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

2 Answers
Nov 15, 2016

(1+x^2)^(-1/2)=sum_(k=0)^oo (-1)^k (Pi_(j=1)^k(1/2+j-1))/(k!)x^(2k)

for absx le 1

Explanation:

We know that

(1+y)^n = 1+ny+(n(n-1))/(2!)y^2+cdots+(Pi_(j=1)^k(n+j-1))/(k!) y^k+cdots

so here

(1+x^2)^(-1/2)=sum_(k=0)^oo (-1)^k (Pi_(j=1)^k(1/2+j-1))/(k!)x^(2k)

The five first terms are

(1+x^2)^(-1/2) approx 1 - x^2/2 + (3 x^4)/8 - (5 x^6)/16 + (35 x^8)/128

Of course this series is convergent only for abs x le 1

Nov 15, 2016

The series is =1-x^2/2+(3x^4)/8-(5x^4)/16

Explanation:

The binomial theorem is (a+b)^n
=a^n+n*(a^(n-1)b)/1+ (n(n-1)a^(n-2)b^2)/(1*2)+(n(n-1)(n-2)a^(n-3)b^(3))/(1*2*3)

Rewriting f(x)=1/sqrt(1+x^2)=(1+x^2)^(-1/2)

So, a=1

b=x^2

and n=-1/2

Therefore, (1+x^2)^(-1/2)=1+(-1/2)x^2+(-1/2)(-3/2)x^4/(1*2)+(-1/2)(-3/2)(-5/2)x^6/(1*2*3)

=1-x^2/2+(3x^4)/8-(5x^6)/16