# Thread: Maclaurin series for square root (1+x)

1. ## Maclaurin series for square root (1+x)

I attempted to find the maclaurin series for the function Square root of 1+x.

F(0)=1 first term= 1
F'(0)=1/2 second term= (1/2)x
F''(0)=-1/4 Third term (-1/4)x^2
F'''(0)=3/8 fourth term (3/8*3!) x^3
F''''(0)=15/16 fifth (-15/16*4!) x^4
F'''''(0)105/32 six (105/32*5!) x^5
Therefore,
f(x)= 1+ (1/2)x + (-1/4)x^2 + (3/8*3!) x^3 + (-15/16*4!) x^4 + (105/32*5!) x^5

The problem is to find generalize term .
I have ( ((-1)^(n-1)) * something *x^n) / ((2^n) * (n!))

I cannot find that "something". because it exists as 1 for the first, second, and the third term , but then it increases to 3,15, 105
so it increases by factor of 3,5,7....
Help..

2. The binomial series is in general defined as...

$\displaystyle (1+x)^{\alpha} = \sum_{n=0}^{\infty} \binom{\alpha}{n}\cdot x^{n}$ (1)

.... where...

$\displaystyle \binom{\alpha}{n}= \frac{\alpha\cdot (\alpha-1)\cdot (\alpha-2)\dots (\alpha-n+1)}{n!}$ (2)

For $\displaystyle \alpha = \frac{1}{2}$ we have...

$\displaystyle (1+x)^{\frac{1}{2}} = 1 + \frac{1}{2}\cdot x - \frac{1}{2\cdot 4}\cdot x^{2} + \frac{1\cdot 3}{2\cdot 4\cdot 6}\cdot x^{3} - \dots$ (3)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

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### sqrt (1 x) maclaurin's series

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