Express #f(x) = (3x)/ [(1 +x)(1+2x^2)]# in partial fractions?

let #f(x) = (3x)/ [(1 +x)(1+2x^2)]#
(i) Express f(x) in partial fractions.
(ii) Hence obtain the expansion of f(x) in ascending powers of x, up to and including the term in #x^3#

2 Answers
Jul 25, 2018

The partial fraction is #f(x)=(-1)/(1+x)+(2x+1)/(1+2x^2)#. The expansion is #=3(x-x^2-x^3)+o(x^3)#

Explanation:

Perform the decomposition into partial fractions

#(3x)/((1+x)(1+2x^2))=A/(1+x)+(Bx+C)/(1+2x^2)#

#=(A(1+2x^2)+(Bx+C)(1+x))/((1+x)(1+2x^2))#

The denominators are the same, compare the numerators

#3x=A(1+2x^2)+(Bx+C)(1+x)#

Let #x=0#, #=>#, #0=A+C#

Let #x=-1#, #=>#, #-3=3A#, #=>#, #A=-1#

#=>#, #C=-A=1#

Coefficients of #x^2#

#0=2A+B#, #=>#, #B=-2A=2#

Therefore,

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

The Taylor expansions are

#-1/(1+x)=-1+x-x^2+x^3 + o(x^3)#

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

You can also obtain theses expansions by performing a long division

Therefore,

#f(x)=-1/(1+x)+(2x+1)/(1+2x^2)=0+3x-3x^2-3x^3+o(x^3)#

#{3x}/{(1+x)(1+2x^2)}=\frac{-1}{1+x}+\frac{2x+1}{1+2x^2}#

Explanation:

Let

#{3x}/{(1+x)(1+2x^2)}=\frac{A}{1+x}+\frac{Bx+C}{1+2x^2}#

#3x=(2A+B)x^2+(B+C)x+(A+C)#

Comparing the corresponding coefficients on both the sides, we get

#2A+B=0 \ ......(1)#

#B+C=3\ ............(2)#

#A+C=0\ .......(3)#

Subtracting (2) from (3), we get

#A-B=-3\ ......(4)#

Adding (1) & (4) we get

#2A+B+A-B=0-3#

#3A=-3#

#A=-1 #

setting #A=-1# in (4), we get

#-1-B=-3#

#B=2#

Setting #A=-1# in (3), we get

#-1+C=0#

#C=1#

Now, setting the values of #A, B, C#, we get following partial fractions

#{3x}/{(1+x)(1+2x^2)}=\frac{-1}{1+x}+\frac{2x+1}{1+2x^2}#