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Application on the limit definition of e

  1. Oct 27, 2013 #1
    Application on the limit definition of "e"

    Hi, I have known that:
    (i)[itex] (1+\frac{a}{n})^n=((1+\frac{a}{n})^\frac{n}{a})^a\to e^a[/itex]
    (ii)[itex] (1-\frac{1}{n})^n=(\frac{n-1}{n})^n=(\frac{1}{\frac{n}{n-1}})^{(n-1)+1}=(\frac{1}{1+\frac{1}{n-1}})^{(n-1)}\cdot (\frac{1}{1+\frac{1}{n-1}}) \to \frac{1}{e}\cdot 1[/itex]

    With above two facts, I wanted to show [itex](\frac{1}{1-\frac{t}{\sqrt{\frac{n}{2}}}})^\frac{n}{2} \to e^{\sqrt{\frac{n}{2}}t}\cdot e^\frac{t^2}{2}[/itex] as n goes to infinity, for a fixed positive real t.

    However, I am continuously getting [itex]e^{\sqrt{\frac{n}{2}}t}\cdot e^{t^2}[/itex] instead of above result and could not find the reason on the following my argument:

    [itex](\frac{\sqrt{\frac{n}{2}}}{\sqrt{\frac{n}{2}}-t})^\frac{n}{2}=(\frac{(\frac{\sqrt{\frac{n}{2}}}{t}-1)+1}{\frac{\sqrt{\frac{n}{2}}}{t}-1})^\frac{n}{2}=(1+\frac{1}{\frac{\sqrt{\frac{n}{2}}}{t}-1})^{(\frac{\sqrt{\frac{n}{2}}}{t}-1)\sqrt{\frac{n}{2}}t+\sqrt{\frac{n}{2}}t}=(1+\frac{1}{\frac{\sqrt{\frac{n}{2}}}{t}-1})^{(\frac{\sqrt{\frac{n}{2}}}{t}-1)\sqrt{\frac{n}{2}}t}\cdot (1+\frac{1}{\frac{\sqrt{\frac{n}{2}}}{t}-1})^{(\frac{\sqrt{\frac{n}{2}}}{t}-1)t^2+t^2}[/itex]
    [itex]=(1+\frac{1}{\frac{\sqrt{\frac{n}{2}}}{t}-1})^{(\frac{\sqrt{\frac{n}{2}}}{t}-1)\sqrt{\frac{n}{2}}t}\cdot (1+\frac{1}{\frac{\sqrt{\frac{n}{2}}}{t}-1})^{(\frac{\sqrt{\frac{n}{2}}}{t}-1)t^2}\cdot (1+\frac{1}{\frac{\sqrt{\frac{n}{2}}}{t}-1})^{t^2} \to e^{\sqrt{\frac{n}{2}}t}\cdot e^{t^2}\cdot 1[/itex] as n goes to infinity.

    It would be very appreciative if you let me know my mistake.
    Thank you very much.
     
  2. jcsd
  3. Oct 28, 2013 #2

    CompuChip

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    Hi fblues. How can there be an ##n## on the right hand side, after you have taken the limit of ##n \to \infty##?
     
  4. Oct 28, 2013 #3

    Office_Shredder

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    I think the statement that you want is something like

    [tex] \lim_{n\to \infty} \left(\frac{1}{1-t/\sqrt{n/2}} \right)^{n/2} e^{-\sqrt{n/2} t} = e^{t^2/2} [/tex]

    (I don't know if this is the correct statement, but is what your statement should look similar to).
     
  5. Oct 28, 2013 #4
    To. CompuChip:
    Thank you for letting me know. I tried to split the part that I don't know from the original problem and made a mistake during this procedure. BTW, it seems Office_Shredder knows the original one.

    To. Office_Shredder:
    Yes. The problem is from "the mgf of Chi_sq(n) becomes the mgf of normal(0,1) as n goes to infinity." I think the general approach is use of Taylor expansion. But, I tried to employ the limit definition of e. Do you have an idea for this?
     
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