Maclaurin series for square root (1+x)

  • #1
hangainlover
83
0

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..
 

Answers and Replies

  • #2
PhaseShifter
276
1
1=1*1
3=3*1
15=5*3
105=7*15

notice a pattern?
 
  • #3
hangainlover
83
0
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
 
  • #4
vela
Staff Emeritus
Science Advisor
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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.
 
Last edited:
  • #5
PhaseShifter
276
1
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.
 

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