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

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Answer 1 · 2016-08-04T13:46:43

#1/sqrt(2)sum_{n=0}^oo(-1)^n/(n!2^{2n} )(Pi_{k=0}^{n-1}(2k+1))x^(n+2)#

Expanding for #x = 0# the function #f(x) = (x+2)^{-1/2}# gives

#f(x) = f(0) + f^'(0)/(1!)x+f^('')(0)/(2!)x^2+cdots + f^((n))(0)/(n!)x^n +cdots#

but

#(d^n)/(dx^n)f(x) = (-1/2)(-1/2-1)cdots(-1/2-n+1)(x+2)^{-1/2-n}#

or

#(d^n)/(dx^n)f(x) = (-1)^n Pi_{k=0}^{n-1}(2k+1)/2(x+2)^{-1/2-n}#

then

# f^((n))(0)=(-1)^n Pi_{k=0}^{n-1}((2k+1)/2) 2^{(-1/2-n)}#
#f^((n))(0)=1/sqrt(2)(-1)^n/2^{2n} Pi_{k=0}^{n-1}(2k+1)#

So the series for #f(x)# is

#f(x) = 1/sqrt(2)sum_{n=0}^oo(-1)^n/(n!2^{2n} )(Pi_{k=0}^{n-1}(2k+1))x^n#

The series representation for #x^2(x+2)^{-1/2}# is given by

#1/sqrt(2)sum_{n=0}^oo(-1)^n/(n!2^{2n} )(Pi_{k=0}^{n-1}(2k+1))x^(n+2)#