How do you use the binomial series to expand #f(x)= sqrt(1+x^2)#?

1 Answer
Nov 25, 2016

Answer:

The series is #==1+x^2/2-x^4/8+x^6/16....#

Explanation:

The binomial theorem is
#(a+b)^n=((n),(0))a^n+((n),(1))a^(n-1)b+((n),(3))a^(n-2)b^2+((n),(4))a^(n-3)b^3+......#

#=a^n+na^(n-1)b+((n)(n-1))/(1*2)a^(n-2)b^2+((n)(n-1)(n-2))/(1*2*3)a^(n-3)b^3+....#

Normally #n in NN#

But there is an extension for #(1+x)^k# where #∣x∣<1# and #k# any number

Let's rewrite #f(x)=(1+x^2)^(1/2)#

So, #(1+x^2)^(1/2)=1+(1/2)x^2+(1/2)(-1/2)(1/2)(x^2)^2+(1/2)(-1/2)(-3/2)(1/6)(x^2)^3+......#

#=1+x^2/2-x^4/8+x^6/16.....#