Write #(9x^2+26x+20)/((1+x)(2+x)^2)# as partial fraction, then expand as binomials up to and including #x^3#. Show the expansion is roughly #5-(7x)/2+Bx^2+Cx^3#?
For partial fractions I got:
#3/((1+x))+2/((2+x))+4/(2+x)^2#
For the expansion, I got: #5-(9x)/2+4x^2-(27x^3)/8# , I'm not sure why I'm getting #-(9x)/2# and not #-(7x)/2#
For partial fractions I got:
For the expansion, I got:
1 Answer
partial fraction decomposition is:
# (9x^2+26x+20)/((1+x)(2+x)^2) -= 3/(1+x) + 6/(2+x) - 4/(2+x)^2 #
The binomial expansion is:
# (9x^2+26x+20)/((1+x)(2+x)^2) = 5 -7/2x+3x^2 -23/8x^3 + ...#
Leading to
Explanation:
The partial fraction decomposition will be of the form:
# (9x^2+26x+20)/((1+x)(2+x)^2) -= A/(1+x) + B/(2+x) + C/(2+x)^2 #
# " " = ( A(2+x)^2 + B(2+x)(2+x) + C(1+x) ) / ((1+x)(2+x)^2)#
Leading to the identity:
# 9x^2+26x+20 -= A(2+x)^2 + B(1+x)(2+x) + C(1+x) #
Where
Put
# x = -1 => 9-26+20=A=> A = 3#
Put# x = -2 => 36-52+20=-C=>C=-4#
Coeff#(x^0) => 20=4A+2B+C=> B=6#
So we can now write:
# (9x^2+26x+20)/((1+x)(2+x)^2) -= 3/(1+x) + 6/(2+x) - 4/(2+x)^2 #
Now we can write the as
# S = 3/(1+x) + 6/(2+x) - 4/(2+x)^2 #
# \ \ = 3/(1+x) + 6/(2(1+x/2)) - 4/(2(1+x/2))^2 #
# \ \ = 3(1+x)^(-1) + 3(1+x/2)^(-1) - (1+x/2)^(-2) #
The Binomial Series tells us that:
# (1+x)^n = 1+nx + (n(n-1))/(2!)x^2 (n(n-1)(n-2))/(3!)x^3 + ...#
# S_1 = 3(1+x)^(-1) #
# \ \ \ \ = 3{1+(-1)x+((-1)(-2))/(2)x^2 + ((-1)(-2)(-3))/6x^3 + ... } #
# \ \ \ \ = 3{1-x+x^2 -x^3 + ... } #
# \ \ \ \ = 3-3+3x^2 -3x^3 + ... #
# S_2 = 3(1+x/2)^(-1) #
# \ \ \ \ = 3{1+(-1)(x/2)+((-1)(-2))/(2)(x/2)^2 + ((-1)(-2)(-3))/6(x/2)^3 + ... }#
# \ \ \ \ = 3{1-1/2x+1/4x^2 -1/8x^3+ ... }#
# \ \ \ \ = 3-3/2x+3/4x^2 -3/8x^3+ ... #
# S_3 = (1+x/2)^(-2) #
# \ \ \ \ = 1+(-2)(x/2)+((-2)(-3))/2(x/2)^2+((-2)(-3)(-4))/6(x/2)^3+... #
# \ \ \ \ = 1-x+3/4x^2-1/2x^3+... #
Combining these results:
# S= {3-3x+3x^2 -3x^3 } + {3-3/2x+3/4x^2 -3/8x^3} - { 1-x+3/4x^2-1/2x^3}+O(x^4) #
# \ \ = (3+3-1) + (-3-3/2+1)x +(3+3/4-3/4)x^2 + (-3-3/8+1/2)x^3#
# \ \ = 5 -7/2x+3x^2 -23/8x^3 + ...#