A shifted cotangent is analytic at zero

(i) Given that y=cot⁡(2x+π4)y=\cot(2x+\frac\pi4), find an equation relating dydx\frac{dy}{dx} and yy. marks

Hence show that d3ydx3=k[yd2ydx2+(dydx)2]\frac{d^3y}{dx^3}=k[y\frac{d^2y}{dx^2}+(\frac{dy}{dx})^2], where kk is a constant to be found. marks

(ii) Find the Maclaurin’s series for yy up to and including the term in x3x^3 if xx is sufficiently small for powers of xx higher than x3x^3 to be neglected. marks

(iii) Using the Maclaurin’s expansion for yy, estimate the value of csc⁡2(13π50)\csc^2(\frac{13\pi}{50}), giving your answer in the form a+bπ+cπ2a+b\pi+c\pi^2, where a,b,ca,b,c are constants to be determined. marks

(iv) Using your result from part (ii), state the equation of the tangent to the curve y=cot⁡(2x+π4)y=\cot(2x+\frac\pi4) at x=0x=0. mark

(v) Explain why a Maclaurin series for y=cot⁡2xy=\cot2x cannot be found. mark

[modified 2009/ACJC/Prelim/I/11]

Total marks: 12

Solution

(i)

A differential relation

y=cot⁡(2x+π4)y=\cot\left(2x+\frac\pi4\right). Find an equation relating dydx\frac{dy}{dx} and yy, and show d3ydx3=k[yd2ydx2+(dydx)2]\frac{d^3y}{dx^3}=k\left[y\frac{d^2y}{dx^2}+\left(\frac{dy}{dx}\right)^2\right].

Hint

  1. ddxcot⁡u=−csc⁡2u⋅u′\frac d{dx}\cot u=-\csc^2u\cdot u'.

  2. Use csc⁡2u=1+cot⁡2u\csc^2u=1+\cot^2u.

Answer · (i)

y′=−2(1+y2)y\prime=-2(1+y^2); k=−4k=-4.

(ii)

Maclaurin series

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

Explore this idea

Hint

  1. At x=0x=0: y=cot⁡π4=1y=\cot\frac\pi4=1.

  2. Use the relation and its derivatives.

Answer · (ii)

1−4x+8x2−64x3/3+O(x4)1-4x+8x^2-64x^3/3+O(x^4).

Explore the idea

cot(2x + π/4) 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 = 0.10
Reset the controls
Function
0.66291
P₃(x)
0.65867
Error
-0.00424

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∣<π8|x|<\frac\pi8: the nearest pole of cot⁡(2x+π4)\cot(2x+\frac\pi4) is at x=−π8x=-\frac\pi8.

Ideas to try
Slide xx towards −π8≈−0.393-\frac\pi8\approx-0.393 where the function has a pole.
(iii)

Using a series to estimate a value

Estimate csc⁡2(13π50)\csc^2\left(\frac{13\pi}{50}\right) in the form a+bπ+cπ2a+b\pi+c\pi^2.

Explore this idea

Hint

  1. Differentiate the series: y′=−2csc⁡2(2x+π4)y'=-2\csc^2\left(2x+\frac\pi4\right).

  2. Choose xx with 2x+π4=13π502x+\frac\pi4=\frac{13\pi}{50}.

Answer · (iii)

2−π/25+π2/12502-\pi/25+\pi^2/1250; a=2,b=−1/25,c=1/1250a=2,b=-1/25,c=1/1250.

Explore the idea

csc²(2x + π/4) 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 = 0.02
Try an example
Function
1.88189
P₂(x)
1.88229
Error
0.00039

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 tiny xx means the quadratic series is almost exact.

Ideas to try
At x=π200≈0.0157x=\frac\pi{200}\approx0.0157 read the function and the degree-2 polynomial.
(iv)

Tangent from a series

State the equation of the tangent to y=cot⁡(2x+π4)y=\cot\left(2x+\frac\pi4\right) at x=0x=0.

Explore this idea

Hint

  1. The first two terms are the tangent.

Answer · (iv)

y=1−4xy=1-4x.

Explore the idea

Degree 1 is the tangent

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 = 0.10
Try an example
Function
0.66291
P₁(x)
0.6
Error
-0.06291

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 tangent at 0 is y=1−4xy=1-4x.

Ideas to try
Set degree 1 and slide xx to see where the tangent leaves the curve.
(v)

Existence of a Maclaurin series

Explain why a Maclaurin series for y=cot⁡2xy=\cot2x cannot be found.

Explore this idea

Hint

  1. A Maclaurin series needs the function and all its derivatives to exist at x=0x=0.

Answer · (v)

cot⁡2x\cot2x has a pole at 0, so no ordinary Maclaurin series exists.

Explore the idea

cot 2x has an asymptote exactly at x = 0

A Maclaurin series needs a finite value at x = 0.

Switch on a feature to add it to the sketch and see how it is found. A full-mark sketch shows all of them.

  • Y = cot 2x
  • Asymptote x = 0
Ideas to try
Shift the graph left by π8\frac\pi8 in your mind: what value does it have at x=0x=0?