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

2 Answers
Nov 15, 2016

Answer:

#(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

Answer:

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#