Question

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

Answer 1

`(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`

Answer 2

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`