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 Exercise [06.05] 
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Post Exercise [06.05]
e^x=\sum _{n=0}^{\infty } \frac{x^n}{n!}
e^x=1+\sum _{n=1}^{\infty } \frac{x^n}{n!}

Differentiating each term gives

\frac{d}{dx}e^x=\sum _{n=1}^{\infty } \frac{n x^{n-1}}{n!}=\sum _{n=1}^{\infty } \frac{ x^{n-1}}{(n-1)!}

Let k=n-1

\frac{d}{dx}e^x=\sum _{k=0}^{\infty } \frac{x^k}{k!}
\frac{d}{dx}e^x=e^x

de^x=e^xdx

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Last edited by Shaun Culver on 18 Aug 2008, 02:57, edited 2 times in total.

26 Apr 2008, 18:31
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Post Re: Exercise [6.5]
Something doesn't feel right about this. Please do point out what I've left out.

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26 Apr 2008, 18:35
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Post Re: Exercise [6.5]
Look Ok to me Shaun. You've just relabeled the infinite sum which is what you should do.


28 Apr 2008, 11:44
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Post Re: Exercise [6.5]
Kurdt wrote:
Look Ok to me Shaun. You've just relabeled the infinite sum which is what you should do.


It seems too informal/glib. I suppose it's enough to get the idea across.

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28 Apr 2008, 15:18
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Post Re: Exercise [6.5]
Shaun Culver wrote:
Kurdt wrote:
Look Ok to me Shaun. You've just relabeled the infinite sum which is what you should do.


It seems too informal/glib. I suppose it's enough to get the idea across.

Well I'm a physicist so it works for me. I don't know what Laura would think. :p


28 Apr 2008, 15:37

Joined: 11 May 2008, 20:07
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Post Re: Exercise [6.5]
I think it would be better if there was the same index under and on the sum (it's just for your first line).


12 May 2008, 20:37
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Post Re: Exercise [6.5]
chag-art wrote:
I think it would be better if there was the same index under and on the sum (it's just for your first line).

I agree.

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12 May 2008, 20:45
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Post Re: Exercise [06.05]
Quote:
e^x=\sum _{n=0}^{\infty } \frac{x^n}{n!}

Differentiating each terms gives
\frac{d}{dx}e^x=\sum _{n=0}^{\infty } \frac{n x^{n-1}}{n!}=\sum _{n=0}^{\infty } \frac{ x^{n-1}}{(n-1)!}

Let k=n-1
\frac{d}{dx}e^x=\sum _{k=0}^{\infty } \frac{x^k}{k!}
\frac{d}{dx}e^x=e^x
de^x=e^xdx


Moving from \sum _{n=0}^{\infty } \frac{n x^{n-1}}{n!}\ to\ \sum _{n=0}^{\infty } \frac{ x^{n-1}}{(n-1)!} is not valid for n=0 since it implies that \frac{0}{0!}=\frac{1}{(-1)!} or, given that 0! is 1, that \frac{0}{1}=\frac{1}{(-1)!}which is clearly false.

This stems from the fact that:

n!=n\times(n-1)! is only valid for n\ge 1

Also when substituting k=n-1 the summation over k would have to start at k=-1 since when n=0 k=-1.

It is better to write the proof as follows I think:

e^x=1+\sum _{n=1}^{\infty } \frac{x^n}{n!} {or to proceed as you did but say \sum _{n=0}^{\infty } \frac{n x^{n-1}}{n!}=\sum _{n=1}^{\infty } \frac{ n x^{n-1}}{n!} since \frac{nx^{n-1}}{n!} = 0when n=0, =\sum _{n=1}^{\infty } \frac{ x^{n-1}}{(n-1)!} }

Differentiating the series gives:

\frac{d}{dx}e^x=\sum _{n=1}^{\infty } \frac{n x^{n-1}}{n!}=\sum _{n=1}^{\infty } \frac{ x^{n-1}}{(n-1)!}

Letting k=n-1 gives:

\frac{d}{dx}e^x=\sum _{k=0}^{\infty } \frac{x^k}{k!}
\frac{d}{dx}e^x=e^x

de^x=e^xdx


Last edited by vasco on 03 Jul 2008, 13:11, edited 1 time in total.

03 Jul 2008, 12:02
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Post Re: Exercise [06.05]
I agree. Thank you.

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Our imagination is stretched to the utmost, not, as in fiction, to imagine things which are not really there, but just to comprehend those things which are there.


03 Jul 2008, 12:45
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