Got this limit I’d thought I should share it. So here you go:
$$\lim_{x\to\infty}{\bigg(\prod_{i=1}^{x}{\big(1+\frac{i}{x}\big)}\bigg)^{\frac{1}{x}}}$$
I have worked on this one for a while, with different angles and ended up with two different answers. Would like to see how you handled it.
Answer
Hint: Its logarithm is a Riemann sum.
It is known that $0^0$ despite being an indeterminate limit form, is usually defined to be equal to $1$. I wonder whether similar conventions exist for some other "indeterminate forms" in the context of two-point compactifications of real numbers. It would be great if someone showed that some authors used these conventions.
Particularly I am interested to know about usage of the following conventions:
$1^\infty=1$
$0 \cdot \infty=0$
$\infty^0=1$
$\frac 00=0$
I also would be interested whether any author proposed distinguishing between "definable" indeterminate forms (those which can be conveniently defined to have certain value, like $0^0$, $1^\infty$) and those which are more problematic, like $\infty-\infty$ or $\frac\infty\infty$ which cannot be conveniently defined.
I know it evaluates to $\frac{rt-\frac{1}{2}\sigma^2t}{2\sigma\sqrt{t}}$ but how to get there is the problem.
Using L'Hôpital you find $$\lim_{n\rightarrow\infty}\frac{e^{rt/n+\sigma\sqrt{t/n}}-1}{e^{2\sigma\sqrt{t/n}}-1} = \frac{1}{2}$$But L'Hôpital doesn't work on the whole thing. The only limit calculator that could figure it out was wolfram but that couldn't give me the steps to get there.
using Taylor series expansion gives me:$$\lim_{n\to\infty}\sqrt{n}(\frac{\sum_{k=0}^{\infty}(rt/n+\sigma\sqrt{t/n})-1}{\sum_{k=0}^{\infty}(2\sigma\sqrt{t/n})^k-1}-\frac{1}{2})$$But I fail to see how to go further from here, any help would be appreciated
Answer
$\displaystyle n:=\frac{t}{x^2}>0$
$$\lim\limits_{n\rightarrow \infty} \sqrt{n}\left(\frac{e^{rt/n+\sigma\sqrt{t/n}}-1}{e^{2\sigma\sqrt{t/n}}-1}-\frac{1}{2}\right)= \lim\limits_{x\rightarrow 0} \frac{\sqrt{t}}{2\sigma x}\left((rx+\sigma)\frac{e^{rx^2+\sigma x}-1}{rx^2+\sigma x}\frac{2\sigma x}{e^{2\sigma x}-1}-\sigma\right)$$
$$= \lim\limits_{x\rightarrow 0} \frac{\sqrt{t}}{2\sigma} (r+(rx+\sigma)\frac{rx+\sigma}{2!}\frac{(2\sigma x)^0}{0!}B_0+(rx+\sigma)\frac{(rx^2+\sigma x)^0}{1!}\frac{2\sigma }{1!}B_1+x\cdot …)$$
$$=\frac{\sqrt{t}}{2\sigma} (r-\frac{\sigma^2}{2})$$
$B_n$ are the Bernoulli numbers and we need here $B_0=1$ and $\displaystyle B_1=-\frac{1}{2}$ .
I need to prove that any irreducible polynomial $f$ of degree $d\,\big|\,n$ over $\mathbb F_p$ devides $x^{p^n} - x$. I know that the splitting field of the latter is the finite field with $p^n$ elements, and that if $\alpha$ is the root of $f$, then $[\mathbb F_p(\alpha):\mathbb F_p] = d$. I don't see, why all the roots of $f$ must lie in the splitting field of $x^{p^n} - x$. In fact, the splitting field of $f$ may have degree $d\,!$ over $\mathbb F_p$, which may well be greater than $n$.
Thank you.
Answer
Q: Why does $\mathbb{F}_{p^n}$ contain all the roots of an irreducible polynomial $f(x)\in\mathbb{F}_{p}[x]$ of degree $d$, $d\mid n$?
One way of seeing this is to observe that the extension $\mathbb{F}_{p^n}/\mathbb{F}_p$ is a Galois extension because it has $n$ distinct automorphisms (powers of the Frobenius automorphism). As a Galois extension it is normal, meaning exactly that any irreducible polynomial with coefficients in the prime field and a root in the bigger field splits there into a product of linear factors.
A possibly more concrete way of seeing the same thing is to observe that if $\alpha$ is a root of $f(x)$, then so are $\alpha^p$, $\alpha^{p^2}$ et cetera. The conjugates $\alpha^{p^i}$ must start repeating at some point, so we get, first $\alpha^{p^i}=\alpha^{p^j}$ for some integers $i,j$ such that $0
At this point we know $\ell$ roots of $f(x)$, namely $\alpha^{p^i},0 \le i<\ell.$ Consider
the polynomial
$$
g(x)=\prod_{i=0}^{\ell-1}(x-\alpha^{p^i})=x^\ell+\sum_{i=0}^{\ell-1}a_ix^i
$$
for some coefficients $a_i\in\mathbb{F}_p[\alpha]$. Because $F:z\mapsto z^p$ is an automorphism of $\mathbb{F}_p[\alpha]$, we see that
$$
g^F(x):=x^\ell+\sum_{i=0}^{\ell-1}F(a_i)x^i=\prod_{i=0}^{\ell-1}(x-F(\alpha^{p^i}))
=\prod_{i=0}^{\ell-1}(x-(\alpha^{p^i})^p)=\prod_{i=1}^{\ell}(x-\alpha^{p^i})=g(x).
$$
Here in the last step we used the fact $\alpha^{p^\ell}=\alpha$. Therefore $F(a_i)=a_i$ for all $i$, in other words $g(x)\in\mathbb{F}_p[x]$. Because $g(x)$ is a non-trivial factor of $f(x)$ with coefficients in the prime field, we must have $g(x)=f(x)$, and hence also that $\ell=d$. This means that all the roots $\alpha_i$ of $f(x)$ are of the form $\alpha^{p^i}$ for some $i$. In particular they all satisfy the equation $\alpha_i^{p^d}=\alpha_i$. Consequently we also have $\alpha_i^{p^n}=\alpha_i$ for all $i$, which is what we wanted to show.
Question about solving congruence. I've worked out how to solve them for the most part except for the following problem I'm having:
$$45x \equiv 15 \pmod{78}$$
By the euclidean algorithm, I work out that the gcd of 45 and 78 = 3 which means there exists 3 solutions.
I divide through the congruence by the gcd, 3, to get a new congruence:
$$15x \equiv 3 \pmod{26}$$
The gcd of this is 1, which means there exists 1 unique solution. By extended euclidean algorithm I work out that the solution is $21 \pmod{26}$
But now I'm asked to find the solution in terms of the original modulus, mod 78.
In my understanding to do this all you need to do is take the solution of the new congruence, which is 21, and keep adding 26 two more times to get 3 different solutions (which works for other problems I've done), which gets me solutions: 21, 47 and 73 (mod 78) but this is incorrect.
The correct solutions are 9, 35, 61.
What am I doing wrong?
Answer
When you divide through by 3, the resulting congruence should be
$$15x\equiv 5\pmod{26},$$
not
$$15x\equiv 3\pmod{26}.$$
Today I had an exam and I mixed up the integration by parts formula.
The question was to integrate
$$ \int\nolimits_0^\infty \frac{e^{-at} - e^{-bt}}{t} \text{d}t $$
I will try solve this again with the right formula when I arrive home. I would appreciate if somebody could tell me the solution so I can double check and maybe give a hint to another way of solving this instead of integration by parts (if possible).
$$f(t)=\frac{\sin^3(\pi t)}{(\pi t)^3}$$
I want to calculate the Fourier transform.
I can't calculate this integral:
$$\int_0^\infty\frac{\sin^3(\pi t)}{(\pi t)^3}\cos(ut)\,\mathrm{d}t$$
