# Thread: Find the taylor series

1. ## 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

2. 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

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

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

find taylor at u=1
$\cos u$

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

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

$f(u)=\cos u= \cos (1) - \sin (1)\cdot (u-1)-\frac{\cos (1) \cdot (u-1)^2 }{2} + \frac{\sin (1) \cdot (u-1)^3 }{3!}$ $+ \frac{\cos (1) \cdot (u-1)^4 }{4!}- \frac{\sin (1) \cdot (u-1)^5}{5!} + ...$

$f(u) = \cos u = \sum_{k=0}^{\infty} \frac{(-1)^k \cos (1) (u-1)^{2k}}{(2k)!} + \sum_{k=0}^{\infty} \frac{(-1)^{k+1} \sin (1) (u-1)^{2k+1} }{(2k+1)!}$

sub u value

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

$f(x) = (x+1)^{\frac{1}{3}}$ at x=0

$f(0) =1$
$f'(x) = \frac{(x+1)^{\frac{-2}{3}}}{3}\Rightarrow f'(0)= \frac{1}{3}$
$f''(x) = \frac{-2\cdot (x+1)^{\frac{-5}{3}}}{3^2 } \Rightarrow f''(1) = \frac{-2}{3^2}$
$f'''(x) = \frac{-2\cdot -5 \cdot (x+1)^{\frac{-8}{3}}}{3^3}\Rightarrow \frac{2\cdot 5 }{3^3}$
$f''''(1) = -\frac{2\cdot 5 \cdot 8 }{3^4}$

$f(x) = (x+1)^{\frac{1}{3}} = 1 + \frac{x}{3\cdot 1!} - \frac{2x^2}{3^2 \cdot 2!}$ $+ \frac{2\cdot 5\cdot x^3 }{3^3 \cdot 3!} - \frac{2\cdot 5 \cdot 8 \cdot x^4}{3^4 \cdot 4!} + ...$

you can use $\prod$ to collect $2.5.8...$
can you find it now ?

3. I make a mistake in cos sqrt{x+1} the Taylor expansion should not contains roots so it will come like this

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

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

but

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

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

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

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

This solution for my prof not me