Factor a logarithm argument

(i) Given that f(x)=ln⁡(2x+1)f(x)=\ln(2x+1), find f(0),f′(0),f′′(0),f′′′(0)f(0),f'(0),f''(0),f'''(0). Hence obtain the first three non-zero terms in the Maclaurin’s series for f(x)f(x). marks

(ii) Hence, or otherwise, show that the first three non-zero terms in the expansion of ln⁡(2x2+3x+1)\ln(2x^2+3x+1) are ax+bx2+ax3ax+bx^2+ax^3, where aa and bb are constants to be found. marks

[2010/MI/Prelim/I/7]

Total marks: 8

Solution

(i)

Maclaurin from derivatives

f(x)=ln⁡(2x+1)f(x)=\ln(2x+1). Find f(0)f(0), f′(0)f'(0), f′′(0)f''(0), f′′′(0)f'''(0) and the first three non-zero terms of the Maclaurin series.

Explore this idea

Hint

  1. Differentiate three times.

  2. Substitute x=0x=0 and use f(0)+f′(0)x+f′′(0)2!x2+⋯f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\cdots.

Answer · (i)

f(0)=0, f′(0)=2, f′′(0)=−4, f(3)(0)=16f(0)=0,\ f\prime(0)=2,\ f\prime\prime(0)=-4,\ f^{(3)}(0)=16; f=2x−2x2+8x3/3+O(x4)f=2x-2x^2+8x^3/3+O(x^4).

Explore the idea

ln(2x + 1) 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 = 0.20
Try an example
Function
0.33647
P₃(x)
0.34133
Error
0.00486

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 series is valid for −12<x≤12-\frac12<x\le\frac12.

Ideas to try
Slide xx past 12\frac12: the series stops converging.
(ii)

Factor and add

Hence show that the first three non-zero terms in the expansion of ln⁡(2x2+3x+1)\ln(2x^2+3x+1) are ax+bx2+ax3ax+bx^2+ax^3, and find aa and bb.

Explore this idea

Hint

  1. 2x2+3x+1=(2x+1)(x+1)2x^2+3x+1=(2x+1)(x+1).

  2. ln⁡(AB)=ln⁡A+ln⁡B\ln(AB)=\ln A+\ln B.

  3. Add the series of ln⁡(2x+1)\ln(2x+1) and ln⁡(1+x)\ln(1+x).

Answer · (ii)

3x−5x2/2+3x3+O(x4)3x-5x^2/2+3x^3+O(x^4); a=3,b=−5/2a=3,b=-5/2.

Explore the idea

ln(2x² + 3x + 1) 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 = 0.20
Try an example
Function
0.51879
P₃(x)
0.524
Error
0.00521

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

Intersect the logarithm-series intervals: −12<x≤12-\frac12<x\le\frac12. The positive endpoint is included because both logarithm series converge there.

Ideas to try
Check that the degree-3 polynomial is 3x−2.5x2+3x33x-2.5x^2+3x^3 at x=0.2x=0.2: it should be 0.524.