Two routes to the same square-root series

It is given that y=excos⁡xy=\sqrt{e^x\cos x}.

(i) Show that 2ydydx=y2−exsin⁡x2y\frac{dy}{dx}=y^2-e^x\sin x. marks

(ii) By further differentiation of the result in part (i), find the Maclaurin series for yy, up to and including the term in x2x^2. marks

(iii) Using standard series from the List of Formulae (MF26), expand excos⁡x\sqrt{e^x\cos x} as far as the term in x2x^2 and verify that the same result is obtained in part (ii). marks

[2020/MI/Prelim/I/6]

Total marks: 9

Solution

(i)

Implicit differentiation

y=excos⁡xy=\sqrt{e^x\cos x}. Show that 2ydydx=y2−exsin⁡x2y\dfrac{dy}{dx}=y^2-e^x\sin x.

Hint

  1. Square: y2=excos⁡xy^2=e^x\cos x.

  2. Differentiate both sides with the product rule.

Answer · (i)

2yy′=y2−exsin⁡x2yy\prime=y^2-e^x\sin x.

(ii)

Maclaurin series from a differential identity

By further differentiation, 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. Use the identity to find y′(0)y'(0).

  3. Differentiate it once more to find y′′(0)y''(0).

Answer · (ii)

y=1+x/2−x2/8+O(x3)y=1+x/2-x^2/8+O(x^3).

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 = 2
Test value xx = 0.40
Reset the controls
Function
1.1722
P₂(x)
1.18
Error
0.0078

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 captures the curve near 0; the real branch near zero reaches y=0y=0 at x=π2x=\frac\pi2 where cos⁡x=0\cos x=0.

Ideas to try
Keep degree 2 and slide xx: where does the quadratic stay within 0.01?
(iii)

Verifying with standard series

Using standard series, expand excos⁡x\sqrt{e^x\cos x} to x2x^2 and verify the result in (ii).

Explore this idea

Hint

  1. Write excos⁡x=ex/2(cos⁡x)1/2\sqrt{e^x\cos x}=e^{x/2}(\cos x)^{1/2}.

  2. Expand each.

Answer · (iii)

ex/2(cos⁡x)1/2=1+x/2−x2/8+O(x3)e^{x/2}(\cos x)^{1/2}=1+x/2-x^2/8+O(x^3).

Real square-root branch near 0, where cos⁡x>0\cos x>0.

Explore the idea

Check against the 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
1.1722
P₂(x)
1.18
Error
0.0078

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

Two independent methods give the same three coefficients.

Ideas to try
Does the dashed curve at degree 2 equal 1+x2−x281+\frac x2-\frac{x^2}8 at x=0.4x=0.4? (1.18)