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  1. #1
    Member pflo's Avatar
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    limit proof

    One way determine the deriveative of e^x is to use inverse functions (natural logs). Another way is to use the definition of a derivative. Thus, \frac{d}{dx}e^x=\lim_{h\rightarrow0}\frac{e^{x+h}-e^x}{h}. When simplified, this results in the equation \frac{d}{dx}e^x=e^x \lim_{h\rightarrow0}\frac{e^h-1}{h}.

    My questions is: what is a proof that \lim_{h\rightarrow0}\frac{e^h-1}{h}=1?
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    Quote Originally Posted by pflo View Post
    One way to determine the derivative of e^x is to use inverse functions (natural logs). Another way is to use the definition of a derivative. Thus, \frac{d}{dx}e^x=\lim_{h\rightarrow0}\frac{e^{x+h}-e^x}{h}. When simplified, this results in the equation \frac{d}{dx}e^x=e^x \lim_{h\rightarrow0}\frac{e^h-1}{h}.

    My question is: what is a proof that \lim_{h\rightarrow0}\frac{e^h-1}{h}=1?
    \displaystyle\ e^h=1+h+\frac{h^2}{2!}+\frac{h^3}{3!}+.......

    \displaystyle\ e^h-1=h+\frac{h^2}{2!}+\frac{h^3}{3!}+.....

    \displaystyle\frac{e^h-1}{h}=1+\frac{h}{2!}+\frac{h^2}{3!}+....

    then take the limit as h\rightarrow\ 0
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  3. #3
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    Quote Originally Posted by Archie Meade View Post
    \displaystyle\ e^h=1+h+\frac{h^2}{2!}+\frac{h^3}{3!}+.......

    \displaystyle\ e^h-1=h+\frac{h^2}{2!}+\frac{h^3}{3!}+.....

    \displaystyle\frac{e^h-1}{h}=1+\frac{h}{2!}+\frac{h^2}{3!}+....

    then take the limit as h\rightarrow\ 0
    To get the above you use the Taylor's expansion where :

    f'(0) = \frac{d(e^x)}{dx} at x=0 ,which is 1

    f"(0) = \frac{d^2(e^x)}{dx^2} at x=0 ,which is again 1 e.t.c .....e.t.c

    Thus admitting that \frac{d(e^x)}{dx}=e^x

    But that is what you want to prove
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    Quote Originally Posted by alexandros View Post
    To get the above you use the Taylor's expansion where :

    f'(0) = \frac{d(e^x)}{dx} at x=0 ,which is 1

    f"(0) = \frac{d^2(e^x)}{dx^2} at x=0 ,which is again 1 e.t.c .....e.t.c

    Thus admitting that \frac{d(e^x)}{dx}=e^x

    But that is what you want to prove
    If the Taylor expansion is used, it can be considered "circular reasoning"
    from one perspective.

    Instead if we use the Binomial Expansion (Newton's Generalised Binomial Theorem) on

    \displaystyle\left(1+\frac{1}{x}\right)^x

    we obtain

    e=\binom{x}{0}+\binom{x}{1}\frac{1}{x}+\binom{x}{2  }\frac{1}{x^2}+.....

    from which we can write the power series expansion of e^h
    Last edited by Archie Meade; December 3rd 2010 at 02:16 AM.
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    Is h a real or a natural No??
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  6. #6
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    Quote Originally Posted by alexandros View Post
    Is h a real or a natural No??
    h\rightarrow\ 0

    \displaystyle\ e=\lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\rig  ht)^n

    \displaystyle\ e^h=\lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\r  ight)^{nh}

    \displaystyle\ =\lim_{n\rightarrow\infty}\left[\binom{nh}{0}+\binom{nh}{1}\frac{1}{n}+\binom{nh}{  2}\frac{1}{n^2}+..........\right]

    \displaystyle\ =\lim_{n\rightarrow\infty}\left[1+h+\frac{(nh)(nh-1)}{2!}\;\frac{1}{n^2}+........\right]

    \displaystyle\ =\lim_{n\rightarrow\infty}\left[1+h+\frac{n^2h^2-nh}{2!}\;\frac{1}{n^2}+......\right]

    \displaystyle\ =\lim_{n\rightarrow\infty}\left[1+h+\frac{h^2}{2!}-\frac{nh}{n^2}+....\right]

    and all the terms with higher power of "n" in the denominator positions go to zero.
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    Quote Originally Posted by Archie Meade View Post
    h\rightarrow\ 0

    \displaystyle\ e=\lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\rig  ht)^n

    \displaystyle\ e^h=\lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\r  ight)^{nh}
    .

    For that you assume the following theorem:

    If ........... lim_{n\to\infty(x_{n})=x\Longrightarrow lim_{n\to\infty}(x_{n})^h=x^h
    Where h is a real No

    BUT is there such a theorem
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  8. #8
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    Quote Originally Posted by alexandros View Post
    .

    For that you assume the following theorem:

    If ........... lim_{n\to\infty(x_{n})=x\Longrightarrow lim_{n\to\infty}(x_{n})^h=x^h
    Where h is a real No

    BUT is there such a theorem
    Please research Newton's Generalised Binomial theorem.
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    Quote Originally Posted by Archie Meade View Post
    Please research Newton's Generalised Binomial theorem.

    It IS NOT a question of Newton,s Generalised Binomial, where the exponent is always natural.

    It is a question whether the theorem i mention holds or not ,so that step 2 in your proof can be justified
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  10. #10
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by alexandros View Post
    It IS NOT a question of Newton,s Generalised Binomial, where the exponent is always natural.

    It is a question whether the theorem i mention holds or not ,so that step 2 in your proof can be justified
    If you're asking whether x_n\to x\implies x_n^\alpha\to x^\alpha the answer is yes, by continuity of f(y)=y^\alpha.
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  11. #11
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    Quote Originally Posted by alexandros View Post
    It IS NOT a question of Newton,s Generalised Binomial, where the exponent is always natural. ????

    It is a question whether the theorem i mention holds or not ,so that step 2 in your proof can be justified
    Why can't you do some research???

    Newton generalised the Binomial Theorem for real numbers a long time ago!

    Also..

    e=2.71828.....\Rightarrow\ e^h=(2.71828....)^h

    \displaystyle\ e=\left[\lim_{n \to \infty}\left(1+\frac{1}{n}\right)^n\right]\Rightarrow\ e^h=\left[\lim_{n \to \infty}\left(1+\frac{1}{n}\right)^n\right]^h

    a=b\Rightarrow\ a^h=b^h
    Last edited by Archie Meade; December 22nd 2010 at 11:31 AM.
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  12. #12
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    Quote Originally Posted by Archie Meade View Post
    h\rightarrow\ 0

    \displaystyle\ e=\lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\rig  ht)^n

    \displaystyle\ e^h=\lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\r  ight)^{nh}

    \displaystyle\ =\lim_{n\rightarrow\infty}\left[\binom{nh}{0}+\binom{nh}{1}\frac{1}{n}+\binom{nh}{  2}\frac{1}{n^2}+..........\right]

    \displaystyle\ =\lim_{n\rightarrow\infty}\left[1+h+\frac{(nh)(nh-1)}{2!}\;\frac{1}{n^2}+........\right]

    \displaystyle\ =\lim_{n\rightarrow\infty}\left[1+h+\frac{n^2h^2-nh}{2!}\;\frac{1}{n^2}+......\right]

    \displaystyle\ =\lim_{n\rightarrow\infty}\left[1+h+\frac{h^2}{2!}-\frac{nh}{n^2}+....\right]

    and all the terms with higher power of "n" in the denominator positions go to zero.
    So we have that: \left(1+\frac{1}{n}\right)^{nh} =
    \left[1+h+\frac{(nh)(nh-1)}{2!}\;\frac{1}{n^2}+........\right]

    Does this equation work for n=1 and h= 1/2 ?
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  13. #13
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    Quote Originally Posted by alexandros View Post
    So we have that: \left(1+\frac{1}{n}\right)^{nh} =
    \left[1+h+\frac{(nh)(nh-1)}{2!}\;\frac{1}{n^2}+........\right]

    Does this equation work for n=1 and h= 1/2 ?
    What did the research indicate ?

    You could compare it to...

    \displaystyle\sqrt{2}=\sum_{k=0}^{\infty}(-1)^{k+1}\frac{(2k-3)!!}{(2k)!!}=1+\frac{1}{2}-\frac{1}{2(4)}+\frac{3}{2(4)6}-\frac{3(5)}{2(4)6(8)}+.....

    What does

    \displaystyle\ 1+h+\frac{(nh)(nh-1)}{2!}+\frac{nh(nh-1)(nh-2)}{3!}+\frac{nh(nh-1)(nh-2)(nh-3)}{4!}+.....

    equal for those values of n and h ?
    Last edited by Archie Meade; December 12th 2010 at 07:27 AM.
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  14. #14
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    Can research show us that:

    \sqrt 2 = \frac{3}{2}??
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  15. #15
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by pflo View Post
    One way determine the deriveative of e^x is to use inverse functions (natural logs). Another way is to use the definition of a derivative. Thus, \frac{d}{dx}e^x=\lim_{h\rightarrow0}\frac{e^{x+h}-e^x}{h}. When simplified, this results in the equation \frac{d}{dx}e^x=e^x \lim_{h\rightarrow0}\frac{e^h-1}{h}.

    My questions is: what is a proof that \lim_{h\rightarrow0}\frac{e^h-1}{h}=1?
    Your question is fully justified!.. effectively the limit \displaystyle \lim_{x \rightarrow 0}\frac{e^x-1}{x}=1 is a 'fundamental limit' from which You can derive all the properties of the exponential function. To demonstrate it we can start from the well known limit...

    \displaystyle \lim_{r \rightarrow \infty} (1+\frac{1}{r})^{r}= e (1)

    Now if x \ne 0 You can set \displaystyle e^{x} = 1 + \frac{1}{r} so that is \displaystyle x= \ln (1+\frac{1}{r}) and therefore...

    \displaystyle \lim_{x \rightarrow 0} \frac{e^{x}-1}{x} = \lim_{r \rightarrow \infty} \frac{\frac{1}{r}}{\ln (1+\frac{1}{r})} = \lim_{r \rightarrow \infty} \frac{1}{\ln (1+\frac{1}{r})^{r}} = \frac{1}{\ln e} = 1 (2)



    Merry Christmas from Italy

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