Maclaurin series for square root (1+x)

[hangainlover]

Homework Statement

Maclaurin series for square root (1+x)

Homework Equations

The Attempt at a Solution

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 the previous term increases by factor of 3,5,7... (somewhat recursive?)
Help..

 

[PhaseShifter]

1=1*1
3=3*1
15=5*3
105=7*15

notice a pattern?

 

[hangainlover]

yeah that part is easy but it goes like this
1,1,1,3,15,105...
how do you account for the first two 1s

 

[vela]

Two of the ones you'd figure would come from the n=0 and n=1 cases. It's the other one that's a bit difficult. What general form did you get for the "something" so far?

By the way, the third term in your original post is missing the 2! in the denominator.

 

[PhaseShifter]

Try following the sequence backwards:
105/7=15
15/5=3
3/3=1
1/1=1
1/-1=-1

Not surprising, since the function is proportional to (1+x)^1/2, and the derivatives are proportional to (1+x)^(-1/2), (1+x)^(-3/2), (1+x)^(-5/2),etc.