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5

Use the Stirling formula: $$\frac{(2n - 2)(2n - 4)\cdots 4 \cdot 2}{(2n - 3)(2n - 5) \cdots 3 \cdot 1} =\frac{(2n - 2)^2(2n - 4)^2\cdots 4^2 \cdot 2^2} {(2n - 2)!} \\= 2^{2n-2} \frac{(n - 1)^2(n - 2)^2\cdots 2^2 \cdot 1^2} {(2n - 2)!} \\=2^{2n-2} \frac{((n - 1)!)^2}{(2n - 2)!} \sim 2^{2n-2}\frac{(n-1)^{2n-2}e^{-2n+2}2\pi n} {(2n - ...


5

You may find : This curious near-identity was apparently noticed almost simultaneously around 1988 by N. J. A. Sloane, J. H. Conway, and S. Plouffe, but no satisfying explanation as to "why" $e^\pi-\pi \simeq 20$ is true has yet been discovered. after some googling.


4

One way to define $e^x$ is via $$ e^x = \lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^n $$ By re-arranging your forumula a bit, you get $$ \frac{1}{2N}\left(1 - \frac{1}{2N}\right)^t = \frac{1}{2N}\underbrace{\left(1 + \frac{\frac{-t}{2N}}{t}\right)^t}_{\text{Compare to $e^x$ for $x=\frac{-t}{2N}$}} \approx \frac{1}{2N}e^{-\frac{-t}{2N}} \text{,} ...


3

I read this post a while ago, and since then I found an extremely good approximation for $e^x$ $e^x=1+\cfrac{2x}{2-x+\cfrac{x^2}{6+\cfrac{6x^2}{10+\cfrac{10x^2}{14+...}}}}$ Truncated here, this gives an error of less than $10^{-7}$ for $x\in(-1.27,1)$. Every additional row increases this range by about 0.4 on both sides, but by using the repeated squaring ...


3

Rewriting the sums a bit, you're asking how to show that $$ \sum_{k=2}^{\infty} \frac{\ln k}{k^2} - \sum_{k=n+1}^{\infty} \frac{\ln k}{k^2} \approx -\ln n \sum_{k=n+1}^{\infty} \frac{1}{k^2} + C. $$ Looking at this one might guess that we'll probably have $C = \sum_{k=2}^{\infty} \frac{\ln k}{k^2}$, so we would just need to prove that $$ ...


3

The proximity to $\pi$ is a pure coincidence. Such coincidences are numerous. For example the root of the equation $x^4+x^5=e^6$ is $x=\pi+0.000 000 029$. A lot of much more accurate coincidences are shown on this page : http://www.contestcen.com/pi.htm For each one of these coïncidences, on could be surprised and ask : "Why is it so close ? ". This is a ...


2

The actual probability of getting the outcome at least once over $n$ trials can be computed pretty easily: it is one minus the probability that the outcome never occurs. So if the outcome has a probability of $p$ of occurring in an individual trial, then over $n$ trials the probability is $$ 1 - (1 - p)^n $$ Going from here to the approximation in your book ...


2

I have used your $\sqrt{x^2+\epsilon}$ function once before, when an application I was working on called for a curve with a tight radius of curvature near $x=0$. Whether it is best for your purpose might depend on your purpose. (Side note: The derivative of $|x|$ does not exist at $x=0$. In physics, we sometimes cheat and write ${d\over dx}|x| = ...


2

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} ...


2

We are asked to find the value of $x$ where: $$\int_0^x \dfrac{1}{\sqrt{2 \pi}} e^{-t^2/2}~dt = 0.45$$ We need two numerical approaches here. One to find the zeros of the function $f(x)$, Newton's Method, and one to estimate the integral, Composite Simpson and Composite Trapezoidal, above where: $$f(x) = \int_0^x \dfrac{1}{\sqrt{2 \pi}} e^{-t^2/2}~dt - ...


2

Your last integral formula for the hypergeometric function is off; according to Wikipedia it should be $$ {}_2F_1(1,1+a,2+a,z) = (1+a) \int_0^1 \frac{t^a}{1-tz}\,dt. $$ Now if we make the change of variables $tz = s$ we get $$ {}_2F_1(1,1+a,2+a,z) = \frac{1+a}{z^{1+a}} \int_0^z \frac{s^a}{1-s}\,dt. \tag{1} $$ Heuristically, the integrand is largest when ...


1

Let $P$ be the product; then $$\log{P} = \sum_{i=1}^{k-1} \log{\left (1-\frac{i}{2 N} \right )} $$ When $N$ is sufficiently large, then we may use the approximation $\log{(1-y)} \approx -y$ and get $$\log{P} \approx -\frac1{2 N} \sum_{i=1}^{k-1} i = -\frac1{2 N} \frac{k (k-1)}{2} = -\frac1{2 N} \binom{k}{2}$$ Thus, $$P \approx e^{-\frac1{2 N} ...


1

Let us compare Taylor expansions of $\left( 1-\frac{1}{x} \right)^t$ and $e^{\frac{-t}{x}}$ around $t=0$. For the first one, we have $$\left( 1-\frac{1}{x} \right)^t \simeq 1+t \log \left(1-\frac{1}{x}\right)+\frac{1}{2} t^2 \log ^2\left(1-\frac{1}{x}\right)+\frac{1}{6} t^3 \log ^3\left(1-\frac{1}{x}\right)+O\left(t^4\right)$$ while for the second ...


1

You can approximate the binomial by a Gaussian: Binomial[n, s]/2^n ~ Exp[-(n/2 - s)^2 /2/(n/4)]/Sqrt[2 Pi n/4] $ \frac{\binom{n}{\frac{n}{2}+\frac{s}{2}}}{2^n}$ to $\frac{\exp \left(-\frac{\left(\frac{n}{2}-\left(\frac{n}{2}+\frac{s}{2}\right)\right)^2}{\frac{2 n}{4}}\right)}{\sqrt{\frac{2 \pi n}{4}}}$ and the right side simplifies to ...


1

By the mean value theorem, 2. 3. 4. together imply 1. So all you have to do is approximate $f'$ by smooth functions $g_n$ vanishing at $0$ (e.g. using convolution with symmetric test functions), then perturb very slightly $g_n$ so as to ensure $g_n(0)>0$ without ruining the convergence to $f'$, and finally set $f_n(t)=\int_0^t g_n(s) \,\mathrm{d}s$.


1

1) A Matroid problem is just a greedy basis algorithm. With graphs, the idea of independence (as it deals with Matroids) comes down to acyclicity. So is a cycle created? The way the greedy basis algorithm works in Matroids is that we sort the ground set $X$. Then create set $A = \emptyset$. Then for each $x \in X$, we check if $A \cup \{x \} \in I$, the ...


1

As @d.k.o.'s comment hints at, your expression is: $$ \frac{1}{1 + \exp(b - a) + exp(c - a) + \exp(d - a)} $$ If the resulting $b - a$, $c - a$, $d - a$ are still (very) negative (small exponentials), a few terms of the geometric series should be accurate enough. If some turn out positive, call them $A$, $B$, $C$ in order of decreasing value: $$ \frac{1}{1 + ...


1

This is the typical situation for a binomial distribution. Let's compute the probability of the complementary, that is, the probability of that the event doesn't occur in $n$ trials. This is $$(1-P)^n$$ Now, the probability that you want to compute is $$1-(1-P)^n=1-\sum_{n=0}^\infty\binom nj (-P)^j$$ Since the first term of the sum is $1$, this make ...


1

Okay, let's start with getting the tangent line. The tangent line will be in the form of: $$y=mx+b$$ Where $m$ is the slope, or $f'(3)$. Find the derivative, then: $$f'(x)=2x-3$$ Hence $f'(3)=3$. Now we have: $$y=3x+b$$ Plug in $x/y$ to find $b$. Well, we can use $x=3$ and $y=3^2-3(3)+5=5$, so $b$ equals: $$5=3(3)+b$$ $$b=-4$$ Hence the tangent line's ...



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