Consider the function $f(x)=e^{x}$.
a. Differentiate the Taylor series about 0 of $f(x)$.
b. Identify the function represented by the differentiated series.
c. Give the interval of convergence of the power series for the derivative.
Answer
Final Answer: \[\boxed{\begin{align*} a. & \text{ The differentiated Taylor series about 0 of } f(x)=e^x \text{ is } f'(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots.\\ b. & \text{ The function represented by the differentiated series is } f(x)=e^x.\\ c. & \text{ The interval of convergence of the power series for the derivative is all real numbers.} \end{align*}}\]
Step 1 ：The Taylor series of a function \(f(x)\) about \(x=a\) is given by the formula: \[f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots\]
Step 2 ：For the function \(f(x)=e^x\), the derivative of \(e^x\) is also \(e^x\), and \(e^0=1\), so the Taylor series of \(f(x)\) about \(x=0\) is: \[f(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots\]
Step 3 ：Differentiating this series, we get: \[f'(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots\]
Step 4 ：This is the same as the original series, so the function represented by the differentiated series is also \(f(x)=e^x\).
Step 5 ：The power series for \(e^x\) converges for all \(x\), so the interval of convergence of the power series for the derivative is also all real numbers.
Step 6 ：Final Answer: \[\boxed{\begin{align*} a. & \text{ The differentiated Taylor series about 0 of } f(x)=e^x \text{ is } f'(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots.\\ b. & \text{ The function represented by the differentiated series is } f(x)=e^x.\\ c. & \text{ The interval of convergence of the power series for the derivative is all real numbers.} \end{align*}}\]
