How do you find a power series representation for # (x/(2-x))^3# and what is the radius of convergence?

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Answer 1 · 2016-10-21T17:25:16

#(x/(2-x))^3 equiv sum_(k=2)^oo(k(k-1))/2^(k+2)x^(k+1)# which is convergent for #abs(x) < 2#

We have #(x/(2-x))^3=(x/2)^3(1/(1-x/2))^3#

Calling #y = x/2# we have

#(x/(2-x))^3 equiv y^3(1/(1-y))^3#

but

#d^2/dy^2(1/(1-y))=2(1/(1-y))^3#

Now, for #abs y < 1# we have

#1/(1-y)=sum_(k=0)^ooy^k# then

#y^3(1/(1-y))^3 = y^3/2d^2/dy^2 sum_(k=0)^ooy^k=y^3/2 sum_(k=2)^ook(k-1)y^(k-2)# or

#y^3(1/(1-y))^3=1/2sum_(k=2)^ook(k-1)y^(k+1)#

but #y=x/2# so finally

#(x/(2-x))^3 equiv sum_(k=2)^oo(k(k-1))/2^(k+2)x^(k+1)# which is convergent for #abs(x/2) < 1# or #abs(x) < 2#