# Question ba4a6

Jan 19, 2017

$\sqrt{1 + h} = 1 + \frac{h}{2} - {h}^{2} / 8 + {h}^{3} / 16. . .$

#### Explanation:

$f \left(x\right) = \sqrt{x} = {x}^{\frac{1}{2}}$ so $f \left(1\right) = 1$
$f p r i m e \left(x\right) = \frac{1}{2} {x}^{- \frac{1}{2}}$ so $f p r i m e \left(1\right) = \frac{1}{2}$
$f p r i m e p r i m e \left(x\right) = \left(- \frac{1}{2}\right) \left(\frac{1}{2}\right) {x}^{- \frac{1}{2}}$ so $f p r i m e p r i m e \left(1\right) = - \frac{1}{4}$
$f p r i m e p r i m e p r i m e \left(x\right) = \left(- \frac{3}{2}\right) \left(- \frac{1}{2}\right) \left(\frac{1}{2}\right) {x}^{- \frac{5}{2}}$ so $f p r i m e p r i m e p r i m e \left(1\right) = \frac{3}{8}$

Taylor series:
f(a+h)=f(a)+(h f'(a))/(1!) + (h^2 f'(1))/(2!) +(h^3 f(a))/(3!)...
Setting $a = 1$ and using the derivatives above:
$\sqrt{1 + h} = 1 + \left(\frac{1}{2}\right) h + \left(- \frac{1}{4}\right) \left(\frac{1}{2}\right) {h}^{2} + \left(\frac{3}{8}\right) \left(\frac{1}{6}\right) \ldots$
$= 1 + \frac{h}{2} - {h}^{2} / 8 + {h}^{3} / 16. . .$

Jan 19, 2017

sqrt(x) = 1+ sum_(n=1)^oo (-1)^(2n+1)/2^(n+1)(1*3*...*(2n-1))/(n!) (x-1)^n

#### Explanation:

The formula for the Taylor series of a function $f \left(x\right)$ indefinitely differentiable in the point $x = a$ is:

f(x) = sum_(n=0)^oo (f^((n))(a))/(n!) (x-a)^n

We have:

$f \left(x\right) = \sqrt{x} = {x}^{\frac{1}{2}} \implies {f}^{\left(0\right)} \left(1\right) = 1$

then we need to find the derivatives of all orders:

${f}^{\left(1\right)} \left(x\right) = \frac{d}{\mathrm{dx}} \sqrt{x} = \frac{1}{2} {x}^{- \frac{1}{2}}$

${f}^{\left(2\right)} \left(x\right) = {d}^{2} / \left({\mathrm{dx}}^{2}\right) \sqrt{x} = - \frac{1}{4} {x}^{- \frac{3}{2}}$

and we can easily see that in general for $n > 1$:

${f}^{\left(n\right)} \left(x\right) = {d}^{n} / \left({\mathrm{dx}}^{n}\right) \sqrt{x} = {\left(- 1\right)}^{n + 1} \frac{1}{2} \left(\frac{1}{2} - 1\right) \left(\frac{1}{2} - 2\right) \ldots \left(\frac{1}{2} - n\right) {x}^{\frac{1}{2} - n}$

${f}^{\left(n\right)} \left(1\right) = {\left(- 1\right)}^{n - 1} \frac{1}{2} \left(\frac{1}{2} - 1\right) \left(\frac{1}{2} - 2\right) \ldots \left(\frac{1}{2} - n\right) = {\left(- 1\right)}^{n + 1} / {2}^{n + 1} \left(1 - 2\right) \left(1 - 4\right) \ldots \left(1 - 2 n\right) = {\left(- 1\right)}^{2 n + 1} / {2}^{n + 1} \left(1 \cdot 3 \cdot \ldots \cdot \left(2 n - 1\right)\right)$

Substituting this in the series expression:

sqrt(x) = 1+ sum_(n=1)^oo (-1)^(2n+1)/2^(n+1)(1*3*...*(2n-1))/(n!) (x-1)^n

Jan 19, 2017

Use the Binomial Theorem and claim that the Taylor Series is necessarily the same.

#### Explanation:

${\left(1 + x\right)}^{\frac{1}{2}}$
=1+(1/2)x+((1/2)(1/2-1))/(2!)x^2+((1/2)(1/2-1)(1/2-2))/(3!)x^3...#
$= 1 + \frac{x}{2} - {x}^{2} / 8 + \frac{3 {x}^{3}}{16.} . .$