How do you use the binomial series to expand #f(x)=1/(sqrt(4+x^2))#?

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Answer 1 · 2018-07-03T11:30:15

The answer is #=1/2-1/16x^2+3/256x^4+.....#

#(a+b)^n=sum_(k=0)^n((n),(k))a^(n-k)b^k#

#=((n),(0))a^nb^0+((n),(1))a^(n-1)b+((n),(2))a^(n-2)b^2+.....#

#=a^n+((n)(n-1))/(1*2)a^2b+((n)(n-1)(n-2))/(1*2.3)a^3b^2+...#

Where,

#((n),(k))=(n!)/((n-k)!(k!))#

Also,

#(1+b)^n=1+((n)(n-1))/(1*2)b+((n)(n-1)(n-2))/(1*2.3)b^2+..#

Here, we have

#1/sqrt(4+x^2)=1/(2(1+(x/2)^2)^(1/2))#

Here

#n=-1/2#

#a=1#

#b=(x/2)^2#

Therefore,

#1/sqrt(4+x^2)=1/2(1+(x/2)^2)^(-1/2)#

#=1/2(((1/2),(0))1^(1/2)(x^2/2)^0+((1/2),(1))1^(1/2-1)(x^2/2)^2+((1/2),(2))1^(1/2-2)(x/2)^4+......)#

#=1/2(1+(-1/2)(x/2)^2+((-1/2)(-3/2))/(1*2)(x/2)^4+..)#

#=1/2-1/16x^2+3/256x^4+.....#