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SolomonZelman
 one year ago
Some algebraic tasks w/ Stirling's approximation, with a relevantly easy limit.
SolomonZelman
 one year ago
Some algebraic tasks w/ Stirling's approximation, with a relevantly easy limit.

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SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1\(\large\color{blue}{\displaystyle\lim_{k \rightarrow \infty}\left(1+\frac{a}{k}\right)^k=e^a}\) We will use Stirling's approximation \(\large\color{royalblue}{\displaystyle x!=\sqrt{2 \pi x}~\left(\frac{x}{e}\right)^x}\) \(\large\color{royalblue}{\displaystyle x!=\sqrt{2 \pi x}~\frac{x^x}{e^x}}\) \(\large\color{royalblue}{\displaystyle e^xx!=\sqrt{2 \pi x}~x^x}\) \(\large\color{royalblue}{\displaystyle \frac{e^xx!}{\sqrt{2 \pi x}}=x^x}\) this is what we want to get:) Now lets perform some manipulations with the limit. (Just follow along with me.) \(\large\color{blue}{\displaystyle\lim_{k \rightarrow \infty}\left(\frac{k}{k}+\frac{a}{k}\right)^k}\) \(\large\color{blue}{\displaystyle\lim_{k \rightarrow \infty}\left(\frac{k+a}{k}\right)^k}\) \(\large\color{blue}{\displaystyle\lim_{k \rightarrow \infty}\frac{\left(k+a\right)^k}{k^k}}\) \(\large\color{blue}{\displaystyle\lim_{k \rightarrow \infty}\frac{\left(k+a\right)^k~\left(k+a\right)^a~\left(k+a\right)^{a}}{k^k}}\) \(\large\color{blue}{\displaystyle\lim_{k \rightarrow \infty}\frac{\left(k+a\right)^{k+a}}{k^k~\left(k+a\right)^a}}\) Now we will apply what I wrote about \(\large x^x\) based on Stirling's approximation. (Although Stirling's approximation is smaller than n!, and thus \(x^x\) is smaller than its equivalent that I derivaed from Stirling's approximation, HOWEVER, since I apply it to both numerator and denominator, it should be reasonable.) \(\large\color{blue}{\displaystyle\lim_{k \rightarrow \infty}\frac{{\LARGE \frac{e^{k+a}(k+a)!}{\sqrt{2 \pi (k+a)}}}}{{\LARGE \frac{e^kk!}{\sqrt{2 \pi k}}}~~\left(k+a\right)^a}}\) the next manipulation can be understood either intuitively, or if not, I am going to let you know that I am multiplying on top and bottom times the following fraction: \(\sqrt{2\pi k}{\bf~/} \sqrt{2 \pi (k+a)}\) \(\large\color{blue}{\displaystyle\lim_{k \rightarrow \infty}\frac{e^{k+a}(k+a)!\sqrt{2 \pi k}}{e^k~k!~\sqrt{2 \pi (k+a)}~\left(k+a\right)^a}}\) \(\large\color{blue}{\displaystyle\lim_{k \rightarrow \infty}\left[\frac{e^{k+a}}{e^k}\times \frac{(k+a)!}{k!~\left(k+a\right)^a}\times \frac{\sqrt{2 \pi k}}{\sqrt{2 \pi (k+a)}} \right]}\) \(\large\color{blue}{\displaystyle\left[\lim_{k \rightarrow \infty}\frac{e^{k+a}}{e^k} \right] \times \left[\lim_{k \rightarrow \infty} \frac{(k+a)!}{k!~\left(k+a\right)^a} \right]\times \left[\lim_{k \rightarrow \infty} \frac{\sqrt{2 \pi k}}{\sqrt{2 \pi (k+a)}} \right]}\) \(\large\color{blue}{\displaystyle e^a \times \left[\lim_{k \rightarrow \infty} \frac{(k+a)!}{k!~\left(k+a\right)^a} \right]\times \left[\lim_{k \rightarrow \infty} \frac{\sqrt{ k}}{\sqrt{ k+a}} \right]}\) \(\large\color{blue}{\displaystyle e^a \times \left[\lim_{k \rightarrow \infty} \frac{(k+a)!}{k!~\left(k+a\right)^a} \right]\times \sqrt{ \lim_{k \rightarrow \infty} \frac{k\color{white}{\Large }}{ k+a}} }\) the last limit is 1. One more limit left. \(\large\color{blue}{\displaystyle e^a \times \left[\lim_{k \rightarrow \infty} \frac{(k+a)!}{k!~\left(k+a\right)^a} \right] }\) The following (\(\Downarrow\)) works for integer "a", but I am pretty sure this limit will be true numerically for any a. \(\large\color{blue}{\displaystyle e^a \cdot \lim_{k \rightarrow \infty} \frac{k!(k+1)(k+2)...(k+a)}{k!~\left(k+a\right)^a} }\) \(\large\color{blue}{\displaystyle e^a \cdot \lim_{k \rightarrow \infty} \frac{(k+1)(k+2)...(k+a)}{\left(k+a\right)^a} }\) \(\large\color{blue}{\displaystyle e^a \cdot \lim_{k \rightarrow \infty} \frac{(k+1)(k+2)...(k+a)}{\left(k+a\right)\left(k+a\right)...\left(k+a\right)} }\) Both top and bottom of the limit contain same number of factors, (a factors). When we take the derivative (a+1) times by l'Hospital's, since the leading terms on top and bottom are \(k^a\), we get 1/1, and thus the limit is equal to 1. So I have to conclude that: (At least for integer a) \(\large\color{red}{\displaystyle\lim_{k \rightarrow \infty}\left(1+\frac{a}{k}\right)^k=e^a}\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1I wrote every little thing out, so it seems very long. But it is not that much to it.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1It is good that my latex didn't get crushed. That sometimes happens.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1Yes, it is \(\forall a\) apparently: http://www.wolframalpha.com/input/?i=lim+k%E2%86%92%E2%88%9E+%281%2Ba%2Fk%29%5Ek

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1I will maye further investigate if I find any fault in this.

Empty
 one year ago
Best ResponseYou've already chosen the best response.0Well, it does seem kind of circular, doesn't the derivation of Stirling's approximation come from using the power series of \(e^x\)?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1If I am correct, then e is the consequence of: \(\displaystyle \lim_{n \rightarrow \infty} \left(1+\frac{1}{n}\right)^n=e\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1I just now looked on wiki, so they use either ln(n!)=ln(2)+ln(3)....+ln(n) or the integral definition of the gamma function.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1I would certainly think \(e^x\) is related, (at least if x=1) \(\large\displaystyle \sum_{n=0}^{\infty}\frac{1}{n!} = \lim_{n\rightarrow \infty}\left(1+\frac{1}{n}\right)^n\) So, the convergence of that series is same as convergence of the sequence on the right side. Perhaps that is where I can look at.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1because both sides are equivalent to e.

Empty
 one year ago
Best ResponseYou've already chosen the best response.0What I mean is, when you use this, where does this come from? \[x! = \sqrt{2 \pi x } \frac{x^x}{e^x}\]

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1I thought it is from: \(\ln\left({\rm x}!\right)=\ln(1)+\ln(2)+\ln(3)+...+\ln({\rm x}1)+\ln({\rm x})\)

Empty
 one year ago
Best ResponseYou've already chosen the best response.0\[\ln x! = \sum_{t=1}^x \ln t \approx \int_1^x \ln t dt \approx x \ln x x \] This is the simple form of Stirling's approximation (and arguably the best in terms of large numbers since the difference is small) but this involves integrating \(\ln x\) which is ultimately where you have to assume you already know the derivative of \(e^x\).

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1Well, then everything comes from \(e^x\). I would certainly not argue against that, but that seems a little too far back....

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1But, yeah, I can see that Stirling's approximation is something that is rather something that has to be proved, than an identity which is to use to prove. If you know whgat I mean.

Empty
 one year ago
Best ResponseYou've already chosen the best response.0Yeah I mean I am just nitpicking for fun like for instance if someone says "take the limit without using L'Hopital's rule" and then they use the Taylor series representation then sure you can take it to be something that is, rather than something that has to be proven but I think that's a mistake. I think this is fun though

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1Yeah, I don't think stirling's approximation is that reasonable. Only for very large values of factorial like 9.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1from 9 and on i mean
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