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

1 Answer
Jun 27, 2015

#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. :)