Use one series to recover its companion

Let y=cos⁡[ln⁡(1+x)]y=\cos[\ln(1+x)] for ∣x∣<1|x|<1.

(a) Show that (1+x)2d2ydx2+(1+x)dydx+y=0(1+x)^2\frac{d^2y}{dx^2}+(1+x)\frac{dy}{dx}+y=0. By further differentiation of this result, show that the Maclaurin series for yy is given by 1−12x2+12x3−512x4+⋯1-\frac12x^2+\frac12x^3-\frac5{12}x^4+\cdots. marks

(b) By using the result in part (a), deduce the Maclaurin series for sin⁡[ln⁡(1+x)]\sin[\ln(1+x)] in ascending powers of xx up to and including x3x^3. marks

(c) Hence state the equation of tangent to the curve y=sin⁡[ln⁡(1+x)]y=\sin[\ln(1+x)] at x=0x=0. mark

[2022/DHS/Prelim/I/6]

Total marks: 9

Solution

(a)

Differential identity and series

y=cos⁡[ln⁡(1+x)]y=\cos[\ln(1+x)]. Show that (1+x)2y′′+(1+x)y′+y=0(1+x)^2y''+(1+x)y'+y=0 and that the Maclaurin series is 1−12x2+12x3−512x4+⋯1-\frac12x^2+\frac12x^3-\frac5{12}x^4+\cdots.

Explore this idea

Hint

  1. Differentiate once: (1+x)y′=−sin⁡[ln⁡(1+x)](1+x)y'=-\sin[\ln(1+x)].

  2. Differentiate again to reach the identity.

  3. Differentiate the identity twice more and substitute x=0x=0.

Answer · (a)

(1+x)2y′′+(1+x)y′+y=0(1+x)^2y\prime\prime+(1+x)y\prime+y=0; y=1−x2/2+x3/2−5x4/12+O(x5)y=1-x^2/2+x^3/2-5x^4/12+O(x^5).

Explore the idea

cos(ln(1+x)) 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 = 4
Test value xx = 0.50
Try an example
Function
0.91892
P₄(x)
0.91146
Error
-0.00746

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 differential identity generates successive coefficients; compare the resulting truncations near zero.

Ideas to try
Raise the degree from 2 to 4 and watch the curve at x=0.8x=0.8.
(b)

Deducing the companion series

Deduce the Maclaurin series of sin⁡[ln⁡(1+x)]\sin[\ln(1+x)] up to and including x3x^3.

Explore this idea

Hint

  1. From part (a): sin⁡[ln⁡(1+x)]=−(1+x)y′\sin[\ln(1+x)]=-(1+x)y'.

  2. Differentiate the series of yy and multiply by −(1+x)-(1+x).

Answer · (b)

sin⁡[ln⁡(1+x)]=x−x2/2+x3/6+O(x4)\sin[\ln(1+x)]=x-x^2/2+x^3/6+O(x^4).

Explore the idea

sin(ln(1+x)) 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 = 3
Test value xx = 0.50
Try an example
Function
0.39445
P₃(x)
0.39583
Error
0.00139

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 companion series comes from one differentiation of the cosine series.

Ideas to try
At degree 1 the dashed line is the tangent at 0.
(c)

Tangent from a series

Hence state the equation of the tangent to y=sin⁡[ln⁡(1+x)]y=\sin[\ln(1+x)] at x=0x=0.

Explore this idea

Hint

  1. The first two terms of the series are the tangent.

Answer · (c)

y=xy=x.

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
0.25936
P₁(x)
0.3
Error
0.04064

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 tangent at x=0x=0 is y=xy=x.

Ideas to try
Set degree 1 and slide xx: the line is a good approximation only near 0.