An even function has no linear term

It is given that y=(cos⁡x)1/2y=(\cos x)^{1/2}.

Show that 2yd2ydx2+2(dydx)2+y2=02y\frac{d^2y}{dx^2}+2(\frac{dy}{dx})^2+y^2=0. marks

Find Maclaurin’s series for yy in ascending powers of xx, up to and including the term in x2x^2. marks

By choosing a suitable value of xx, deduce the approximate relation 124≈1+kπ2\frac1{\sqrt[4]2}\approx1+k\pi^2, where kk is a constant to be determined. marks

[2008/AJC/Prelim/I/4]

Total marks: 7

Solution

Differential identity

Implicit differentiation

y=(cos⁡x)1/2y=(\cos x)^{1/2}. Show that 2yd2ydx2+2(dydx)2+y2=02y\dfrac{d^2y}{dx^2}+2\left(\dfrac{dy}{dx}\right)^2+y^2=0.

Hint

  1. Square: y2=cos⁡xy^2=\cos x.

  2. Differentiate twice.

Answer · Differential identity

2yy′′+2(y′)2+y2=02yy\prime\prime+2(y\prime)^2+y^2=0.

Series

Maclaurin series

Find the Maclaurin series of yy up to and including x2x^2.

Explore this idea

Hint

  1. At x=0x=0: y=1y=1.

  2. y′(0)=0y'(0)=0 since 2yy′=−sin⁡0=02yy'=-\sin0=0.

  3. Use the identity for y′′(0)y''(0).

Answer · Series

1−x2/4+O(x4)1-x^2/4+O(x^4).

Explore the idea

√(cos x) 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 = 0.79
Try an example
Function
0.8409
P₂(x)
0.84579
Error
0.00489

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

Only even powers appear, so degree 1 gives the same polynomial as degree 0.

Ideas to try
Compare degree 2 with degree 4: which is better at x=1x=1?
Approximation

Using a series to approximate

Deduce that 124≈1+kπ2\dfrac1{\sqrt[4]2}\approx1+k\pi^2 and find kk.

Explore this idea

Hint

  1. Choose xx so that cos⁡x=2−1/4\sqrt{\cos x}=2^{-1/4}, i.e. cos⁡x=12\cos x=\frac1{\sqrt2}.

Answer · Approximation

x=π/4x=\pi/4: 2−1/4≈1−π2/642^{-1/4}\approx1-\pi^2/64, so k=−1/64k=-1/64.

Explore the idea

The test value x = π/4

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 = 0.79
Try an example
Function
0.8409
P₂(x)
0.84579
Error
0.00489

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

Close to 0 the approximation is accurate, so x=π4x=\frac\pi4 gives a useful estimate of 2−1/42^{-1/4}.

Ideas to try
At degree 2, is the polynomial 0.84580.8458 and the function 0.84090.8409?