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This was something that popped up in class and confused me...

So knowing that the power series may be differentiated term by term inside the interval of convergence, using the Maclaurin series you can derive the differentiation formula for the function $f(x) = \frac{1}{1-x}$.

What do they mean by differentiation formula and how do I get it using Maclaurin series? I'm quite confused xD

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    $\begingroup$ Maybe this is about $\tfrac{1}{1-x} = \sum_{k=0}^∞ x^k$ for $x ∈ (0..1)$ (which follows from $\tfrac{1-x^n}{1-x} = \sum_{k=0}^n x^k$). Then you already know the MacLaurin series for $f(x)$ and you can conclude $\tfrac{f^{(k)}(0)}{k!} = 1$ for all $k ∈ ℕ_0$, so you don’t need to differentiate to get the derivatives of $f$ at $0$. $\endgroup$ – k.stm Mar 15 '14 at 12:30
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The McLaurin formula is just the Taylor expansion calculated in $0$, so: $$ f(x)=f(0)+f'(0)x+f''(0)\frac{x^2}{2}+\dots=\sum_{k=0}^\infty f^{(k)}(0)\frac{x^k}{k!}. $$

In your case: $$ f(0)=1, $$ $$ f'(x)=-\frac{1}{(1-x)^2}\cdot(-1); f'(0)=1, $$ and so on.

It says differentiation formula, as you have to compute all the derivatives of $f(x)$.

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  • $\begingroup$ so basically they are asking for the taylor expansion and that is the differentiation formula? $\endgroup$ – Kennan Mar 14 '14 at 10:00
  • $\begingroup$ @k.stm brilliant interpretation. $\endgroup$ – 7raiden7 Mar 15 '14 at 12:56

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