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SolomonZelman

  • one year ago

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

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  1. SolomonZelman
    • one year ago
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    \(\large\color{blue}{\displaystyle\lim_{k \rightarrow \infty}\left(1+\frac{a}{k}\right)^k=e^a}\) 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) \(\large\color{red}{\displaystyle\lim_{k \rightarrow \infty}\left(1+\frac{a}{k}\right)^k=e^a}\)

  2. SolomonZelman
    • one year ago
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    I wrote every little thing out, so it seems very long. But it is not that much to it.

  3. imqwerty
    • one year ago
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    :)

  4. Empty
    • one year ago
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    Oooh fun

  5. SolomonZelman
    • one year ago
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    It is good that my latex didn't get crushed. That sometimes happens.

  6. SolomonZelman
    • one year ago
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    Yes, it is \(\forall a\) apparently: http://www.wolframalpha.com/input/?i=lim+k%E2%86%92%E2%88%9E+%281%2Ba%2Fk%29%5Ek

  7. SolomonZelman
    • one year ago
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    I will maye further investigate if I find any fault in this.

  8. Empty
    • one year ago
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    Well, it does seem kind of circular, doesn't the derivation of Stirling's approximation come from using the power series of \(e^x\)?

  9. SolomonZelman
    • one year ago
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    If I am correct, then e is the consequence of: \(\displaystyle \lim_{n \rightarrow \infty} \left(1+\frac{1}{n}\right)^n=e\)

  10. SolomonZelman
    • one year ago
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    I 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.

  11. SolomonZelman
    • one year ago
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    I 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.

  12. SolomonZelman
    • one year ago
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    because both sides are equivalent to e.

  13. Empty
    • one year ago
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    What I mean is, when you use this, where does this come from? \[x! = \sqrt{2 \pi x } \frac{x^x}{e^x}\]

  14. SolomonZelman
    • one year ago
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    I thought it is from: \(\ln\left({\rm x}!\right)=\ln(1)+\ln(2)+\ln(3)+...+\ln({\rm x}-1)+\ln({\rm x})\)

  15. Empty
    • one year ago
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    \[\ln x! = \sum_{t=1}^x \ln t \approx \int_1^x \ln t dt \approx x \ln x -x \] 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 \(e^x\).

  16. SolomonZelman
    • one year ago
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    Well, then everything comes from \(e^x\). I would certainly not argue against that, but that seems a little too far back....

  17. SolomonZelman
    • one year ago
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    But, 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.

  18. Empty
    • one year ago
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    Yeah 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

  19. SolomonZelman
    • one year ago
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    Yeah, I don't think stirling's approximation is that reasonable. Only for very large values of factorial like 9.

  20. SolomonZelman
    • one year ago
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    from 9 and on i mean

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