Q 28. - Calculus

Question

Manipulating power series: Find power series representations

for the indicated functions.

$e-\frac{e^3}{3!}+\frac{e^5}{5!}-\frac{e^7}{7!}+\cdots$

Step-by-Step Solution

Answer

The power series representations for the indicated functions

is $e-\frac{e^3}{3!}+\frac{e^5}{5!}-\frac{e^7}{7!}+\cdots =\sin e$

Step 1: Given information

The series is $e-\frac{e^3}{3!}+\frac{e^5}{5!}-\frac{e^7}{7!}+\cdots$

Step 2: Find the values of the series.

The Maclaurin series for the function $f\left(x\right)=\sin x$ is

$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots$

The series $e-\frac{e^3}{3!}+\frac{e^5}{5!}-\frac{e^7}{7!}+\cdots$is the Maclaurin series for $\sin x$at $x=e$

$x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots =\sin x$

Therefore,

$e-\frac{e^3}{3!}+\frac{e^5}{5!}-\frac{e^7}{7!}+\cdots =\sin e$