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[SOLVED] Taylor expansion to express e^x

DeusAbscondus

Active member
Jun 30, 2012
176
Hi folks,

If $e^x= \Sigma_{k=0}^\infty \frac{x^k}{k!}$

what do I evaluate $x$ at?
How does the sigma notation tell me what to do with $x$?

$$e^x= \Sigma_{k=0}^\infty \frac{x^k}{k!}\ = 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!} .... \text {ad infinitum}$$

Sorry, I just realized my error: this formula gives me a way of finding out a given value for $e^x$ with $x=a$ ie: with $x$ set to any value.
Right?

My real problem is actually related, but quite other than this.
And your response, believe it not, gave me pause to ask myself the question again and realize this ie: that I actually have another question.

Thanks !
 
Last edited:

Chris L T521

Well-known member
Staff member
Jan 26, 2012
995
Re: taylor expansion to express $e^x$

Hi folks,

If $e^2= \Sigma_{k=0}^\infty \frac{x^k}{k!}$

what do I evaluate $x$ at?
How does the sigma notation tell me what to do with $x$?

My inadequate understanding of key concepts is evidenced below, as I try to work with the sigma expression:

$$e^x= \Sigma_{k=0}^\infty \frac{x^k}{k!}\ = 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!} .... \text {ad infinitum}$$

The only reason I derive $1$ for the first term is that $k=0$ in that term;
thereafter, where is the instruction to alter the values of $x$?

Thanks for any help,
Deo Abscondito
I don't think I get what you're asking. Are you trying to ask if (a) we know how to figure out what $x$ is if $\displaystyle e^2=\sum_{k=0}^{\infty} \frac{x^k}{k!}$ or (b) if we're given $\displaystyle e^x=\sum_{k=0}^{\infty} \frac{x^k}{k!}$, then what is $e^2$ in terms of a series?