Multiply two binomial series

Find the expansion of (1−x1+2x)1/3\left(\frac{1-x}{1+2x}\right)^{1/3} in ascending powers of xx, up to and including the term in x2x^2. State the range of values of xx for which the expansion is valid. marks

Without performing any calculations, explain why putting x=16x=\frac16 into the result gives a better approximation to 53\sqrt[3]{5} than putting x=−411x=-\frac4{11}. mark

Hence by putting x=16x=\frac16, find an approximation for 53\sqrt[3]{5}, expressing your answer as a fraction in its lowest terms. marks

[2009/IJC/Prelim/I/7]

Total marks: 8

Solution

Expansion

Product of binomials

Find the expansion of (1−x1+2x)1/3\left(\dfrac{1-x}{1+2x}\right)^{1/3} in ascending powers of xx up to x2x^2, and state the range of validity.

Explore this idea

Hint

  1. Write it as (1−x)1/3(1+2x)−1/3(1-x)^{1/3}(1+2x)^{-1/3}.

  2. Expand each with the binomial series.

  3. Multiply and keep up to x2x^2. Validity is the intersection.

Answer · Expansion

1−x+x2+O(x3)1-x+x^2+O(x^3)

∣x∣<1/2|x|<1/2.

Explore the idea

Terms of ((1 − x)/(1 + 2x))^(1/3)

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 = 0.17
Reset the controls
Function
0.85499
P₂(x)
0.86111
Error
0.00612

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 shared standard open validity interval is ∣x∣<12|x|<\frac12, set by (1+2x)−1/3(1+2x)^{-1/3}. Endpoint convergence requires a separate check.

Ideas to try
Slide xx beyond ±12\pm\frac12 and watch the readouts: the series stops being reliable.
Comparison

Choosing substitutions

Without calculation, explain why x=16x=\frac16 gives a better approximation to 53\sqrt[3]5 than x=−411x=-\frac4{11}.

Explore this idea

Hint

  1. Compare the sizes of ∣x∣|x| and check validity.

Answer · Comparison

1/61/6 is closer to zero than −4/11-4/11, reducing higher-power contributions.

A truncation heuristic, not a rigorous bound.

Explore the idea

Compare x = 1/6 and x = −4/11

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 = -0.36
Try an example
Function
1.70981
P₂(x)
1.4958
Error
-0.214

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

Close to 0 the series is very accurate; near the edge of the valid range it is poor.

Ideas to try
Set x=0.1667x=0.1667 then x=−0.3636x=-0.3636 and compare the Error readouts for degree 2.
Approximation

Exact fraction

By putting x=16x=\frac16, find an approximation for 53\sqrt[3]5 as a fraction in lowest terms.

Hint

  1. At x=16x=\frac16: 1−161+26=58\frac{1-\frac16}{1+\frac26}=\frac58.

  2. So the left side is (58)1/3=532\left(\frac58\right)^{1/3}=\frac{\sqrt[3]5}2.

Answer · Approximation

53≈31/18\sqrt[3]5\approx31/18.