Question

Consider the function ƒ(x) = eˣ

(a) Differentiate the Taylor series about 0 of ƒ(x)
(b) Identify the function represented by the differentiated series.
(c) Give the interval of convergence of the power series for the derivative.

Answer 1

Answer:

a) The Taylor series expansion of $f(x)=e^{x}$about$x=0$ is

$f(x)=e^{x}\\\\f(x)=f(0)+f'(0)x+\frac{f''(0)}{2!}x^{2}+\frac{f'''(x)}{3!}x^{3}+...\\\\f(x)=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+...$

Differentiating the expansion with respect to 'x' we get

$f(x)=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+...\\\\f'(x)=0+1+\frac{2x}{2!}+\frac{3x^{2}}{3!}+\frac{4x^{3}}{4!}+...\\\\f'(x)=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+...$

b)

As we can identify from the expansion calculated above we have

$f'(x)=e^{x}$

c)

The interval of convergence of the given power series is

$-\infty < x< +\infty$