Analytic function

**thahachaina** · December 11th 2008, 10:33 PM

Prove that 2^x is analytic on R and find its Maclaurin expansion.

thanks in advance

**o_O** · December 11th 2008, 11:23 PM

Verify for yourself that: $f^{(n)}(x) = \left(\ln 2\right)^n 2^x$

So the Maclaurin series of $f(x) = 2^x$ is given by: $\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!} x^n = \sum_{n=0}^{\infty}\frac{\left(\ln 2\right)^n}{n!} x^n$

This series converges for all $x$ (use the ratio test to verify).

To prove that $f(x) = 2^x$ is equal to the sum of its Maclaurin series, we must show that $\lim_{n \to \infty} R_{n}(x) = 0$ where $R_n(x)$ is the remainder, i.e.:

$R_{n}(x) = \frac{f^{(n+1)}(c)}{(n+1)!}x^{n+1} = \frac{(\ln 2)^n2^c}{(n+1)!}x^{n+1}$ where $c \in (0, x)$

This fact should come in handy: $\lim_{n \to \infty} \frac{x^n}{n!} = 0$

With this, we've shown $f(x)$ is equal to its Taylor series over the reals and thus analytic.

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