Prove that 2^x is analytic on R and find its Maclaurin expansion.
thanks in advance
Verify for yourself that: $\displaystyle f^{(n)}(x) = \left(\ln 2\right)^n 2^x$
So the Maclaurin series of $\displaystyle f(x) = 2^x$ is given by: $\displaystyle \sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!} x^n = \sum_{n=0}^{\infty}\frac{\left(\ln 2\right)^n}{n!} x^n$
This series converges for all $\displaystyle x$ (use the ratio test to verify).
To prove that $\displaystyle f(x) = 2^x$ is equal to the sum of its Maclaurin series, we must show that $\displaystyle \lim_{n \to \infty} R_{n}(x) = 0$ where $\displaystyle R_n(x)$ is the remainder, i.e.:
$\displaystyle R_{n}(x) = \frac{f^{(n+1)}(c)}{(n+1)!}x^{n+1} = \frac{(\ln 2)^n2^c}{(n+1)!}x^{n+1}$ where $\displaystyle c \in (0, x)$
This fact should come in handy: $\displaystyle \lim_{n \to \infty} \frac{x^n}{n!} = 0$
With this, we've shown $\displaystyle f(x)$ is equal to its Taylor series over the reals and thus analytic.