Further numerical methods (Further Pure 1)

Explanation

• Three finite-difference approximations are quoted in the specification and must be known. The forward difference $\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)_n\approx\dfrac{y_{n+1}-y_n}{h}$ steps forward from a single known value.
• The central difference $\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)_n\approx\dfrac{y_{n+1}-y_{n-1}}{2h}$ is more accurate but needs two consecutive starting values.
• The second-derivative formula is $\left(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\right)_n\approx\dfrac{y_{n+1}-2y_n+y_{n-1}}{h^2}$.
• In every case, rearrange to make $y_{n+1}$ the subject and then step forward, keeping more decimal places in the working than the question asks for in the answer.
• The forward difference has error of order $h$, whereas both central formulae have error of order $h^2$, which is why halving the step size reduces their error by about a factor of four.

Explanation

• Simpson's rule estimates $\displaystyle\int_a^b y\,\mathrm{d}x$ by fitting a quadratic through each consecutive group of three ordinates.
• With $n$ strips of width $h=\dfrac{b-a}{n}$ and ordinates $y_0,y_1,\dots,y_n$, the rule is $\displaystyle\int_a^b y\,\mathrm{d}x\approx\frac{h}{3}\left[\left(y_0+y_n\right)+4\left(y_1+y_3+\cdots\right)+2\left(y_2+y_4+\cdots\right)\right]$.
• The number of strips must be even, because the ordinates are consumed three at a time with the outer ones shared, so the strips pair up.
• The end ordinates take coefficient $1$, the odd-numbered ones take $4$ and the even-numbered interior ones take $2$; a quick check is that the coefficients must total $3n$.
• The rule is exact for every polynomial of degree at most three, and its error is of order $h^4$, so halving the strip width reduces the error by roughly a factor of sixteen.