An inverse tangent generates a reciprocal series

If y=2tan⁡−1(x2x+3)y=\sqrt2\tan^{-1}(\frac{x\sqrt2}{x+3}), show that (x2+2x+3)dydx=2(x^2+2x+3)\frac{dy}{dx}=2. mark

By further differentiation of the above result, find the Maclaurin’s series expansion for yy in ascending powers of xx up to and including the term in x3x^3.

Hence find the first three non-zero terms in the expansion of 1x2+2x+3\frac1{x^2+2x+3}. marks

[2008/HCI/Prelim P2/1]

Total marks: 8

Solution

Identity

Differentiating an inverse tangent

y=2tan⁡−1x2x+3y=\sqrt2\tan^{-1}\dfrac{x\sqrt2}{x+3}. Show that (x2+2x+3)dydx=2(x^2+2x+3)\dfrac{dy}{dx}=2.

Hint

  1. ddxtan⁡−1u=u′1+u2\frac d{dx}\tan^{-1}u=\frac{u'}{1+u^2} with u=x2x+3u=\frac{x\sqrt2}{x+3}.

  2. Use the quotient rule for u′u'.

Answer · Identity

(x2+2x+3)y′=2(x^2+2x+3)y\prime=2.

Series

Differential recurrence

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

Explore this idea

Hint

  1. Differentiate (x2+2x+3)y′=2(x^2+2x+3)y'=2 twice.

  2. Evaluate yy, y′y', y′′y'', y′′′y''' at 0.

Answer · Series

2x/3−2x2/9+2x3/81+O(x4)2x/3-2x^2/9+2x^3/81+O(x^4).

Explore the idea

The function 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 = 0.60
Reset the controls
Function
0.32736
P₃(x)
0.32533
Error
-0.00203

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∣<3≈1.73|x|<\sqrt3\approx1.73, set by the complex zeros of x2+2x+3x^2+2x+3.

Ideas to try
Raise the degree and see how far the series can follow the curve.
Reciprocal

Series by differentiation

Hence find the first three non-zero terms of the expansion of 1x2+2x+3\dfrac1{x^2+2x+3}.

Explore this idea

Hint

  1. 1x2+2x+3=12y′\frac1{x^2+2x+3}=\frac12y'.

Answer · Reciprocal

1/3−2x/9+x2/27+O(x3)1/3-2x/9+x^2/27+O(x^3).

Explore the idea

1/(x² + 2x + 3) 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
Try an example
Function
0.271
P₂(x)
0.27
Error
-0.001

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

A reciprocal of a quadratic needs no binomial expansion when you can differentiate a known series.

Ideas to try
Check degree 2 at x=0.3x=0.3: 13−29(0.3)+127(0.09)≈0.27\frac13-\frac2{9}(0.3)+\frac1{27}(0.09)\approx0.27.