thereddevils
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How is
[tex]1+p+\frac{p^2}{2!}+\frac{p^3}{3!}+...=e^p[/tex] ?
[tex]1+p+\frac{p^2}{2!}+\frac{p^3}{3!}+...=e^p[/tex] ?
Understanding the Binomial Expansion and its Relationship to e^p
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I know it has to do with taylor expansions, but I've never studied this so I can't answer your question. I'd also like to see a proof for this so this is like some pointless post I'm making so I can subscribe to this thread 
l'Hôpital
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Depends how you define e^p. You could just define it as the power series. However, I'm assuming you are using something like...
[tex] e^p = \lim_{n \rightarrow \infty} (1 + \frac{p}{n})^n[/tex]
Try using the binomial theorem on the right side, then take the limit.
[tex] e^p = \lim_{n \rightarrow \infty} (1 + \frac{p}{n})^n[/tex]
Try using the binomial theorem on the right side, then take the limit.
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thereddevils said:
Mentallic said:I know it has to do with taylor expansions, but I've never studied this so I can't answer your question. I'd also like to see a proof for this so this is like some pointless post I'm making so I can subscribe to this thread
You just show the remainder upon approximating it with the first n terms goes to zero as n --> infinity. See, for example,
http://en.wikipedia.org/wiki/Taylor's_theorem