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32

You may write, for $N \geq 2$, $$ \begin{align} e^{3/2}\prod_{n=2}^{N}e\left(1-\dfrac{1}{n^2}\right)^{n^2}&=e^{3/2}\times\prod_{n=2}^{N}e\times\prod_{n=2}^{N}\left(1-\dfrac{1}{n^2}\right)^{n^2}\\\\ &=e^{3/2}\times e^{N-1}\times\prod_{n=2}^{N}\dfrac{(n-1)^{n^2}}{n^{n^2}}\dfrac{(n+1)^{n^2}}{n^{n^2}}\\\\ &=e^{3/2}\times ...


21

We have for the sequence $f(n)$ $$\begin{align} f(n)&=\sum_{k=1}^n2^k\\\\ &=2^{n+1}-2\tag 1 \end{align}$$ where we summed a Geometric Progression to arrive at $(1)$. Thus, the continuation of $f$ for real arguments is $$\bbox[5px,border:2px solid #C0A000]{f(x)=2^{x+1}-2}$$ NOTE: We remark that the continuation is not unique inasmuch as we ...


12

One can write the log of the product as $$\sum_{n=2}^{\infty} \left [1+n^2 \log{\left (1-\frac1{n^2} \right )} \right ] $$ Now, $$\log{\left (1-\frac1{n^2} \right )} = -\int_0^1 \frac{du}{n^2-u} $$ So the sum is equal to $$-\int_0^1 du \, u \sum_{n=2}^{\infty} \frac1{n^2-u} $$ $$\sum_{n=-\infty}^{\infty} \frac1{n^2-u} = -\frac{\pi \cot{\pi ...


9

You may just use the fact that, as $x \to 0$, using the Taylor expansion, $$ e^x=1+x+O(x^2) $$ giving, for some $n_0\geq1$, $$ \sum_{n\geq n_0}\left(\exp \left(\frac{(-1)^n}{n}\right)-1\right)=\sum_{n\geq n_0}\frac{(-1)^n}{n}+\sum_{n\geq n_0}O\left(\frac1{n^2}\right). $$ The latter series is absolutely convergent and the series $\displaystyle \sum_{n\geq ...


9

One way is to use 'Feynman's trick', and differentiate under the integral. Note $$\int_0^\infty x^4e^{-\alpha x^2}=\int_0^\infty \frac{\partial^2}{\partial \alpha^2}e^{-\alpha x^2}=\frac{\partial^2}{\partial \alpha^2}\int_0^\infty e^{-\alpha x^2}$$ Calculate the inner definite integral first, then differentiate it twice with respect to $\alpha$.


7

so start with for $a>1$ and $\forall n$ big enough we have $$n^a < a^n \Leftrightarrow \log(n^a)<\log(a^n)\Leftrightarrow a\log(n)<n\log(a)\Leftrightarrow 1<\frac{\log(a)}{a}\frac{n}{\log(n)} $$ and we fixed $a$, so $\frac{\log(a)}{a}=c>0$ is just a constant. So in fact we have to show, that $$ 1<c\frac{n}{\log(n)} $$ holds for all ...


6

HINT : $$4.8=1000^{\frac 1x},\ \ \ 0.8=1000^{\frac 1y}$$ So, $$1000^{\frac 1x-\frac 1y}=\frac{1000^{1/x}}{1000^{1/y}}=\frac{4.8}{0.8}$$


6

I've checked my answer with Mathematica and it gives me the same, it's much more complicated as you thought, so your answer isn't good! $$z = \frac{1 + i}{\sigma \delta \left(1 - e^{-(1 + i)t/\delta} \right)}= \frac{1 + i}{\sigma \delta \left(1 - e^{-\frac{(1+i)t}{\delta}} \right)}$$ Assuming $\sigma,\delta,t\in \mathbb{R}$: $$\Re\left(\frac{1 + ...


6

I don't think this has anything to do with the complex logarithm function. If the series $\sum a_i$ converges, then the left-hand side will exist (because $\exp$ is continuous): $\exp (\sum \limits _{i=1} ^\infty a_i) = \exp (\lim \limits _{N \to \infty} \sum \limits _{i=1} ^N a_i) = \lim \limits _{N \to \infty} \exp (\sum \limits _{i=1} ^N a_i) = \lim ...


6

This is incorrect as written. E.g., for $x = 0, y = 1$, $$LHS = 1 - e^{-1/2} \approx 0.393 \color{red}{>} \frac{1}{4} = RHS$$ However it is true that $$\left|e^{-x/2} - e^{-y/2}\right| \leq \frac{|x-y|}{2} \quad\text{ for all } x, y \geq 0$$ To see this, write $f(x) = e^{-x/2}$. Then by the Mean Value Theorem, for all $0 \leq x < y$, ...


6

You may write $$ \begin{align} \sum_1^{\infty} \frac{n(n+1)}{2(n!)}&=\sum_1^{\infty} \frac{n(n-1)+2n}{2(n!)}\\\\ &=\sum_2^{\infty} \frac{n(n-1)}{2(n!)}+\sum_1^{\infty} \frac{2n}{2(n!)}\\\\ &=\frac12\sum_2^{\infty} \frac{1}{(n-2)!}+\sum_1^{\infty} \frac{1}{(n-1)!}\\\\ &=\frac12\sum_0^{\infty} \frac{1}{n!}+\sum_0^{\infty} \frac{1}{n!}\\\\ ...


5

Let $F(a)$ be defined as $$F(a)=\int_0^{\pi}e^{i(x+a\cos x)}\sin x\,dx \tag 1$$ Now, it is easy to show that the real part of $F$ is identically $0$. To do this, we write the real part of $(1)$ as $$\text{Re}\lbrace F(a)\rbrace=\int_0^{\pi/2}\cos(x+a\cos x)\sin x\,dx+\int_{\pi/2}^{\pi}\cos(x+a\cos x)\sin x\,dx \tag 2$$ Then, enforcing the substitution ...


5

Another way to look at it (that probably isn't simpler) : Let $u=x^2$, then the integral becomes: $$\frac{1}{2} \int_0^\infty u^{3/2}e^{-\alpha u}du$$ which is the Laplace transform of $u^{3/2}$, times ${1}/{2}$. We know that the Laplace transform of $u^n$ is $n!*\alpha^{-n-1}$ or generally $\Gamma(n+1)*\alpha^{-n-1}$. Here, $n=3/2$ so $$\Gamma(3/2 ...


5

HINT: $$\sum_{i=1}^n 2^i=2(2^n-1)$$


5

The other answers show you how to derive the function $f(x) = 2^{x+1} - 2$, but I'm just going to show how the problem is approached (in complete unrigorous terms). I know from experience how magical it seems that the above answers just managed to come up with the right function all of a sudden. So, we look at the problem and we can see that it's forming a ...


5

You could either use numerical methods such as interval bisection, Newtons Method of finding roots applied to $f(x) = 2^x + x - n$, as in Winther's solution. Secondly, you could solve this equation manually to get the solutions as $x=0$ for $n=1$ and $$x=n-\frac{W_{c}\left(2^n \ln 2\right)}{\ln 2}$$ where we let $c$ range over the integers. If you are ...


5

Since $$\lim_{x \to 2^{+}} \frac{3}{2(2-x)} = -\infty$$ then $$\lim_{x \to 2^{+}}\exp\left(\frac{3}{2(2-x)}\right) = \exp\left( \lim_{x \to 2^+} \frac{3}{2(2-x)}\right) = 0$$


5

Hint: Factor it as $$(k+1)2^x=8$$ and divide by $k+1$ to obtain $$2^x=\frac{8}{k+1}$$ provided that $k\neq-1$. What can we say about above equation?


5

Take natural logs of both and use Stirling. $$\ln(k^m) =m \ln(k) \text{ and } \ln(n!) \approx \frac12 \ln(2 \pi)+(n+\frac12)\ln(n) - n. $$ Comparing these should be no problem.


4

Hint: Put $-\alpha x^2=u \Rightarrow -2\alpha x \mathrm dx=\mathrm du$,then $x\mathrm dx=\frac {-1}{2\alpha}\mathrm du$. you will have, $$\displaystyle\int xe^{-\alpha x^2}\,dx=\frac {-1}{2\alpha}\displaystyle\int e^u \mathrm du$$


4

Taking logs and using the power series for $\log(1-x)$, we get $$ \begin{align} \frac32+\sum_{n=2}^\infty\left[1+n^2\log\left(1-\frac1{n^2}\right)\right] &=\frac32-\sum_{n=2}^\infty\sum_{k=1}^\infty\frac1{(k+1)n^{2k}}\\ &=\frac32-\sum_{k=1}^\infty\frac{\zeta(2k)-1}{k+1}\tag{1} \end{align} $$ Transcribing $(5)$ from this answer: $$ \begin{align} ...


4

The standard approach here is to use the CDF of your distribution, which is defined on $\mathbb{R}$ (regardless of the support of your distribution). For any finite distribution, you generally use cumsum to calculate the CDF; for an infinite distribution, you'll need an explicit formula for $F:\mathbb{R}\rightarrow[0,1]$, where $F(x)=\mathbb{P}[X\leq x]$. ...


4

It's a local minimum. In order to establish that it's a global minimum, you'll need to do more than take the derivative and apply the first or second derivative test. One theorem you can use is Rolle's theorem (or the more general version, the mean value theorem). It states that if you have a function $f : [a, b] \rightarrow \mathbb{R}$ that is continuous, ...


4

You are correct up till $$\int e^{-x} +2 + e^x \, \mathrm{d}x$$ This evaluates to $$e^x + 2x - e^{-x} + c.$$ This is due to the "rule" that $$\int e^{\alpha x} \, \mathrm{d}x = \frac{e^{\alpha x}}{\alpha} + c.$$ In the case that $\alpha = 1$ we get $\int e^x \, \mathrm{d}x = e^x + c$. For $\alpha = -1$ we get $\int e^{-x} \, \mathrm{d}x = \frac{e^{-x}}{-1} ...


3

We can show that the only $nice$ solutions to $$f(x^y)=f(x)^{f(y)}\qquad(1)$$ are the identity function, and the constant functions $1$ and $-1$. There are also a few $ugly$ solutions. To avoid problems with $0^0$, let's first find all functions $f$ such that $f:(0,+\infty)\to(-\infty,0)\cup(0,+\infty)$. (Note that because the question is about nonnegative ...


3

Newton's method is usually the best numerical method for such problems: it's very simple, converges very fast and it works most of the time. To use it define a function $f(x) = 2^x +x - n$ so that the desired solution to your problem is the solution to $f(x) = 0$. Pick a guess $x_0$, say $x_0 = 0$, and the itterate via $$x_{n+1} = x_n - ...


3

For first, get rid of the extra parameter by setting $c=\frac{b}{a^2}$: $$I= \int_{\mathbb{R}}e^{ibx^2}\frac{\sin(ax)}{x}\,dx = \int_{\mathbb{R}}e^{icx^2}\frac{\sin x}{x}\,dx=\text{Im PV}\int_{\mathbb{R}}e^{icx^2+ix}\frac{dx}{x} $$ then translate the $x$ variable in order to get: $$ I = ...


3

If you have an $f: \mathbb{N} \rightarrow [0,1]$ function such that $n$ is a side and $f(n)$ the probability of getting that side, and if we can produce a function $g: \mathbb{N} \rightarrow [0,1]$ such that $g$ is invertible and: $$ g(n) = \sum_{i=1}^n f(n) $$ Then you could just produce a random float $ k\in [0,1]$ and return $g^{-1}(k)$ Edit: As was ...


3

Following what wythagoras did in his answer, we arrive at $$2^x = \frac{8}{k+1}$$ Now, the function $2^x$ is strictly increasing and is always positive. (*) So, we must have $$\frac{8}{k+1} > 0$$ and $k > -1$. (*) To see what I mean, look at the graph below: the graph is the function $2^x$, as you can see the function hits every value greater than ...



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