Integrate a chain-rule derivative first

It is given that dydx=etan⁡−1x1+x2\frac{dy}{dx}=\frac{e^{\tan^{-1}x}}{1+x^2}, where tan⁡−1x\tan^{-1}x denotes the principal value.

(i) Find an expression for yy in terms of xx given that y=1y=1 when x=0x=0. marks

(ii) Show that (1+x2)d2ydx2+(2x−1)dydx=0(1+x^2)\frac{d^2y}{dx^2}+(2x-1)\frac{dy}{dx}=0. marks

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

(iv) Hence, deduce the series expansion for (a) etan⁡−1x(1+x)2\frac{e^{\tan^{-1}x}}{(1+x)^2} (b) e2x+tan⁡−1xe^{2x+\tan^{-1}x}, up to and including the term in x2x^2. marks

[2010/SAJC/Prelim P1/9]

Total marks: 12

Solution

(i)

Reverse chain rule

Given dydx=etan⁡−1x1+x2\dfrac{dy}{dx}=\dfrac{e^{\tan^{-1}x}}{1+x^2} and y=1y=1 when x=0x=0, find yy.

Hint

  1. ddxtan⁡−1x=11+x2\frac d{dx}\tan^{-1}x=\frac1{1+x^2}, so the integrand is f′eff'e^f with f=tan⁡−1xf=\tan^{-1}x.

Answer · (i)

y=etan⁡−1xy=e^{\tan^{-1}x}.

(ii)

Differential equation

Show that (1+x2)d2ydx2+(2x−1)dydx=0(1+x^2)\dfrac{d^2y}{dx^2}+(2x-1)\dfrac{dy}{dx}=0.

Hint

  1. From the given equation (1+x2)y′=y(1+x^2)y'=y.

  2. Differentiate.

Answer · (ii)

(1+x2)y′′+(2x−1)y′=0(1+x^2)y\prime\prime+(2x-1)y\prime=0.

(iii)

Maclaurin series

By further differentiation, find the Maclaurin series of yy up to and including x3x^3.

Explore this idea

Hint

  1. Differentiate the result of (ii) once.

  2. Evaluate at 0.

Answer · (iii)

1+x+x2/2−x3/6+O(x4)1+x+x^2/2-x^3/6+O(x^4).

Explore the idea

e^(tan⁻¹x) 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
1.58986
P₃(x)
1.60417
Error
0.0143

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 ∣x∣<1|x|<1, because tan⁡−1x\tan^{-1}x has radius 1.

Ideas to try
Raise the degree from 1 to 3.
(iv)(a)

Multiplying series

Deduce the series of etan⁡−1x(1+x)2\dfrac{e^{\tan^{-1}x}}{(1+x)^2} up to x2x^2.

Explore this idea

Hint

  1. Multiply by (1+x)−2=1−2x+3x2−⋯(1+x)^{-2}=1-2x+3x^2-\cdots.

Answer · (iv)(a)

1−x+3x2/2+O(x3)1-x+3x^2/2+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
0.79194
P₂(x)
0.835
Error
0.04306

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 (1+x)−2(1+x)^{-2} limits the validity to ∣x∣<1|x|<1.

Ideas to try
Slide towards x=−1x=-1 where the function has a pole.
(iv)(b)

Product of exponentials

Deduce the series of e2x+tan⁡−1xe^{2x+\tan^{-1}x} up to x2x^2.

Explore this idea

Hint

  1. e2x+tan⁡−1x=e2x⋅etan⁡−1xe^{2x+\tan^{-1}x}=e^{2x}\cdot e^{\tan^{-1}x}.

Answer · (iv)(b)

1+3x+9x2/2+O(x3)1+3x+9x^2/2+O(x^3).

Explore the idea

The product 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 = 0.40
Try an example
Function
3.25602
P₂(x)
2.92
Error
-0.33602

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

Multiplying two series is how you build the series of a product.

Ideas to try
Compare degrees 2 and 3 near x=0.8x=0.8.