Compose sine and logarithm carefully

It is given that y=ln⁡(2+sin⁡2x)y=\ln(2+\sin2x).

(i) Show that eyd2ydx2+ey(dydx)2=−4sin⁡2xe^y\frac{d^2y}{dx^2}+e^y(\frac{dy}{dx})^2=-4\sin2x. marks

(ii) By further differentiation of the above results, find the Maclaurin series for yy, up to and including the term in x3x^3. marks

(iii) Verify that the series for ln⁡(2+sin⁡2x)\ln(2+\sin2x) is the same as the result obtained in part (ii), if the standard series from the List of Formulae (MF26) are used. marks

(iv) Hence or otherwise, find the series expansion for ln⁡(2+sin⁡2x)1−x\frac{\ln(2+\sin2x)}{\sqrt{1-x}}, up to and including the term in x2x^2. marks

[2022/ACJC/Prelim/I/7]

Total marks: 10

Solution

(i)

Implicit differentiation

y=ln⁡(2+sin⁡2x)y=\ln(2+\sin2x). Show that eyy′′+ey(y′)2=−4sin⁡2xe^yy''+e^y(y')^2=-4\sin2x.

Hint

  1. Write ey=2+sin⁡2xe^y=2+\sin2x.

  2. Differentiate twice.

Answer · (i)

ey[y′′+(y′)2]=−4sin⁡2xe^y[y\prime\prime+(y\prime)^2]=-4\sin2x.

(ii)

Maclaurin series from derivatives

Find the Maclaurin series of yy up to and including x3x^3.

Explore this idea

Hint

  1. Differentiate the identity once more.

  2. Find y(0)y(0), y′(0)y'(0), y′′(0)y''(0), y′′′(0)y'''(0).

Answer · (ii)

ln⁡2+x−x2/2−x3/3+O(x4)\ln2+x-x^2/2-x^3/3+O(x^4).

Explore the idea

ln(2 + sin 2x) 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 = 0.40
Try an example
Function
0.99966
P₃(x)
0.99181
Error
-0.00785

The shading is a local comparison window, not a convergence or accuracy guarantee.

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
  • A local comparison window, not a convergence or accuracy guarantee
Key idea

The curve rises through (0,ln⁡2)(0,\ln2) with gradient 1.

Ideas to try
Compare degree 2 and 3 near x=0.5x=0.5.
(iii)

Verifying with standard series

Verify the same series using standard results.

Hint

  1. 2+sin⁡2x=2[1+x−2x33+⋯ ]2+\sin2x=2\left[1+x-\frac{2x^3}3+\cdots\right].

  2. Use ln⁡(1+u)\ln(1+u).

Answer · (iii)

ln⁡2+ln⁡(1+x−2x3/3+⋯ )\ln2+\ln(1+x-2x^3/3+\cdots) gives the same cubic series.

(iv)

Multiplying series

Hence find the series for ln⁡(2+sin⁡2x)1−x\dfrac{\ln(2+\sin2x)}{\sqrt{1-x}} up to x2x^2.

Explore this idea

Hint

  1. (1−x)−1/2=1+x2+3x28+⋯(1-x)^{-1/2}=1+\frac x2+\frac{3x^2}8+\cdots.

  2. Multiply by the series from (ii).

Answer · (iv)

ln⁡2+(1+12ln⁡2)x+38ln⁡2 x2+O(x3)\ln2+(1+\tfrac12\ln2)x+\tfrac38\ln2\,x^2+O(x^3).

Explore the idea

The quotient 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
1.12569
P₂(x)
1.12051
Error
-0.00518

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 factor 11−x\frac1{\sqrt{1-x}} limits the valid range to ∣x∣<1|x|<1.

Ideas to try
Raise the degree and slide xx towards 1.