Read a tangent from the first two coefficients

Find the first three terms in the expansion of 1+x23−2x\frac{1+x^2}{\sqrt{3-2x}} in ascending powers of xx.

State the range of values of xx for which the expansion is valid. marks

Deduce that the equation of the tangent to the curve y=2+2x23−2xy=\frac{2+2x^2}{\sqrt{3-2x}} at the point where x=0x=0, is given by 9y=23x+639y=2\sqrt3x+6\sqrt3. marks

[modified DHS/2008/Prelim/I/5]

Total marks: 5

Solution

Expansion

Binomial expansion of a product

Find the first three terms in the expansion of 1+x23−2x\dfrac{1+x^2}{\sqrt{3-2x}} in ascending powers of xx.

Explore this idea

Hint

  1. (3−2x)−1/2=13(1−2x3)−1/2(3-2x)^{-1/2}=\frac1{\sqrt3}\left(1-\frac{2x}3\right)^{-1/2}.

  2. Expand, then multiply by 1+x21+x^2.

Answer · Expansion

13+x33+7x263+O(x3)\frac1{\sqrt3}+\frac{x}{3\sqrt3}+\frac{7x^2}{6\sqrt3}+O(x^3).

Explore the idea

Terms of (1 + x²)/√(3 − 2x)

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 = 0.40
Reset the controls
Function
0.78207
P₂(x)
0.7621
Error
-0.01997

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 are already close to the curve near 0.

Ideas to try
Slide towards x=32x=\frac32: what happens to the function and to the series?
Validity

Range of validity

State the range of values of xx for which the expansion is valid.

Hint

  1. The binomial series in (1−2x3)−1/2\left(1-\frac{2x}3\right)^{-1/2} needs ∣2x3∣<1\left|\frac{2x}3\right|<1.

Answer · Validity

∣x∣<3/2|x|<3/2.

Tangent

Tangent from a series

Deduce that the tangent to y=2+2x23−2xy=\dfrac{2+2x^2}{\sqrt{3-2x}} at x=0x=0 is 9y=23x+639y=2\sqrt3x+6\sqrt3.

Explore this idea

Hint

  1. This curve is twice the expansion above: y=23+2x33+⋯y=\frac2{\sqrt3}+\frac{2x}{3\sqrt3}+\cdots.

  2. The first two terms are the tangent.

Answer · Tangent

9y=23 x+639y=2\sqrt3\,x+6\sqrt3.

Explore the idea

Degree 1 is the tangent

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 = 1
Test value xx = 0.30
Try an example
Function
1.40718
P₁(x)
1.27017
Error
-0.13701

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 first two terms of the series give the tangent line.

Ideas to try
  1. Set the degree to 1: the dashed line touches the curve at x=0x=0.
  2. Raise the degree to see the curvature added.