Use partial fractions for distant coefficients

Express 2x2+3x−1(x−1)(x2+1)\frac{2x^2+3x-1}{(x-1)(x^2+1)} in partial fractions.

Hence find the series expansion of 2x2+3x−1(x−1)(x2+1)\frac{2x^2+3x-1}{(x-1)(x^2+1)} in ascending powers of xx, up to and including the term in x4x^4. Find the coefficient of x2008x^{2008}. marks

[2008/HCI/Prelim/I/6]

Solution

Partial fractions

Partial fractions

Express 2x2+3x−1(x−1)(x2+1)\dfrac{2x^2+3x-1}{(x-1)(x^2+1)} in partial fractions.

Hint

  1. Ax−1+Bx+Cx2+1\frac A{x-1}+\frac{Bx+C}{x^2+1}.

  2. Substitute values of xx and compare coefficients.

Answer · Partial fractions

2x−1+3x2+1\frac2{x-1}+\frac3{x^2+1}.

Expansion

Geometric series

Hence find the series expansion up to and including x4x^4.

Explore this idea

Hint

  1. 2x−1=−2(1−x)−1\frac2{x-1}=-2(1-x)^{-1} and 3x2+1=3(1+x2)−1\frac3{x^2+1}=3(1+x^2)^{-1}.

  2. Use (1−x)−1=1+x+x2+⋯(1-x)^{-1}=1+x+x^2+\cdots and (1+x2)−1=1−x2+x4−⋯(1+x^2)^{-1}=1-x^2+x^4-\cdots.

Answer · Expansion

1−2x−5x2−2x3+x4+O(x5)1-2x-5x^2-2x^3+x^4+O(x^5)

∣x∣<1|x|<1.

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 = 4
Test value xx = -0.50
Reset the controls
Function
1.06667
P₄(x)
1.0625
Error
-0.00417

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 series is valid for ∣x∣<1|x|<1, because both partial fractions have radius 1.

Ideas to try
Slide xx towards −1-1 and past it: where does the series stop working?
Coefficient

Pattern in coefficients

Find the coefficient of x2008x^{2008}.

Explore this idea

Hint

  1. Every xkx^k gets −2-2 from the first series.

  2. Only even powers get ±3\pm3 from the second.

Answer · Coefficient

[x2008]=1[x^{2008}]=1.

Explore the idea

The coefficients repeat every 4 steps

Switch on the coefficients of x⁰ to x¹¹: 1, −2, −5, −2, then again.

Switch on a feature to add it to the sketch and see how it is found. A full-mark sketch shows all of them.

  • Coefficients equal to 1
  • Coefficients equal to −5
  • Coefficients equal to −2
Ideas to try
Continue the pattern to x2008x^{2008}: which row of dots does it land in?