Question

Find the Taylor series expansion of the function cosh(x) centered at x = 0.

(a) \(1-\frac{x^2}{2!}+\frac{x^4}{4!}+….\infty\)

(b) \(\frac{x}{1!}+\frac{x^3}{3!}+\frac{x^5}{5!}….\infty\)

(c) \(1+\frac{x^2}{2!}+\frac{x^4}{4!}+….\infty\)

(d) \(1+\frac{x}{1!}+\frac{x^2}{2!}+….\infty\)

I had been asked this question in an internship interview.

My question is based upon Taylor Mclaurin Series topic in section Differential Calculus of Engineering Mathematics

Answer 1

The correct option is (c) \(1+\frac{x^2}{2!}+\frac{x^4}{4!}+….\infty\)

The explanation: We know the general expression for the expansion of the Taylor series

\(\tau[f(x)]=f(a)+\frac{x.f^{(1)}(a)}{1!}+\frac{x^2.f^{(2)}(a)}{2!}+…..\infty\)

Given a=0 we substitute in the equation to get

\(\tau[f(x)]=f(0)+f^{(1)}(0)\times \frac{x}{1!}+f^{(2)}(0)\times \frac{x^2}{2!}+….\infty\)

Now the n^th derivatives can be calculated as

\(f^{(n)}(x)=(\frac{e^x+e^{-x}}{2})^{(n)}\)

\(=\frac{e^x+(-1)^ne^{-x}}{2}\)

Substituting x=0 yields the final expansion

\(f^{(n)}(x)=\frac{1+(-1)^n}{2}\)

We get

\(\tau[f(x)]=1+(0)\times \frac{x}{1!}+(1)\times \frac{x^2}{2!}+(0)\times \frac{x^3}{3!}+…..\infty\)

\(\tau[f(x)]=1+\frac{x^2}{2!}+\frac{x^4}{4!}+…\infty\)