Normalise before expanding

It is given that f(x)=4−3xf(x)=\sqrt{4-3x}.

(a) Find the binomial expansion for f(x)f(x), up to and including the term in x2x^2. Give the coefficients as exact fractions in their simplest form, and state the range of values of xx for which this expansion is valid. marks

(b) By taking x=14x=\frac14, show that 13≈1847512\sqrt{13}\approx\frac{1847}{512}. marks

(c) We may also substitute x=113x=\frac1{13} to obtain an approximation of 13\sqrt{13}. Without further calculation, explain why this leads to a better approximation of 13\sqrt{13} than the value shown in (b). mark

[2022/EJC/Prelim/II/1]

Total marks: 8

Solution

(a)

Binomial expansion

Find the binomial expansion of f(x)=4−3xf(x)=\sqrt{4-3x} up to and including x2x^2 (exact fractions), and state the range of xx for which it is valid.

Explore this idea

Hint

  1. Write 4−3x=2(1−3x4)1/2\sqrt{4-3x}=2\left(1-\frac{3x}4\right)^{1/2}.

  2. Apply (1+u)1/2=1+u2−u28+⋯(1+u)^{1/2}=1+\frac u2-\frac{u^2}8+\cdots with u=−3x4u=-\frac{3x}4.

  3. Validity: ∣u∣<1|u|<1.

Answer · (a)

2−34x−964x2+O(x3)2-\frac34x-\frac9{64}x^2+O(x^3)

−4/3<x<4/3-4/3<x<4/3.

Explore the idea

Adding terms to 2(1 − 3x/4)^(1/2)

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 = 0.25
Reset the controls
Function
1.80278
P₂(x)
1.80371
Error
0.00094

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

Inside −43<x<43-\frac43<x<\frac43 extra terms improve the approximation; the displayed truncated polynomial approximates the function there.

Ideas to try
  1. Slide xx towards 43\frac43: does the polynomial still approach the curve?
  2. Move xx beyond 43\frac43 and see what happens to the readouts.
(b)

Approximating a surd

By taking x=14x=\frac14, show that 13≈1847512\sqrt{13}\approx\frac{1847}{512}.

Explore this idea

Hint

  1. Substitute x=14x=\frac14 into the expansion: the left side is 4−34=132\sqrt{4-\frac34}=\frac{\sqrt{13}}2.

Answer · (b)

13≈1847/512\sqrt{13}\approx1847/512.

Explore the idea

Test value x = 1/4

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 = 0.25
Try an example
Function
1.80278
P₂(x)
1.80371
Error
0.00094

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

Three terms give 1847/512≈3.607421847/512\approx3.60742, compared with 13≈3.60555\sqrt{13}\approx3.60555: the error is about 0.00187.

Ideas to try
  1. Set the degree to 2 and read the polynomial at x=0.25x=0.25: is it 18471024=1.8037\frac{1847}{1024}=1.8037?
  2. How much does the error shrink if you keep one more term?
(c)

Choosing a better value

Explain why substituting x=113x=\frac1{13} gives a better approximation than the value in (b).

Explore this idea

Hint

  1. Higher powers of xx are omitted: they matter less when xx is closer to 0.

Answer · (c)

x=1/13x=1/13 has smaller expansion input; use 13=7/f(1/13)\sqrt{13}=7/f(1/13).

Smaller omitted powers support this particular estimate; no universal error bound.

Explore the idea

Test value x = 1/13

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 = 0.08
Try an example
Function
1.94147
P₂(x)
1.94149
Error
0.00002

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

For these two substitutions, the smaller input reduces the omitted higher powers. Recover the target with 13=7/f(1/13)\sqrt{13}=7/f(1/13); this is not a universal error bound.

Ideas to try
Compare the error at x=113x=\frac1{13} with the error at x=14x=\frac14 for the same degree.