Answer to Question #224771 in Calculus for Israr

Question

Find the Maclaurin series or Taylor series centered at a, interval of

convergence and the radius of convergence of 1

√1−𝑥

.

Answer 1

Solution;

Maclaurin series is given by;

Since a=0;

"f(x)=\\displaystyle\\sum_{n=0}^\\infin\\frac{f^n(0)x^n}{n!}=f(0)+f'(0)x+f"(0)\\frac{x^2}{2!}+f'''(0)\\frac{x^3}{3!}+..."

Given;

$f(x)=\sqrt{1-x}$

$f(0)=\sqrt{1-0}=1$

By differentiation;

$f'(x)=-\frac{1}{2\sqrt{1-x}}$

$f'(0)=-\frac{1}{2\sqrt{1-0}}=-\frac12$

$f''(x)=-\frac{1}{4(1-x)^{\frac32}}$

$f''(0)=-\frac{1}{4(1-0)^{\frac32}}=-\frac14$

$f'''(x)=-\frac{3}{8(1-x)^{\frac52}}$

$f'''(0)=-\frac{3}{8(1-0)^{\frac52}}=-\frac38$

By substitution;

$f(x)=1-\frac12x-\frac18x^2-\frac1{16}x^3+...$

From the series;

$f(x)=\displaystyle\sum_{n=1}^{\infin}(-1)\frac{x^n}{2(n^{2n-2})}$

$a_n=-\frac{1}{2(n^{2n-2}}$

$a_{n+1}=\frac{-1}{2(n+1)^{2n}}$

Using root test;

$L=|x|\displaystyle{lim_{n\to\infin}}|\frac{a_{n+1}}{a_n}|$

$L=|x|lim_{n\to\infin}|\frac{n^{2n-2}}{(n+1)^{2n}}|$

$L=|x|$

The interval of convergence will be ;

$1<x<1$

Hence Radius of convergence is;

$R=1$