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How do I write this infinite series using the sigma notation?

$$1+f'(n)m+\frac{f''(n)}{2!}m^2f(n)+\frac{f'''(n)}{3!}m^3(f(n))^2+...$$

My attempt:

$$\sum_{k=0}^\infty\frac{f^{(k)}(n)}{k!}m^k(f(n))^{k-1}$$

But this yields an initial value of $(f(n))^{-1}$ instead of $1$

So we could represent this as:

$$1-(f(n))^{-1}+\sum_{k=0}^\infty\frac{f^{(k)}(n)}{k!}m^k(f(n))^{k-1}$$

Is there a neater way with just a sigma sign followed by a function?

Sam Houston
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    Actually the initial value it gives is $f(n)(f(n))^{-1} = 1$, because $f^{(0)}$ (the zeroth derivative of $f$) is just $f$. So your attempt looks right to me. But what is this from (that's not the Taylor expansion)? –  May 03 '15 at 12:54
  • Oh so the attempt was correct? – Sam Houston May 03 '15 at 12:55

1 Answers1

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Since $0!$ is usually defined as $0!=1$ the first term in your sum for $k=0$ is given by $$f(n)f(n)^{-1}m^0=1$$ hence your attempt was correct!

EDIT: ups Bye_World already told you and I didn't see it. sorry!

the.polo
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