# Find the taylor series

• Nov 14th 2009, 07:14 AM
gichfred
Find the taylor series
Hello,
I have been trying to find the taylor series for the follwing problems for a couple of hours but I have made no progress....please show how I can solve them......

1. f(x)=cos((x+1)^1/2) at x=0
2. f(x)=(x+1)^1/3 at x=0
• Nov 14th 2009, 11:47 AM
Amer
Quote:

Originally Posted by gichfred
Hello,
I have been trying to find the taylor series for the follwing problems for a couple of hours but I have made no progress....please show how I can solve them......

1. f(x)=cos((x+1)^1/2) at x=0
2. f(x)=(x+1)^1/3 at x=0

taylor

$\displaystyle \cos (\sqrt{1+x})$ at x=0

let $\displaystyle u= \sqrt{x+1}$

find taylor at u=1
$\displaystyle \cos u$

$\displaystyle f(u) = \cos u$
$\displaystyle f(1) = \cos (1)$
$\displaystyle f'(u) = -\sin u \Rightarrow f'(1) =-\sin (1)$
$\displaystyle f''(u) = -\cos u \Rightarrow f''(1) = -\cos (1)$
$\displaystyle f'''(u) = \sin u \Rightarrow f'''(1) = \sin (1)$

$\displaystyle f(u)=\cos u = f(1) + f'(1)(u-1) + \frac{f''(1)(u-1)^2}{2} + \frac{f'''(1)(u-1)^3}{3!} + ...$

$\displaystyle f(u)=\cos u= \cos (1) - \sin (1)\cdot (u-1)-\frac{\cos (1) \cdot (u-1)^2 }{2} + \frac{\sin (1) \cdot (u-1)^3 }{3!}$$\displaystyle + \frac{\cos (1) \cdot (u-1)^4 }{4!}- \frac{\sin (1) \cdot (u-1)^5}{5!} + ... \displaystyle f(u) = \cos u = \sum_{k=0}^{\infty} \frac{(-1)^k \cos (1) (u-1)^{2k}}{(2k)!} + \sum_{k=0}^{\infty} \frac{(-1)^{k+1} \sin (1) (u-1)^{2k+1} }{(2k+1)!} sub u value \displaystyle \cos \sqrt{x+1} =\sum_{k=0}^{\infty} \frac{(-1)^k \cos (1) (\sqrt{x+1}-1)^{2k}}{(2k)!} + \sum_{k=0}^{\infty} \frac{(-1)^{k+1} \sin (1) (\sqrt{x+1}-1)^{2k+1} }{(2k+1)!} \displaystyle f(x) = (x+1)^{\frac{1}{3}} at x=0 \displaystyle f(0) =1 \displaystyle f'(x) = \frac{(x+1)^{\frac{-2}{3}}}{3}\Rightarrow f'(0)= \frac{1}{3} \displaystyle f''(x) = \frac{-2\cdot (x+1)^{\frac{-5}{3}}}{3^2 } \Rightarrow f''(1) = \frac{-2}{3^2} \displaystyle f'''(x) = \frac{-2\cdot -5 \cdot (x+1)^{\frac{-8}{3}}}{3^3}\Rightarrow \frac{2\cdot 5 }{3^3} \displaystyle f''''(1) = -\frac{2\cdot 5 \cdot 8 }{3^4} \displaystyle f(x) = (x+1)^{\frac{1}{3}} = 1 + \frac{x}{3\cdot 1!} - \frac{2x^2}{3^2 \cdot 2!}$$\displaystyle + \frac{2\cdot 5\cdot x^3 }{3^3 \cdot 3!} - \frac{2\cdot 5 \cdot 8 \cdot x^4}{3^4 \cdot 4!} + ...$

you can use $\displaystyle \prod$ to collect $\displaystyle 2.5.8...$
can you find it now ?
• Nov 15th 2009, 02:30 AM
Amer
I make a mistake in cos sqrt{x+1} the Taylor expansion should not contains roots so it will come like this

$\displaystyle \cos \sqrt{x+1} = \sum_{k=0}^{\infty} \frac{(-1)^k (\sqrt{x+1})^{2k}}{(2k)!}$

$\displaystyle \cos \sqrt{x+1} = \sum_{k=0}^{\infty} \frac{(-1)^k (x+1)^k}{(2k)!}$

but

$\displaystyle (x+1)^k = \sum_{i=0}^{k} com(k,i) x^i$

$\displaystyle com(k,i)= \frac{k!}{i! (k-i)!}$

$\displaystyle \cos \sqrt{x+1} = \sum_{k=0}^{\infty} \frac{(-1)^k \cdot \sum_{i=0}^{k} com(k,i) x^i }{(2k)!}$

simplify it, sub k values and take $\displaystyle x,x^2,x^3,x^4,...$ as factors

This solution for my prof not me