Proof of Chain Rule and Intuition about what it Does I looked up a proof of the chain rule here https://web.williams.edu/Mathematics/lg5/A37W12/Chain.pdf

which made sense from a computational standpoint. However, I do not understand what intuitive details about the process of the chain rule or why it works, from this proof.

Let $h(x) = f(g(x))$. Intuitively, $h'(x)$ is a measure of how much $h$ changes per unit $x$ for a small change in $x$:

A small change $\delta x$ in $x$ causes the change $g'(x) \delta x$ in the value of g(x). The change in the value of $h(x) = f(g(x))$ as a result of the change in $g(x)$ is then given by $f'(g(x))$ multiplied by the change in $g(x)$ which was $g'(x) \delta x$. So the total change in $h$ is given by $f'(g(x))g'(x) \delta x$.

Divide both sides by $\delta x$ and take the limit of $\delta x \to 0$ (where the approximations become exact) and you have your chain rule: $h'(x) = f'(g(x)) g'(x)$.

Here is a proof of the chain rule. If $h(x)=f(g(x))$ then:
$$h(x + Δx) = f(g(x + Δx))$$
The gradient equation is of course:
$$f(x + Δx) = f(x) + f'(x)Δx$$
where $f(x)$ is any smooth function. Therefore we can say:
\begin{align*}
h(x + Δx) &= f[g(x) + g'(x)Δx]\\
&=f(g(x)) + f'(g(x))g'(x)Δx
\end{align*}
This is a direct consequence of the gradient equation i.e. it applies to both finite differences and calculus itself without any other treatment e.g. taking the standard part (or limit) or neglecting infinitesimals (which is unusual). Consequently $Δx→dx$, so substituting and rearranging:
$$\frac{h(x + dx) - h(x)}{dx} = f'(g(x))g'(x)$$
Therefore, the chain rule is:
$$h'(x) = f'(g(x))g'(x)$$
The above functional form can be directly converted to the differential form:
$$\frac{dh}{dx} = \frac{df}{dg}\frac{dg}{dx}$$
Note that the first numerators on the two sides are different. Usually they are written the same, but that may be a case of a bad convention. However, the equation still balances - $df$ is the differential as a function of $g$. This is implied by the denominator $dg$ so it doesn't have to be stated. If we want to express the LHS in terms of $f$ it should be:
$$\frac{d(f(g))}{dx} = \frac{df}{dg}\frac{dg}{dx}$$

EDIT: This was the first answer.


Find all natural number $n$ for which $3^9+3^{12}+3^{15}+3^n$ is a perfect cube. Find all natural number $n$ for which $3^9+3^{12}+3^{15}+3^n$ is a perfect cube.

What I have tried.

$3^9+3^{12}+3^{15}+3^n$

$=3^9(757+3^{n-9})$

Let $757+3^{n-9}=a^3$

Taking modulo $3$:

$a^3\equiv 1 \pmod 3$

$\implies a\equiv 1 \pmod 3$

Also, $a^3>757>729=9^3$

$\therefore a =10+3k$

$a=10$ gives $n=14$

Now, how can I know if there is any other $n>14$ satisfying the given condition.

PS: Please not use computer programmes to answer. I want pure mathematical solution.

Check manually if there exist any solution with $n \le 9$.
Now assume $n>9$, and look at the equation $3^m+757=a^3$ modulo 7 (with $m=n-9$):
$3^m+1 \equiv 3^m+757 \equiv a^3 \equiv \{0,-1,1\} \pmod 7$
$3^m \equiv \{0,-1,-2\} \pmod 7$
$n \equiv \{3,5\} \pmod 6$
If $m=6k+3$,
$ 757 = a^3-(3^{2k+1})^3 = (a-b)(a^2+ab+b^2)$
where $b=3^{2k+1}$. Factoring 757, you can check there is no solutions.
Else, if $m=6k+5$,
$757 = a^3-3^5(3^{2k})^3 = a^3-3^5b^3$
where $b=3^{2k}$. This is a Thue equation, effectively solvable:
using PARI/GP

tnf = thueinit(x^3-243)
thue(tnf, 757)
[[10, 1]]

you can check the only solution is $(a,b)=(10,1)$, hence $(a,n)=(10,14)$

Only long comment.

Let $n-9=x$, then we have equation $757+3^x=a^3$ for positive integers $a,x$.

$2\mid a$ and $a\equiv 1\pmod3$, then some prime $2<p\mid a$ and $p\equiv 2\pmod3$.

Find suitable primes $p$ and related $x$ using descrete logarithm, multiplicative order and CRT in pari/gp.

First, $x$ is odd, becose $8\mid a^3$, $ord_8(3)=2$ and $zlog_{3_{8}}(-757)=1$:

? m=Mod(3,8);
?
? h=znorder(m)
%1 = 2
?
? znlog(-757,m,h)
%2 = 1

gp-code for finding $p$ and $x$:

 forprime(p=5,100, if(p%3==2,
  m=Mod(3,p);
  h=znorder(m);
  z=znlog(-757,m,h);
  if(z,
   c=iferr(chinese(Mod(1,2),Mod(z,h)), E, 0);
   if(c,
    m=Mod(3,p^2);
    h=znorder(m);
    z=znlog(-757,m,h);
    if(z,
     c=chinese(c,Mod(z,h));
     m=Mod(3,p^3);
     h=znorder(m);
     z=znlog(-757,m,h);
     if(z,
      c=chinese(c,Mod(z,h));
      j=c.mod; c=lift(c);
      print("p = "p"    x = "c" + k*"j);
     )
    )
   )
  )
 ))

Output:

p = 5    x = 5 + k*100
p = 23    x = 8169 + k*11638
p = 29    x = 11719 + k*23548
p = 47    x = 28121 + k*101614
p = 71    x = 332919 + k*352870

I.e. if $23\mid a$, then $x=8169 + k\cdot 11638$, where $k$ is 0,1,2,....


Proof $f$ has a minimum if $f$ is continuous and $\lim_{x \rightarrow \infty}(f(x))=\infty=\lim_{x \rightarrow -\infty}(f(x))$ I have a question about the validity of this proof I wrote:
Claim: If $f$ is continuous on $\mathbb{R}$ and $\lim_{x \rightarrow \infty}(f(x))=\infty=\lim_{x \rightarrow -\infty}(f(x))$, then $f$ has a minimum.
My proof:
Lower Bound Theorem) If $g$ is continuous on $[a,b]$, then $\exists z \in [a,b]: \forall x \in [a,b]: f(z)\leq f(x)$
(1) $\forall N_1: \exists n_1>0: \text{if } x>n_1 \text{ then } f(x)>N_1$ (Definition)
(2) $\forall N_2: \exists n_2<0: \text{if } x<n_2 \text{ then } f(x)>N_2$ (Definition)
(3) Given $N_1, N_2$: Let $I = [n_2, n_1]$
(4) $\therefore f$ is continuous on $I$
(5) $\therefore \exists y \in I: \forall x \in I:f(y) \leq f(x)$ (Using Lower Bound Theorem)
(6) If $x \in \mathbb{R}$ but $x \notin I$, then $x < n_2$ or $x>n_1$.
(7) $\therefore f(x) > N_1$ or $f(x) > N_2$
(8) Choose $N_1, N_2: f(y)<N_1$ and $f(y)<N_2$
(9) $\therefore \forall x \in \mathbb{R}:f(y) \leq f(x)$
Thanks for reading. I'm not sure about line 8. I already defined $N_1$ and $N_2$ in line 3, and constructed $y$ based on that in line 5. So if I redefine $N_1$ and $N_2$ based on $y$ in line 8, that could potentially redefine $y$ in line 5, which could redefine $N_1$ and $N_2$ in line 8, which could redefine $y$ in line 5, etc. etc. I'm pretty sure either this is just a little technicality that can be fixed easily but I'm not sure how, or it's not a problem at all and I'm overthinking. Can someone help?

As an alternative, let consider the restriction $f:[a,\infty) \to \mathbb R$ then if $f(a)$ is not a minimum for the restriction by IVT $\exists b>a$ such that $f(b)=f(a)$ and $f(x)\ge f(b) \,\forall x\ge b$, then by EVT the restriction attains a minimum $f(c)$ at $c\in [a,b]$.
We can use the same argument for the restriction $f:(-\infty,a] \to \mathbb R$ to show that the restriction attains a minimum $f(c')$ at $c'\in [b',a]$.
Then $$\min (f(x))=\min(f(c),f'(c))$$

Let $x_0$ be some point in $\mathbb{R}$. Since $\lim_{x\to\pm\infty} f(x)=\infty$ and $f$ is  continuous, there exists an $M>0$ such that $$f(x_1)>f(x_0)$$ for any $x$ satisfying $|x|>M$. Since any minimizer $x^*$ of $f$ over $\mathbb{R}$ satisfies $f(x^*)\leq f(x_0)$, it follows that the set of minimizers of $f$ over $\mathbb{R}$ is the same as the set over $\mathbb{R}\cap B_{|\cdot|}[0,M]$, which is compact and non-empty. This is because both $\mathbb{R}, B_{|\cdot|}[0,M]$ are closed and $B$ is bounded and contains $x_0$. Therefore by Weierstrass Theorem for closed functions, there exists a minimizer of $f$ over $\mathbb{R}\cap B_{|\cdot|}[0,M]$ and hence also over $\mathbb{R}$.
P.S.
The above holds for every coercive function $f$ and closed non-empty set $S$.


How do I evaluate this trigonometric limit involving infinite product? Tried a lot but couldn't solve it. How do I begin approaching this limit?

$ \displaystyle\lim _{n\to\infty} \displaystyle\prod_{k=3}^{n} 1 - \tan^4\left( \frac{\pi}{2^k}\right)$

There isn't that many trick to create a horrible looking and yet doable limit.

In school, if you are asked to evaluate a series or product which involve trigonometry functions whose arguments contain terms like $\frac{(\cdots)}{2^k}$.  One thing you should do is lookup the double-angle formula for various trigonometry functions and see whether you can use them to turn your series/product into a telescoping one.

In this case, it do work.

Let $t = \tan\theta$, we have

$$1 - t^4 = (1-t^2)^2\frac{1+t^2}{1-t^2}
= \left(\frac{2t}{\frac{2t}{1-t^2}}\right)^2\frac{1+t^2}{1-t^2}$$

Recall
$$\frac{2t}{1-t^2} = \tan(2\theta)\quad\text{ and }\quad
\frac{1-t^2}{1+t^2} = \cos(2\theta) = \frac12\frac{\sin(4\theta)}{\sin(2\theta)}
$$

We obtain
$$1 - t^4 = 8\left(\frac{\tan(\theta)}{\tan(2\theta)}\right)^2\frac{\sin(2\theta)}{\sin(4\theta)}
=  8\frac{\tan^2\theta\sin(2\theta)}{\tan^2(2\theta)\sin(4\theta)}
$$
The product is telescoping and

$$\begin{align}\prod_{k=3}^n\left(1 - \tan^4\frac{\pi}{2^k}\right)
&= 8^{n-2} \frac{\tan^2\frac{\pi}{2^n}\sin \frac{\pi}{2^{n-1}}}{\tan^2\frac{\pi}{4}\sin\frac{\pi}{2}}\\
&= \frac{1}{32} \left(2^n\tan \frac{\pi}{2^n}\right)^2\left(2^{n-1}\sin\frac{\pi}{2^{n-1}}\right)\end{align}
$$

Using
$$\lim_{N\to\infty} N\tan\frac{\pi}{N} = \lim_{N\to\infty} N\sin\frac{\pi}{N} = \pi$$
We obtain
$$\lim_{n\to\infty} \prod_{k=3}^n\left(1 - \tan^4\frac{\pi}{2^k}\right) = \frac{\pi^3}{32}$$

I wouldn't quickly have an idea either, but WolframAlpha gives that $\prod_{k=3}^\infty\left(1-\tan^4\left(\frac{\pi}{2^k}\right)\right)\approx0.968946$ (https://www.wolframalpha.com/input/?i=product+of+(1-tan(pi%2F2%5En)%5E4)+from+3+to+infty).


Showing the equality of $\lim_{s \to 1^+}\int_{n}^{n+1}\frac1{x^s}dx = \int_{n}^{n+1}\frac1xdx$ I want to show that$\displaystyle\lim_{s \to 1^+}\int_{n}^{n+1}\cfrac{1}{x^s}dx = \int_{n}^{n+1}\cfrac{1}{x}dx$.

I find the left hand side as:
$\lim_{s \to 1^+}(s-1)(\cfrac{1}{n^s}-\cfrac{1}{(n+1)^s})$. However, if I am not wrong, the right hand side is equal to $\ln(n+1)-\ln(n)$.

Do I miss something? How can I show this equality?

My main goal is to understand the equality $ \lim_{s \rightarrow 1^+} \left( \zeta(s) - \frac{1}{s-1} \right) = \gamma$ that is asked at Limit of Zeta function.

Edit:
Thank you all for your valuable comments and answers.  We clearly have $ \lim_{s \rightarrow 1^+}  \frac{1}{s-1} =\infty$. Since  $\frac{1}{s-1}$ diverges, I can not seperate the limits. Now I would like to conclude that $ \lim_{s \rightarrow 1^+}  \zeta(s)   = \infty$. I need some hints because, unfortunately, I can not see how to get the result.

For $s>1$, an antiderivative of $x^{-s}$ is
$$
\frac{x^{-s+1}}{-s+1}
$$
so the integral on the left-hand side is
$$
\frac{(n+1)^{1-s}}{1-s}-\frac{n^{1-s}}{1-s}=
\frac{(n+1)^{1-s}-n^{1-s}}{1-s}
$$
(this shows you computed the integral wrongly).

Thus you need
$$
\lim_{s\to 1^+}\frac{(n+1)^{1-s}-n^{1-s}}{1-s}=
\lim_{t\to 0^-}\frac{(n+1)^{t}-n^t}{t}
$$
which is the derivative at $0$ of the function
$$
f(t)=(n+1)^{t}-n^t
$$
Since
$$
f'(t)=(n+1)^{t}\ln(n+1)-n^t\ln n
$$
you have
$$
f'(0)=\ln(n+1)-\ln n=\int_{n}^{n+1}\frac{1}{x}\,dx
$$

The easiest way to see this doesn't involve any fancy algebraic tricks like the paths you tried. Rather, since the argument of the integral approaches 'nicely' enough, so must its argument: in particular, here, the integrand is uniformly continuous and monotonic, which are both very nice properties for taking limits. Merely verifying that the integral is well-defined at all s>1 is then enough. I don't cite a particular theorem from real analysis because there are many different ones that could apply, depending on which properties you particularly choose to use.

I do find it easiest to imagine graphically, though: if you have the 'regions' in the xy plane for each of the s>1, then taking their (nicely measurable) intersection yields the region for the right-hand-side integral.


Proving the Bourbaki–Witt Theorem using Recursion. I am trying to prove the Bourbaki–Witt theorem, which states:

  Let $\left(X,\leq\right)$ be a partially ordered set and let
  $f:X\rightarrow X$ be a function. Suppose that $f\left(x\right)\geq x$
  for all $x\in X$ and that every chain in $X$ has a supremum.  Then for
  all $x\in X$, there exists an element $y\in Y$ such that $y\geq x$ and
  $f\left(y\right)=y$.

There is more than one way to prove this theorem. The method I want to use involves Hartogs Lemma (there exists an ordinal which does not inject to a given set) and transfinite recursion. The proofs I have seen start by recursively defining a function as follows:

This is from the following proofwiki article: Bourbaki–Witt theorem
.

First of all, I am uncomfortable with how they defined $g$. It's not rigorous enough for me. My goal is to define $g$ more rigorously/precisely using the recursion theorem. This is the version of the recursion theorem that I will use:

  Let $\alpha$ be an ordinal, let $Y$ be a non-empty set, and let $a\in Y$.
  Let $g:Y\rightarrow Y$ be a function and let
  $h:\wp\left(Y\right)\rightarrow Y$ be a function. Then there exists a
  unique function $\varphi:\alpha\rightarrow Y$ such that

  • $\varphi\left(\emptyset\right)=a$,

  •
  $\varphi\left(\beta^{+}\right)=g\left(\varphi\left(\beta\right)\right)$
  for all $\beta\in\alpha$ with $\beta^{+}\in\alpha$, and

  • $\varphi\left(\gamma\right)=h\left(\varphi\left[\gamma\right]\right)$
  for every limit ordinal $\gamma\in\alpha$.

By $\beta^{+}$, I mean the immediate successor of the ordinal $\beta$ and by $\wp\left(Y\right)$, I mean the power set of $Y$. Also, $\varphi\left[\gamma\right]$ is the set $\left\{ \varphi\left(\beta\right):\beta\in\gamma\right\} $.

Is this version of the Recursion Theorem sufficient to define $g$? I know that there are many different versions of the recursion theorem that may be more appropriate than this one. Now using this recursive definition, I will try to define the function $g$ used in the proofwiki article, which I will just label  $\varphi$:

  Let $x\in X$ and let $\alpha$ be an ordinal such that there does not exist an injection from $\alpha$ to $X$. Let
  $h:\wp\left(X\right)\rightarrow X$ be any function such that for all
  $A\in\wp\left(X\right)$, if $A$ is a chain in $X$, then $h\left(A\right)=\sup A$
   . By the Recursion Theorem above, there exists
  a unique function $\varphi:\alpha\rightarrow X$ such that

  • $\varphi\left(\emptyset\right)=x$,

  •
  $\varphi\left(\beta^{+}\right)=f\left(\varphi\left(\beta\right)\right)$
  for all $\beta\in\alpha$ with $\beta^{+}\in\alpha$, and

  • $\varphi\left(\gamma\right)=h\left(\varphi\left[\gamma\right]\right)$
  for every limit ordinal $\gamma\in\alpha$.

This is how I started the proof. Now evidently, the function $\varphi$ is apparently not the function required. I need to show that if $\gamma\in\alpha$ is a limit ordinal, then $\varphi\left[\gamma\right]$ is a chain in $X$. That will guarantee that $\varphi\left[\gamma\right]$ will have a supremum. I will then have obtained the function in the proofwiki article. This is the part that I am having trouble with. The proofwiki article stated that indeed this is the case and follows from the fact that $f\left(x\right)\geq x$
 for all $x\in X$. However, this is not at all obvious to me. How can I show that $\varphi\left[\gamma\right]$ is a chain in $X$ for every limit ordinal $\gamma$?

$f$ is known as inflationary iff for any $x, f(x)\geqslant x$. So it is obvious that $f$ inflationary implies $\varphi(\beta)$ increasing (as mentioned in proofwiki ) for
$$
\varphi(\beta^+)=f(\varphi(\beta))\geqslant \varphi(\beta)
$$
Also $\varphi\left[\gamma\right]$ is clearly a chain for $\gamma_1<\gamma_2$
$$
\varphi[\gamma_1]=\{\varphi(\beta):\beta<\gamma_1\} \subset \{\varphi(\beta):\beta<\gamma_2\}=\varphi[\gamma_2]
$$
If $\alpha$ is limit ordinal, then for $\gamma_1<\gamma_2<\cdots<\alpha$
$$
\varphi[\gamma_1]\subset\varphi[\gamma_2]\subset\cdots\subset\bigcup_{\gamma<\alpha}\varphi[\gamma]=\sup_{\gamma<\alpha}\varphi[\gamma]
$$
So we can define $\quad h(\varphi[\gamma])=\sup_{\gamma<\alpha}\varphi[\gamma]$.


Determine whether sequence of independent rvs satisfy SLLN Determine if the following sequences of independent rvs satisfy the SLLN:

  $(a)$ $P(X_n=1) = \frac{1}{n^2},\, P(X_n=0) = 1-\frac{1}{n^2}$
  $(b)$ $P(X_n=n) = \frac{1}{n^2},\, P(X_n=0) = 1-\frac{1}{n^2}$
  $(c)$ $P(X_n=n) = \frac{1}{n}, \, P(X_n=0) = 1-\frac{1}{n}$

My attempt: Let $S_n = X_1 + X_2+\ldots + X_n$.

$(a)$: $E(X_n) = \frac{1}{n^2} = E(X_n^2)$.  We need to verify: $\displaystyle \lim_{n\rightarrow \infty} \frac{S_n - E(S_n)}{n}\rightarrow 0$ almost surely. Now, $E(S_n) = \sum_{k=1}^{n} \frac{1}{k^2}$, so as $n\rightarrow \infty$, $E(S_n)\rightarrow \frac{\pi^2}{6}$, which implies $\frac{E(S_n)}{n}\rightarrow 0$. However, $S_n\leq n$ for every $n$, thus as $n\rightarrow \infty$, $\frac{S_n}{n} = \frac{n-k}{n} \rightarrow 1$, unless all $X_i$'s $=0$ (which is an extreme case, since $X_n$ are random). Thus $X_n$ does not satisfy SLLN.
$(b)$: $E(X_n) = \frac{1}{n}$ and $E(X_n^2) = 1$. Thus $E(S_n)$ diverges as $n\rightarrow \infty$, while $S_n = kn$ for some nonnegative integers $k$. Thus, we could apply the L'Hospital rule (variable is $n$) to compute $\lim_{n\rightarrow \infty} \frac{S_n - E(S_n)}{n} = \lim_{n\rightarrow \infty} (\sum_{k=1}^{n} \frac{1}{k^2}) = \frac{\pi^2}{6}\neq 0$, so $X_n$ does not satisfy SLLN.
$(c)$: $E(X_n) = 1$, so $E(S_n)=  n$. Thus, $S_n - E(S_n) = (k-1)n$ for some nonnegative integers $k$. Thus, $\lim_{n\rightarrow \infty} \frac{S_n - E(S_n)}{n} = k-1 = 0$ if and only if $k=1$. So, $\lim_{n\rightarrow \infty} \frac{S_n - E(S_n)}{n}$ does not converge to $0$ almost surely (is this correct?). Thus, $X_n$ does not satisfy SLLN.

My question: I'm quite skeptical with my solution above since the way I tried to express $S_n$ seems to be weird. Could anyone please help me with these problems in case my solutions above are incorrect? Any thoughts would really be appreciated.

The SLLN by Kolmogorov for independent sequence $\left(X_n\right)_{n\in \mathbb N}$, states that if

$Var(X_n)<\infty, \mathbb E[X_n]=μ_n$ for every $n$,
$\sum_{n=1}^\infty\dfrac{Var(X_n)}{n^2}<\infty$

then $\frac1n(S_n-\mathbb ES_n)\overset{a.s.}\longrightarrow 0$. These are sufficient but not necessary meaning, that every sequence that satisfies these (actually the second is the one to check) satisfies the SLLN, but if a sequence does not satisfy these, then this does not mean that it cannot satisfy the SLLN.

So, in your case:

$(a)$: $\mathrm{Var}[X_n]\le \mathbb E[X_n^2]=\frac1{n^2}$ and so $$\sum_{n=1}^{\infty}\frac{Var(X_n)}{n^2}\le\sum_{n=1}^\infty\frac1{n^4}<\infty$$
and so, it does satisfy the SLLN.
$(b)$: $\mathrm{Var}[X_n]\le \mathbb E[X_n^2]=1$ and so, as in $(a)$
$$\sum_{n=1}^{\infty}\frac{Var(X_n)}{n^2}\le\sum_{n=1}^\infty\frac1{n^2}<\infty$$
and so, it does satisfy the SLLN.
$(c)$: Here $Var(X_n)=n-1$, hence $$\sum_{n=1}^{\infty}\frac{Var(X_n)}{n^2}=\sum_{n=1}^\infty\frac{n-1}{n^2}=\infty$$
and so, this way does not work! (But as mentioned above, it does not mean that we are done). However, if we consider the independent events $E_n:\{ω:X_n(ω)=n\}$, we have that $$\sum_{n=1}^\infty P(E_n)=\sum_{n=1}^\infty\frac1n=\infty$$ and hence by the second Borel-Cantelli lemma: $$P(\lim\sup_nE_n)=P(X_n=n \text{ occurs i.o.})=1$$ This, however, still does not allow you to show that \begin{align}P\left(\lim_{n}\frac{S_n-\mathbb E[S_n]}{n}\neq 0\right)&=P\left(\lim_{n}\frac{S_n}{n}\neq 1\right)\\[0.2cm]&\ge P\left(\lim_{n}\frac{S_n}{n}> 1\right)= P\left(\lim\sup E_n\right)=1\end{align}
since there are unbounded sequences that are cesaro-summable.


Domain of definition. What is the difference between domain of a function $f$ and its domain of definition? I searched it in my books but did't got the answer. I am thinking like this

If $f:[0,1]\rightarrow\mathbb{R}$ be a function (say) $f(x)=x^{2}$ then domain of $f$ is $ [0,1]$ and its domain of definition is $\mathbb{R}$.

Please give me the right definition. Thanks in advance.

domain of f is referring to the specified given cases.For example in real life, we always have bound for different function.(populationzxac cost, profit)

domain of definition only lies on the definition. It is a general case ignoring the bound.


Use the definition of compact set to prove the uniform continuity on compact sets. The question is: Recall the theorem

  A function that is continuous on a compact set $K$ is uniformly continuous on $K$.

Provide a proof by the definition

  $K\subset R$ is compact if every open cover of $K$ has a finite subcover.

Can someone tell me how to solve this question?

This is a well known theorem, whose proof you can find in any analysis textbook, so I'll just give a sketch.

Suppose $f : A \to \mathbb{R}$ is a continuous function and that $A \subset \mathbb{R}$ is compact.

To use compactness, we need an open cover. Fix $\epsilon > 0$. For each $x \in A$, choose $\delta(x) > 0$ such that $y \in A$ and $|x-y| < \delta(x)$ implies $|f(x) - f(y)|< \epsilon$.

Define $N(x) = \{y \in A : |x - y| < \delta(x)\}$. These $N(x)$ form an open cover of $A$. Use compactness to choose a subcover $N(x_i)$. Take $\delta = \frac{1}{2} \min_i \delta(x_i)$. Suppose two points $x,y$ are $\delta$ close. Argue that $x$ is $\delta$ close to some $x_i$, and that $y$ has to be $\delta(x_i)$ close to that same $x_i$. What can you conclude about the distance between $f(x), f(y)$ and $f(x_i)$?


Deciding if a recursively enumerable language is finite A language $L \subseteq X^{\ast}$ is recursively enumerable if there exists a Turing machine enumerating the strings in $L$, here after a word from $L$ is written we do not know if it will be the last one, and the machine runs forever without printing anything new (or just repeats words already written). Equivalently a lanuage is recursively enumerable if there exists a Turing machine which accepts inputs, and halts if the input word is in $L$, and has an unspecified behaviour otherwise. Or similar if there exists a partial computable function $f : L \to \{1\}$.

  Given a recursively enumerable language, is there an effective way to decide if the language is finite?

I guess not, but how to prove this?

To try to determine whether a language is finite, we have to have some information about the language. So, to make the problem precise, it's necessary to specify how that information will be provided. Perhaps the most typical way to pose this sort of question is the following:

  Given a program $e$ for a Turing machine, is the language accepted by machine $e$ finite?

So we are asking whether that problem can be solved by a computable function:

  Is there a computable function $f$ such that, given a program $e$ for a Turing machine, $f(e) = 1$ if the language accepted by machine $e$ finite, and $f(e) = 0$ otherwise?

The answer to that is "no". There is a general result, Rice's theorem, which applies. It shows that, when we ask questions like this one about the language accepted by a Turing machine (i.e. not questions about the machine itself), no nontrivial property of languages can be decided by a computable function that takes as input the machine $e$.

The general proof method is by reducing the halting problem to the question above. If we want to tell whether a particular Turing machine $e$ halts when started on an empty tape, we can make another Turing machine $e'$ which accepts a finite language if and only if $e$ halts. The function taking $e$ to $e'$ is computable. So if $f$ above was computable, we could solve the halting problem by taking input $e$ and asking whether $f(e') = 1$. That would let us solve the halting problem with a computable function, which is impossible. We conclude that $f$ is not computable, because the map taking $e$ to $e'$ is.

You can feed all words of the language to the second version Turing machine and it will halt for each word. If the whole run is finished in finite time, the given language was finite.

The question is how to do this practically. We can not start a TM with some maybe infinite language as input. We rather need to remove an element from the set and repeat until the set is exhausted. This needs removal and test for emptiness. Or remembering the enumerated words and a means of comparison.

In the other view one could let the enumerating TM list all words of the language and compare with the given language. If all words of the given language are listed within finite time, the given language is finite.

Again, how to do this in practice?


Calculating pi manually Hypothetically you are put in math jail and the jailer says he will let you out only if you can give him 707 digits of pi. You can have a ream of paper and a couple pens, no computer, books, previous pi memorization or outside help.

What is the best formula to use? Where best will result in the least margin of error (most important) and is decently quick (second importance).

707 probably seems arbitrary, but I got it from here: http://en.wikipedia.org/wiki/William_Shanks

Also I don't really understand how he could use that formula because I thought you need a table to get arctan values or it would probably take a long time to make values of arctan to use in the formula, but that isn't too important.

I asked this question today because it celebrates the most accurate pi day for the next 100 years :)

As metioned in Wikipedia's biography, Shanks used Machin's formula
$$ \pi = 16\arctan(\frac15) - 4\arctan(\frac1{239}) $$

The standard way to use that (and the various Machin-like formulas found later) is to compute the arctangents using the power series

$$ \arctan x = x - \frac{x^3}3 + \frac{x^5}5 - \frac{x^7}7 + \frac{x^9}9 - \cdots $$

Getting $\arctan(\frac15)$ to 707 digits requires about 500 terms calculated to that precision. Each requires two long divisions -- one to divide the previous numerator by 25, another to divide it by the denominator.

The series for $\arctan(\frac1{239})$ converges faster and only needs some 150 terms.

(You can know how many terms you need because the series is alternating (and absolutely decreasing) -- so once you reach a term that is smaller than your desired precision, you can stop).

The point of Machin-like formulas is that the series for $\arctan x$ converges faster the smaller $x$ is. We could just compute $\pi$ as $4\arctan(1)$, but the series converges hysterically slowly when $x$ is as large as $1$ (and not at all if it is even larger). The trick embodied by Machin's formula is to express a straight angle as a sum/difference of the corner angles of (a small number of different sizes of) long and thin right triangles with simple integer ratios between the cathetes.

The arctangent gets easier to compute the longer and thinner each triangle is, and especially if the neighboring side is an integer multiple of the opposite one, which corresponds to angles of the form $\arctan\frac{1}{\text{something}}$. Then going from one numerator in the series to the next costs only a division, rather than a division and a multiplication.

Machin observed that four copies of the $5$-$1$-$\sqrt{26}$ triangle makes the same angle as an $1$-$1$-$\sqrt2$ triangle (whose angle is $\pi/4$) plus one $239$-$1$-$\sqrt{239^2+1}$ triangle. These facts can be computed exactly using the techniques displayed here.

Later workers have found better variants of Machin's idea, nut if you're in prison without reference works, it's probably easiest to rediscover Machin's formula by remembering that some number of copies of $\arctan\frac1k$ for some fairly small $k$ adds up to something very close to 45°.

A pretty simple method i would use if i had no access to books would be to just calculate the alternating sum of reciprocals of consecutive odd numbers and multiply by four, since
$$\sum_{k=0}^\infty\frac{(-1)^k}{2k+1}=\frac \pi 4$$
It would take a very very very long time though, but it is a fairly simple method.

Another method would be to make some kind of small, circular pool, measure the diameter, then pour water from a measure cup into the pool, keeping in mind the amount of water you have pured. Once the pool is filled to a certain height, measure the height and, since the volume of water is given by $h*d*\pi$, then with a big (diameter-wise) enough pool we could be able to work out the digits accurately enough. This approach is less time consuming but needs a lot more resources.


multivariable limit problem involving sine This is a homework problem, but I am having lots of trouble figuring this out. I know homework problems aren't well accepted on this website, but I couldn't find a better place to ask this.
$$\lim_{(x,y)\to (0,0)}\sin\left(\frac xy\right)(ax+by)$$

Is there a way to bring the $ax+by$ inside the sine? I'm not so good with trig. But, really the $a,b$ are throwing me off. I'm not even sure if they are integers or what. The book doesn't say anything.

Note that
$$-|ax+by|\leq \sin\left(\frac{x}{y}\right)(ax+by)\leq|ax+by|$$

Therefore, by the Squeeze Theorem, $$\lim_{(x,y)\to (0,0)}\sin\left(\frac {x}{y}\right)(ax+by)=0$$

$a$ and $b$ are arbitrary real numbers in your problem.

One way to approach this is to see what happens if you approach $(0,0)$ along some arbitrary line.  If the answer for the limit is not independent of the direction of that line, then the limit does not exist.  So write your line as $y=mx$ (no constant; it's got to go through the origin).  (We will have to deal with the case of the line $x=0$ separately to make our set of lines complete.)

Then you have $$f(x,y) = (ax+bmx)\sin \left( \frac{x}{mx} \right)
= (ax+bmx) \sin \left( \frac{1}{mm} \right)$$ which clearly approaches zero as $x \to 0$.

But now look at the line $x=0$, that is, approaching the origin along the $y$ axis.
$$f(0,y) = (by)\sin \left( \frac{0}{y} \right) $$ which of course is zero everywhere along that line.

So it looks like your limit is zero.


Exponential Random Variables QUESTION:
Let $X$ and $Y$ be exponentially distributed random variables with parameters $a$ and $b$ respectively. Calculate the following probabilities:
(a) $P(X>a)$
(b) $P(X>Y)$
(c) $P(X>Y+1)$
ANSWER:
(a) since $X$ is exponentially distributed, we have $P(X>a) = 1 - P(X \leq a) = 1 - (1 - e^{-a^2}) = e^{-a^2}$ .
(b) Not sure
(c) Not sure

Any hints/answers would help!!

ad a): As noted by others,$P(X>a)=1-P(X\leq a)$, if you have troubles with this use the expected value of the indicator functions on sets like $\{X>a\}$.

ad b) and c): The link below discusses the same problem. pdf of the difference of two exponentially distributed random variables

$$
\mathbb P(X>Y)
=
\frac b{a+b}ae^{-ax}
$$


How to define a bijection between $(0,1)$ and $(0,1]$? How to define a bijection between $(0,1)$ and $(0,1]$?
  Or any other open and closed intervals?

If the intervals are both open like $(-1,2)\text{ and }(-5,4)$ I do a cheap trick (don't know if that's how you're supposed to do it):
I make a function $f : (-1, 2)\rightarrow (-5, 4)$ of the form $f(x)=mx+b$ by
\begin{align*}
-5 = f(-1) &= m(-1)+b \\
4 = f(2) &= m(2) + b
\end{align*}
Solving for $m$ and $b$ I find $m=3\text{ and }b=-2$ so then $f(x)=3x-2.$

Then I show that $f$ is a bijection by showing that it is injective and surjective.

Choose an infinite sequence $(x_n)_{n\geqslant1}$ of distinct elements of $(0,1)$. Let $X=\{x_n\mid n\geqslant1\}$, hence $X\subset(0,1)$. Let $x_0=1$. Define $f(x_n)=x_{n+1}$ for every $n\geqslant0$ and $f(x)=x$ for every $x$ in $(0,1)\setminus X$. Then $f$ is defined on $(0,1]$ and the map $f:(0,1]\to(0,1)$ is bijective.

To sum up, one extracts a copy of $\mathbb N$ from $(0,1)$ and one uses the fact that the map $n\mapsto n+1$ is a bijection between $\mathbb N\cup\{0\}$ and $\mathbb N$.

Note that:

If A is a countable infinite set and a is any element in A, then: there's a bijection for A \ {a} and A;

Or, equivalently if A is contained strictly in B, b is in B \ A, A is countably infinite then: there's a bijection for A $\bigcup$ {b} and A;

To prove the second claim, for example: denote A = {an| n is in $N$}, let: b = f(a0), and an = f(an + 1) for n >= 0. then f is a bijection.

(Similar to a previous answer for your question)

If A is uncountable, you can choose a subset in A as A' that is countable, and do a similar thing to add element to A'. (similar proof for removing element from A, by choosing the element you wish to remove in A')

By this, you can add \ remove as many finite element to an infinite set A as possible, and still have a bijection between them. This is a famous question, called "Hilbert's Hotel", to the name of German Mathematician David Hilbert, you can see more on wiki.

(In fact, (1) if A is countably many, you can even add countably infinite elements to A, and there's still a bijection; and sometimes, you can also do this for removing (2) if A is uncountable, you can always do this for both add/remove a coutably many set to/from A)


Prove $e=\lim\limits_{x \to \infty}(1+\frac1x)^x$ The same exercise requires us to prove that $\lim_{x \to \infty}x\log(1+\frac1x)=1$ just before proving the identity in the title. I can use this fact as follows:
$$\lim_{x \to \infty}\left(1+\frac1x\right)^x=\lim_{x \to \infty}e^{x\log(1+\frac1x)}=e$$
Is this proof valid? I'm asking because the author is using a complicated $\varepsilon-\delta$ proof which could have been avoided, if I'm not mistaken.

If you have
$\log(1+x)
=\int_0^x \dfrac{dt}{1+t}
$,
then,
if $x > 0$,
$\log(1+x)
=\int_0^x \dfrac{dt}{1+t}
\lt\int_0^x dt
=x$
and
$\log(1+x)
\gt \int_0^x \dfrac{dt}{1+x}
= \dfrac{x}{1+x}
= \dfrac{x+x^2-x^2}{1+x}
= x-\dfrac{x^2}{1+x}
$
so
$0
\gt \dfrac{\log(1+x)}{x}-1
\gt -\dfrac{x}{1+x}
$
so
$\lim_{x\to 0}\dfrac{\log(1+x)}{x}
=1
$.
This is probably your proof.


Probability that there are exactly two wrongly addressed envelopes There are four Envelopes with letters. Two are chosen Randomly and opened and found that they are wrongly addressed. Find the Probability that there are exactly two wrongly addressed envelopes.

My Try: Let the Envelopes be $E_1$,$E_2$,$E_3$ and $E_4$ and Corresponding Letters be $L_1$,$L_2$,$L_3$ and $L_4$ Since two opened are found wrongly addressed,implies there are minimum of two wrongly addressed envelopes.So Favorable cases are :

$1$.There are Exactly two wrongly addressed Envelopes i.e.,Remaining two are correctly addressed and this can happen in $\binom{4}{2}=6$ ways

$2$.There are exactly three wrongly addressed envelopes i.e., remaining one has correctly addressed and this an happen in $\binom{4}{3}\left(3!-1\right)=20 $ways

$3.$There are exactly four wrongly addressed envelopes and this an happen in $\left(4!-1\right)=23 $ways, so Required Probability is

$\frac{6}{6+20+23}=\frac{6}{49}$. I am not sure whether i have done in a right way, please help me if i am wrong.

If two randomly chosen envelopes are wrongly addressed, then the other two envelopes can only be correctly addressed if the two persons who receive the randomly chosen envelopes can get the letter that is meant for them by switching. Person $A$ must have received the letter for person $B$ and vice versa. If not then person $C$ or $D$ will receive a letter meant for $A$ or $B$ and more than $2$ letters are wrongly addressed.

Let's say that order $\left(1,2,3,4\right)$ stands for a fully correct
addressing. The first two letters are wrongly addressed if
we are dealing with for instance $\left(2,4,.,.\right)$. Counting
tells you that there are $14$ of such cases. The first two letters
are 'switched' in the cases $\left(2,1,3,4\right)$ and $\left(2,1,4,3\right)$
and in only case $\left(2,1,3,4\right)$ there are exactly $2$ wrongly
addressed letters. If the first two letters are not switched then
there are more than $2$ wrongly addressed letters as argued above. We find
a probability of $\frac{1}{14}$ that there are exactly $2$ letters
are wrongly addressed under condition that the first two are wrongly
addressed.

In a pack of four envelopes,

one pair of envelopes can be wrongly addressed in only 1 way. But these two envelopes can be chosen from four envelopes in $^{4}C_{2}$ ways. Hence exactly two envelopes can be wrongly addressed in $^{4}C_{2} * 1 = 6$ ways.
one triplet of envelopes can be wrongly addressed in 2 ways. Suppose 1-2-3-4 is the correctly addressed sequence and supposing that envelope number 4 is always correctly addressed, the 2 ways of wrongly addressing the remaining envelopes are 3-1-2-4 and 2-3-1-4. But any randomly chosen three envelopes can be wrongly addressed. Hence exactly three envelopes can be wrongly addressed in $^{4}C_{3} * 2 = 8$ ways.
similarly, one quartet of envelopes can be wrongly addressed in 9 ways. Suppose 1-2-3-4 is the correctly addressed sequence, the envelopes can be wrongly addressed as 2-1-4-3, 2-3-4-1, 2-4-1-3 and so on. But four envelopes can be randomly chosen in $^{4}C_{4}$ ways. Hence exactly four envelopes can be wrongly addressed in $^{4}C_{4} * 9 = 9$ ways.

Therefore, the probability that there are exactly two wrongly addressed envelopes = $\frac{6}{6 + 8 + 9}$ i.e. 26.09$\%$.


Stopping Time Subset Proof My probability textbook has a really crappy proof for the following result.  Suppose $S$ and $T$ are stopping times, with $S(\omega) \le T(\omega)$ for all $\omega$.  Prove that $\mathcal{F}_S \subset \mathcal{F}_T$.  Can anyone give me a direct proof for this?

Take $A \in \mathcal{F}_S$.  Then $A \cap \{S = n\} \in \mathcal{F}_n$ for all $n$.  Since $\mathcal{F}_m \subset \mathcal{F}_n$ for all $m \le n$, we also have $A \cap \{S \le n\} \in \mathcal{F}_n$, we know $A \cap \{T = n\} = A \cap \{S \le n\} \cap \{T = n\} \in \mathcal{F}_n$.  Therefore, $A \in \mathcal{F}_T$ as well.

Starting from a probability space $(\Omega,\mathcal{F},P)$ with a filtration $\{\mathcal{F}_t\}$ we have the following definitions:

$$\mathcal{F}_S:=\{A\in\mathcal{F}:A\cap\{S\le t\}\in\mathcal{F}_t, \forall t\ge 0\}$$

The proof $\mathcal{F}_S\subset\mathcal{F}_T$ is straightforward. Let $A\in\mathcal{F}_S$ and $t\ge 0$. All we need to show is $A\cap\{T\le t\}\in\mathcal{F}_t,\forall t\ge 0$:

Since $T$ is a stopping time we know $\{T\le t\}\in\mathcal{F}_t$,$\forall t$. Hence

$$A\cap\{T\le t\}=A\cap \{S\le t\}\cap \{T\le t\}\in\mathcal{F}_t$$

Since both, $A\cap \{S\le t\}$ and $\{T\le t\}$ are elements of $\mathcal{F}_t$, so is their intersection. The equality is true since $\{S\le t\}\subset\{T\le t\}$.


Cauchy-Riemann equations, $f(z) = \bar{z} \sin z$ Find all $z \in \mathbb{C}$ such that the function $f(z) = {\bar z}\sin(z)$ satisfies the Cauchy-Riemann equations.

Own work:
Let $z = x+iy$ where $x$ and $y$ are real numbers

Thus,
${\bar z} = x-iy$, and
$\sin(z) = \frac{\exp(iz)-\exp(-iz)}{2i}$

Hence $f(x+iy)= (x-iy)\frac{\exp(i(x+iy))-\exp(-i(x+iy))}{2i}$

Im struggling to simplify this in order to do the partial derivatives to see that it satisfies the Cauchy-Riemann equations.

Any help will be appreciated.

If you know the complex Cauchy–Riemann equation namely:
$$\frac{\partial f}{\partial \bar{z}}=0,$$ you have
$$\frac{\partial }{\partial \bar{z}} (\bar{z} \sin z) = \left(\frac{\partial }{\partial \bar{z}} \bar{z}\right) \sin z + \bar{z} \left(\frac{\partial}{\partial \bar{z}} \sin z\right)=\sin z$$ as $\sin z$ is holomorphic. Hence $f$ satisfies the Cauchy-Riermann equations when $\sin z$ vanishes.

sin(z)=sin(x+iy)=sin(x)cos(iy)+cos(x)sin(iy)

now multiplying with z bar and taking partial derivatives becomes easier


Plotting solutions to first order system The question gives the initial value problem $\dfrac {\Bbb d^2y} {\Bbb dt^2} + y = 0$ with the initial conditions $y(0) = 0, y'(0) = v(0)=1$, and that $y(t) = \sin t$ is a solution.

The question asks to plot the solution curve to the solution. Do I just draw the $\sin$ curve? Is it asking for a closed system shape?

Draw the three curves approximately as function of time $t$.

$ Sin, Cos, -Sin $

The question wants you to recognize that graphs 1 and 3 are same except sign!


Linear Algebra - Dimensions of Subspace How do I go about finding the dimension of the subspace:
$$$$S:={p($x$) ∈ $P_4$: p($x$)= 2p($x$) for  all $x\in\mathbb{R}$} of $P_4$

My textbook says $dim(P_n)=n+1$, but this does not give me the correct answer. All help is appreciated.

Let $n=4$, then $p \in S \leq P_n$ has the form of $\sum_{j=0}^{n}a_jx^j$ for some $a_j \in R$.  Therefore $p = 2p$ implies that for all $x \in R$,
$$\sum_{j=0}^{n}a_jx^j = 2\sum_{j=0}^{n}a_jx^j \iff 0 = \sum_{j=0}^{n}(2-1)a_jx^j = p(x).$$
What's the dimension of a space where every element has this property?

Now that we are given $R = \mathbb R$, it's even easier: which element of $P_n \leq \mathbb R[x]$ has infinitely many roots?

I'm extremely confused by the equivalence drawn in linear algebra between vectors in Rn and polynomials. Can someone please point out the flaws in my understanding?

For Rn, a vector (1, 2, 3) has components in three different dimensions (along three different axes), so adding these together is impossible, right?

And when we express a system of linear equations in matrix form, say

2x + 3y + 10z = 45

we treat the variables x, y, z as living in 3 different dimensions that cannot be directly added together, just like the different dimensions ("axes") of the vector (1, 2, 3) in Rn, right?

My understanding (please correct me if I'm wrong) is that x y z somehow correspond to DIFFERENT dimensions, just as 1, 2, 3 lie on three different axes, so you can't just add them together like (1,2,3) = 1+2+3 = 6, nor can you do x+y+z = 3x, for example.

But in linear algebra I see people treat the same variable x* raised to different degrees, as different variables, exactly as in the previous case. So for

2x^3 + 3x^2 + 10x = 45

you would make a matrix with three columns, corresponding to the x^3, x^2, and x, exactly as if it had been

2x + 3y + 10z = 45

This feels totally wrong to me, because x^3 and x^2 all come from x!

I assume that since you can't just add variables of a different power together in a linear way, that's why they can be slotted into the matrix in this way?

i.e.)

2x^2 + 3x^2 can be comined into 5x^2 but 2x^2 + 3x^3 cannot be combined.

But it still feels like a stretch to me and is completely baffling conceptually. How can you say that the same variable raised to different degrees lives in different dimensions, in the same way as the components of a vector in Rn??? For one, x^2 can be calculated if you know x, but a value living in dimension 2 of Rn, can never be calculated by just knowing a value in dimension 1 - they have totally different meanings.

If you graphed y = x and y = x^2, the two different graphs would look kind of different, but still be in the same dimension on the graph paper, right? One wouldn't be 3D or sticking out of the paper, right? So how do you draw this equivalency, that the 3 dimensional vector (1,2,3) corresponds to (x, x^2, x^3)?


This is meant mostly as a joke, but partly not:

We clearly have $2^{15}-2^3=32{,}760$ divides $a^{15}-a^3$ when $a=0$, $\pm1$, and $\pm2$, so that leaves only $32{,}755$ other cases to check....

(The part that's not meant as a joke is the implicit question, where do you draw the line between proving something by brute force computation versus spending time looking for a clever theory.)


Math without pencil and paper For someone who is physically unable to use a pencil and paper, what would be the best way to do math?

In my case, I have only a little movement im my fingers. I can move a computer mouse and press the left button. Currently I do very little math and when I do I use MS Office but this is a very slow way of doing it and will break my thinking process.

EDIT:
Thank you for the answers. I have a lot to try. I won't accept an answer since it's dependent on the disability but thank you all.

(Background: I have a chronic pain condition, and it is extremely painful for me to do any sort of repetitive fine motor activity, including writing and typing.  I earned an undergraduate degree in math with quite little handwriting, and I'm currently a graduate student.)

Unfortunately, I've never been able to find any one solution or approach to replace the work that many people do with pencil and paper, but there are lots of partial solutions that work for different situations and types of mathematics.  Here are some of the ones that I use and which, based on your description, might be applicable.

Getting really proficient at mental math.  I think you've already done a lot of this, so all I'm going to say is that I can do a lot more math in my head than I would've believed was possible pre-disability.  It just takes a lot of practice.  Of course, it doesn't work for everything; some calculations just require remembering too many things at once.
Speech-to-text.  If you have clear speech, then a speech-to-text system is potentially very helpful.  (I use Dragon NaturallySpeaking.)  Maybe you already use one for general writing and computer use.  If so, then you have almost certainly noticed that they are not designed for mathematics (or, for that matter, computer science), so some additional software is necessary in order to do math. I use a system based on NatLaTeX to dictate all of my formal mathematics, including anything in that I'm going to turn in for my coursework.  Basically, NatLaTeX defines a speakable form of many common LaTeX commands, including everything you need for most mathematical expressions.  Using a custom vocabulary in Dragon, I can dictate a plain text file containing this NatLaTeX source.  I then use in scripts from the NatLaTeX project to transform my dictated text into actual LaTeX source, which I can then compile into nicely typeset mathematics using a standard LaTeX compiler.  (Actually, I use a batch file to automate the process.)  Just as a note, I have made several modifications to NatLaTeX in order to optimize it for mathematics (the original author was a physicist) and to adjust for changes in LaTeX.  Feel free to contact me if you want a copy of the modified scripts.  I do eventually intend to post them somewhere online, but I need to spend some time updating the documentation first (and that's really hard to justify spending time on it while I'm preparing for comprehensive exams!).

The big advantage to a dictation system like this is that you get to go through the process of formally presenting your work, which is really helpful for checking understanding and practicing proof writing.  Disadvantages include a steep learning curve and not being able to see your work (typeset, at least) in real time.  You also have very limited choices in text editors because Dragon will only "play nice" with a couple of them.
Programming languages and computer algebra systems.  When you actually need to do calculations, plot functions, etc., it is hard to beat a good computer algebra system.  There are lots of choices out there, and I think the choice of which one to use ultimately comes down to personal preference and perhaps compatibility with whatever assistive technology you use.  Of course, you still face the problem of how to handle getting input to that computer algebra system.  Typing with one finger on an on-screen keyboard sounds rather slow and tedious.  Here are a few alternatives you might want to look into.

If you have head and neck movement, a head pointer is one option.  I sometimes use one that actually marketed as a gaming device called TrackIR.  (Gaming peripherals are much less expensive than assistive technology peripherals!)  You can use this to type on an on-screen keyboard or to interact with the input panels found in many computer algebra systems.  For mouse click, you could use a switch with your finger or a dwell/hover click.
Eye tracking is a technology that has recently become much more affordable due to relatively new consumer-level devices marketed for gaming applications.  Just a couple months ago, I got an Tobii EyeX eye tracker, and it has been great!  I use it with a simple mouse emulation script and Dasher to write code in Sage.  Dasher is really cool, by the way, and quite good for text entry with low-precision input devices like eye trackers.  It's also much faster than most on-screen keyboards.  Just as a warning, the EyeX is not intended as an assistive device so you do have to do a some software configuration in order to make it do what you want it to do.  But you're studying computer science, so I don't think that should be a problem for you.  (FreePIE, OptiKey, and Dasher are all good pieces of free software to look up and consider for use with the EyeX.)

The big advantage to computer algebra systems is that you can do all of the numerical calculations and tedious symbolic manipulations without needing to ever write down a single number.  The disadvantage is that hiding all of the details can sometimes hinder your conceptual understanding.

Those are the main strategies I use.  I hope something here can help you out, too.

EDIT:
There has been a request for some examples of what NatLaTex can do, so here are a couple of different examples pulled files on my hard drive.

Discrete Math Example:
NatLaTeX Input (dictated with Dragon)
Given a poset "(P, precedes)", a collection of linear extensions "{calligraphy R } equals left curly brace precedes sub one, precedes sub two, low dots, precedes sub k right curly brace" is called a ``realizer'' of "P" if "precedes equals intersection of sub {i equals one } to the k precedes sub k", where each relation "precedes sub i" is interpreted as a set of ordered pairs and "intersection of" is set intersection. Equivalently, "{calligraphy R }" is a realizer of "P" if, for all "p, q in P", "p precedes q" if and only if "p precedes sub i q" for all "one less than or equal to i less than or equal to k".

LaTeX Output
Given a poset \( ( P , \prec )\), a collection of linear extensions \(
{ \mathcal R } = \{ \prec_1 , \prec_2 , \ldots , \prec_k \}\) is called
a ``realizer'' of \( P\) if \( \prec = \bigcap_{ i = 1 }^k \prec_k\),
where each relation \( \prec_i\) is interpreted as a set of ordered
pairs and \( \bigcap\) is set intersection. Equivalently, \( { \mathcal
R }\) is a realizer of \( P\) if, for all \( p , q \in P\), \( p \prec
q\) if and only if \( p \prec_i q\) for all \( 1 \leq i \leq k\).

Analysis Example:
NatLaTeX Input (dictated with Dragon)
begin theorem [Monotone Convergence Theorem]
Let "left curly brace f sub n right curly brace sub {n equals one } to the infinity" be a sequence of nonnegative measurable functions with "f sub one less than or equal to f sub two less than or equal to low dots less than or equal to f sub n less than or equal to f sub {n +1 } less than or equal to low dots" and "limit of sub n f sub n equals f" (pointwise). Then, "f" is measurable and
@begin{equation}
limit of sub {n right arrow infinity } integral f sub n d Greek mu equals integral limit of sub {n right arrow infinity } f sub n d Greek mu equals integral f d Greek mu
@end{equation}
end theorem

LaTeX Output
\begin{theorem}[Monotone Convergence Theorem]
Let \( \{ f_n \}_{ n = 1 }^\infty\) be a sequence of nonnegative
measurable functions with \( f_1 \leq f_2 \leq \ldots \leq f_n \leq f_{ n
+ 1 } \leq \ldots\) and \( \lim_n f_n = f \) (pointwise). Then, \( f \) is
measurable and
\begin{equation}
\lim_{ n \rightarrow \infty } \int f_n d \mu = \int \lim_{ n
\rightarrow \infty } f_n d \mu = \int f d \mu
\end{equation}
\end{theorem}

Though not impaired, I recently began working with Accessibility, so my computer and I can communication audibly.  Was very challenging at first, but re-running the "improve" process helps, plus this week I found a note and downloaded WSRMacros, for Windows Speech Recognition.
Essentially I use it to make boilerplates ahead of time, set one up so I'd say, for instance, "quadratic" and it might output Ax^2 + Bx + C, which appears fairly useless, except that you can then turn around and by voice "select A" and replace it, repeat for B and C.  Tedious, but each new macro adds more functionality, and the symbols are integrated into the macro, rather than somehow spoken into the equation for each instance.  Maybe you can combine this with the other ideas above, like using Google voice to get a good basic equation on your clipboard.
Also, MathML is interesting, it's like writing webpages in HTML, but with specific math tags.
These are free and/or open source tools, or included with Windows.


Two definitions of Measurable Functions Let $\mathfrak M_n$ the Lesbegue-Measurable $\sigma$-algebra on ${\mathbb {R}}^n$. Let $f: E \longrightarrow \mathbb R$ with $E \in \mathfrak M_n$.

f is said to be measurable in the "first form" if $\forall A $ open subset of $\mathbb{R}$ it implies $f^{-1}(A) \in \mathfrak M_n$.
f is said to be measurable in the "second form" if $\forall A \in \mathfrak M_1$ it implies $f^{-1}(A) \in \mathfrak M_n$.

Is $ 1 \Rightarrow 2$ true?

There exists a continuous fuction $f$ and a Lebesgue measurable set $E$ such that $f^{-1} (E)$ is not measurable. Thus such a function $f$ satsfies the condition $1$ but not satisfies the condition $2.$


f measurable function then $E=\left\{x\in X:f(x)=0\right\}$ is measurable set. Let $(X,m)$ measure space.
Let $f:X\to \mathbb{R}$ measurable function.
Show that $E=\left\{x\in X: f(x)=0\right\}$ is a measurable set.

I have this definition.
If $X$ is a measurable space, $Y$ is a topological space, and $f$ is a mapping
of $X$ into Y, then f is said to be measurable provided that $f^{-1}(V)$ is a
measurable set in X for every open set $V$ in $Y$.

Now, $E=f^{-1}(\left\{0\right\})$ but $\left\{0\right\}$ is not open...
Why is E measurable?

$\{0\}$ is not open, but it is measurable in the Borel $\sigma$-algebra on $\mathbb R$.

Let $(E, \mathcal E)$ and $(F, \mathcal F)$ be measure spaces. Then $f: E \to F$ is measurable iff $f^{-1}(A) \in \mathcal E$ for all $A \in \mathcal F$. This is the most general definition of a measurable function: note it doesn't require any topology on either set. But checking this property for every single element of a $\sigma$-algebra is hard. A key theorem is the following:

If $\mathcal T$ generates $\mathcal F$, i.e. $\sigma(\mathcal T) = \mathcal F$, then $f$ is measurable as long as $f^{-1}(A) \in \mathcal E$ for all $A \in \mathcal T$.

This is a simple application of the fact that $f^{-1}$ preserves unions and complements. In particular, the Borel $\sigma$-algebra on a topological space $(X, \mathcal T)$ is generated by the open sets $\mathcal T$, hence the definition in Rudin. However the more general definition still holds. To see this explicitly in your case

$$ f^{-1}(\{0\}) = f^{-1}((\mathbb R \setminus \{0\}^C)) = \left( f^{-1}(\mathbb R \setminus \{0\}) \right)^c $$
hence is measurable as the complement of a measurable set. A similar proof proves the theorem.

Note that,
E = {0} $f$ $^{-1}$ = ($\bigcap_{n=1}^{\infty}(-\frac{1}{n},\frac{1}{n})$) $f^{-1}$ = $\bigcap_{n=1}^{\infty}$ (($-\frac{1}{n},\frac{1}{n})$$f^{-1}$).
But $f$ is measurable, and ($-\frac{1}{n},\frac{1}{n})$ is open in $\mathbb{R}$, for all n $\in$ $\mathbb{N}$. Then,
$\bigcap_{n=1}^{\infty}$ (($-\frac{1}{n},\frac{1}{n})$$f^{-1}$)  is measureble in X, namely, $E$ is measurable in X.


Sum to infinity of An and Sk Consider a series $\sum_{n=1}^\infty A_n$ for which $S_k = \frac{k + 1}{k}$ $\forall k
\in \mathbb N$. Find $A_n$ and the value of $\sum_{n=1}^\infty A_n$.

I think i have worked out that $\sum_{n=1}^\infty A_n$ is $1$. But i don't know how to work out $A_n$ or show fully how i got my answer for the sum to infinity.

$S_k=\frac{k+1}{k}$

hence $\lim_{k\to \infty} S_k=1$.

which implies that $\sum_{n=1}^\infty A_n$ is $1$.

By definition,
$$
\sum_{n=1}^\infty a_n = \lim_{n \to +\infty} \sum_{k=1}^n a_k.
$$
If you have really proved that
$$
S_n = \sum_{k=1}^n a_k = \frac{n+1}{n}
$$
for every $k$, then your conclusion is perfectly legitimate, since $\lim_{n \to +\infty} \frac{n+1}{n}=1$.


How do I know if a matrix is diagonalizable by congruence? Let V be a unitary space over $\mathbb{C}^n$ with an orthonormal basis $K$.
Let $A: V \times V \rightarrow \mathbb{C}$ be a sesquilinear form.

If there exists a basis $B$ in $V$ such that $[A]_B$ is diagonal, what does that mean for the matrix $[A]_K$? What are the necessary conditions for $B$ to exist? And is there a general algorithm to find one?

I know that if $[A]_K$ is hermitian, or more generally normal, there always exists an orthonormal base $B$ which satisfies this. But what if $[A]_K$ isn't normal?

Just to be clear, I'm asking about the diagonalization by congruence, rather than by similarity. The diagonal matrix $[A]_B$ has to meet this condition:
$$ [A]_K = R^+ \; [A]_B \; R, $$
where $R$ is a regular matrix and $R^+$ its conjugate transpose.

If $[A]_B$ is diagonalisable by congruence, of course $[A]_K$ is also diagonalisable by congruence, because the matrices of a sequilinear form are related by congruences.

So, the basic question is, when is a complex square matrix diagonalisable by congruence? Note that this is not always possible. E.g. if $e_1e_2^T=\pmatrix{0&1\\ 0&0}$ is diagonalisable by congruence, there will exist an invertible $P$ and some $z\ne 0$ such that
$$
(P^\ast e_1)(P^\ast e_2)^\ast=P^\ast e_1e_2^TP=\pmatrix{z\\ &0}=ze_1e_1^T\ne0,
$$
meaning that both $P^\ast e_1$ and $P^\ast e_2$ are scalar multiples of $e_1$. But this is impossible because $P$ is invertible. Thus $e_1e_2^T$ is not diagonalisable by congruence.

In general, let $M=H_1+iH_2$, where $H_1$ and $H_2$ are Hermitian. Then $M$ diagonalisable by congruence if and only if $H_1$ and $H_2$ are simultaneously diagonalisable by congruence.

Here is a known result. When at least one of $H_1$ and $H_2$ is nonsingular, they are simultaneously diagonalisable by congruence (and hence $M$ is diagonalisable by congruence) if and only if $H_k^{-1}H_{3-k}$ is similar to a real diagonal matrix. For details, see the section on "Congruence and simultaneous diagonalization" or more specifically, theorem 4.5.15 of Horn and Johnson, Matrix Analysis, first edition, 1985, p.228 (theorem or page numbers may be different in the second edition).


Efficiently finding congruencies All variables are integers with a value somewhere between 0-99, inclusive.
ax + b % 100 = n and you are given everything except x.
Example: 65x + 6 is congruent with 81 mod 100.
I am trying to find a way to determine the potentially valid values of x without resorting to brute force.  Is there a known method more efficient than brute force for a problem like this?  If so, can anyone explain how the congruencies are found and/or possibly point me in the direction of any tutorials, books, or videos?


Must the dimensions of $V$ and $U$ be equal for $f: U \mapsto V$ to be a diffeomorphism? Suppose we have $U, V \subset \mathbb{R}^{n}$, and a map $f: U \mapsto V$. If $f$ is a diffeomorphism, must the dimensions of $U$ and $V$ be equal?

I'm thinking this is true since the tangent map $Df_{x}: T_{x}U \mapsto T_{f(x)}V$ has to be a linear isomorphism if $f$ is a diffeomorphism.

You are right, the explanation you gave is also the right one.

If $f$ is a diffeomorphism, then there exists $g\colon V\rightarrow U$ such that $g\circ f=\operatorname{id}_{U}$ and using the chain rule:
$$Dg_{f(x)}\circ Df_x=\operatorname{id}_{T_xU},$$
so that $Df_x$ is an invertible linear map.

Let $n=2$, $U$ be the $x$-axis, $V$ be $\mathbb{R}^2$, and $f:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ be the identity.  Notice $\left. f \right|_U$ is a diffeomorphism $U \rightarrow V$.


Moment generating function and $E(2^X)$ A moment-generating function of $X$ is given by$$
M(t) = 0.3e^t + 0.4e^{2t} + 0.2e^{3t} + 0.1e^{5t}.
$$
  Find the pmf of $X$.

My solution
$$
x-f(x)\\
1-0.3\\
2-0.4\\
3-0.2\\
5-0.1\\
$$
(correct?)
The next question asks to calculate $E(2^X)$, which I am tottally unsure about. I calculated the mean as $0.22$ and variance as $5.28$.

Anyone understand what to do?

You are given the value of $M(t)=E[\mathrm e^{tX}]$ for every $t$ and you are looking for the value of $E[2^X]$. Well, note that $E[2^X]=M(t)$ for $t=$ $____$.

Sanity check: As it happens, $E[2^X]$ is an integer.

hint: $E[g(X)] = \sum g(x)P (X = x)$ Can you solve now?


What is the meaning of percentile? I am confused by the term percentile.
Once my teacher told me that percentile means the percentage with respect to the score of the highest achiever.

This means that if in a competition I got $80$ out of $100$ and the highest score in that competition was 90 out of 100 then my percentile would be $\frac{80}{90}*100=88.89$.

So I got $80\%$ and $88.89$ percentile.

I was believing that my above concept was right.

But when I see the definition of percentile on Wikipedia then I got something new (but I don't understand this definition) and then I thought that what my teacher told me was wrong.

Kindly tell me if my teacher right or wrong.

If your memory is correct, your teacher was wrong.

The percentile tells you the value below which a certain percentage of the observations fall.

Thus, if the median score on the test was $80$, a score of $80$ would put you at the fiftieth percentile regardless of what the high score was.

First, consider the possibility that you misunderstood your teacher or that your words above do not quite capture what (s)he meant. That said, the "percentage with respect to the score of the highest achiever" has nothing to do with percentiles, and I am not sure there is a single word to express "I got within 89% of the best".

Percentiles (or, more generally, quantiles) are about relative positions compared to everyone else. It is not required that whatever you are looking at has numerical values, only that the values can be ordered. Taking an example in the spirit of @Nick's first link, consider a test where

60 people got a D
you and no one else got a C
39 people got a B
no-one got an A or worse than D.

How many % of a B are a C? However, one can still make statements in terms of quantiles, for example:

Everyone in the lower half got a D. (Since some in the upper half did so, too, the median grade is a D, too.)
Everyone in the highest (fourth) quartile got a B.
Everyone in the sixth decile got a D. (= those who scored better than the bottom 50%, but worse than the top 30%.)
Those in the 61st percentile got a C.
Those in the 62nd percentile got a B.

Thus, quantiles are best imagined as lining up everyone in (ascending) order and putting (ascendingly numbered) separators at n equidistant steps (n=100 for percentiles, 10 for deciles, 4 for quartiles)

Your percentile rank is then the number of the marker closest to you towards the "worse" side. (Note that percentiles make most sense if the considered set has >>100 elements. See the aforementioned link on how handle sets with < 100 elements.)


Difference between Perpendicular, Orthogonal and Normal It seems to me that perpendicular, orthogonal and normal are all equivalent in two and three dimensions.  I'm curious as to which situations you would want to use one term over the other in two and three dimensions.

Also... what about higher dimensions?  It seems like perpendicular and normal would not have a nice meaning whereas orthogonal would as it is defined in terms of the dot product.

Can someone give me a detailed breakdown as to the differences in their meanings, their uses and the situations for which each should be used?

In two or three dimensions, I agree, perpendicular is more natural than orthogonal.

In higher dimensions, or if the dimension is represented by an unknown, both are correct, but I think orthogonal is preferable.

Here's an excerpt from Wikipedia (https://en.wikipedia.org/wiki/Orthogonality):

  In mathematics, orthogonality is the generalization of the notion of perpendicularity to the linear algebra of bilinear forms.

Normal can be used in any dimension, but it usually means perpendicular to a curve or surface (of some dimension).

Perpendicular lines may or may not touch each other. Orthogonal lines are perpendicular and touch each other at junction.


Expected measure of a ball in a probability space with a metric Assume we are given a probability space $(\mathbb{X}, \mathcal{X}, \mathbb Q)$ and a measurable distance function defined on it $d:\mathbb{X}\times \mathbb{X}\to \mathbb{R}^+\cup\{0\}$ that conforms to the usual definition of distance on metric spaces. I am trying to understand (in lay terms) the "expected value of the measure of a ball with fixed radius $r> 0$ centered at a randomly chosen point in $\mathbb{X}$ with measure $\mathbb{Q}$". That is more formally
\begin{equation}
\int_{\mathbb{X} }q(x)\mathbb{Q}(\mathrm{d}x)
\end{equation}
where $q(x) = \mathbb{Q}(B(x,r))$ and $B(x,r) = \{y\in \mathbb{X}: d(x,y)\leq r\} $. The first thing to ask here is if $q(x)$ is measurable but for now I assume $\mathcal{X}$ is large enough that it is.
My problem originates from this question: "find a lower bound for the probability that a randomly chosen iid set of points in a space are close-by". I am wondering what assumptions are needed from $\mathbb{Q}$ and $d$ such that this can be lower bounded. Of course I would not expect a bound valid for all $\mathbb{Q}$ and $d$ but any tip about how to treat this integral (or you know if anybody thought about this before and gave it a name so I can search) would be helpful. The trivial bound $\inf_{x\in\mathbb X} \mathbb{Q}(B(x,r))$ is not useful for my purposes.

How is it defined the probability density function $\mathbb{Q}$? Is it discrete or continuous?
If it's continuous then $\lim_{r\rightarrow0}\int_{B_r}q(x)dx=0$
And if it's discrete then $\lim_{r\rightarrow0}\int_{B_r}q(x)dx=q(y)$


Is $V$ a simple $\text{End}_kV$-module? Let $V$ be a finite-dimensional vector space over $k$ and $A = \text{End}_k V$. Is $V$ a simple $A$-module?

A detailed answer was requested, so here it is.

The answer is yes (unless $V$ is the zero vectorspace). And we don't need finite-dimensionality of $V$, at least if we accept the axiom of choice. First, a convention: whenever $V$ is a $k$-vectorspace, write $\hat{V}$ for $V$ viewed as an $\text{End}_k V$-module. Let us now unpack the definitions.

  Observation A. $\hat{V}$ is simple iff:

  $\hat{V}$ has a non-zero submodule, and
  every non-zero submodule $X$ of $\hat{V}$ has the property that $X=\hat{V}.$

That's just the definition. But with a bit of thought, we see that this can be written in terms of $V$ only.

In particular:

  Observation B. $\hat{V}$ is simple iff:

  $V$ has a non-zero subspace, and
  every non-zero subspace $X$ of $V$ has the property that if for all endomorphisms $\varphi$ of $V$ we have $\varphi(X) \subseteq X$, then $X=V$.

To see that the two condition 1's are equivalent, notice firstly that since every submodule of $\hat{V}$ is a subspace of $V$, hence the forward direction is easy. The backward direction takes a tiny bit more work; start with a non-zero subspace $X$ of $V$, deduce that $V$ is non-zero, or in other words that $\hat{V}$ is non-zero, and hence that $\hat{V}$ is a non-zero submodule of itself.

The two condition 2's are seen to be equivalent simply by unpacking the definitions.

Clearly, unless $V$ has a non-zero subspace, it cannot be the case that $\hat{V}$ is simple.

So, let us conjecture:

  Proposition. If $V$ has a non-zero subspace, then $\hat{V}$ is simple.

Proof.

We need to show (1) and (2) from Observation B. Condition (1) is immediate. So suppose we're given a non-zero subspace $X$ of $V$ such that for every endomorphism $\varphi$ of $V$, we have $\varphi(X)\subseteq X.$ The goal is to show that $X=V$. Since $X \subseteq V$ is obvious, we need to show $V \subseteq X$. So consider arbitrary $v \in V$. The goal is to show that $v \in X$.

Since $X$ is non-zero, we can find non-zero $x \in X$. I claim that, to achieve our goal, it suffices to find an endomorphism $\varphi$ of $X$ such that $\varphi(x)=v$. For suppose such an endomorphism exists. Then since $x \in X$, we have $\varphi(x) \in \varphi(X)$, so $\varphi(x) \in X$, so $v \in X$.

In summary, the set up is:

  Assume

  $V$ is a vectorspace
  $x$ is a non-zero element of $V$
  $v \in V$

  Show

  there exists an endomorphism $\varphi$ of $V$ such that $\varphi(x)=v$.

Now since $k$ is a field and $x$ is non-zero, hence $\{x\}$ is linearly independent. Hence by a fairly deep theorem of linear algebra, the set $\{x\} \subseteq V$ extends to a basis $B$ of $V$. This doesn't assume finite-dimensionality of $V$ (as long as we accept the axiom of choice; otherwise, we have to be more careful.)

Now define a function $f : B \rightarrow V$ as follows: $$f(b) = \begin{cases}x & b = x \\ 0 & b \neq x\end{cases}$$ Then there exists a unique linear transform $\varphi : V \rightarrow V$ that agrees with $f$ on $B$. We have: $$\varphi(x) = f(x) = \left(\begin{cases}x & x = x \\ 0 & x \neq x\end{cases}\right) = \left(\begin{cases}x & \mathrm{True} \\ 0 & \mathrm{False}\end{cases}\right) = x$$

Ergo, there exists an endomorphism $\varphi$ of $V$ with $\varphi(x)=v$. This completes the proof.

Simply look at the definition: Recall that a module $M$ over a ring $R$ is simple if the only non-zero submodule of $M$ is $M$ itself. Equivalently, for any $x\neq 0$ in $M$, the submodule $Rx=\left\{rx:r\in R\right\}$ generated by $x$ is $M$.

Now in your case: Take a nonzero $x\in V$ and any $y\in V$. Is it true that there exists $T\in A=\operatorname{End}(V)$ with $Tx=y$?


What is the value of $i^0$? I have to solve the following question -

$$\sum_{n=0}^{1000} i^n$$

where $i = \sqrt{-1}$

To be able to solve the problem, I need to know the value of $i^0$. What is the value of $i^0$? Is it 0 or indeterminate or something else?

In every multiplicative group $G$ we define
$$
g^0=e
$$
for every $g\in G$. In this case we have $G=\mathbb C^*$

One can use the definition $z^a=e^{a\log(z)}$, in this case we defined $\log(z)$ for all complex numbers except negative real numbers. And $e^z$ is defined for every $z\in\mathbb C$, so by this definition we have
$$
i^0=1
$$

$$i^0=i^{1-1}=\frac{i^1}{i^1}=\frac ii$$
Now, $\frac ii$ is asking us for the number $c$ such that $c\times i=i$. It's not hard to show that $c=1$ is the only one that works. Thus, $\frac ii=1$, and:
$$i^0=1$$
(This exact argument shows that $z^0=1$ for any nonzero number $z$.)


Phase portrait of ODE in polar coordinates Given the system of ODEs in polar coordinates, $$r' = r(1-r^2)(4-r^2)$$ $$\theta'=2-r^2,$$

one can determine its equilibrium points and limit cycles as follows:

$\gamma_1:= \begin{cases} r = 0,\\ \theta = 2t\end{cases}$, $\gamma_2:= \begin{cases} r = 1,\\ \theta = t\end{cases}$, $\gamma_3:= \begin{cases} r = 2,\\ \theta = -2t\end{cases}$.

$\gamma_1$ corresponds to $(0,0)$ in the $xy$-plane, and $\gamma_2$ and $\gamma_3$ correspond to circles.

Now I need to sketch the phase portrait of this system and, based on this sketch, determine the stability of the equilibrium points and limit cycles.

Does one need to solve this system explicitly in order to sketch the phase portrait, or is there a neater way to do it, without solving the system?

I've also tried with solving for the ODE in terms of $\frac{d\theta}{dt}$, but it doesn't appear to be an equation which is easy to plot either.

Also, do you think that I've found all the possible limit cycles, or maybe missed something? I'm new to this kind of analysis.

Hints: I would have made this a comment if I could.

Here is the polar plot.

Here is the transformed system, if you want to validate the rest of your analysis using other approaches.

$$x' = x \left(-x^2-y^2+1\right) \left(-x^2-y^2+4\right)-\frac{y \left(-x^2-y^2+2\right)}{\sqrt{x^2+y^2}} \\ y' = y \left(-x^2-y^2+1\right) \left(-x^2-y^2+4\right) + \frac{x \left(-x^2-y^2+2\right)}{\sqrt{x^2+y^2}}$$

Update Reduced the plot area to see further detail in the areas of interest.

EDIT 1&2

Before coming to phase portrait, I thought of gaining understanding of LC loci numerically.

The limit cycle LC (above) are running  in $ r<1, r=1 - 2, r >2 $ three regions. In the central region there is a a turnaround at radial direction of velocity at $r= \sqrt 2 $. When started inside $r<1$ it cannot cross $r =1$. Starting inside the annulus must remain there between the two circular asymptotes . Motion $r>2 cannot enter the annulus at all.

Without solving you can plot  first PP $ (r, r^{'} ) $ but one cannot plot second PP $ (\theta, \theta ^{'}) $ without solving the full system either analytically or numerically. There are two circle boundaries at $ r= (1,2)$ and a radial tangent at $ \theta^{'} =0, r = \sqrt 2.$

Where exactly the outer radial asymptotic comes about $ $ r>r2$  ... is yet to be understood (by me).


how to calculate 2^1.4 So I have got a very basic question but it didn't come up as a google search so I am posting it here.

I want to know how to easy calculate

2^1.4 = 2.6390...

Using log and antilogs i.e not easy approach ?

i.e.

log y =  log m^n

log y = n log m

log y = 1.4 log 2

log y = 1.4 * 0.301

log y = 0.4214

antilog (log y) = antilog 0.4214

y= antilog 0.4212 (look for this in a table, should give result)

What I found on internet ,

turn the decimal into a fraction

2^(1.4)
=> 2^(1 + 2/5)

now given a^(n + m) = a^n * a^m
=> 2^1 * 2^(2/5)
=> 2 * 2^(2/5)

you can stop there if you want to.

however a^(nm) = (a^n)^m

=> 2 * (2^2)^(1/5)

  //I was ok till here what to do next to get the right result !

which would be read as 2 times the 5th root of 2 squared and would be the exact answer. [the nth root of a = a^(1/n)]

You can use Newtons approximation method.

We want to find $x=2^{1.4}$, or equivalently, $x^5=\left(2^{1.4}\right)^5=2^7=128$

Define $f(x)=x^5-128$

We want to find the root of $f(x)$

as noted in my comment, $x\approx \sqrt{8} \approx 2.828$

So we start with this guess of $2.828$.

We put it into the formula,

$$x'=x-\frac{f(x)}{f'(x)}$$

and $x'$ will be a more accurate guess.

$$x'= 2.828 - \frac{(2.828)^5-128}{5(2.828)^4}$$

with a bit of hand calculation we get $x'=2.66\ldots$ which is very close to the actual value.

2^1.4 =(1+1)^1.4   approximately
           =1+1×1.4     approximately
           =1+1.4         approximately
           =2.4             approximately
For the above I used binomial theorem


set and subset question Given  a nonempty set S, we select a fixed subset $S_0\subseteq S$ and use it to define two functions on the set of all subsets of $S$:

$f(A)=A\cup S_0$
$g(A)=A\setminus S_0$

For what choices of $S_0$ do these functions commute?

Only for $S_0=\varnothing$? or
Only for $S_0=S$? or
For all $S_0$?

HINT:

Recall that the functions commute if for every $A$, $f(g(A))=g(f(A))$, that is $$(A\cup S_0)\setminus S_0=(A\setminus S_0)\cup S_0.$$

The choices for which the functions don't commute can be found by testing the case where $S=\{a,b,c\}$. The choice where the functions do commute is easily provable.

$gf(A)=g(A\cup S_0)=(A\cup S_0)\setminus S_0=A\setminus S_0$

$fg(A)=(A\setminus S_0)\cup S_0=A$

1 -true


Computing a conditional expectation for uniform RVs Suppose $X_1, ..., X_n \sim U[0, 1]$ are iid uniform RVs. How would I go about computing $E[X_n | X_{(n)}]$ where $X_{(n)}$ is the nth order statistic, i.e. $\max\{X_1, ..., X_n\}$ ?

I'm stuck without any ideas on this.

A joint independent uniform distribution remains "flat" even after conditioning on what region the values lie in. Thus, your conditional expectation is $c/n + ((n-1)/n)E(Y_c)$ where $c$ is the observed maximum value (which could have been $X_n$), and $Y_c$ is the uniform random variable restricted to $[0,c]$. Thus your conditional expectation is $c/n + ((n-1)/n)c/2$, where $c$ is the given maximum of the $n$ samples.

Assume $Y = X_{(n)}.$ The pdf of $Y$ is $f(y) = ny^{n-1}$.  The joint pdf of $Y$ and $X_n$ is
$$
f(y,x_n)= \begin{cases} x_n^{n-1}\delta(y-x_n) & x_n = y\\ (n-1)y^{n-2} & x_n < y \end{cases}
$$
The conditional pdf of $X_n$ given $Y$ is
$$
f(x_n/y) = \begin{cases} \frac{x_n^{n-1}}{ny^{n-1}}\delta(y-x_n) & x_n = y\\ \frac{(n-1)}{ny} & x_n < y \end{cases}
$$
So we have
$$
E[X_n/Y] = \frac{n+1}{2n}Y
$$


How to do a double sum in MATLAB given the explicit values of the function I have a double summation which I am trying to compute as follows

$$\sum_i^3\sum_j^3 h_{i,j} w_{i,j}$$

which I am trying to implement in MATLAB.. My application gives me the explicit values of both $h_{i,j} $ and $w_{i,j}$ and gives them as vector such as

For example $${\bf h} = [ h_{11}\,\, h_{12}\,\, h_{13} \,\,h_{21}\,\, h_{22}\,\, h_{23} ]^T$$
$${\bf w} = [ w_{11}\,\, w_{12}\,\, w_{13} \,\,w_{21}\,\, w_{22}\,\, w_{23} ]^T$$

where $T$ means transpose.
In order to find the double summation, I do the following in MATLAB

   for l =1: 6
    v(i)=  h(i)*w(i)
  end

for i = 1:6
summation = sum_initial+ v(i)
end

Am I doing things correctly here?

Thanks,

you defined your vectors as two rows, then

just $h.'*w$ or if $h$ is real then $h'*w$ enough, this is a row vector times column vector, i.e. a dot\inner product

however, if you want them to be matrices (as it looks from your mathematical definition), then $h(:).'*w(:)$ will do the trick.

Your code could work if you fix it as following

for l =1: 6
   v(i)=  h(i)*w(i);
end

summation=0;
for i = 1:6
   summation = summation + v(i);
end

Assuming h and w are 3x3 matrices:

  h(:).'*w(:)


A question about atlas in smooth differentiable I know this is definitely an easy question. However, I just cannot figure out.

I am using Do Carmo's Riemannian Geometry to teach myself Differential Geometry.

In his book, he gave a definition of differentiable manifold:

A differentiable manifold of dimension $n$ is a set $M$ and a family of injective mappings $x_\alpha:U_\alpha\subset \mathbb{R}^n\to M$ of open sets $U_\alpha$ of $\mathbb{R}^n$ into $M$ such that:

(1)$\bigcup_\alpha x_\alpha(U_\alpha)=M$

(2)for any pair $\alpha,\beta$, with $x_\alpha(U_\alpha)\cap x_\beta(U_\beta)=W\neq\emptyset$, the sets $x_\alpha^{-1}(W)$ and $x_\beta^{-1}(W)$ are open sets in $\mathbb{R}^n$ and the mappings $x_\beta^{-1}\circ x_\alpha$ are differentiable.

(3) The family $\{(U_\alpha,x_\alpha)\}$ is maximal relative to the conditions (1) and (2).

Furtuhermore, he define $A\subset M$ to be an open set in $M$ iff $x_\alpha^{-1}\left(A\cap x_\alpha(U_\alpha)\right)$ is an open set in $\mathbb{R}^n$ for every $\alpha$.

I can show that this is a topology, and that $x_{\alpha}(U_\alpha)$ is open and $x_\alpha$ is continuous. However, I have no idea how to prove that $x_\alpha$ is actually a homeomorphism. It suffices to prove that $x_\alpha$ is a open mapping, but I have no idea.

Any help will be appreciated. Thanks.

We check that $x_{\alpha}:U_{\alpha}\rightarrow M$ is an open mapping. So given $U\subset U_{\alpha}$ open, we only need to see that $x_{\alpha}(U)\subset M$ is open according to the definition you gave.

For any $\beta$, we have
\begin{align}x_{\beta}^{-1}\left(x_{\alpha}(U)\cap x_{\beta}(U_{\beta})\right)&=(x_{\beta}^{-1}\circ x_{\alpha})(U)\cap (x_{\beta}^{-1}\circ x_{\beta})(U_{\beta})=(x_{\alpha}^{-1}\circ x_{\beta})^{-1}(U)\;\cap\; U_{\beta}\end{align}

which is  open because $x_{\alpha}^{-1}\circ x_{\beta}$ is differentiable.

Hint: The $x_\alpha$ are quotient maps by definition, but they are also injective.


Prove that the sum is less than $6/5$. How to prove that $$\sum_{n=1}^{15} \frac{1}{n^3}\lt\frac 65$$

I tried to compare this sum to the infinite sum, but Apery's constant is just above $1.2$ so this approach doesn't work.

Then I typed this into wolfy and the sum seems as if it is just under $6/5$.

However, I was wondering if there was a better way of proving this result rather than summing all the terms.

You can estimate the remainder of the series by comparing it to the integral.

$\displaystyle \sum\limits_{n=N}^{\infty}\frac 1{n^3}\ge\int_{N}^{\infty}\frac{dt}{t^3}=\bigg[\frac{-1}{2t^2}\bigg]_{N}^{+\infty}=\frac{1}{2N^2}$

$\displaystyle \sum\limits_{n=1}^{15}\frac 1{n^3}\le\zeta(3)-\frac 1{512}\simeq 1.2001... > \frac 65$

$\displaystyle \sum\limits_{n=1}^{14}\frac 1{n^3}\le\zeta(3)-\frac 1{450}\simeq 1.1998... < \frac 65$

Unfortunately the integral comparison cannot tell with certitude if the proper $N$ is $15$, so unless there are other ideas, I think the effective computation is required.

$\sum_{N+1}^{15} (1/x^3) $ < $\int_N^{15} (1/x^3) dx $  Try N=1 ,if it doesn't work,try N=2 etc this comes from the integral test . In fact you should   increase the upper limit of the integral from 15 to  +infinity  to make it easier .I believe N=3 might work


$H, K$ conjugate subgroups of $G$, then $G/H$ and $G/K$ are isomorphic as G-sets Let $H, K$ be conjugate subgroups of $G$. Then we can regard $G/H, G/K$ as G-sets defining $g'(gH) := (g'g)H$ and similarly for $K$.

Two G-sets $X, Y$ are said to be isomorphic if there exists some bijection $\phi: X \rightarrow Y$ such that

$$\phi(gx) = g\phi(x) \; \forall x \in X, g \in G$$

My intuition seems to be that $\phi(H) = K$ and so $\phi(gH) = gK$, but I'm having trouble trying to show $\phi$ is well defined, since $\phi(ghH) = ghK =_{?} gK = \phi(gH)$ does not look right.

I've found this question here in this site, but the only answer is rather brief and I don't fully understand what is going on, as seems to be the case for the OP. Anyway, it was never restated or further detailed, so I'd really appreciate if someone would care to give some insight into it.

I will try to explain better the answer which appears in the post you mentioned.

To define a $G$-map $\phi \colon G/H \to G/K$ is equivalent to fixing some $g \in G$ such that $\phi(1H)=gK$ (note that $g$ does not need to be $1$). Indeed, then we have automatically $\phi(xH)=x\phi(1H)=xgK$. But $\phi$ is well-defined iff for all $h\in H$ we have $hgK=\phi(hH)=\phi(1H)=gK$, which is equivalent to $g^{-1}hgK = K$ for all $h$, which in turn is equivalent to $g^{-1}Hg \subseteq K$.

Now we show that the other inclusion is instead equivalent to injectivity of $\phi$. Note that $xgK=gK$ is equivalent to $x \in gKg^{-1}$. Now, injectivity is equivalent to requiring that all such $x$ are in $H$, i.e. $gKg^{-1}\subseteq H$.

Surjectivity is automatic for $\phi$, because if we want to find $x$ such that $\phi(xH)=xgK=qK$, we just need to set $x:=qg^{-1}$. Hence we conclude:

$\phi\colon G/H \to G/K, xH \mapsto xgK$ is an isomorphism of $G$-sets iff $H$ and $K$ are $g$-conjugated.

Second isomorphism theorem: Let G be a group and HK < G. If H  Δ G, then K/HnK  ≅ HK / H

Define $\phi : K \to H k/H$  by $\phi(a) = aH$. I state that $\phi$ is well-defined homomorphism and endomorphism at the same time, this means that $\operatorname{Ker}f = HnK$


Simplify $\frac{x}{c} - \frac{x}{c-d}$ There's a long time that I don't solve questions like this one. I'm having problems to simplify this one:

$$\frac{x}{c} - \frac{x}{c-d}$$

First recall that
\[
\frac{a}{b}-\frac{c}{d}= \frac{ad-bc}{bd}
\]
Then
\[
\frac{x}{c}-\frac{x}{c-d}= x\left(\frac{1}{c}-\frac{1}{c-d}\right)= x\left(\frac{c-d-c}{c(c-d)}\right)= \frac{-xd}{c^2-cd}
\]
I attempted to be as clear as possible.

Multiply the first term by $\frac{c-d}{c-d}$ and the second term by $\frac{c}{c}$ Then you will have $\frac{x(c-d)-x(c)}{c(c-d)} = \frac{-xd}{c(c-d)}$


Finding $\frac{d}{dx}(\sqrt{3\cos^3x})$ I am attempting to find the derivative of $\sqrt{3\cos^3 x}$.

Using the chain rule, I got:

$$\frac{1}{2} \sqrt{(3\cos x)^3} × 3(3\cos x)^2 × (-3\sin x).$$

Apparently this is where I went wrong, the correct answer stated that instead of the derivative of the inner function should have been $-\sin x$ instead of $-3\sin x$. To my understanding the derivative of $3\cos x$ is $-3\sin x$. Is my reasoning incorrect?

$$\begin{align*}
\frac{\mathrm{d}}{\mathrm{d}x}\sqrt{3\cos^3x} &= \sqrt{3}\cdot\frac{\mathrm{d}}{\mathrm{d}x}\cos^\frac32x\\
&= \sqrt{3}\cdot\frac32\cos^{\frac32 - 1}x\cdot\frac{\mathrm{d}}{\mathrm{d}x}\cos x\\
&= \sqrt{3}\cdot\frac32\cos^\frac12x\cdot(-\sin x) \\
&= -\frac{3^\frac32}{2}\sin x\sqrt{\cos x}
\end{align*}$$

Here is a correct implementation of chain rule

$\frac{d\sqrt{3cos^3(x)}}{dx} = \frac{1}{2\sqrt{3cos^3(x)}}*\frac{d( {3cos^3(x)})}{dx}$ = $\frac{1}{2\sqrt{3cos^3(x)}}*3*3*cos^2(x)*\frac{d({cos(x)})}{dx}$

$\qquad \qquad  =  \frac{1}{2\sqrt{3cos^3(x)}}*9*cos^2(x)*(-sin(x)) = -\frac{3^{\frac{3}{2}}}{2}*sin(x)*\sqrt{cos(x)}$


Inverse of prime counting function The prime counting function $ \pi (x) \approx \dfrac {x} {\ln(x-1)} $.
This function returns the number of primes less than $x$.

Note: $x-1$ gives a better estimate than $x$.

How to find $x$ given $ \pi(x) $ ?

Instead of an analytical Expression or 1-Step Approximation use a Fixed Point Iteration that converges relatively fast to the Inverse of x/ln(x).

8 Iterations yield the Result e.g. x = 4_210_563_857 for y = 190_000_000:

    public static double PI_INVERSE(double y) {
        double xOld = 0;
        double x = y*Math.Log(y);
        const double maxDiff = 0.5;
        for (; x - xOld > maxDiff;) { //strictly monotonous
            xOld = x;
            x = y*Math.Log(x);
        }
        return x;
    }

Look up
Meissel's prime-counting function.

One reference is here:

https://en.wikipedia.org/wiki/Prime-counting_function


Does taking expectation of numerator and denominator respectively preserve monotonic structure ? Suppose $f(x,\mu)$ and $g(x,\mu)$ are both positive functions, and for any $x$,
$\frac{f(x,\mu)}{g(x,\mu)}$ increases with $\mu$.

Let $X$ denote an arbitrary random variable, is it true that
$$\frac{\mathbb{E}[f(X,\mu)]}{\mathbb{E}[g(X,\mu)]}$$
also increases with $\mu$?

For my purpose we can assume $f$ and $g$ are differentiable, and that we can change order of differentiation and expectation in any step of calculation.

Being $E$ a linear and monotone operator, you could prove $E\biggl[\frac{f(X,\mu)}{g(X,\mu)}\biggr]$ is increasing in $\mu$. But as $E$ doesn't function nicely with quotients, you wouldn't be able to prove that $\frac{E[f(X,\mu)]}{E[g(X,\mu)]}$ increases with $\mu$.


How does $\tan^{-1}(x-\sqrt{1+x^2})=\frac{1}{2}\tan^{-1}x+C$ directly? I'm teaching baby calculus recitation this semester, and I meet a problem to calculate the derivative of
$$y=\tan^{-1}(x-\sqrt{1+x^2})$$
Just apply the chain rule and after some preliminary algebra, I find
$$\frac{dy}{dx}=\frac{1}{2(1+x^2)}$$
What surprises me is that the result implies
$$y=\frac{1}{2}\tan^{-1}x+C$$
Can anyone tell me how to see that directly?

Consider the following figure:

Triangles $\triangle ABC$ and $\triangle ACD$ are right triangles, and $AB\cong BD$, so $\triangle ABD$ is isosceles with

$$\angle BAD=\angle BDA={1\over2}(\pi-\angle ABD)={1\over2}\left(\pi-\left({\pi\over2}-\angle BAC \right)\right)={\pi\over4}+{1\over2}\angle BAC$$

where all angles are taken in a positive sense.  We also have

$$\angle CAD=\angle BAD-\angle BAC={\pi\over4}-{1\over2}\angle BAC$$

Finally, taking care with minus signs, we have

$$\angle CAD=-\tan^{-1}(x-\sqrt{1+x^2})\quad\text{and}\quad\angle BAC=\tan^{-1}x$$

which gives the desired identity

$$\tan^{-1}(x-\sqrt{1+x^2})={1\over2}\tan^{-1}x-{\pi\over4}$$

Remark (added later):  The picture and proof here assume $x\ge0$.  To cover the case $x\lt0$ requires either its own picture or an appeal to some kind of analytic continuity.

First, you have $\arctan (\tan y)=y$ for $y\in]-\pi/2, \pi/2[$.

And for all $y\neq \pi/4+k\pi/2$,
$$\tan (2y)=\frac{2\tan y}{1-\tan^2 y}$$

Thus, for all $z\in]-1,1[$, $y=\arctan z$ is in $]-\pi/4,\pi/4[$, and you can apply the above:

$$\tan(2\arctan z)=\frac{2z}{1-z^2}$$

And since $2\arctan z \in ]-\pi/2,\pi/2[$, you can also apply the first identity to get

$$\arctan \tan (2\arctan z)=2\arctan z=\arctan \left(\frac{2z}{1-z^2}\right)$$

Bear in mind that this is proved only for $z\in]-1,1[$, and it actually is wrong outside this interval.

Now, for $x>0$, $x-\sqrt{1+x^2}<0$ since $1+x^2>x^2$ implies $\sqrt{1+x^2}>x$.

But it's also $>-1$, since

$$(1+x)^2=x^2+1+2x>x^2+1$$

And taking square root, $1+x>\sqrt{1+x^2}$

Thus, for $x>0$, $-1<x-\sqrt{1+x^2}<0$.

It's thus a good candidate to apply the formula above with arctangent, which is true for $|z|<1$:

That is, for $x>0$,

$$2\arctan (x-\sqrt{1+x^2})=\arctan \left(\frac{2(x-\sqrt{1+x^2})}{1-(x-\sqrt{1+x^2})^2}\right)$$

And

$$\frac{2(x-\sqrt{1+x^2})}{1-(x-\sqrt{1+x^2})^2}=\frac{2(x-\sqrt{1+x^2})}{-2x(x-\sqrt{1+x^2})}=-\frac{1}{x}$$

Thus, still for $x>0$,

$$2\arctan (x-\sqrt{1+x^2})=-\arctan\frac{1}{x}$$

And since $\arctan x+\arctan \frac1x=\pi/2$ for $x>0$, you have

$$2\arctan (x-\sqrt{1+x^2})=\arctan (x) -\pi/2$$

Now, we have to prove the same for $x<0$.

Then, $x-\sqrt{1+x^2}<-1$, so we can't apply our formula with arctangent.
However, its inverse is in $]-1,0[$, and for $x<0$,

$$\arctan (x-\sqrt{1+x^2})+\arctan \left(\frac{1}{x-\sqrt{1+x^2}}\right)=-\pi/2$$

$$\arctan (x-\sqrt{1+x^2})-\arctan (x+\sqrt{1+x^2})=-\pi/2$$
$$\arctan (x-\sqrt{1+x^2})=\arctan (x+\sqrt{1+x^2})-\pi/2$$

Then, still for $x<0$, $x+\sqrt{1+x^2} \in ]0,1[$ and

$$2\arctan (x+\sqrt{1+x^2})=\arctan \frac{2(x+\sqrt{1+x^2})}{1-(x+\sqrt{1+x^2})^2}$$

$$2\arctan (x+\sqrt{1+x^2})=\arctan \frac{2(x+\sqrt{1+x^2})}{-2x(x+\sqrt{1+x^2})}=-\arctan \frac1x$$

And $\arctan x+\arctan \frac1x=-\pi/2$ (remember that $x<0$), hence

$$2\arctan (x+\sqrt{1+x^2})=\arctan (x)+\pi/2$$

And finally

$$2\arctan (x-\sqrt{1+x^2})=2\arctan (x+\sqrt{1+x^2})-\pi=\arctan (x)-\pi/2$$

This formula is now proved for $x>0$ and for $x<0$. It's also true for $x=0$, since then it amounts to $2\arctan (-1)=-\pi/2$.


Power of Hypothesis Test for Variance Ratio from Two Normal Samples Consider independent random samples from two normal distributions, $X_i\sim N(\mu_1,\sigma_1^2)$ and $Y_j\sim N(\mu_2,\sigma_2^2)$ for $i=1,...,n_1$ and $j=1,...,n_2$.  Let $n_1=n_2=9,\bar{x}=16,\bar{y}=10,s_1^2=36,$ and $s_2^2=45$.
For testing $H_0:\sigma_2^2/\sigma_1^2\le1$ vs $H_1:\sigma_2^2/\sigma_1^2>1$, what is the power of this test if in fact $\sigma_2^2/\sigma_1^2=1.33?$  Use a significance level of $\alpha=0.05$.

I know that the power of a test is the probability of rejecting $H_0$ given that $H_a$ is actually true, but computing this probability is where I'm getting stuck; essentially, I need to compute:
$$P(S_1^2/S_2^2>F_\alpha(8,8)|_{\sigma_2^2/\sigma_1^2=1.33})$$
Where $F$ is just the fisher percentile and the bar represents "given (conditional probability)."  How do I eliminate this conditional probability so that I can actually calculate a numerical value here?

The answer is easy. The variance ratio, $V2/V1 = 1.333$, so $L = V1/V2 = 3/4 = 0.75, UT = F^{-1}(1-0.05, 8, 8) = 3.4381, \beta = F(UT/L, 8, 8, True) = 0.899$ and Power $= 1 - \beta = [0.101].$

Minitab confirmed this since $n_1=n_2$ and a VisualBasic $6$ Monte-Carlo simulation showed $0.101$ after $30$ million iterations using the Box-Muller generator.

I don't know why this simple formula for $F$ distribution Power is not well published like the $\chi^2$ power formula is.


Derivative of vector consisting of euclidean distances I have $g: \Bbb R^2 \to \Bbb R^4$ given by $g(x) = (\|c_1 - x\|, \dots, \|c_4 - x\|)$, where $c_1, \dots, c_4 \in \Bbb R^2$.

I want to find $\left( \dfrac {\partial g} {\partial x_1}, \dfrac {\partial g} {\partial x_2} \right)$. Any hints?

If my understanding is correct, you have

$$
g(x) =  f_1(x) f_12(x) f_13(x) f_14(x)
$$

with $f_i(x) = || \bf c_i - x ||$

Evaluating $\frac{\partial g}{\partial x_i}$ then is simply a matter of using the nested chain rule.


Understanding the Graph of a Double Integral I am trying to understand the graph of $\int_0^1\int_{8y}^1dxdy$. Upon solving the double integral, I get $\int_0^1{1-8y}dy\Rightarrow-3.$ Which I believe is correct, but I'm mainly putting this work in case it matters for the graph or I did something wrong. But upon graphing the boundary equations, I get the following graph https://www.desmos.com/calculator/xofiu5ggqs.

This is what I don't understand. I have not seen this particular case. I'm not sure what happens when a horizontal bound "cuts off" (for lack of a better phrase),  a vertical bound like this. So does this double integral solve for the red region or black region or something entirely different? Also, to my understanding, a double integral will only be negative if the region is below the x-axis as well so perhaps my work is incorrect as well. So there's a lot I don't understand about this graph.

Edit: I saw some posts that were deleted by their poster that stated that the result of $-3$ comes from the difference of the triangle formed by the intersection of $x=8y, y=1$ and the red region. Is it proper for this to happen? In other words, is it not acceptable for a double integral to represent the area of two regions or is it simply a unique case?

As you can see from Desmos the boundary of your integral does not describe any valid region since it is doing the black triangle plus the negative of a triangle to the right of the red region. You'll have to ask your professor to correct the typo.


Probability $2$ people have a birthday in the same month out of $7$ What is the probability that $2$ people in the group have a birthday in the same month out of $7$ people?

I know the answers $88.85\%$ however I want to know how to work it out using factorials instead of fractions.

HINT: Please do tell about your approach next time.

Use exclusion:

$P[$ atleast $2$ people have birthday in same month$ ]= 1-P[$no two people have birthday on same month$]$
$= 1-\frac{12.11.10.9.8.7.6}{12^7}=0.88$


Find all polynomials that fix $\mathbb Q$ and the irrationals Problem: Describe all polynomials $\mathbb{R}\rightarrow\mathbb{R}$ with coefficients in $\mathbb C$ which send rational numbers to rational numbers and irrational numbers to irrational numbers.

Take a polynomial $f=a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0$, and assume $f$ satisfies the problem conditions.

Since $f(1),f(2),f(3),...,f(n)$ are all rationals, we can use them to get $n$ independent equations in the variables $a_i$. Then, we can solve for all the $a_i$ with linear algebra. Linear algebra tells us that this equation-solving uses only +,-,*, and /, so all coefficients $a_i$ of the polynomial must be rational.

Now, multiply $f$ by all the denominators of all the $a_i$ to get a new polynomial $g=b_nx^n+b_{n-1}x^{n-1}+...+b_1x+b_0$ with integer coefficients. Clearly $g$ satisfies the problem conditions as well.

Consider all values $x$ for which $g(x)$ is an integer. By the Rational Root Theorem, any such $x$ can be expressed as $\frac{k}{b_n}$ for some integer $k$. From this we deduce that any two values $x$ for which $g(x)$ is an integer are spaced at least $\frac{1}{b_n}$ apart.

But if the polynomial $g(x)$ has degree at least $2$, it attains arbitrarily large slope as $x$ goes to $\infty$. In particular, we can pick $x$ so large that $g(x+\frac{1}{b_n})>g(x)+100$. But then $g$ attains many integer values between $x$ and $x+\frac{1}{b_n}$, violating the deduction in the previous paragraph. Therefore, $g$ has degree at most $1$.

After proving this, we are done: the only polynomials $f$ that work are the ones in the form $a_1x+a_0$, where both $a_1$ and $a_0$ are rational.

Any polynomial of the form $\prod (x-r_i)$, where the $r_i$ are rational, will do it. If a polynomial has an irrational root (i.e., the real or imaginary part is irrational), then it maps that irrational number to zero, which is rational.

Therefore, the polynomial must be of the form stated above.

This seems too easy and simple. So, what is wrong?