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I'm sorry for my language. English is not my first language.

I'm trying to find $$f^{(10)}(0)\;\;\text{when}\;\;f(x)=\frac{1}{2+x}$$ by using MacLaurin. The answer is: $$f^{(10)}(0)=\frac{10!}{2^{11}}$$

Wolfram Alpha gives the formula: $$\frac{1}{2+x}=\sum_{n=0}^n x^n (-1)^n2^{-1-n}.$$ How do I get from $\dfrac{1}{2+x}$ to $\displaystyle \sum_{n=0}^n x^n (-1)^n2^{-1-n}$?

Thanks for your help!

Robert Z
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user9060784
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2 Answers2

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$$\frac {1}{(2+x)}= \frac{1}{2} \frac {1}{(1+x/2)}= \frac{1}{2} \frac {1}{(1-(-x/2))}$$

G Cab
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The series is a geometric one: for any $x$ such that $|x|<2$, $$\frac{1}{2+x}=\frac{1}{2}\cdot \frac{1}{1-(-x/2)}=\frac{1}{2}\sum_{k=0}^{\infty}(-x/2)^k$$ On the other hand, the given function is analytic in $|x|<2$ with MacLaurin expansion $$\frac{1}{2+x}=\sum_{k=0}^{\infty}\frac{f^{(k)}(0)}{k!}x^k.$$ Hence, it follows that $$\frac{f^{(k)}(0)}{k!}=\frac{(-1/2)^k}{2}\implies f^{(k)}(0)=\frac{(-1)^k k!}{2^{k+1}}.$$

Robert Z
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  • I appreciate you taking the time to reply, thank you!

    I get the first part, but i dont understand how this work. Do you have time to explain it?

    Hence, it follows that$$ $\frac{f^{(k)}(0)}{k!}=\frac{(-1/2)^k}{2}\implies f^{(k)}(0)=\frac{(-1)^k k!}{2^{k+1}}.$$

     – user9060784 Oct 25 '19 at 20:57
  • Do you mean this: $\frac{(-1/2)^k}{2}=\frac{(-1)^k}{2\cdot 2^k}$? – Robert Z Oct 25 '19 at 21:05