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NST First Year Maths

Some unofficial solutions to Cambridge NST IA maths papers

Fourier series

2016 Paper 2 Question 7

11th Oct 20201st Nov 2020nstmathsupervisorLeave a comment

(a)

The ‘trick’ is to re-write 2 \sin x \cos x

f(x) = \sin x + \sin 2x + \sin 3x

which is the required Fourier series.

(b)

The derivative of f(x) is

f^{'}(x) = \cos x + 2\cos 2x + 3\cos 3x

which again is the required Fourier series.

2016, Paper 2Fourier series

2011 Paper 1 Question 14

8th Oct 20201st Nov 2020nstmathsupervisorLeave a comment

\sqrt3 \frac{dy}{dx} + y = 4 \sin x \qquad(*)

(a)

f(x)=\frac{1}{2}a_0 + \sum\limits_{r=1}^\infty a_r \cos \left(\frac{2\pi rx}{L}\right) + \sum\limits_{r=1}^\infty b_r \sin \left(\frac{2\pi rx}{L}\right)

The period L=2\pi so

f(x)=\frac{1}{2}a_0 + \sum\limits_{r=1}^\infty a_r \cos \left( r x\right) + \sum\limits_{r=1}^\infty b_r \sin \left(r x \right)

(b)

Assuming such an expansion exists for y

\frac{dy}{dx} = \sum\limits_{r=1}^\infty a_r (-)\sin \left( r x\right)r +  b_r \cos \left(r x \right) r

Substitute in (*)

\frac{1}{2}a_0 + \sum\limits_{r=1}^\infty \left( \sqrt3 r b_r + a_r \right) \cos{(rx)} + \left( b_r -  \sqrt3 r a_r \right) \sin{(rx)} = 4 \sin{x}

This yields the constraints on the coefficients, by comparing constants, coefficients of \cos and \sin

\begin{aligned}a_0 &= 0 \\ \sqrt3 r b_r + a_r &= 0 \\  b_r -  \sqrt3 r a_r &= 4\delta_{1,r}  \end{aligned}

Here, \delta_{1,r} is the Kronecker delta.

2011, Paper 1Fourier series

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