Repeated poles and general coefficients

Express 9+4x2(1−2x)2(2+x)\frac{9+4x^2}{(1-2x)^2(2+x)} in partial fractions. marks

Hence find in its simplest form,

(a) the first four terms in the expansion of 9+4x2(1−2x)2(2+x)\frac{9+4x^2}{(1-2x)^2(2+x)} in ascending powers of xx and state the set of values of xx for which the expansion is valid.

(b) the coefficient of xrx^r in the expansion. marks

[2009/SAJC/JCT/8]

Total marks: 9

Solution

Partial fractions

Repeated linear factors

Express 9+4x2(1−2x)2(2+x)\dfrac{9+4x^2}{(1-2x)^2(2+x)} in partial fractions.

Hint

  1. A1−2x+B(1−2x)2+C2+x\frac A{1-2x}+\frac B{(1-2x)^2}+\frac C{2+x}.

Answer · Partial fractions

4(1−2x)2+12+x\frac4{(1-2x)^2}+\frac1{2+x}.

(a)

Series of a sum

Find the first four terms of the expansion and the set of values of xx for which it is valid.

Explore this idea

Hint

  1. 4(1−2x)−2+12(1+x2)−14(1-2x)^{-2}+\frac12\left(1+\frac x2\right)^{-1}.

  2. Expand each and add. Validity is the intersection.

Answer · (a)

92+634x+3858x2+204716x3+O(x4)\frac92+\frac{63}4x+\frac{385}8x^2+\frac{2047}{16}x^3+O(x^4)

∣x∣<1/2|x|<1/2.

Explore the idea

Terms of the series

Compare the function with a finite polynomial. Change the highest included power and test x, then use the error plot to see how the approximation changes at that x.

Highest included powerxⁿ, n = 3
Test value xx = 0.20
Reset the controls
Function
11.56566
P₃(x)
10.5985
Error
-0.96716

Inside the stated open interval. A finite truncation can still have a noticeable error.

Teal marks the function, amber marks the polynomial, and the red segment joins their values at the selected x.

Does keeping more powers help at this x?

Highest included power n runs horizontally; signed error Pₙ(x) − f(x) runs vertically. Zero means agreement at this x. A missing power leaves the polynomial unchanged; individual steps need not improve the error.

  • Function
  • Series truncated at the chosen power
  • Stated open interval. Endpoint convergence requires a separate check
Key idea

The pole at x=12x=\frac12 limits the valid range to ∣x∣<12|x|<\frac12.

Ideas to try
Slide xx towards 12\frac12: the function blows up, while a fixed-degree polynomial stays finite and becomes a poor approximation.
(b)

General coefficient

Find the coefficient of xrx^r.

Explore this idea

Hint

  1. Use (1−z)−2=∑(r+1)zr(1-z)^{-2}=\sum(r+1)z^r and (1+w)−1=∑(−w)r(1+w)^{-1}=\sum(-w)^r.

Answer · (b)

[xr]=(r+1)2r+2+(−1)r/2r+1[x^r]=(r+1)2^{r+2}+(-1)^r/2^{r+1}

Integer r≥0r\ge0.

Explore the idea

The coefficient pattern

Compare the function with a finite polynomial. Change the highest included power and test x, then use the error plot to see how the approximation changes at that x.

Highest included powerxⁿ, n = 4
Test value xx = 0.35
Reset the controls
Function
44.86998
P₄(x)
26.1956
Error
-18.67437

Inside the stated open interval. A finite truncation can still have a noticeable error.

Teal marks the function, amber marks the polynomial, and the red segment joins their values at the selected x.

Does keeping more powers help at this x?

Highest included power n runs horizontally; signed error Pₙ(x) − f(x) runs vertically. Zero means agreement at this x. A missing power leaves the polynomial unchanged; individual steps need not improve the error.

  • Function
  • Series truncated at the chosen power
  • Stated open interval. Endpoint convergence requires a separate check
Key idea

Each extra term contributes [xr]xr[x^r]x^r; the closed form lets you write them all down.

Ideas to try
Raise the degree one step at a time and watch how the polynomial improves.