Expert: Sherry Wallin - 8/17/2009

Question
QUESTION: I)Express (2-x+8x^2)/((1-x)(1+2x)(2+x)) in partial fractions. II) Hence obtain the expansion of (2-x+8x^2)/((1-x)(1+2x)(2+x)) in ascending powers of x, up to & including the term in x^2. How to do the ii) part? I know how to do the i) part only. Please help me. Thanks a lot

ANSWER: Warm~
    What did you get for part I?

Math Prof

---------- FOLLOW-UP ----------

QUESTION: I got this (1/(1-x))+(2/(1+2x))-(4/(2+x))

Answer
Hi Warm~
    You did the partial fraction decomposition correctly...
To use the binomial expansion expand one term at a time and since you are suppose to do expand up through x^2 expand out to the x^2 term in each and then combine like terms and write the result in ascending powers of x (which is how each expansion is written anyway).

The binomial expansion form looks like (out to the x^2 term): (1+x)^n = 1 + nx + [n(n-1)/2!]x^2

1/(1-x) = (1+(-x))^-1 = 1 +(-1)(-x) + [(-1)(-1-1)/2!)x^2 = 1 + x + x^2 (here x is (-x) and n = -1)

2/(1+2x)= 2[(1+2x)^-1) = 2[1 +(-1)(2x)+[(-1)(-1-1)/2!](2x)^2] = 2(1 + (-2x) + [(-1)(-2)/2]4x^2)
= 2(1 + (-2x) + [2/2]4x^2) = 2 -4x + 8x^2  (x here is (2x) and again n = -1)

4/(2+x) = 4(2+x)^-1 = 4[2(1+(1/2)x)^-1] = 4(2^-1)(1+(1/2)x)^-1 = 4(1/2)(1+(1/2)x)^-1 = 2((1+(1/2)x)^-1 (x here is (1/2)x and n = -1 again)

so the expansion for 4/(2+x) is:
2[1 + (-1)(1/2)x + [(-1)(-1-1)/2!]((1/2)x)^2 = 2(1 - (1/2)x +(1/4)x^2) = 2 -x +(1/2)x^2

Now combine all the expansion terms:
1 + x + x^2 + 2 -4x + 8x^2 -[2 -x +(1/2)x^2] = 1 + x + x^2 + 2 -4x + 8x^2 -2 + x -(1/2)x^2
= 1 -2x +(17/2)x^2

So there is your answer to part II. If you have any further questions about this problem feel free to write and ask.

Math Prof