Check an approximation by its error function

(i) Using appropriate expansions from the List of Formulae (MF26), show that the series expansion of ln⁡(1+e2x)\ln(1+e^{2x}) in ascending powers of xx, up to and including the term in x2x^2, is ln⁡2+x+x22\ln2+x+\frac{x^2}2. marks

(ii) Find the set of values of xx such that the value for the above expansion for ln⁡(1+e2x)\ln(1+e^{2x}) is within ±0.3\pm0.3 of the value of ln⁡(1+e2x)\ln(1+e^{2x}). marks

(iii) Deduce an expansion for 21+e−2x\frac2{1+e^{-2x}} up to and including the term in xx. marks

[2018/DHS/Prelim/I/2]

Total marks: 7

Solution

(i)

Composition of series

Show that the series of ln⁡(1+e2x)\ln(1+e^{2x}) up to x2x^2 is ln⁡2+x+x22\ln2+x+\frac{x^2}2.

Explore this idea

Hint

  1. Expand e2xe^{2x} first, then ln⁡(1+⋅)\ln(1+\cdot).

  2. Factor the 2 out so the bracket starts with 1.

Answer · (i)

ln⁡2+x+x2/2+O(x3)\ln2+x+x^2/2+O(x^3).

Explore the idea

ln(1 + e^(2x)) and its degree-2 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 = 1.00
Try an example
Function
2.12693
P₂(x)
2.19315
Error
0.06622

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 quadratic is most accurate near 0; its error reaches 0.3 at about x=±1.55x=\pm1.55.

Ideas to try
Keep degree 2 and slide xx: where is the gap about 0.3?
(ii)

Error within a tolerance

Find the set of xx for which the value of the expansion is within ±0.3\pm0.3 of ln⁡(1+e2x)\ln(1+e^{2x}).

Explore this idea

Hint

  1. Define the error E(x)=ln⁡(1+e2x)−(ln⁡2+x+x22)E(x)=\ln(1+e^{2x})-\left(\ln2+x+\frac{x^2}2\right).

  2. Solve ∣E∣<0.3|E|<0.3 with a graphing calculator.

Answer · (ii)

−α<x<α-\alpha<x<\alpha, where ln⁡cosh⁡α−α2/2=−0.3\ln\cosh\alpha-\alpha^2/2=-0.3 and α≈1.549476\alpha\approx1.549476 (about 1.55).

Strict error less than 0.3; alpha is the exact positive boundary and rounded values are illustrative.

Explore the idea

The error of the quadratic approximation

E(x) = ln(1+e^(2x)) − (ln 2 + x + x²/2). The approximation is acceptable where E stays above −0.3.

Height of the horizontal linek = -0.3
Try an example
Test boundary heights

Buttons use full-precision boundary values; numbers shown here are rounded.

Number of distinct crossings
2

Two intersections, at opposite x-values. Between them the error has smaller magnitude.

All listed intersections are within the original drawing window. A touching point counts once; hollow endpoints are excluded.

Coordinates are rounded to 6 significant figures. Inspect a row to highlight that intersection and bring it into view.
Pointxy
-1.54948-0.3
1.54948-0.3
  • Error E(x)
  • Y = k
  • Crossings
  • Intersection selected in the coordinate table
Key idea

The quadratic is within the strict 0.3 tolerance for ∣x∣<α|x|<\alpha, where E(α)=−0.3E(\alpha)=-0.3 and α≈1.549476\alpha\approx1.549476. Rounded values are not exact boundaries.

Ideas to try
Set k=−0.3k=-0.3 and read the two crossings: they are at x≈±1.549x\approx\pm1.549.
(iii)

Deducing a new series

Deduce an expansion for 21+e−2x\dfrac2{1+e^{-2x}} up to and including the term in xx.

Explore this idea

Hint

  1. Differentiate the series of ln⁡(1+e2x)\ln(1+e^{2x}).

  2. ddxln⁡(1+e2x)=2e2x1+e2x=21+e−2x\frac d{dx}\ln(1+e^{2x})=\frac{2e^{2x}}{1+e^{2x}}=\frac2{1+e^{-2x}}.

Answer · (iii)

2/(1+e−2x)=1+x+O(x3)2/(1+e^{-2x})=1+x+O(x^3).

Explore the idea

2/(1 + e^(−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 = 1
Test value xx = 0.50
Try an example
Function
1.46212
P₁(x)
1.5
Error
0.03788

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

21+e−2x=1+tanh⁡x=1+x−x33+⋯\frac2{1+e^{-2x}}=1+\tanh x=1+x-\frac{x^3}3+\cdots.

Ideas to try
Set degree 1: the dashed line is the tangent y=1+xy=1+x.