Question

The Mclaurin Series expansion of sin(e^x) is?

(a) sin(1)+\(\frac{x.cos(1)}{1!}+\sum_{n=2}^{\infty}\sum_{a=0}^{\infty}\frac{x^n.(-1)^a}{n!}\times\frac{(2a+1)^n}{(2a+1)!}\)

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

(c) \(-\frac{e^x}{1!}+\frac{e^{3x}}{3!}-\frac{e^{5x}}{5!}…\infty\)

(d) \(\sum_{n=2}^{\infty}\sum_{a=0}^{\infty}\frac{x^n.(-1)^a}{n!}\times \frac{(2a+1)^n}{(2a+1)!}\)

This question was addressed to me in an international level competition.

Asked question is from Taylor Mclaurin Series topic in portion Differential Calculus of Engineering Mathematics

Answer 1

Correct option is (a) sin(1)+\(\frac{x.cos(1)}{1!}+\sum_{n=2}^{\infty}\sum_{a=0}^{\infty}\frac{x^n.(-1)^a}{n!}\times\frac{(2a+1)^n}{(2a+1)!}\)

The best explanation: We know the series expansion for sin(x) is

sin(t)=\(\frac{t}{1!}-\frac{t^3}{3!}+\frac{t^5}{5!}…\infty\)

Substituting t=e^x we have

sin(e^x)=\(\frac{e^x}{1!}-\frac{e^3x}{3!}+\frac{e^5x}{5!}…\infty\)

Now using

\(e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+…\infty\)

We have

sin(e^x)=\(\frac{1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+…\infty}{1!}-\frac{1+\frac{3x}{1!}+\frac{(3x)^2}{2!}+\frac{(3x)^3}{3!}+…\infty}{3!}+\frac{1+\frac{5x}{1!}+\frac{(5x)^2}{2!}+\frac{(5x)^3}{3!}+…\infty}{5!}\)

Grouping terms with same power we have

\(=(\frac{1}{1!}-\frac{1}{3!}+\frac{1}{5!}…\infty)+x(\frac{1}{1!}-\frac{3}{3!}+\frac{5}{5!}…\infty)\)

+\(\frac{x^2}{2!}(\frac{1}{1!}-\frac{3^2}{3!}+\frac{5^2}{5!}…\infty)+…\infty\)

We can rewrite the last terms of the series as a double series we have

sin(1)+\(\frac{x.cos(1)}{1!}+\sum_{n=2}^{\infty}\sum_{a=0}^{\infty}\frac{x^n.(-1)^a}{n!}\times\frac{(2a+1)^n}{(2a+1)!}\)