Question

How do you find the 1st 4 nonzero terms in the Taylor series expansion about x=0 for `f(x)= sqrt(1+x)`?

Answer 1

`f(x)~~1 + x/2-x^2/(8)+(3x^3)/(48)` (for x close to 0)

Explanation:

For a general function `f(x)`, we can do a Taylor series expansion about `x=0`, (called the Maclaurin Series) by doing the following:

`f(x) = f(0) + x*f'(0) + (x^2*f''(0))/(2!) + (x^3*f^((3))(0))/(3!) + ...`

This can be written more concisely with summation notation as

`f(x) = sum_(n=0)^(\oo)(x^n f^((n))(0))/(n!)`

In this case, we have `f(x)=sqrt(1+x)`
Note that
`f'(x)=1/(2sqrt(1+x))`

`f''(x)=-1/(4(1+x)^(3/2))`

`f^((3))(x)=3/(8(1+x)^(5/2))`

This gives:

`f(0) = 1`
`f'(0)=1/2`
`f''(0)=-1/4`
`f^((3))(0)=3/8`

So the first 4 nonzero terms of the Taylor expansion about `x=0` for `f(x)=sqrt(1+x)` are

`f(x)~~1 + x/2-x^2/(8)+(3x^3)/(48)`

PS: I hope there are no errors. :)