Recover parameters and a general coefficient

It is given that the first 3 terms in the expansion of (2−ax)−b(2-ax)^{-b} in ascending powers of xx are 18+916x+cx2+⋯\frac18+\frac9{16}x+cx^2+\cdots, where a,b,ca,b,c are positive constants.

(i) Find the values of a,b,ca,b,c. marks

(ii) Show that the coefficient of xrx^r is in the form p(r+1)(r+2)qrp(r+1)(r+2)q^r, where pp and qq are constants to be determined. marks

[ACJC/2009/Prelim/II/2]

Total marks: 11

Solution

(i)

Matching coefficients

(2−ax)−b=18+916x+cx2+⋯(2-ax)^{-b}=\frac18+\frac9{16}x+cx^2+\cdots with a,b,c>0a,b,c>0. Find aa, bb, cc.

Explore this idea

Hint

  1. Normalise: (2−ax)−b=2−b(1−ax2)−b(2-ax)^{-b}=2^{-b}\left(1-\frac{ax}2\right)^{-b}.

  2. Match the constant term first: it gives bb.

  3. Then match xx and x2x^2.

Answer · (i)

a=3, b=3, c=27/16a=3,\ b=3,\ c=27/16.

Explore the idea

(2 − 3x)⁻³ and its 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 = 2
Test value xx = 0.30
Reset the controls
Function
0.75131
P₂(x)
0.44562
Error
-0.30569

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 constant 1/8, the slope 9/16 and the quadratic coefficient c=2716c=\frac{27}{16} come directly from the first three terms.

Ideas to try
Slide xx towards 23≈0.667\frac23\approx0.667 and beyond: where does the series fail?
(ii)

General term

Show that the coefficient of xrx^r is p(r+1)(r+2)qrp(r+1)(r+2)q^r and find pp and qq.

Explore this idea

Hint

  1. Use (−3r)(−z)r=(r+22)zr\binom{-3}r(-z)^r=\binom{r+2}2z^r for (1−z)−3(1-z)^{-3} with z=3x2z=\frac{3x}2.

Answer · (ii)

[xr]=116(r+1)(r+2)(3/2)r[x^r]=\frac1{16}(r+1)(r+2)(3/2)^r; p=1/16, q=3/2p=1/16,\ q=3/2.

Integer r≥0r\ge0; ∣x∣<2/3|x|<2/3.

Explore the idea

The coefficients grow like (r+1)(r+2)(3/2)^r/16

Switch on the first coefficients and compare them with the smooth curve.

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

Ideas to try
Check r=3r=3: 116⋅4⋅5⋅(32)3=13532\frac1{16}\cdot4\cdot5\cdot\left(\frac32\right)^3=\frac{135}{32}.