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Let $A(z)=\sum\limits_{n=0}^\infty a_nz^n$ be a power series with radius of convergence $R>0$.

Given that $\forall\ x\in[0,R):A(x)=e^x-2e^{-x}$, Prove that $R=\infty$ and find $A(\pi i)$.

My attempt:

$$A(x)=e^x-2e^{-x}=\sum\limits_{n=0}^\infty\frac{x^n}{n!}-2\sum\limits_{n=0}^\infty\frac{(-x)^n}{n!}=\sum\limits_{n=0}^\infty\frac{(1+2(-1)^{n+1})x^n}{n!}$$

now by using the Cauchy–Hadamard theorem we can compute that $R=\infty$.

But I don't know how to find $A(\pi i)$, can I just assign the value to $A(x)$ without any explanation?

please help and correct my mistakes.

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    To compute $A(\pi i)$, use the formula $e^{i\theta}=\cos(\theta)+i\sin(\theta)$ for $\theta\in\mathbb R$. – 2017-02-07
  • 0
    I know this identity but can I use it (since the function is Real)? – 2017-02-07
  • 0
    How do you know the function is real? what makes you say the function is real? – 2017-02-07

1 Answers 1

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The power series $A(z)=\sum\limits_{n=0}^\infty\frac{(1+2(-1)^{n+1})z^n}{n!}$ converges for all $z \in \mathbb C$ !

From $e^{i \pi}=-1$ we get

$A(i \pi)=-1-2\frac{1}{-1}=1$.