limits - Find $lim_{n to infty} frac{(-1)^n n^{frac{3}{2}}}{5e^n}$

Question: Find $\lim_{n \to \infty} \frac{(-1)^n n^{\frac{3}{2}}}{5e^n}$

My attempt: Since $(-1)^n$ can be written as $\cos(\pi n)$ we can use the squeeze theorem as such: $ -1 \leq \cos(\pi n) \leq 1 \implies \frac{-n^{\frac{3}{2}}}{5e^n} \leq \frac{\cos(\pi n)\cdot n^{\frac{3}{2}}}{5e^n}\leq \frac{n^{\frac{3}{2}}}{5e^n}$ and so we can take the limit as $n$ approaches zero and because exponentials grow faster than polynomials we get: $0\leq \frac{\cos(\pi n)\cdot n^{\frac{3}{2}}}{5e^n}\leq 0$ so $\lim_{n \to \infty} \frac{(-1)^n n^{\frac{3}{2}}}{5e^n} =0 $.

Is this correct? If so what other ways could I have solved this limit?

Answer

alternative method:-$|a_n|=\frac{n^{\frac{3}{2}}}{5e^n}$. Consider the series $\sum_{n=0}^{\infty}|a_n|$. Use cauchy's nth root test, we get $\sum_{n=0}^{\infty}|a_n|$ converges. so $\lim_{n\to\infty}|a_n|=0.$ Hence, $\lim_{n\to\infty}a_n=0$.

probability - Calculate $int_{-infty}^infty x^2e^{-x^2} dx$ using Gaussian random variables and the properties of pdfs

I've been asked the following question in an exam of probability and statistics undergraduate course. Help is appreciated.

Calculate the numeric value of $\int_{-\infty}^\infty x^2e^{-x^2} dx$ using Gaussian random variables and the properties of pdfs (probability density functions).

Note: The normal cdf table was given

Here is how I tried to solve it:

• Take the normal pdf and use the fact that pdfs are equal to 1 when integrated from $-\infty$ to $\infty$.

$\int_{-\infty}^\infty \frac{1}{\sigma \sqrt{2\pi}} e^{-\frac{(x-\mu)^2}{2\sigma^2}} dx = 1$

• Let $\mu = 0$ and $\sigma^2 = \frac{1}{2}$. And the integral simplifies to this.

$\int_{-\infty}^\infty e^{-x^2} dx = \sqrt{\pi}$

• Then I've tried to integrate the main integral by parts.

$\int x^2e^{-x^2} dx = x^2 \int e^{-x^2} dx - \int2x(\int e^{-x^2} dx) dx $

• Plug in $\sqrt{\pi}$ where needed and simplify.

$= x^2 \sqrt{\pi} - \int 2x( \frac{\sqrt{\pi}}{2} erf(x) ) dx $

• Further simplifying I get the following and this is where I stuck.

$= x^2 \sqrt{\pi} - \sqrt{\pi} \int x \, erf(x) dx $

• The functions don't converge, so I can't proceed. Also, I don't know how to integrate the error function.

If you know how to proceed, or if you think this should've been solved in another way, please answer. Note that I need a solution involving Gaussian random variables and the properties of pdfs (as required.)

Answer

The step where you "plug in $\sqrt{\pi}$ as needed" is incorrect. On the preceding line you have indefinite integrals, where you haven't kept track of the bounds of integration. Turning it into definite integrals would also affect the factor $x^2$, for example.

Also, it's much better to do the integration by parts like this, and not involve the error function:
$$
\int_a^b (x e^{-x^2}) \cdot x \, dx = \Bigl[(-\tfrac12 e^{-x^2} ) \cdot x \Bigr]_a^b - \int_a^b (-\tfrac12 e^{-x^2} ) \cdot 1 \, dx

,
$$
and then see what happens as $a \to -\infty$ and $b \to \infty$.

(I know this isn't a probabilistic solution, but it's just to comment on your attempt.)

real analysis - Show that $sum_{n=1}^{infty}{frac{1}{n^2}}=frac{pi^2}{6}$

Show that $$\sum_{n=1}^{\infty}{\frac{1}{n^2}}=\frac{\pi^2}{6}$$ Anyone can help ?

Answer

This is known as the Basel problem and was first solved by Euler. His derivation (shown at above link) does some clever manipulations with the power series expansion of $\frac{\sin(x)}{x}$.

A more advanced proof uses Fourier transforms and Parseval's identity for the function $f(x)=x$.

The link also gives a rigorous but elementary proof, requiring only trigonometric identities and binomial coefficients, together with the "pinching lemma", but no calculus.

calculus - How to find $lim_{x to 0}frac{1-cos(2x)}{sin^2{(3x)}}$ without L'Hopital's Rule.

How would you find $\displaystyle\lim_{x \to 0}\frac{1-\cos(2x)}{\sin^2{(3x)}}$ without L'Hopital's Rule?

The way the problem is set up, it makes me think I would try and use the fact that

$\displaystyle\lim_{x \to 0}\frac{1-\cos(x)}{x}=0$ or $\displaystyle\lim_{x \to 0}\frac{\sin(x)}{x}=1$

So one idea I did was to multiply the top and bottom by $2x$ like so:

$\displaystyle\lim_{x \to 0}\frac{1-\cos(2x)}{\sin^2{(3x)}}\cdot\frac{2x}{2x}$.

Then I would let $\theta=2x$:

$\displaystyle\lim_{\theta \to 0}\frac{1-\cos(\theta)}{\sin^2{(\frac{3}{2}\theta)}}\cdot\frac{\theta}{\theta}$

which would let me break it up:

$\displaystyle\lim_{\theta \to 0}\frac{1-\cos(\theta)}{\theta}\cdot \frac{\theta}{\sin^2(\frac{3}{2}\theta)}$

So I was able to extract a trigonometric limit that is zero or at least see it. The second part needed more work. My immediate suspicion was maybe the whole limit will go to zero, but when checking with L'Hopital's Rule, I get $\frac{2}{9}$..... :/

Answer

Also try this one: with
$$1-\cos2t=2\sin^2t$$
then
$$\lim_{x \to 0}\frac{1-\cos(2x)}{\sin^2{(3x)}}=\lim_{x \to 0}\frac{2\sin^2x}{\sin^23x}\times\dfrac{(3x)^2}{2(x)^2}\times\dfrac{2}{9}=\lim_{x \to 0}\frac{2\sin^2x}{2(x)^2}\times\dfrac{(3x)^2}{\sin^23x}\times\dfrac{2}{9}=\color{blue}{\dfrac{2}{9}}$$

functions - Show that for any subset $Csubseteq Y$, one has $f^{-1}(Ysetminus C) = X setminus f^{-1}(C)$

Let $f: X\rightarrow Y$ be a map

Show that for any subset $C\subseteq Y$, one has

$f^{-1}(Y\setminus C) = X \setminus f^{-1}(C)$

In this case $f^{-1}$ refers to preimage

I started off with trying to show $f^{-1}(Y\setminus C) \subseteq X \setminus f^{-1}(C)$

Let $ x\in f^{-1}(Y\setminus C) \Rightarrow x \in f^{-1}(Y), x\notin f^{-1}(C)\Rightarrow x\in X, x\notin f^{-1}(C) \Rightarrow x\in X \setminus f^{-1}(C) \Rightarrow f^{-1}(Y\setminus C) \subseteq X \setminus f^{-1}(C) $

Then I tried to show $f^{-1}(Y\setminus C) \supseteq X \setminus f^{-1}(C)$

Let $ x \in X\setminus f^{-1}(C) \Rightarrow x \in X, x \notin f^{-1}(C)$ and since $ C \subseteq Y$ ,if $ x\notin f^{-1}(C)$ , $x$ must be in $f^{-1}(Y\setminus C)$ , so $f^{-1}(Y\setminus C) \supseteq X \setminus f^{-1}(C)$.

Could anyone tell me if this is the correct way to answer this question? It almost seems like I'm repeating the same argument and it looks too simple. Would appreciate if anyone could point out any mistakes or if i should be more vigorous in my working. Thank you!

Answer

It's correct. I would only be careful with the "obvious" parts. For instance, when you say "$x$ must be in $f^{-1}(Y\setminus C)$". This is because $$x\in X \land x\notin f^{-1}(C)\implies f(x)\in Y\land f(x)\notin C \implies f(x)\in Y\setminus C\implies x\in f^{-1}(Y\setminus C)$$

integration - Calculus Question: Improper integral $displaystyleint_{-infty}^{infty} x^{2}e^{x-e^{2x}}dx$

I am curious about evaluation of the following integral

$$\int_{-\infty}^{\infty} x^{2}e^{x-e^{2x}}dx$$
Is it possible to evaluate it? This not my homework but I will share my attempt. I tried standard technique, integration by part but without any success. I also couldn't find any suitable substitution. The integral seems as if it were evaluating the expected value or moment generating function of a certain distribution but I couldn't find any pdf like the integrand in my textbook table.

Answer

Hint: Show that $\displaystyle\int_{-\infty}^\infty e^{ax-e^x}~dx=\Gamma(a).~$ Then, after substituting $x=2t$, differentiate twice with

regard to a, and let $a=\dfrac12$.

quadratics - Finding all pairs of integers that satisfy a bilinear Diophantine equation

The problem asks to "find all pairs of integers $(x,y)$ that satisfy the equation $xy - 2x + 7y = 49$.

So far, I've got

\begin{align}
xy - 2x + 7y &= 49 \\
x\left(y - 2\right) + 7 &= 49
\\
y &\leq 49
\end{align}

I can't get any further. Any help?