Q. 63 - Calculus

Question

Q. 

Let \(f(x) = \begin{cases} & 0,\text { if } x=0 \\ & e^{\frac{-1}{x^{2}}},\text{ if } x\neq 0 \end{cases}\)

(a) Use the definition of the derivative at a point to show that f has derivatives of all orders at \(x = 0\).

(b) Show that \(f^{(k)}(0)=0\) for every nonnegative integer \(k\).

(c) What is the Maclaurin series for \(f\) ?

(d) Explain why the Maclaurin series does not converge to \(f(x)\).

Step-by-Step Solution

Answer

The answers are

(c) Maclaurin series is the constant function 0.

(d) The Maclaurin series does not converge to the actual function (other than at \(x=0\)).

Step 1. Given function.

\(f(x) = \begin{cases} & 0,\text { if } x=0 \\ & e^{\frac{-1}{x^{2}}},\text{ if } x\neq 0 \end{cases}\) 

Step 2. Definition of derivative at any point \(a\).

\(f^{'}(a)=\displaystyle \lim_{h \to 0}\frac{f(a+h)-f(a)}{h}\)

Step 3. Part (a): Derivative at point \(x=0\)

\(f^{'}(0)=\displaystyle \lim_{h \to 0}\frac{f(0+h)-f(0)}{h}\)

\(f^{'}(0)=\displaystyle \lim_{h \to 0}\frac{f(h)-f(0)}{h}\)

\(f^{'}(0)=\displaystyle \lim_{h \to 0}\frac{e^{\frac{-1}{h^{2}}} -0}{h}\)

It is \(\frac{0}{0}\) form, Using L'Hopital's rule:

\(f^{'}(0)=\displaystyle \lim_{h \to 0} \frac {\frac{2e^{\frac{-1}{h^{2}}}}{3}h}{h}\)

\(f^{'}(0)=\displaystyle \lim_{h \to 0} \frac{2 e^{\frac{-1}{h^{2}}}}{3}\)

\(f^{'}(0) =\frac{2 e^{-\propto }}{3}\)

\(f^{'}(0)=0\) 

By induction we get that,\(f^{(k)}(0)=0\) for every non-negative integer \(k\) 

Thus \(f\) has derivatives of all orders at \(x = 0\). 

Part (b): From Part (a) we get,

\(f^{(k)}(0)=0\) for every non-negative integer \(k\)

Step 4. Part (c): Maclaurin Series

In order to have a Maclaurin series, a function \(f\) must have a derivative of every order at

\(x = 0\). 

We know Maclaurin series is of the form \(\sum_{k=0}^{\propto } \frac{f^{(k)}(0)}{k!}x^{k}\)

Here we have,  \(f^{(k)}(0) =0\)

Thus, our Maclaurin series is the constant function \(0\).

Step 5. Part(d): Convergence of Maclaurin's series.

The Maclaurin series is the constant function 0. The series is made of infinitely many zeroes which converges to zero for every value of \(x\), but the function \(f(x)=0\) if and only if \(x=0\). That is why it is not convergent to \(f(x)\)