# How do you write the taylor series for f(x)=sqrt(x) at a=16 and find the radius of convergence.?

May 28, 2017

$f \left(x\right) = 4 + \frac{1}{8} \left(x - 16\right) - \frac{1}{512} {\left(x - 16\right)}^{2} + \frac{1}{16384} {\left(x - 16\right)}^{3} + \cdots$

$L = \left\mid 1 - \frac{x}{16} \right\mid < 1$

#### Explanation:

A Taylor series is a series expansion of a function about a point. A one-dimensional Taylor series is an expansion of a real function $f \left(x\right)$ about a point $x = a$ is given by

f(x)=f(a)+f'(a)(x-a)+(f''(a))/(2!)(x-a)^2+(f^(3)(a))/(3!)(x-a)^3+cdots+(f^(n)(a))/(n!)(x-a)^n+cdots

We are given
$f \left(x\right) = \sqrt{x}$ and $a = 16$

Here are the first four derivatives
$f ' \left(x\right) = \frac{1}{2 \sqrt{x}}$

$f ' ' \left(x\right) = - \frac{1}{4 {x}^{3 / 2}}$

${f}^{\left(3\right)} \left(x\right) = \frac{3}{8 {x}^{5 / 2}}$

${f}^{\left(4\right)} \left(x\right) = - \frac{15}{16 {x}^{7 / 2}}$

Plug these derivatives into the Taylor series expansion above

f(x)=sqrt(a)+1/(2sqrt(a))(x-a)-(1)/(2!4a^(3//2))(x-a)^2+(3)/(3!8a^(5//2))(x-a)^3-(15)/(4!16a^(7//2))(x-a)+cdots

Plug in $a = 16$

f(x)=sqrt(16)+1/(2sqrt(16))(x-16)-(1)/(2!4(16)^(3//2))(x-16)^2+(3)/(3!8(16)^(5//2))(x-16)^3-(15)/(4!16(16)^(7//2))(x-16)+cdots

Simplify $\sqrt{16} = 4$

f(x)=4+1/(2(4))(x-16)-(1)/(2!4(4)^(3))(x-16)^2+(3)/(3!8(4)^(5))(x-16)^3-(15)/(4!16(4)^(7))(x-16)+cdots

Simplify the fractions

$f \left(x\right) = 4 + \frac{1}{8} \left(x - 16\right) - \frac{1}{512} {\left(x - 16\right)}^{2} + \frac{1}{16384} {\left(x - 16\right)}^{3} - \frac{1}{2097152} {\left(x - 16\right)}^{4} + \cdots$

Finally, the radius of convergence is found when

$L = \left\mid 1 - \frac{x}{16} \right\mid < 1$