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Exercise [06.05]
Author	Message
Shaun Culver Site Admin Joined: 25 Feb 2008, 13:32 Posts: 105 Location: Cape Town, South Africa	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 $\frac{d}{dx}e^x=\sum _{k=0}^{\infty } \frac{x^k}{k!}$ $\frac{d}{dx}e^x=e^x$ _________________ “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.” Last edited by Shaun Culver on 18 Aug 2008, 02:57, edited 2 times in total.
26 Apr 2008, 18:31

Shaun Culver Site Admin Joined: 25 Feb 2008, 13:32 Posts: 105 Location: Cape Town, South Africa	Re: Exercise [6.5] Something doesn't feel right about this. Please do point out what I've left out. _________________ “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.”
26 Apr 2008, 18:35

Kurdt Joined: 19 Mar 2008, 14:09 Posts: 36	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

Shaun Culver Site Admin Joined: 25 Feb 2008, 13:32 Posts: 105 Location: Cape Town, South Africa	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. _________________ “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.”
28 Apr 2008, 15:18

Kurdt Joined: 19 Mar 2008, 14:09 Posts: 36	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

chag-art Joined: 11 May 2008, 20:07 Posts: 11 Location: france	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

Shaun Culver Site Admin Joined: 25 Feb 2008, 13:32 Posts: 105 Location: Cape Town, South Africa	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. _________________ “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.”
12 May 2008, 20:45

vasco Supporter Joined: 07 Jun 2008, 08:21 Posts: 182	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 $\frac{d}{dx}e^x=\sum _{k=0}^{\infty } \frac{x^k}{k!}$ $\frac{d}{dx}e^x=e^x$ 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 $\frac{nx^{n-1}}{n!} = 0$ when 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 gives: $\frac{d}{dx}e^x=\sum _{k=0}^{\infty } \frac{x^k}{k!}$ $\frac{d}{dx}e^x=e^x$ Last edited by vasco on 03 Jul 2008, 13:11, edited 1 time in total.
03 Jul 2008, 12:02

Shaun Culver Site Admin Joined: 25 Feb 2008, 13:32 Posts: 105 Location: Cape Town, South Africa	Re: Exercise [06.05] I agree. Thank you. _________________ “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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Exercise [06.05]

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