyle{\lim_{s\to\infty}} \sqrt{s_n} = \sqrt{s}$ for some sequence $s_n$.
We must note that $s_n > 0$ for all $n$ for the limit to be defined in $\mathbb{R}$.
I've put together this proof, but I'm not sure if it is valid:
Let $\epsilon >0$. Since $s=\lim{s_n}$, there exists an $N$ such that $n > N$ implies $|s_n-s| < \epsilon^2$. Thus, for an arbitrary $n>N$, we have:
$\sqrt{|s^n - s|} < \epsilon$
Case 1 ($s_n > s$):
$\sqrt{s_n-s} < \epsilon$
$\sqrt{s_n} - \sqrt{s} < \sqrt{s_n-s} < \epsilon$
$|\sqrt{s_n} - \sqrt{s}| < \epsilon$
Case 2 ($s > s_n$):
$\sqrt{s-s_n} < \epsilon$
$\sqrt{s_n} - \sqrt{s} < \sqrt{s-s_n} < \epsilon$
$|\sqrt{s} - \sqrt{s_n}| < \epsilon$ $\blacksquare$
Couple of things that are worrying me:
I'm not sure if the $\sqrt{a}-\sqrt{b} < \sqrt{a-b}$ actually holds up in this case. For proving it, I though $a-b > a-b-2\sqrt{ab}$ if $a, b > 0$, which is true in this case since $s, s_n > 0$.
Does this whole casework idea make sense?
AI: I was going to post this as a comment, but it started to get a bit long.
So if both $a>b$, you're argument establishes that $\sqrt{a}-\sqrt{b}<\sqrt{a-b}$ (since $f(x)=x^2$ is strictly increasing on $[0,\infty)$).
Then, Case 1 works fine, since $s_n>s$ implies that $\sqrt{s_n}-\sqrt{s}>0$ (since $f(x)=\sqrt{x}$ is also increasing on $[0,\infty)$) which combined with $\sqrt{s_n}-\sqrt{s}<\varepsilon$ gives you that $|\sqrt{s_n}-\sqrt{s}|<\varepsilon$.
Case 2 requires minor tweaking since you haven't actually bound from below $\sqrt{s_n}-\sqrt{s}$. However, if you use the same argument as in Case 1 you get that
$$0<\sqrt{s}-\sqrt{s_n}<\sqrt{s-s_n}<\varepsilon$$
which gives you the desired $|\sqrt{s}-\sqrt{s_n}|<\varepsilon$. |
H: Fourier transform of a function over finite group
Let $G$ be finite abelian group and $\hat G$ be its character group.
The Fourier transform of a function $f:G \to \mathbb C$, is the function $\hat{f}:\hat{G}\to \mathbb C$ defined by $\hat{f}(\chi)=\sum_{a\in G}f(a)\chi(-a)$.
I need hint proving that $\hat{\hat{f}}(a)=|G|f(-a)$.
AI: Hint
$$\hat{\hat{f}}(a)=\sum_{\chi\in \hat{G}}\hat{f}(\chi)\chi(-a)=\sum_{\chi\in \hat{G}}\sum_{b\in G}f(b)\chi(-b)\chi(-a)=\sum_{ b \in G}f(b)\left(\sum_{\chi\in \hat{G}} \chi(-a-b)\right) \,.$$
For $c \in G$ what is
$$\sum_{\chi\in \hat{G}} \chi(c) ?$$ |
H: Integral curve for vector field tangent to sphere
Let $S^1$ be the unit sphere $x_1^2+x_2^2=1$ in $\mathbb{R}^2$ and let $X=S^1\times S^1\in\mathbb{R}^4$ with defining equations $f_1=x_1^2+x_2^2-1=0, f_2=x_3^2+x_4^2-1=0$. The vector field $$w=x_1\frac\partial{\partial x_2}-x_2\frac\partial{\partial x_1}+\lambda\left(x_4\frac\partial{\partial x_3}-x_3\frac\partial{\partial x_4}\right)$$ ($\lambda\in\mathbb{R}$) is tangent to $X$ and hence defines by restriction a vector field $v$ on $X$. What are the integral curves of $v$?
At a point $p=(x_1,x_2,x_3,x_4)\in X$, we have $v(p) = (p, (-x_2,x_1,\lambda x_4,-\lambda x_3))$.
The definition of an integral curve is the following:
A $C^{\infty}$ map $\gamma:(a,b)\rightarrow X$ is an integral curve of $v$ if, for $c\in (a,b)$ and $p=\gamma(c)$, $$v(p)=\left(p,\dfrac{d\gamma}{dt}(c)\right)$$
So we must have $$\dfrac{d\gamma(c)}{dt}=(-x_2,x_1,\lambda x_4,-\lambda x_3)$$
How can we solve for $\gamma$?
AI: First, pick an initial point $\gamma(0)$. Your formula then also gives you the initial tangent of the curve $\gamma'(0)$.
Write $\gamma(t) = [\gamma_1(t) ,\gamma_2(t), \gamma_3(t), \gamma_4(t)].$ Then
$\frac{d}{dt}\gamma_1 = -\gamma_2$ and $\frac{d}{dt}\gamma_2 = \gamma_1$, so by substitution
$$\frac{d^2}{dt^2}\gamma_1 = -\gamma_1.$$
Can you solve this ODE, given the two initial conditions $\gamma_1(0)$ and $\gamma_1'(0) = -\gamma_2(0)$?
Can you work out the other three components? |
H: If a Banach space $X$ is isometric to its first dual $X^*$, must $X$ be reflexive?
Suppose that $X$ is a Banach space such that there exists a linear isometry $X \rightarrow X^*$. Must $X$ be reflexive?
Of course, this implies that $X$ is isometric with its second dual $X^{**}$. But with this alone it is not possible to conclude that $X$ is reflexive, James space is the famous counterexample for this. So a negative answer to my question should be at least as difficult as finding an example like the James space.. so probably not very easy.
AI: No, consider $J\oplus_2 J^*$, where $J$ is a James space |
H: Properties of Lie derivative
Let's have Lie derivative:
$$
L_{V}\varphi = V^{\mu}\partial_{\nu}\varphi , \quad L_{V}A_{\mu} = V^{\nu}\partial_{\nu}A_{\mu} + (\partial_{\mu}V^{\nu})A_{\nu}.
$$
How to show that for scalar and vector fields
$$
L_{V}L_{U} - L_{U}L_{V} = L_{[U, V]}, [U, V]^{\nu} = U^{\mu}\partial_{\mu}V^{\nu} - V^{\mu}\partial_{\mu}U^{\nu}?
$$
An attempt to prove equality for vector field.
$$
L_{[U, V]}A_{\mu} = [U, V]^{\nu}\partial_{\nu}A_{\mu} + (\partial_{\mu}[U, V]^{\nu})A_{\nu} =
$$
$$
=[U, V]^{\nu}\partial_{\nu}A_{\mu} + (\partial_{\mu}(U^{\alpha}\partial_{\alpha}V^{\nu} - V^{\alpha}\partial_{\alpha}U^{\nu}))A_{\nu}. \qquad (.1)
$$
By the other way [edited],
$$
(L_{U}L_{V} - L_{V}L_{U})A_{\mu} = L_{U}(V^{\nu}\partial_{\nu}A_{\mu} + \partial_{\mu}(V^{\nu})A_{\nu}) - L_{V}(U^{\nu}\partial_{\nu}A_{\mu} + \partial_{\mu}(U^{\nu})A_{\nu})=
$$
$$
= U^{\alpha}\partial_{\alpha}[V^{\nu}\partial_{\nu}A_{\mu} + \partial_{\mu}(V^{\nu})A_{\nu}] + \partial_{\mu}(U^{\alpha})[V^{\nu}\partial_{\nu}A_{\alpha} + \partial_{\alpha}(V^{\nu})A_{\nu}] -
$$
$$
-V^{\alpha}\partial_{\alpha}[U^{\nu}\partial_{\nu}A_{\mu} + \partial_{\mu}(U^{\nu})A_{\nu}] - \partial_{\mu}(V^{\alpha})[U^{\nu}\partial_{\nu}A_{\alpha} + \partial_{\alpha}(U^{\nu})A_{\nu}]=
$$
$$
=[U, V]^{\nu}\partial_{\nu}A_{\mu} + \partial_{\mu}(U^{\alpha})V^{\nu}\partial_{\nu}A_{\alpha} + U^{\alpha}\partial_{\alpha}(\partial_{\mu}(V^{\nu})A_{\nu}) + \partial_{\mu}(U^{\alpha})\partial_{\alpha} (V^{\nu})A_{\nu} -
$$
$$
- \partial_{\mu}(V^{\alpha})U^{\nu}\partial_{\nu}A_{\alpha} - V^{\alpha}\partial_{\alpha}(\partial_{\mu}(U^{\nu})A_{\nu}) - \partial_{\mu}(V^{\alpha})\partial_{\alpha}(U^{\nu})A_{\nu}. \qquad (.2)
$$
But I don't know how to make $(.2)$ equal to $(.1)$.
Edit.
By rewriting the second summand in $(.1)$ as
$$
\partial_{\mu}(U^{\nu})\partial_{\nu}(V^{\alpha})A_{\alpha} + \partial_{\mu}(U^{\nu})V^{\alpha}\partial_{\nu}A_{\alpha} + U^{\nu}\partial_{\nu}(\partial_{\mu}(V^{\alpha})A_{\alpha}) - \partial_{\mu}(V^{\nu})\partial_{\nu}(U^{\alpha})A_{\alpha} - \partial_{\mu}(V^{\nu})U^{\alpha}\partial_{\nu}A_{\alpha} -V^{\nu}\partial_{\nu}(\partial_{\mu}(U^{\alpha})A_{\alpha})
$$
I got
$$
[L_{[U, V]} - (L_{U}L_{V} - L_{V}L_{U})]A_{\mu} =
$$
$$
=\partial_{\mu}(U^{\nu})V^{\alpha}\partial_{\nu}A_{\alpha} - \partial_{\mu}(V^{\nu})U^{\alpha}\partial_{\nu}A_{\alpha} - \partial_{\mu}(U^{\alpha})V^{\nu}\partial_{\nu}A_{\alpha} + \partial_{\mu}(V^{\alpha})U^{\nu}\partial_{\nu}A_{\alpha},
$$
which isn't equal to zero.
AI: On scalars
$$L_V(L_U(\phi)):=v^{\rho}\partial_\rho u^{\nu}\partial_\nu\phi+
v^{\rho}u^{\nu}\partial^2_{\rho\nu}\phi; $$
$$L_U(L_V(\phi)):=u^{\rho}\partial_\rho v^{\nu}\partial_\nu\phi+
u^{\rho}v^{\nu}\partial^2_{\rho\nu}\phi=(\text{repeated indices})=\\
u^{\nu}\partial_\nu v^{\rho}\partial_\rho\phi+
u^{\nu}v^{\rho}\partial^2_{\rho\nu}\phi.$$
$$L_{[U,V]}(\phi):=[U,V]^{\rho}\partial_\rho \phi=
u^{\nu}\partial_\nu v^{\rho}\partial_\rho\phi
-v^{\nu}\partial_\nu u^{\rho}\partial_\rho\phi=(\text{repeated indices})=\\
u^{\nu}\partial_\nu v^{\rho}\partial_\rho\phi
-v^{\rho}\partial_\rho u^{\nu}\partial_\nu\phi$$
i.e. $L_U\circ L_V-L_V\circ L_U=L_{[U,V]}.$
On vector fields
$$L_{[U,V]}(A_\mu)=u^{\rho}\partial_\rho v^{\nu}\partial_\nu A_\mu-
v^{\rho}\partial_\rho u^{\nu}\partial_\nu A_\mu+
\partial_\mu(u^{\rho}\partial_\rho v^{\nu}-v^{\rho}\partial_\rho u^{\nu})A_\nu,$$
$$L_U(L_V(A_\mu)):=u^{\rho}\partial_\rho B_{\mu}+
\partial_\mu u^\rho B_\rho,$$
with $B_\mu=v^\nu\partial_\nu A_\mu+\partial_\mu v^\nu A_\nu$ and similarly for $B_\rho$. Similarly
$$L_V(L_U(A_\mu)):=v^{\rho}\partial_\rho C_{\mu}+
\partial_\mu v^\rho C_\rho,$$
with $C_\mu=u^\nu\partial_\nu A_\mu+\partial_\mu u^\nu A_\nu$ and similarly for $C_\rho$.
The identity on vector fields is then proven by a bit long but straightforward computation.
Edit: on the OP computations
Two little remarks. In the computations leading to .2) the last term in the r.h.s of the second equality has a wrong sign (it should be a minus).
The first term in the last line of the same sequence of computations, i.e. $-V^{\alpha}\partial_{\alpha}(U^{\nu}\partial_{\nu}A_{\mu}) $ should not be.
Removing such term, we arrive at
$$L_{[U,V]}(A_\mu)-L_U(L_V(A_\mu))-L_V(L_U(A_\mu))=
U^{\alpha}\partial_{\mu}V^{\nu}\partial_\alpha A_{\nu} - V^{\alpha}\partial_{\mu} U^{\nu}\partial_{\alpha}A_{\nu}+
\\
\partial_{\mu}U^{\alpha}V^{\nu}\partial_\nu A_{\alpha}
-\partial_{\mu}V^{\alpha} U^{\nu}\partial_{\nu}A_{\alpha};$$
the r.h.s. of the above equality is equal to $0$: one needs to exchange the repeated indices $\nu\leftrightarrow \alpha$ in the last 2 terms to finish the proof. |
H: Constructing a cochain complex out of a chain complex
Let $(C,\partial)$ be a chain complex where $C_i$ is an $R$-module ($R$ is a given ring) , we can always construct a cochain complex out of the chain complex $(C,\partial)$ in the following way: We construct the $i$th $R$-module of the cochain complex as $C^i=Hom_R(C_i,R)$ this is the $R$-module of $R$-module homomorphisms from $C_i$ to $R$. The $R$-module homomorphism $\delta_i:C^i\rightarrow C^{i+1}$ sends a morphism $f:C_i\rightarrow R$ to the morphism $f\circ \partial_{i+1}:C^{i+1}\rightarrow R$.
Question 1 : What if we take another functor other than $Hom_R(-, R)$ and get another cochain complex out of the original chain complex, do we obtain isomorphic quotients $ker/Im$ in the two cochain complexes?
Question 2 : Given a cochain complex $(C,\delta)$, how can we construct a chain complex out of $(C,\delta)$? I think we can still use the $Hom(-,R)$ functor to the cochain complex to get a chain complex, Is this correct? and why this question seems to be not interesting as i can't find anything about this converse construction, it seems like we always need to get a cochain complex out of a chain complex but not a chain complex out of a cochain complex?
AI: For Question 1, the answer is no. The simplest counterexample is to start with any chain complex whose homology is not trivial and to take the functor sending all modules to the zero module. A less trivial example is to take the functor $M\mapsto \text{Hom}_R(-,X)$ for some $R$-module $X$ other than $R$.
For Question 2, the answer is yes, you can go from cochain to chain complexes the same way. The reason people usually go from chain to cochain complexes is probably that in the original applications to topology, it was easy and natural to define the desired chain complexes directly (singular, simplicial, and cell complexes, for example) and then one needed a Hom construction to get cochain complexes. But in other situations, one can define a cochain complex directly (e.g., the deRham complex for a smooth manifold), and then one could use Hom to get a chain complex. |
H: Norm in $L^2$ goes to zero for function away from zero
Let $f\in L^2(\mathbb{R})$ and $M>0$. Define $f_M(x)=f(x)$ for $|x|\leq M$ and $f_M(x)=0$ for $|x|>M$. Show that $\|f_M-f\|_2\rightarrow 0$ as $M\rightarrow \infty$.
Well, we have $$\|f_M-f\|_2^2=\int_{|x|>M}|f|^2dx$$
Since $f\in L^2(\mathbb{R})$, we know that $$\int_{x\in\mathbb{R}}|f|^2dx<\infty$$ But why does this imply that $$\int_{|x|>M}|f|^2dx\rightarrow 0 \text{ as } M\rightarrow\infty?$$
AI: Because
$$\int_\mathbb{R} |f|^2 = \sum_{n=1}^\infty A_n\ ,$$
where $A_n = \int_{\{n< |x|\leq n+1\}} |f|^2$. Then $\int_{|x|>M} |f|^2 = \sum_{n=M}^\infty A_n \to 0$. |
H: $Z$ score probability
I was given a question where I was supposed to find the probability of obtaining $y$ between two scores, however when I input my answer it tells me that I'm wrong, the question is given below along with my answer to the question:
Question
For a normal distribution with sample mean $= -19$ and standard deviation $= 6.85$ find $p( -16.15 \leq y \leq -15.27 )$, where $y$ is a random draw from the normal distribution.
Round to $4$ decimal places.
My answer
I obtained the $z$ scores for both the $y$ values $-15.27$ and $-16.15$ and their
respective $z$ scores are $0.5445$ and $0.4161$. The probability under the curve with a $z$ score of $0.5445$ is $0.7054$ and the probability under the curve with a $z$ score of $0.4161$ is $0.6628$. With those two probabilities, I obtained the difference (getting $0.0426$) and I put in that answer and was told that I was wrong. Can someone give me insight as to what I'm doing wrong?
AI: The problem is your CDF table for the normal distribution only allows you to get a couple decimals, so you're saying $\Phi(.5445) \approx \Phi(.54) = .7054$ which (apparently) isn't a good enough approximation.
$\Phi(.5445)- \Phi(.4161) \approx .706951 - .661332 = .045619 \approx .04562$
I found these values using Wolfram Alpha. |
H: Prove that $\sqrt{x}$ is continuous on its domain $[0, \infty).$
Prove that the function $\sqrt{x}$ is continuous on its domain $[0,\infty)$.
Proof.
Since $\sqrt{0} = 0, $ we consider the function $\sqrt{x}$ on $[a,\infty)$ where $a$ is real number and $a \neq 0.$ Let $\delta=2\sqrt{a}\varepsilon.$ Then, $\forall x \in \mathit{dom},$ and $\left | x-x_0\right | < \delta \Rightarrow \left| \sqrt{x}-\sqrt{x_0}\right| = \left| \frac{x-x_0}{ \sqrt{x}+\sqrt{x_0}} \right| < \left|\frac{\delta}{2\sqrt{a}}\right|=\varepsilon.$
Can I do this?
AI: I think you mean a≠0. As you have it written now, you still have to show $\sqrt{x}$ is continuous on $[0,a)$, but you are on the right track. Consider when $x=0$, let $\delta = \epsilon^2$ and it follows. Then, for $x>0$, at every point $x_0 \geq x$, $\sqrt{x}>0$, so $\frac{1}{\sqrt{x}+\sqrt{x_0}}<1$, then your inequality follows. |
H: Elementary Properties of cyclic groups
Homework Problem from Group Theory:
Prove the following:
For any cyclic group of order n, there are elements of order k, for every integer, k, which divides n.
What I have so far..
Take G as a cyclic group generated by a. >>>> G=, a^(n)=e, where e is the indentity.
I know that if G is indeed cyclic, it must be generated by a single element, a. So, if another element, say b, of order k is in the group, it must be some multiple of n.
ord(b)=k
ord(a)=n
We are given that k divides n, which I know is the key here, but I am unsure how to start this.
I have tried
a^(n)=(b^(n))^k
which is then equal to
e=e^k, which is simply e.
I feel like I'm on the right track, but am missing something here.
AI: You've almost got it. Since $k | n$, there exists an $l$ for which $kl = n$. Now define an element by
$$h = g^l$$
Then we have
$$h^k = (g^l)^k = g^{lk} = g^n = e$$
so the order of $h$ is at most $k$. The order is at least $k$, and I'll leave that for you to show. So $h$ is the desired element. |
H: Using Plancherel formula to compute exponential integral
Let $c>0$. Use the Plancherel formula to compute the integral $$\int_{-\infty}^\infty \dfrac{1}{c^2+y^2}dy$$
We note that the Fourier transform of the function $f(x)=e^{-cx}$ for $x\geq 0$ and $0$ for $x<0$ is $\dfrac{1}{c+iy}$. (This follows from direct computation.)
The Plancherel formula says that $\|\hat{f}\|_2^2=2\pi\|f\|_2^2$. Writing out fully, this means $$\int_{-\infty}^\infty \hat{f}(y)^2dy=2\pi\int_0^\infty f(x)^2dx$$ Plugging in $f$ and $\hat{f}$, this yields
$$\int_{-\infty}^\infty \dfrac{1}{(c+iy)^2}dy=2\pi\int_0^\infty e^{-2cx}dx$$ I don't see how this helps with the original integral.
Edit.. thanks to Daniel Fischer's comment, it should be $$\int_{-\infty}^\infty \dfrac{1}{|c+iy|^2}dy=2\pi\int_0^\infty e^{-2cx}dx$$
The integral on the right is $\dfrac{1}{-2c}e^{-2cx}|_0^\infty = \dfrac{\pi}{c}$
AI: It's not just the square of $\hat{f}$, since $\hat{f}$ takes on complex values; you have to multiply by the complex conjugate: Recall that the $L^2$ inner product is given by $$(f, g) = \int f \overline{g}$$
So in this case, the complex conjugate is $1 / (c - iy)$, so we actually get
$$\int_0^{\infty} e^{-2cx} = \int_{-\infty}^{\infty} \frac{1}{(c - iy)(c + iy)} = \int_{-\infty}^{\infty} \frac{1}{c^2 + y^2}$$ |
H: Hemitian operator inequality
I am trying to find two Hermitian operators $A$ and $B$ (whose representations are $2 \times 2$ complex matrices) for which neither $A \leq B$ nor $A \geq B $ holds.
Note that $A \geq B$ iff $\langle(A-B)x, x \rangle\geq 0$ where $\langle~{,}~\rangle$ is an inner product.
Any help would be appreciated.
AI: How about:
$$ A = \left(
\begin{array}{cc}
1 & 0 \\ 0 & 0
\end{array}\right) $$
$$ B = \left(
\begin{array}{cc}
0 & 0 \\ 0 & 1
\end{array}\right)
$$ |
H: 15 distinguishable balls into five distinguishable boxes, Assume there is no restriction on which box gets one ball, which box gets two balls, etc.
I am preparing for an exam and I came across this problem. I am a little confused.
Give the expression of ways to distribute 15 distinguishable balls into five
distinguishable boxes so that the boxes have one, two, three, four, and five balls in
them, respectively. Assume there is no restriction on which box gets one ball, which
box gets two balls, etc.
How do I approach such a problem?
EDIT: Would the answer to this be $15!/(1!\cdot2!\cdot3!\cdot4!\cdot5!)$? I'm confused about the assumption provided at the end.
AI: Choose the lucky person who will end up with $5$ balls. This can be done in $\binom{5}{1}$ ways. Then choose the balls the person will get. This can be done in $\binom{15}{5}$ ways.
Now choose the person who will get $4$ balls. This can be done in $\binom{4}{1}$ ways. Choose the balls the person will get. This can be done in $\binom{10}{4}$ ways. And so on. We end up with
$$\binom{5}{1}\binom{15}{5}\binom{4}{1}\binom{10}{4}\binom{3}{1}\binom{6}{3}\binom{2}{1}\binom{3}{2}\binom{1}{1}\binom{1}{1}.$$
The last two entries are a bit superfluous!
Remark: A little simpler is to divide the balls into $5$ piles, with $5,4,3,2,1$ in the piles. We can write down the number of such divisions as a multinomial coefficient, or as $\binom{15}{5}\binom{10}{4}\binom{6}{3}\binom{3}{2}$. Then multiply by $5!$ for the ways to distribute the piles among the $5$ people.
The answer you provided probably used the sort of reasoning described above. For if you express the binomial coefficients in terms of factorials and simplify, you will find that $\binom{15}{5}\binom{10}{4}\binom{6}{3}\binom{2}{1}=\frac{15!}{5!4!3!2!1!}$. This answer should be multiplied by $5!$. |
H: Ring of continuous functions on $\mathbb{R}$, maximal ideal, quotient
Let $I(S) = \{f \in \mathcal{C}(\mathbb{R}) \ | \ \ \forall x \in S: f(x)=0\}$
I've already proven that it is an ideal in the ring $\mathcal{C}(\mathbb{R})$.
However, I have troubles proving that if $I(S)$ is maximal, then it must have the form $I(c), \ c \in \mathbb{R}$. The second problem is proving that $\mathcal{C}(\mathbb{R})/I(S) \cong \mathcal{C}(S)$. I know that an ideal $M$ of a ring $R$ is maximal $\iff$ $R/M$ is a field. But that doesn't help me here, does it?
AI: Since the zero set of a continuous function is closed, we have $I(S) = I(\overline{S})$ for every subset $S \subset \mathbb{R}$. Therefore, it is advisable (and common practice) to only consider closed $S$ for the vanishing ideals.
Now, it is easy to see that $S_1 \subset S_2 \Rightarrow I(S_2) \subset I(S_1)$. In particular, $I(S) \subset I(\{c\})$ for every $c \in S$. Since $I(\{c\}) \neq \mathcal{C}(\mathbb{R})$, if $I(S)$ is maximal, then we must have $I(S) = I(\{c\})$ for every $c\in S$. But if $c\neq d$, then $I(\{c\}) \neq I(\{d\})$, for example $x \mapsto (x-c)$ lies in $I(\{c\})$, but not in $I(\{d\})$. That means, a necessary condition for $I(S)$ to be maximal is that $S$ is a singleton set. That this condition is also sufficient is not hard to see.
It is, however, worth pointing out that not all maximal ideals of $\mathcal{C}(\mathbb{R})$ are of this form. |
H: Is there ever a requirement to change the limits of integration?
I don't have issues with doing integration problems, but occasionally I see the solution changing the limits of integration whenever a $u$-substitution is done.
I obviously don't have a problem doing this, and I just recently noticed my book doing this under the chapter involving "area of surface of revolution."
My question is did I develop a bad habit by never changing the limits of integration, or is it a best practice to always change limits of integration?
My teacher said on a test, and in general, if we do not change the limits of integration then we should be signifying this by labeling our limits of integration $x=$ lower-limit and $x=$upper-limit.
EDIT EXAMPLE INCLUDED
After further investigation, my confusion is because of the two below equations:
Find the exact area of the surface obtained by rotating the curve about the x-axis
$$y=\sqrt{1+4x}, 1\le x\le 5$$
The limits of integration were changed in the solution to this problem.
The given curve is rotated about the y-axis. Find the area of the resulting surface.
$$y=x^\frac{1}{3}, 1\le y \le 2$$
The limits of integration were NOT changed in the solution to this problem.
AI: Sometimes we are doing an indefinite integral by $u$-substitution, and then the form of integral involving $u$ is easily converted back into $x$ or whatever the original variable was at the end of the process.
However when a definite integral is involved, you have a choice of either converting the limits of integration from (say) $x$ limits to $u$ limits, or considering the $u$-substitution as a means to obtaining the final indefinite integral in terms of $x$ and using the original limits of integration.
The former has the advantage of skipping the substitution back into $x$, but at the cost of figuring out how to change the limits of integration into terms of $u$. This was a bad habit you learned, or more precisely, a useful habit you failed to learn. Converting the limits from $x$ to $u$ is ordinarily just a matter of using the $x$ limits in the expression for $u$ in terms of $x$. |
H: Is there a compact complex manifold with trivial $H_2$?
I don't believe that every complex manifold should have nontrivial $H_2$, otherwise we would easily prove the Chern's conjecture... But the problem is I don't have any counterexample. The Kähler manifold will have nontrivial $H_2$ and so do Riemann surface. Hence I guess there would be a complex surface with trivial $H_2$?
AI: Yes, consider the Hopf torus, $({\mathbb C}^2-0)/{\mathbb Z}$, where the generator of ${\mathbb Z}$ acts by a dilation, say, $(x,y)\to 2(x,y)$. |
H: Showing that $x^me^{-ax}$ is bounded
Let $a>0$, $m\geq 0$ and let $f:[0,\infty)\rightarrow\mathbb{R}$ be defined as $f(x)=x^me^{-ax}$. It should be true that this function is bounded, because near $0$ both terms are bounded, and far from $0$ the term $e^{-ax}$ decreases more rapidly than the term $x^m$ increases. What is a nice way to show this rigorously?
AI: Since
$$
\lim_{x\to \infty}f(x)=0,
$$
there exists $r>0$ such that
$$
f(x) \le 1 \quad \forall x \ge r.
$$
Thus for every $x \ge 0$ we have
$$
f(x) \le \max\{\max_{0 \le x \le r}f(x),1\}.
$$ |
H: understanding the proof of even-odd handshake problem
I´d like to know if I understand correctly the argument behind the even-odd handshake problem. Basically the theorem says that the number of persons who have shaken an odd number of hands is even.
My reasoning goes like this. Suppose that the number of people who have shaken an odd number of hands is odd. Then the expression for number of handshakes in that group is : 2a+1 + 2b+1 + 2c+1 + ... 2z+1 which has an odd number of terms. So we have an odd number of odd numbers and that gets us that the number of handshakes in the group is an odd number. But! We also know that a number of handshakes in any group has to be even (since every handshake gets counted twice.) Contradiction.
Btw I didn´t come up with the proof, I am just trying to understand a proof which I found on-line. Is my understanding sound?
AI: The total number of handshakes overall has to be even.
Suppose the number of people who have shaken an odd number of hands is odd. Odd times odd is odd. So the total handshakes in this group is odd.
The number of people who have shaken an even number of hands can odd or even. But the total hands shaken by them will always be even times even or even times odd = even.
Adding them up gives a contradiction that says "odd + even = even".
Edit: You need to consider the sum of hands shaken in both groups instead of just the "odd handshake group" because the "odd group" could have shaken hands with people outside of their group too and so is allowed to have shaken an odd number of hands.
i.e. It is not true that the total number of hands shaken by a subset of the group has to be even. |
H: Question on one theorem for uniform continuity.
In my text book, one theorem states this. A real-valued function $f$ on $(a,b)$, is uniformly continuous on $(a,b)$ if and only if it can be extended to a continuous function $g$ on $[a,b].$ And the book gives two examples. The first example says that the function $f(x) = x \sin (\frac{1}{x})$ for $x \in (0,\frac{1}{\pi}]$ is uniformly continuous because it can be extended to
$$g(x) = \begin{cases}
x \sin(\frac{1}{x}) & 0<x\leq \frac{1}{\pi}\\
0 & x=0.
\end{cases}$$
The second example is for the function $h(x)=\sin(\frac{1}{x})$, and it says that it can be extended to a closed interval but not uniformly continuous. And what does it mean a function can be extended to other function? Can anyone explain why is this true?
AI: In this example extending a function means that you can include the endpoints in the domain by defining values for the function at the endpoints as you did for $f$, extending the domain to include $a$ and $b$ ($0$ and $\frac{1}{\pi}$). For $g$ this works because you get continuity at $x=0$, and uniform continuity on the domain. For $h(x)$ this will not work. Consider what you would extend the function $h(x)$ to be at $x=0$. Or, what would you set $h(0)$ equal to to make it continuous at $x=0$? This is impossible. You can set $h(0)=0$, but then the function is not continuous at $0$. This relates to the topologist sine curve. As $x\rightarrow 0$, $h(x)$ oscillates infinitely rapidly so the limit is not well defined.
Regarding your question about extending a function, note we can define any function to anything we wish as long as it is well defined. For example, consider $q(x) = x$ on $[0,1]$. I can extend $q$ to also be defined on $(400,1200)$ by saying
$q(x) = \begin{cases}
x & x\in[0,1]\\
\sin(x+40\pi) & x\in (400,1200),
\end{cases}$
but you don't have continuity between the two intervals, only inside of them. |
H: homework combinatorics carousel
I was asked this question in my homework:
How many different combinations are there to paint a carousel with $n$ seats, in $r$ different colors, such that any combination that you can get via rotating is considered the same combination?
Example: $red -> green -> blue$ is the same arrangement as $blue -> red -> green$
a) first question n=9
b) second question n=15
c) last question n=7
What I did:
I know the answer for n=7 (not because i understand it, but because someone told me) and it's a special case because 7 is a prime number...
I would love someone to explain to my why 7 is a special case, why does it matter if n is prime or not, and help me solve the question for n=9 and n=15.
The answer for n=7 is $\frac{r^7-r}{7}+r$
AI: If it is not prime (for example 6) you can break in up into repeating sections: aka
$$a b c a b c$$
or
$$ababab$$
you can only do this for 7 if they are all the same color. Understanding this should help answer the homework problem.
I realise this might not be enough of a hint so I will add to it.
The number of ways to paint the $7$ seats $r$ colors is $r^7$. $r$ of these patterns are all of one color. For the $r^7-r$ patterns where there is no real rotational symmetry, you can define the patterns that matter by their starting seat and find that there are severn duplicates of each pattern when you rotate it. For the $r$ on colors each of these are identical. You therefore end up with:
$$\frac{r^7-r}{7}+r$$ patterns. When you change from 7 to other prime numbers you just change the 7 but if the number is not prime, you have to take into account patterns that have more complex rotation symetry that is not possible for 7. |
H: Show that if $a \sim b$, then $C(a) = C(b)$ where $C(x)$ is the equivalence class containing $x$.
I'm working through a textbook on my own, so I don't want the full answer. I'm only looking for a hint on this problem.
Show that if $a \sim b$, then $C(a) = C(b)$ where $C(x)$ is the equivalence class containing $x$.
I'm reading through Paul Sally's Tools of the Trade, and he gives this problem as an example in a theorem. An equivalence class is defined like this: If $a \in X$, then we write $C(a) = \lbrace b \in X \mid b \sim a \rbrace$. The proof is as "Transitivity" and that's it, but I'm not sure how to use transitivity to prove it. Here's what I have so far:
Let $a, b \in X$ be arbitrary elements of $X$. If $a \sim b$, then $a \in C(b)$ by definition. But, if $a \sim b$, then $b \sim a$ by symmetry, so $b \in C(a)$. This is where I get stuck. Since $a, b$ are arbitrary elements in $X$, can I just immediately conclude that $C(a) = C(b)$ for any $a,b \in X$ where $a \sim b$. I'm not sure how to use transitivity here.
AI: Let us introduce some more variety in the letters used. For $x \in X$, define $C(x)$ by:
$$C(x) = \{y \in X\mid y \sim x\}$$
Hence:
$$C(a) = \{y \in X \mid y \sim a\}, \quad C(b) =\{y \in X\mid y \sim b\}$$
What does it mean for these to be equal? How does transitivity come into play when proving this? |
H: What's the proper proof to show that H is a subspace of V?
Let $V$ be the vector space, which is defined as $$V=\left\{f \colon \mathbb{R} \to\mathbb{R} \right\}.$$ and let H be the set of functions which satisfy $f{(0)}=0$, i.e. $$H=\left\{ f\colon \mathbb{R} \to \mathbb{R}\colon f{(0)}=0 \right\}.$$ How would one prove that H is a subspace of V? I understand that you must prove that $H$ contains the zero vector, $H$ is closed under addition, and that $H$ is closed under scalar multiplication. But how would you prove this for ALL functions? Many thanks in advance.
AI: By definition of $H$ it contains the zero vector. Consider adding two functions $g(x)$ and $h(x)$ both in $H$. So $(g+h)(0) =g(0)+h(0)= 0 + 0$ because they are in $H$. I will leave it to you to show how it works for scalar multiplication. Because $V$ defines all functions from $\mathbb{R}$ to $\mathbb{R}$ and $H$ describes just those that meet one criteria, than the functions in $H$ are a subset of those in $V$ by definition, thus $H$ is a subspace of $V$. |
H: Ensuring positive definite matrix defined by variables in Matlab
I am working with a piece of Matlab where I have:
n = 4;
Mbar = zeros(n,n);
Mbar(1,1) = M;
Mbar(2,2) = Ixx;
Mbar(3,3) = Iyy;
Mbar(4,4) = Izz;
L = zeros(n,n);
L = chol(Mbar)';
An error message said that it didn't like M, Ixx, Iyy, and Izz. So I changed the code to:
syms M Ixx Iyy Izz;
assume(M>0);
assume(Ixx>0);
assume(Iyy>0);
assume(Izz>0);
M = vpa(Mbar(1,1));
Ixx = vpa(Mbar(2,2));
Iyy = vpa(Mbar(3,3));
Izz = vpa(Mbar(4,4));
L = zeros(n,n);
L = chol(Mbar)';
An error message was returned stating that Mbar must be positive definite for the Cholesky decomposition to be performed. Is there a way to define the variables in Mbar such that I am ensured that Mbar is positive definite?
AI: I really have no idea what you are trying to do. All of your code is out of order. I think that this is all that you need, though the result is trivial for a diagonal matrix:
syms M Ixx Iyy Izz;
assume(M>0);
assume(Ixx>0);
assume(Iyy>0);
assume(Izz>0);
Mbar = diag([M Ixx Iyy Izz]);
L = chol(Mbar)'
which returns
L =
[ M^(1/2), 0, 0, 0]
[ 0, Ixx^(1/2), 0, 0]
[ 0, 0, Iyy^(1/2), 0]
[ 0, 0, 0, Izz^(1/2)]
The reason the first case didn't work is because you were using Matlab's default numeric functionality and had not assigned any values to M, Ixx, etc. You got the error in the second case because you didn't create your Mbar matrix other than allocate it as all zeros. |
H: Is my proof correct for: $\sqrt[7]{7!} < \sqrt[8]{8!}$
I have to show that
$$\sqrt[7]{7!} < \sqrt[8]{8!}$$
and I did the following steps
\begin{align}
\sqrt[7]{7!} &< \sqrt[8]{8!} \\
(7!)^{(1/7)} &< (8!)^{(1/8)} \\
(7!)^{(1/7)} - (8!)^{(1/8)} &< 0 \\
(7!)^{(8/56)} - (8!)^{7/56} &< 0 \\
(8!)^{7/56} \left(\left( \frac{7!}{8!} \right)^{(1/56)} - 1\right) &< 0 \\
\left(\frac{7!}{8!}\right)^{(1/56)} - 1 &< 0 \\
\left(\frac{7!}{8!}\right)^{(1/56)} &< 1 \\
\left(\left(\frac{7!}{8!}\right)^{(1/56)}\right)^{56} &< 1^{56} \\
\frac{7!}{8!} < 1 \\
\frac{1}{8} < 1 \\
\end{align}
Did I do this properly? Is this way the best way or is there another much easier way?
Thanks a bunch!
AI: Your proof is correct, but perhaps a bit elaborate.
How about:
\begin{align}
&&\sqrt[7]{7!} &< \sqrt[8]{8!}\\
\iff&&(7!)^8 &< (8!)^7\\
\iff&&(7!)^7 7! &< (8\cdot 7!)^7\\
\iff&&7! &< 8^7
\end{align} |
H: Prove that the improper integrals are equal
Prove that
$$\int_0^{\infty} \frac{\cos{x}}{1+x} dx = \int_0^{\infty} \frac{\sin{x}}{(1+x)^2} dx$$
Things that I tried so far: I tried to create integral (0, infinity) cos x/1+x - sin x/(1+x)^2 and prove that it converges to 0 but it did not work. Another thing that I tried is change variables by x=pi/2 -t , and did not lead me anywhere.
hints..?
AI: All you need to do is integrate by parts:
$$\int_0^{\infty} dx \frac{\cos{x}}{1+x} = \underbrace{\left [ \frac{\sin{x}}{1+x}\right ]_0^{\infty}}_{\text{this}=0} + \int_0^{\infty} dx \frac{\sin{x}}{(1+x)^2}$$ |
H: Question on geometrically reduced, geometrically connected.
I have a question from a book which I am trying to attempt.
Let $k$ be a field not of characteristic 2 and let $a\in k$ be not a square (i.e. for all $b\in k$, $b^{2}\neq a$). I want to show that $X=\mbox{Spec}(k[U,T]/(T^{2}-aU^{2}))$ is geometrically reduced and geometrically connected.
My attempt is as follows: I am hoping to show that if $\overline{k}$ denotes the algebraic closure of $k$, then $X_{\overline{k}}$ is reduced. Since we have $X_{\overline{k}}=\mbox{Spec}(k[U,T]/(T^{2}-aU^{2})\otimes_{k}\overline{k})=\mbox{Spec}(\overline{k}[U,T]/(T^{2}-aU^{2}))$, and I need to show that this is reduced. To do this, I will have to show that localisation of $A=\overline{k}[U,T]/(T^{2}-aU^{2})$ at any prime ideal is a reduced ring. I was actually hoping to show that $A$ is a reduced ring (since then localisation of a reduced ring is also reduced), but algebraically I am not sure how to do it.
I have no idea how to start on proving geometrically connectedness. Glad if someone can give me some hints.
Thanks! (Will update this page if I have ideas on proving geometrically connectedness)
AI: a) Yes, the ring $A=\overline{k}[U,T]/(T^{2}-aU^{2})$ is reduced (and hence the scheme $X_\bar k$ is reduced) since the ideal $(T^{2}-aU^{2})\subset \overline{k}[U,T]$ is equal to its nilradical.
This is because in the prime factorization $T^{2}-aU^{2}=(T-\sqrt a U)(T+\sqrt a U)$ there are no repeated factors.
b) Geometric connectedness is clear geometrically (!) because $\bar X=\operatorname {Spec} A$ is the union of the two intersecting (connected !) lines $V(T-\sqrt a U), V(T+\sqrt a U) \subset \mathbb A^2_\bar k\operatorname ={Spec} (\overline{k}[U,T])$ . |
H: Show $|1-e^{ix}|^2=2(1-\cos x)$
Show $|1-e^{ix}|^2=2(1-\cos x)$
$$|1-e^{ix}||1-e^{ix}|=1-2e^{ix}+e^{2ix}=e^{ix}(e^{-ix}-2+e^{ix})=e^{ix}(2\cos x-2)=-2e^{ix}(1-\cos x)$$
Not sure how they got rid of the $-e^{ix}$ factor. Did I expand the absolute values wrong? thank you
AI: $$|1-e^{i x}|^2 = (1-e^{i x}) (1-e^{-i x}) $$
Then use $e^{i x}+e^{-i x} = 2 \cos{x}$.
Generally, $|z|^2 = z \bar{z}$. |
H: multiplying a non-square number to get a square
I have a theory that the only way how can I get a square from a non-square is to multiply it by some power of itself. For example 3 multiplied by 27 gives 81.
Is this always true? If yes, how would one go about proving that?
AI: No, it is not true: $$3 \cdot 12 = 36 = 6^2$$ but $12$ is not a power of $3$.
What can be said is this, as a consequence of unique factorization:
If $3s$ is a perfect square for some $s$, then it is necessary that $3^{2k + 1}$ is a divisor of $s$ for some $k$. That is, an odd power of $3$ divides $s$.
In the previous case, $k = 0$. |
H: Integral and Summation Exchange
Why is it possible to do the following?
$\int \space[\space \sum_{n=0}^\infty(-1)^n x^n \space]\space dx = \sum_{n=0}^\infty (-1)^n\int x^n dx$
I know that this is legal:
$\int \sum_{n=0}^\infty x^n dx = \sum_{n=0}^\infty\int x^n dx$
But I am not sure why the $(-1)^n$ can be taken out seperately.
AI: $(-1)^n$ is a constant as far as the integral is concerned, since the integral is with respect to $x$, while $(-1)^n$ has no relation to $x$, so it can be factored out. |
H: Unitary invariance
Why is it that for any non-negative matrix $M$ and unitary matrix $U$, we have
$$\sqrt{UMU^\dagger}=U\sqrt{M}U^\dagger$$?
This question has to do with Problem 2c from this sheet. I think I am allowed to assume the "fact" but I'd like to know why.
AI: Suppose $\sqrt{ U M U^\dagger } = B$. Then
$$
B^2 = U M U^\dagger \implies U^\dagger B^2 U = M \implies \left( U^\dagger B U \right) \left( U^\dagger B U \right) = M
$$
This implies
$$
\left( U^\dagger B U \right)^2 = M
$$
Now since $M$ is a positive definite operator, we can take a square root of both sides. We then find
$$
U^\dagger B U = \sqrt{M} \implies B = U \sqrt{M} U^\dagger
$$
Thus
$$
\boxed{ \sqrt{U M U^\dagger} = U \sqrt{M} U^\dagger }
$$ |
H: Inverse Function Theorem - Challenging question
I found the following question on the internet, and I think it would be useful to solve it as part as studying to my midterm:
Let $f:\mathbb{R}^n \to \mathbb{R}^n $ having continuous partial derivatives up to first order , such that:
$ \| f(x)- f(y) \| > \frac{1}{10} \|x-y\| $ for every $x,y\in \mathbb{R}^n $ such that $x\neq y$.
Prove that $J(f) \neq 0 $
Prove that $f$ sends $\mathbb{R}^n $ onto $\mathbb{R}^n $ .
Thanks in advance
AI: Part 1:
Since $f$ is differentiable at $x$, for all $\epsilon>0$ there is some $\delta>0$ such that for all $\|h\| < \delta$, we have $\|f(x+h)-f(x) -DF(x)h\| \le \epsilon \|h\|$. Choose $\epsilon = \frac{1}{20}$.
Then
\begin{eqnarray}
\|Df(x)h\| &=& \|DF(x)h - (f(x+h)-f(x))+(f(x+h)-f(x))\| \\
&\ge& \|f(x+h)-f(x)\| - \|DF(x)h - (f(x+h)-f(x))\| \\
&\ge& \frac{1}{20} \|h\|
\end{eqnarray}
And so $DF(x)$ is invertible and $J(f(x)) \neq 0$.
Part 2:
Let $Y= f(\mathbb{R}^n)$.
Since $DF(x)$ is invertible, the implicit function theorem shows that $f$ has a local inverse, and hence $f$ maps open sets into open sets. Hence $Y$ is open.
Suppose $y_n \in Y$, and $y_n \to y$. Since $y_n \in Y$, there is some $x_n$ such that $y_n = f(x_n)$. Since $\|y_n-y_m \| = \|f(x_n)-f(x_m)\| \ge \frac{1}{10} \|x_n-x_m\|$, we see that $x_n$ is Cauchy, and hence $x_n \to x$ for some $x$. Since $f$ is continuous, we have $f(x) = y$, and so $y \in Y$. This shows that $Y$ is closed.
Since $\mathbb{R}^n$ is connected, and $Y \ne \emptyset$, we have $Y = \mathbb{R}^n$. |
H: Relations in Discrete Math/ tables
Does anyone know how to make this table? I can do a table with normal values but the $x^2$ throws me off.
'Write the relation as a table, the relation $\mathbb{Z}$ on $\{1,2,3,4\}$ by $(x,y) \in \mathbb{Z}$ if $x^2\geq y$.'
Thanks!
AI: $$\begin{matrix}
xRy & 1 & 2 & 3 & 4 \\
1 & T & F & F & F \\
2 & T & T & T & T \\
3 & T & T & T & T \\
4 & T & T & T & T
\end{matrix}$$
You simply need to evaluate the truth value of each relation. $2R1\iff 2^2\geq1$, so $2R1$ is true. Comment if you have questions, there's nothing complicated about this. |
H: Is the set of rational numbers a vector space?
Is the set of all rational numbers $\mathbb{Q}$ a vectorspace?
I assumed no, because if
$$\vec{x} \in \mathbb{Q}$$
$$\pi \vec{x} \notin \mathbb{Q}$$ failing scalar mult closure
This was a question out of my linear algebra book, looking at the solutions says that it is a vectorspace. Am I making an incorrect assumption?
AI: Any question that asks if this or that set is a vector space is a wrong question. To be a vector space, one needs to specify a set, a field, an addition operation and a scalar product operation. Then one needs to check the axioms. Without that, the question is meaningless.
The intention was probably to ask whether $\mathbb Q$ is a vector space over the field $\mathbb Q$ with the ordinary notions of addition and multiplication, which it is. Your argument in the question shows that $\mathbb Q$ is not a vector space over $\mathbb R$ with the usual addition and multiplication. But, if the field is not specified, and the operations are not specified, then really the question is just meaningless. In particular, one can endow $\mathbb Q$ with infinitely many different vector space structures over infinitely many different fields. For that matter, the set $\mathbb N$ can also be endowed with infinitely many such vector space structures. |
H: Conjugacy classes of rotational symmetry tetrahedron
I am struggling with this question, as I don't know how to give a "short proof" as requested by our professor, as a non-graded exercise. Especially, since we have to do this without using any shapes, and just using our head!!
Problem: Let T be the tetrahedral rotation group. Give a short proof that the elements of order 3 are NOT all conjugate to each other in T.
Any help would be appreciated.
Thanks.
AI: Hint: Number of elements in a conjugacy class must divide the order of the group. |
H: Expected value of max(x, y)
We are given a square of unit length which has its centre at $(0,0)$ and its edges are parallel to the axes. How to find the expected value of $\max(x, y)$ where $(x,y)$ is a point in the square.
Edit:
Firstly I thought about just $x$ axis. $x$ axis varies in $[-0.5, 0.5]$. And $y$ will always be $0$. Therefore, $\max(x, y)$ in $-0.5 \le x \lt 0$ will be equal to $0$ and so will be the expected value. For $0 \le x \lt 0.5, \space \max(x, y) = x$. Therefore, expected value will be $\int_0^{0.5} x^2 dx $. But I don't know how to account for $2$ dimensional case.
AI: Draw a picture of the square and overlay the line $y=x$. Anything above $y=x$ has y greater than x and the problem is exactly symmetrical about this line. The value is, therefore, $$2\int_{-\frac{1}{2}}^{\frac{1}{2}} \int_x^\frac{1}{2} y dy dx$$ which would be divided by the area but the area is 1. I am looking for a way to do this without integrals but don't know one off hand. |
H: Riemann Integrable Functions Sequence
I cannot use the definition of limsup for this problem so I'm kind of stuck.
Let $f$ be continuous on $[a,b]$, let $f(x)\geq 0$ for $x\in [a,b]$, and let $M_n := \left(\int_a^b f^n\right)^{1/n}$. Show that $\mathrm{lim}(M_n)=\mathrm{sup}\{f(X): x\in [a,b]\}$.
My attempt: I thought about using the maximum minimum theorem and squeezing it between two functions but it didn't work.
Since $f$ is continuous there are elements $x^*, x_* \in [a,b]$ such that $f(x_*)\leq f \leq f(x^*)$. Then $f(x_*)^n(b-a)\leq \int_a^b f^n \leq f(x^*)^n(b-a)$ or $f(x_*)(b-a)^{1/n}\leq \left(\int_a^b f^n\right)^{1/n} \leq f(x^*)(b-a)^{1/n}$. This is where I'm stuck. Should I try a different approach? Thanks for the assistance.
AI: Note that you can assume without loss of generality that $f(x^*)=1$.
Fix $\varepsilon>0$. Since $f(x^*)=1$, there exists $\delta>0$ with $\delta<1/2$ and $f(x)>1-\frac\varepsilon2$ on $(x^*-\delta,x^*+\delta)$. Then
$$
\left(\int_a^bf^n\right)^{1/n}\geq \left(\int_{x^*-\delta}^{x^*+\delta}f^n\right)^{1/n}\geq(1-\frac\varepsilon2)(2\delta)^{1/n}
$$
As you say, now using $\limsup$ would be very useful. But we can do it this way: fix $n_0$ such that $(2\delta)^{1/n}>1-\varepsilon/2$ for all $n\geq n_0$. Then, for all $n\geq n_0$,
$$
\left|1-\left(\int_a^bf^n\right)^{1/n}\right|=1-\left(\int_a^bf^n\right)^{1/n}\\\leq1-(1-\frac\varepsilon2)(2\delta)^{1/n}=1-(2\delta)^{1/n}+\frac\varepsilon2(2\delta)^{1/n}<\frac\varepsilon2+\frac\varepsilon2=\varepsilon
$$ |
H: If $A^2+A=0$,then $\lambda=1$ cannot be an eigenvalue of A.
Prove the following statement:
If $A^2+A=0$,then $\lambda=1$ cannot be an eigenvalue of A.
I've been struggling on this question for a couple of hours and don't know how to approach it.
AI: If there were an eigenvector $v$ of $A$ with eigenvalue $1$, we'd have
$$(A^2 + A)v = A A v + Av = Av + v = v + v = 2v$$
On the other hand,
$$(A^2 + A)v = 0v = 0$$ |
H: Integration of $c(y^2)(1-y)^4$
Could anyone please help with integrating $f(y)=cy^2(1-y)^4$? Where $c$ is a constant.
AI: Hint: Substitute $u = 1 - y$, giving
$$-c\int (1 - u)^2 u^4 du$$
Now expand $(1 - u)^2$ and integrate term-by-term. |
H: Limit of $\text{ex} (n;P)/ \binom n2$ for the Petersen graph
This question is linked to
For a graph $G$, why should one expect the ratio $\text{ex} (n;G)/ \binom n2$ to converge?
where an argument was given that this specific ratio converges for $n\rightarrow\infty$. I would like to find the specific limit in the case where the graph $G$ is the Petersen graph $P$, so want to find $\lim_{n\rightarrow \infty}\text{ex} (n;P)/ \binom n2$
I do not know if there is an explicit formula for $\text{ex} (n;P)$ but if you have any ideas I would be thankful.
AI: I think it can be followed from the Erdős–Stone Theorem.
Let $T_r(n)$ denotes the Turán graph, it is a 2-partite Turán-graph with $n$ vertices. Petersen graph has chromatic number 3, i.e the graph cannot be colours with 2 colours $\Rightarrow$ $P\subseteq T_2(n)$ $\forall n\in\mathbb N\rightarrow t_2(n)\le ex(n;P)$ where $t_r(n)$ is the number of edges of $T_r(n)$. Now for sufficiently large $t$ is has to be true that $P\subseteq K_2(t)$ where $K_2(t)$ denotes the complete 2 partite grpah with t vertices.
$\Rightarrow$ $\forall \epsilon >0$ the Erdős–Stone Theorem gives us $ex(n;K_2(t))<t_2(n)+\epsilon n^2$
$\Rightarrow$ For large $n$ we get $t_2(n)/\binom n2\le ex(n;P)/\binom n2\le ex(n;K_2(t))/\binom n2\le t_2(n)/\binom n2+\epsilon n^2/\binom n2\le t_2(n)/\binom n2+4\epsilon$
Now $t_{r-1}(n)/\binom n2$ converges to $\frac{r-2}{r-1}$ and we set $r=3$, i.e the chromatic number of the Petersen grpah it follows that $\lim_{n\rightarrow\infty} ex(n;P)/\binom n2=\frac{3-2}{3-1}=1/2$
In my argumentation we can replace the chromatic number $3$ by any other chromatic number which yields to $\lim_{n\rightarrow\infty} ex(n;G)/\binom n2=\frac{\gamma(G)-2}{\gamma(G)-1}$ where $\gamma(G)$ is the chromatic number of $G$ |
H: Number of ways of sorting distinct elements into 4 sets
This was on a test I just had. The first part says:
"A person donates nine antique clocks to four different museums. Supposing all clocks are identical and he can distribute them in any way he chooses, how many ways are there of donating the clocks"?
This is choosing elements with replacement which has a well known formula of $n+r-1\choose r$ where $n=9$ and $r=3$ in our case. The next part is what has me stumped:
"Now suppose each clock is different. In how many ways can they be donated"?
He can still distribute them any way he chooses, so the first museum could get 8, the second 1 and the others none. I still can't figure out how to do this. I thought one would first want to choose an arrangement of the 9 objects of which there are 9! such arrangements. But then I'm not sure how to account for all the possible number of elements that each museum could get.
AI: Every clock goes to one of four museums, so there are $4$ choices for the first clock, $4$ for the second,... for a total of $4^9$ in all by the rule of product for combinatorics. |
H: An application of Fubini-Tonelli
Let $f$ be a nonnegative measurable function which is finite $\mu$ almost everywhere. Suppose that $\mu(E_t) < \infty$ for all $t>0$, where $E_t = \{x: f(x)>t\}$. Let $\lambda$ be another measure, where $\lambda((a,b]) = \mu(E_a) - \mu(E_b)$ for all $a<b<\infty$. It is true that $\lambda$ is a Lebesgue-Stieljes measure though it is not necessary to prove (though the proof is straightforward - right continuity follows from continuity from above which is applicable since $\mu(E_t)$ is finite.
Prove that $\int f\,d\mu = \int_{(0,\infty)} \mu(E_t)\,dm(t) = \int_{(0,\infty)} t\,d\lambda(t)$.
Here $m$ is Lebesgue measure.
I was able to successfully prove the first equality in the case where $f$ is a simple function, and unsuccessfully do anything else. I wasn't able to conclude anything about $f$ in general because it's not clear how any convergence theorem would be applied. So maybe what I need to do is somehow prove the second equality in general and then use it with a convergence theorem?
Any tips appreciated. I will also post a solution if I find one.
Thanks.
AI: For the first integral, you have to use Tonelli:
$$
\int_{(0,\infty)}\mu(E_t)\,dm(t)=\int_{(0,\infty)}\int_{E_t}1\,d\mu(s)\,dm(t)=\int_{\mathbb R}\int_0^{f(s)}1\,dm(t),d\mu(s)=\int_{\mathbb R}f(s)\,d\mu(s).
$$
The only difficulty here is the change of the limits. The region we are considering is initially described as
$$
0\leq t<\infty,\ \ s\in E_t\, (\mbox{i.e. }f(s)>t);
$$
so we re-think of this as
$$
s\in\mathbb R,\ 0\leq t< f(s)
$$
For the second equality we use the same kind of trick:
$$
\int_{(0,\infty)}\mu(E_t)\,dm(t)=\int_{(0,\infty)}\lambda((t,\infty))\,d\lambda(s)\,dm(t)\\
=\int_{(0,\infty)}\int_{(t,\infty)}1\,d\lambda(s)\,dm(t)=\int_{(0,\infty)}\,\int_{(0,s)}\,dm(t)\,\lambda(s)=\int_{(0,\infty)}\,s\,d\lambda(s).
$$
Here again we are writing the region $0\leq t<\infty$, $t<s$ as $0\leq s<\infty$, $0\leq t<s$. We also used that $f$ is finite almost everywhere, as this implies that $\mu(E_b)\to0$ as $b\to\infty$; this implies that $\lambda((t,\infty))=\mu(E_t)$. |
H: Prove anti-symmetric-ness of partial ordered set in lattice.
Prove anti-symmetric-ness of partial ordered set in lattice.
Definition:
If $(A, \le_{A})$ is a lattice and $C$ is a set, $([C \rightarrow A], \le)$ is also a lattice.
And $\rightarrow$ is defined as follows:
$f \le g$ if and only if for any $c \in C$, $f(c) \le_{A} g(c)$
Such lattices are defined as point wise.
And for $([C \rightarrow A], \le)$ as point wise lattice, we can drop $_{A}$.
Then I need to prove that $\le$ of $([C \rightarrow A], \le)$ is anti-symmetric.
And below is my proof based on this posting.
Let $f, g \in [C \to A]$ be arbitrary
such that $f \leq g$ and $g \leq f$.
Then by definition $f(c) \leq_A g(c)$
4, and $g(c) \leq_A f(c)$ for any $c \in C$.
Since $A$ is a lattice, it is a partially ordered set
, therefore anti-symmetric.
We have $f(c), g(c) \in A$
We have $f(c) \leq_A g(c)$ and $g(c) \leq_A f(c)$
, thus $f(c) = g(c)$ for any $c \in C$ by the anti-symmetric-ness of $A$
This means $f = g$
Therefore, $\le$ of $([C \rightarrow A], \le)$ is anti-symmetric.
Is it correct?
Thanks in advance.
AI: In this answer I assume $[C\to A]$ denotes the set of functions from $C$ to $A$.
Note that if $C$ is the empty set, then $[C\to A]=\varnothing$ and thus $\left([C\to A],\leq \right)$ can't be a lattice. Despite this, $\leq$ comes out trivially anti-symmetric.
The rest is essentially correct. On step 8. the justification should be anti-simmetry instead of transitivy. |
H: Find the Least Prime Divisor of $2^{17}-1$
Show that the least prime divisor of $2^{17}-1$ is $2^{17}-1$ itself.
This question is really anoying. Let $N=2^{17}-1$. What I know is that if $q\mid N$, then $q=34k+1$ for some $k \in \Bbb{N}$ and $q \equiv \pm1 \pmod 8$. Therefore,
$$q \equiv 2k+1 \equiv \pm1 \pmod 8$$
We get $k \equiv -1, 0 \pmod 4$. So $q=136x+1$ or $q=136x-33$ for some $x \in \Bbb{N}$.
This is all I know about finding the prime divisor of the form ${2^{p}-1}$, which might be the Mersenne number. But do I have to just substitute each possible value of $q$ into $N$ in order to show that it ultimately doesn't have any prime divisors other than $N$ itself? It takes so much time and effort. And I don't think the problem is just to make students tediously substitute each case one by one(if it is, then I'm so frustrated).
So the question: is there any more time-saving way to distinguish certain types of Mersenne number without goint too far from my basic understading of Number Theory? Or should I just subsitute one by one and that's the best choice I can take?
Thanks!
Edited This was a question that baffled me about few months ago, and after studying the case for perfect numbers, now I do know that $2^{17}-1$ is a Mersenne number(related to the perfect number $8589869056$). But I still think just memorizing the case doesn't give me full insight into this area.
AI: Let $p|2^{17} - 1$, then we have $2^{17} \equiv 1 \pmod p$, but we also have from Fermat's Little Theorem $2^{p-1} \equiv 1 \pmod p$.
From Lagrange Theorem for group order we get that $2^{\gcd(p-1,17)} \equiv 1 \pmod p$.
Obivously the greatest common divisor is $17$, because otherwise (if the gcd were $1$) then $2 \equiv 1 \pmod p$, which is impossible.
Then we have $17 \mid p-1$ and because $p-1$ is even we have $34\mid p-1$. So we need to check all prime numbers of the form $p=34k + 1 \quad \forall k \in \mathbb{N}$ such that $p \in (1,\sqrt{2^{17} - 1}]$. Which reduces the number of divisors to check. Actually there are just $4$ primes in this inteval, those are $103, 137, 239, 307$. You can check that none of them divides $2^{17} - 1$, implying that $2^{17} - 1$ is prime.
Also you can check how Euler proved that $2^{31} - 1$ is prime, and apply the same method on $2^{17} - 1$.
Actualy $2^{17} - 1 = 131071$ isn't that big number, even without knowledge in number theory it isn't so much hord work to do.
And at last as other users have suggested use Lucas-Lehmer Test. |
H: Which step is wrong in this proof
Proof: Consider the quadratic equation $x^2+x+1=0$. Then, we can see that $x^2=−x−1$. Assuming that $x$ is not zero (which it clearly isn't, from the equation) we can divide by $x$ to give
$$x=−1−\frac{1}{x}$$
Substitute this back into the $x$ term in the middle of the original equation, so
$$x^2+(−1−\frac{1}{x})+1=0$$
This reduces to
$$x^2=\frac{1}{x}$$
So, $x^3=1$, so $x=1$ is the solution. Substituting back into the equation for $x$ gives
$1^2+1+1=0$
Therefore, $3=0$.
What happened?
AI: Up to $x^3=1$ everything is fine. This allows you to conclude that $x\in \{z\in \Bbb C\colon z^3=1\}$. Since $\{z\in \Bbb C\colon z^3=1\}=\left\{\dfrac{-1 + i\sqrt 3}{2}, \dfrac{-1 - i\sqrt 3}{2},1\right\}$, then $x$ is one of the elements of this set.
You made a reasoning consisting of logical consequences, not one of logical equivalences. That's why you can tell that $x\in \{z\in \Bbb C\colon z^3=1\}$, but you can't say which one is it.
See this for a similar issue.
An even more simpler version of your mistake is this: suppose $x^2=1$, then $x=1$.
You can convince yourself that this is wrong and that you did the same thing in your question. |
H: Writing the identity permutation as a product of transpositions
I am reading Introduction to Abstract Algebra by Keith Nicholson and ran into a lemma that states:
If the identity permutation $\varepsilon$ can be written as a product of $n \geq 3$ transpositions, then it can be written as a product of $n - 2$ transpositions.
So let us look at $S_4$. So $\varepsilon = (1234) = (12)(23)(34)$ which is the product of 3 transpositions and thus satisfies the criteria for this lemma. So I then tried to write $\varepsilon$ as the product of 1 transposition and I did not see how this was possible. What am I missing here?
Thanks.
AI: If you're using cycle notation in your post (which it seems you are), then the element $\alpha=(1234)$ is not the identity permutation ($\alpha$ has order $4$) - The identity permutation in $S_4$ is given by $\varepsilon=(1)(2)(3)(4)$. |
H: A normed space of continuous functions with norm $\int_{0}^{1}|f(t)|dt$ is not complete
Suppose $E$ is a normed space of all continuous functions on $[0,1]$ with norm $\int_{0}^{1}|f(t)|dt$. Prove that $E$ is not complete
I know that we must do is to find a Cauchy sequence of continous functions that doesn't converge in $E$, but I can't find that sequence. Can any one help me? Thanks.
AI: Consider
$$f_n=\begin{cases}
0 & \text{if}\> 0\le x<1/2\\
n(x-1/2) & \text{if}\> 1/2\le x < 1/2 + 1/n\\
1 & \text{otherwise}
\end{cases}$$
Then each $f_n$ is continuous and $\langle f_n\rangle$ is Cauchy but it does not converge at any continuous function. |
H: Relationship between group actions and homomorphisms
I know that there exist no nontrivial homomorphism from $S_3$ into $Z_5$ as they are groups of co-prime order. I am not looking for an explanation of this but for an explanation concerning the obvious misunderstanding I have between group actions and homomorphisms. As to me $S_3$ acting on $Z_5$ does not seem trivial.
This is how I have been thinking of group actions.
$\begin{array}{ l | c |c|c|c|r } &0 & 1 & 2 &3 &4\\id&1&1&2&3&4\\(12)&0&2&1&3&4\\(13)&0&3&2&1&4\\(23)&0&1&3&2&4\\(123)&0&2&3&1&4\\(213)&0&3&1&2&4 \\ \end{array}$
The left hand column represent an element in $S_3$ and the upper row are element of $Z_5$. How does this now relate to homomorphism? And why is the homomorphism trivial?
AI: You have defined a successful group action on the set $\{0,1,2,3,4\}$, but this does not give a homomorphism to $\mathbb Z_5$. To do this, you would need to assign an element of $\mathbb Z_5$ to each element of $S_3$. So for example if your homomorphism is called $\phi\colon \mathbb S_3\to \mathbb Z_5$, then $\phi((12))$ would have to be an element of $\mathbb Z_5$. Say $\phi((12))=k$. But then $0=\phi(id)=\phi((12)^2)=2k$. But the only element of $\mathbb Z_5$ satisfying $2k=0$ is $0$. Hence $\phi((12))=0$. You can argue similarly for all the other elements of $S_3$. |
H: Decide the dimension of maximal ideals
Let $A=\mathbb C[x,y]/(y^2-x^3,y^5-x^3)$. I want to know the dimension of each maximal ideal over $\mathbb C$. Actually I can't decide it's maximal ideal. And how to decide its dimension?
AI: To find the maximal ideals, use the Nullstellensatz. If $I$ is an ideal in a polynomial ring over an algebraically closed field, then each point of the common zeros of $I$, $\mathcal Z(I)$, is in bijective correspondence with the maximal ideals containing $I$ (equivalently, maximal ideals in the quotient by $I$). |
H: Find a polynomial $p$ of degree $3$ if its value in $4$ points is given
Find a polynomial $p$ of degree $3$ such that
\begin{align*}
p(−4) &= −142, \\
p(1) &= −2, \\
p(−5) &= −242, \\
p(4) &= 10.
\end{align*}
Then use your polynomial to approximate $p(2)$.
\begin{align*}
p(x) &= ?, \\
p(2) &= ?.
\end{align*}
I can't find this sort of question in the textbook so I'm having trouble.
Please teach me how to solve this question and, perhaps, fill in point 2?
Thank you.
AI: You general cubic has the form $p(x) = ax^3 + bx^2 + cx + d$.
Plugging in the four points will give you four equations in the four unknowns $\{a,b,c,d\}$.
For example, $p(1) = -2 \Rightarrow -2 = a + b + c + d$.
Solve that system of equations however you know best! |
H: Solve for an Undetermined Coefficent by using annihilator method.
The problem is
I know that we need to find and
There are three roots and at least two of them are complex.
Edit: Sorry I forgot the D = 1
The is supposed to be . The + right next to the sinx is just a typo.
Now for the . Since there is a on the right hand side of the equation, we need to use . Do I expand the equation to take three derivatives? I know that there's going to be a lot of product rule executions. Is there an easier way to do this problem? I've heard about the annihilator method, but I don't know how to use it. I could do variation of parameters, but it produces too many overwhelming variables.
This is what I got so far for the first derivative of Yp. I figure if I put the a and b aside for a bit and take the product rule for the terms, it may be easier.
This is a screenshot of my first derivative and my second derivative of Yp
AI: The equivalent system is:
$$y''' - 3 y'' + 4 y' -2y = e^x \cos x$$
The homogeneous solution gives us $m^3 - 3m^2 + 4 m -2 = 0 \rightarrow m= 1, 1 \pm i$, so:
$$y_h(x) = e^x(c_1 + c_2 \cos x + c_3 \sin x)$$
Now, we notice that we have an $e^x \cos x$ term in the particular solution, we multiply by an $x$ term, so we choose:
$$y_p(x) = x~e^x(a \cos x + b \sin x)$$
We now form $y''' - 3 y'' + 4 y' -2y = e^x \cos x$ and solve for the constants.
Yes, you need to form the first, second and third derivative, it is a bit of algebra, but there are also many cancellations.
Update
$y'_p = e^x x (b \cos x - a \sin x) + e^x (a \cos x + b \sin x) +
e^x x (a \cos x + b \sin x)$
$y''_p = 2 e^x (b \cos x - a \sin x) + 2 e^x x (b \cos x - a \sin x) +
e^x x (-a \cos x - b \sin x) + 2 e^x (a \cos x + b \sin x) +
e^x x (a \cos x + b \sin x)$
$y'''_p = 6 e^x (b \cos x - a \sin x) + 3 e^x x (b \cos x - a \sin x) +
e^x x (-b \cos x + a \sin x) + 3 e^x (-a \cos x - b \sin x) +
3 e^x x (-a \cos x - b \sin x) + 3 e^x (a \cos x + b \sin x) +
e^x x (a \cos x + b \sin x)$
Now, you have to form $y_p''' - 3 y_p'' + 4 y_p' -2y_p$
Spoiler
$$y(x) = e^x (c_1 + c_2 \cos x + c_3 \sin x -\dfrac{1}{2} x~ \cos x)$$
Notes:
$(1)$ Variation of parameters may actually be easier to use.
$(2)$ You can see some examples of the Annihilator Method
$(3)$ If you want an alternate method to undertermined coeeficients or the Ann Method, see example $2$ of AN ALTERNATIVE METHOD FOR THE UNDETERMINED COEFFICIENTS AND THE ANNIHILATOR METHODS |
H: Find a plane parallel to a line and perpendicular to $5x-2y+z=3$.
Find a plane parallel to the following line: $$x=3t+2, y=-t+1, z=t-1$$ and perpendicular to:
$$5x-2y+z=3.$$
I've tried the following:
The normal vector to the plane above is <5,-2,1>.
The normal/direction vector to line is <3, -1, 1>.
I did the dot product on these two normal vectors, and I got 18. Now I'm not sure where to go from here to get the equation of the plane I'm supposed to be finding.
AI: If a plane is given by $Ax + By + Cz = D$, then the direction of the vector normal to a plane is $\begin{bmatrix} A \\ B \\ C\end{bmatrix}$.
If a line is given by $x = M_1t + B_1, y = M_2t + B_2, z = M_3t + B_3$, then the direction of the line is $\begin{bmatrix} M_1 \\ M_2 \\ M_3\end{bmatrix}$
A line is parallel to a plane (I'm assuming this is what you mean) if it never intersects the plane or always intersects the plane. That means that the normal vector of the plane and the direction of the line are perpendicular.
Two vectors, $\begin{bmatrix} x_1 \\ y_1 \\ z_1\end{bmatrix}$ and $\begin{bmatrix} x_2 \\ y_2 \\ z_2\end{bmatrix}$, are perpendicular if and only if $x_1x_2 + y_1y_2 + z_1z_2 = 0$. |
H: Help with a proof of Analysis (subsequences)
I have a bad time with the next problem. I'd appreciate any help.
Problem: Let $(a_n)_{n=0}^{\infty}$ be a sequence which is not bounded. Show that there exists a subsequence $(b_n)_{n=0}^{\infty}$ of $(a_n)$ such that $(1/b_n)\rightarrow0$.
Scratchwork:
Claim 1: The set $A_j: =\{\, n \in \mathbb{N}: |a_n|\ge j\,\}$ is not empty. Let $j\in \mathbb{N}$ be given. Since $(a_n)$ is not bounded, then there is some $N (j)$ so that $|a_n|\ge j$, in particular $|a_N|\ge j$. Thus $N \in A_j$ which shows that is non-empty.
Claim 2: The function $n: \mathbb{N}\rightarrow \mathbb{N}$, $j\mapsto \text{min} (A_j)$ is well define. This follows immediately from (1), because each $A_j$ is not empty, and the uniqueness of the minimum.
Claim 3: The function $n$, is an ordering preserving map, i.e., $n(j)< n(j+1)$.
Here is where I'm stuck. I can show $\,n(j) \le n(j+1)$ but not $n(j) \not= n(j+1)$
Claim 4: $|a_{n(j)}|\le |a_{n(j+1)}|$. We argue for contradiction, suppose $|a_{n(j)}|>|a_{n(j+1)}|$. So, we have $|a_{n(j)}|>j+1$, and then $n(j)\in A_{j+1}$, which contradicts that $n(j+1)$ is the minimum of $A_{j+1}$, since $n(j) < n(j+1)$ by (2). By induction we can show that $|a_{n(j)}|\le |a_{n(\ell)}|$, whenever $n(j)< n(\ell)$.
We set $b_j:= a_{n(j)}$
Claim 5: $(1/b_j)\rightarrow 0$. Let $\epsilon>0$ be given. Then there is some $N$ such that $|a_n|\ge \left\lfloor{1/\epsilon}\right\rfloor+1$ for infinitely many $n\ge N$. So, if $j \ge N$ then $n(j)\ge N$, which means $|a_{n(j)}|\ge \left\lfloor{1/\epsilon}\right\rfloor+1$ and since $|a_{n(\ell)}|\ge|a_{n(j)}|$ for each $n(j)<n(\ell)$. Thus $|b_j|\ge \left\lfloor{1/\epsilon}\right\rfloor+1$, for each $j\ge N$ and hence $1/|b_j| \le \epsilon$, which proves that $(1/b_j)\rightarrow 0$, as desired.
Using this definition of the function $n$ is possible to show that conserve the strict inequality (because this is the hint which the author gives)? Thanks.
Here are some changes:
We define $n$ recursively as follows: Let $n_1:= \text{min} (A_1)$ and $n_{j+1}:= \text{min } (A_{j+1}-\{n_{k}\})$.
We claim that $n(j) < n(j+1)$. Suppose for the sake of contradiction that $n(j) > n(j+1)$ (notice that $n(j) \not= n(j+1)$ otherwise we reached a contradiction ). Since, $n(j+1) \in A_{j}$ (this is because $|a_{n(j+1)}|\ge j+1>j$), $n(j+1)< n(j)$ contradicts the minimality of $n(j)$.
We thus have $|a_{n(j)}|\ge j$, which implies $|1/a_{n(j)}|\le 1/j$ and so $(1/a_{n(j)})\rightarrow 0$. Setting, $b_j:= a_{n(j)}$ we're done.
AI: You are doing too much work.
Let $n(1) = \min A_1$, $n(k) = \min A_k \cap \{n(k-1)+1,...\}$. Then you always have $n(k+1) > n(k)$, and $|a_{n(k)}| \ge k$.
Then you have $| \frac{1}{a_{n(k)}} | \le \frac{1}{k}$, and so $\lim_{k \to \infty} \frac{1}{a_{n(k)}} = 0$.
Let $b_k = a_{n(k)}$. |
H: Is there a named theorem for the fact that two points define a line, and three points define a quadratic function?
In particular, is there a theorem stating the fact that a polynomial function of degree d is defined by d+1 points?
I'm asking because I want to use this fact in a different proof but I want to be able to cite a theorem for it if one exists.
AI: This is known as the unisolvence theorem. |
H: Normalization of Orthogonal Polynomials?
The generalized Rodrigues formula (Hassani, Mathematical Physics, p. 174) is of the form
$$p_n=K_n\frac{1}{w}\left(\frac{d}{dx}\right)^n(wp^n)$$
The constant $K_n$ is seemingly chosen completely arbitrarily, and I really need to be able to figure out a quick way to derive whether it should be $K_n = \tfrac{(-1)^n}{2^nn!}$ in the case of Jacobi polynomials (which covers by extension the Legendre, Chebyshev and Gegenbauer polynomials), $K_n = \tfrac{1}{n!}$ for Laguerre polynomials, and $K_n = (-1)^n$ for Hermite polynomials. The best I have so far is actually working out the $n$th derivative of $(wp^n)$ in the case of Legendre polynomials, but that method becomes crazy with any of the other polynomials, and as Hassani says the choices are arbitrary so they probably don't work. My question is, how can I derive the constants without any memorization, whether by some nice trick or by the method one uses to arbitrarily choose their values. I'd really appreciate some help.
AI: There is, in general, no general mnemonic for the $K_n $. Those constants are fixed by convention and such conventions are different between the big families of classical OPs, which reflects the different situations they're used in. No instructor should expect you to memorize them. For problem solving, or as a working research phycisist, the usual is to have a copy of Abramowitz & Stegun, or simply the DLMF, to refer to for the precise form of the normalization factors in such formulae.
Edit in response to comment:
To be clear, this is not doable. The conventions are chosen independently for the different families and respond to different pressures; there is no overarching scheme. What Szegö is doing in your quote is reducing the Rodrigues formula to the form $\text{const}\times P_n^{(\alpha,\beta)}$ and then fixing the normalization using the fact that
$$
P_n^{(\alpha,\beta)}(1)=\begin{pmatrix}n+\alpha\\n\end{pmatrix},
$$
which is equally arbitrary.
Have a look at the table with the definitions of the different families in the DLMF (the notation for which is here) and at the table of special values. Some are normalized to a constant norm (e.g. Chebyshev), some are normalized to constant leading coefficient (e.g. Hermite $\rm{He}_n$, Chebyshev), some are normalized to a constant value at $x=0$ or $x=1$ (e.g. Legendre). Others are normalized due to other considerations (like the Hermite $H_n$, which are normalized so $H_n(x)e^{-x^2/2}$ is an eigenfunction of the Fourier transform).
What you want is impossible. If you want the normalization constants at your fingertips, memorize them the hard way, or have the DLMF or the NIST Handbook or Abramowitz and Stegun at your fingertips. |
H: Prove there exists a natural number $n$ such that $p_1+n(p_2-p_1)$ is prime but $p_2+n(p_2-p_1)$ is not, where $p_1 < p_2$ are primes
I am told $n$ should be taken to be the smallest possible n such that $p_2 + n(p_2 - p_1)$ is composite without actually picking a specific value for n and then somehow reach a contradiction, but I cannot for the love of god figure out what to do.
AI: $\textbf{Hint:}$ $p_1 + n(p_2 - p_1) = p_2 + (n-1)(p_2 - p_1)$. |
H: How to tell if an integral can be integrated (has an elementary anti-derivative)?
When dealing with improper integrals I sometimes have to figure out whether or not to use the comparison test. Everything I read says something along the lines of "Also, there will be some integrals that we simply won’t be able to integrate and yet we would still like to know if they converge or diverge."
How do I actually figure out whether or not I can integrate without actually trying to integrate and failing?
AI: In full generality this is an extremely hard problem, in fact it is impossible.
In practice, by learning lots of integration techniques and integration criteria you develop better intuition and have more tools to answer the question. Also, knowing many examples of non-integrable functions helps in quickly spotting more such functions. But, there is not general method. |
H: Find the maximum and minimum values of $A \cos t + B \sin t$
Let $A$ and $B$ be constants. Find the maximum and minimum values of $A \cos t + B \sin t$.
I differentiated the function and found the solution to it as follows:
$f'(x)= B \cos t - A \sin t$
$B \cos t - A \sin t = 0 $
$t = \cot^{-1}(\frac{A}{B})+\pi n$
However, I got stuck here on how to formulate the minimum and maximum points. Any explanation would be appreciated.
AI: Let $\displaystyle C = \sqrt{A^2 + B^2}$.
Then $\displaystyle A \cos t + B \sin t = C \left(\frac{A}{\sqrt{A^2+B^2}} \cos t + \frac{B}{\sqrt{A^2+B^2}} \sin t\right)$, and we can visualize a right triangle with opposite $A$ and adjacent $B$ to notice that we can find an angle $\phi$ such that
$\displaystyle \frac{A}{\sqrt{A^2+B^2}} = \sin \phi$
and
$\displaystyle \frac{B}{\sqrt{A^2+B^2}} = \cos \phi$.
This means that $\displaystyle A \cos t + B \sin t = C (\sin \phi \cos t + \cos \phi \sin t) = C \sin (\phi + t)$. Now we can easily see that the maximum is $C = \sqrt{A^2+B^2}$ and the minimum is $-C = -\sqrt{A^2+B^2}$.
And of course, the maximum is reached when $\displaystyle \phi + t = \frac{\pi}{2}$, and the minimum when $\displaystyle \phi + t = \frac{3\pi}{2}$. |
H: A question regarding $\,3 \times 4$ matrices
Good day, I'm currently studying for an exam and need to learn about matrices too. Well, since I'm not good at English I'll just write what I've done so far. Below is a photo showing the full sheet of paper with the steps I did so far.
The thing I'm wondering about is, that I'd have four unknown variables $(w, x, y, z)$ after writing down the equations, but my matrix has got 3 rows only. Therefor I'm wondering how I'm supposed to find a generic solution for A*x = b
$$A=\begin{bmatrix}0&2&-2&3\\ 1&3&-1&2\\ 2&3&1&0 \end{bmatrix} , b=\begin{bmatrix}-1\\ -2\\ 0 \end{bmatrix}$$
$$A=\begin{bmatrix}1&3&-1&2&-2\\ 0&1&-1&3/2&-1/2\\ 0&0&0&1/2&-3/2 \end{bmatrix} $$
$$z=-3 $$
$$\operatorname{rank}(A)=3 $$
So I actually solved for $z$ but I can't seem to solve for the other unknown variables (or can I?). The thing that confuses me is the fact this matrix is $3 \times 4$ now instead of $3 \times 3$. Any tips would be very much appreciated!
Click image with middle-mouse-button for full size.
AI: We can write this as an augmented matrix:
$$\begin{bmatrix}0&2&-2&3&-1\\ 1&3&-1&2&-2\\ 2&3&1&0&0 \end{bmatrix}$$
Using Gaussian Elimination (Row-Reduced-Echelon-Form), we arrive at:
$$A=\begin{bmatrix}1&0&2&0&12 \\ 0&1&-1&0&-8\\ 0&0&0&1&5 \end{bmatrix}$$
This means our solution can be written as:
$z = 5$
$x = -8 + y$
$w = 12 - 2y$
$y$ is a free variable and this system has an infinite number of solutions.
You can easily test this solution by picking random values for $y$, substituting the values back into the system and validating equality. |
H: Change of variables to derive Fourier series
Let $f\in C^{\infty}(\mathbb{R})$ be a periodic function of period $2L$. Define $$a_n=\dfrac{1}{2L}\int_{-L}^Lf(x)e^{-in\pi x/L}dx$$ Show by change of variables that $$f(x)=\sum_{n=-\infty}^\infty a_ne^{i\pi nx/L}$$
I'm quite confused about what "change of variables" refers to here.
Edit: Following mathematician's hint: the sum can be rearranged as
$$\dfrac{1}{2L}\sum_{n=-\infty}^\infty \left(\int_{-L}^Lf(y)e^{in\pi (x-y)/L}dy\right)$$
But what can we do next? How can we get that this above thing is equal to $f(x)$?
AI: Start with $\sum_{-\infty}^\infty (\frac{1}{2L}\int_{-L}^L f(y)e^{-in\pi x/L} dy ) e^{i\pi nx/L}$ with the goal of showing this equals $f(x)$. The use of two different variables is necessary because the $y$ is being "integrated away" to get the constant $a_n$. Change of variables may be used once you move the infinite sum and $e^{i\pi nx/L}$ inside the integral (which you can do by linearity plus some convergence theorem e.g. the dominated cvgce theorem). |
H: Clean characterization of the matrix of a linear transformation
Does anyone have a clean characterization of the matrix of a linear transformation? I would like one that is concise and clean. This is what I have, although I am not sure if it is correct:
Matrix of the transformation T with respect to B,C:
The matrix M such that $T(x)=Mx$ $\forall x$ in the domain and such that the columns of M are the coordinate vectors of the image of the transformation with respect to C.
Thanks.
AI: Technically, the statement $T(x) = Mx$ doesn't make sense, since the former is an element of a general vector space and the latter is a vector in $R^m.$
To be more precise, you could say that if $T:V \rightarrow W$ is a linear map between spaces of dimension $n,m$ respectively, then $M$ is the unique $m \times n$ matrix such that $[T(x)]_C = M[x]_B$ for every $x \in V$, where $[ \cdot ]_B$ is the coordinate isomorphism in the basis $B.$ |
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