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#### 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

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 !

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 !

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