Towards Maclaurin

Build series for the sine and cosine functions by adding one term at a time, alternately making the approximation too big then too small but getting ever closer.

Age

16 to 18

Challenge level

Exploring and noticing Working systematically Conjecturing and generalising Visualising and representing Reasoning, convincing and proving

Being curious Being resourceful Being resilient Being collaborative

Problem

(1) We know $\cos x \leq 1$ for all $x$. By considering the derivative of the function $$f(x) = x - \sin x$$ prove that $\sin x \leq x$ for $x \geq 0$.

(2) By considering the derivative of the function $$f(x) = \cos x - \left(1 - {x^2\over 2}\right)$$ prove that $\cos x \geq 1 - {x^2\over 2}$ for $x \geq 0$.

(3) By considering the derivative of the function $$f(x) = \left(x - {x^3 \over 3!}\right) - \sin x $$ prove that $\sin x \geq (x - {x^3 \over 3!})$ for $x \geq 0$.

(4) By considering the derivative of the function $$f(x) = \cos x - \left(1 - {x^2 \over 2!} + {x^4\over 4!}\right) $$ prove that $\cos x \leq \left(1 - {x^2\over 2!} + {x^4 \over 4!}\right)$ for $x \geq 0$.

(5) What can you say about continuing this process?

Student Solutions

Thank you for your solutions to this problem to Annie from Newstead Wood, Andre from Tudor Vianu National College, Romania and Ben who did not give the name of his school. This is Ben's solution.

(1) \begin{eqnarray} f(x)&=& x - \sin x \\ f(0) &=& 0 \\ f'(x) &=& 1-\cos x \geq 0\end{eqnarray} As $ \cos x \leq 1$ then $f'(x)$ is always positive so the function $f(x) = x - \sin x$ is always increasing, so $x$ is becoming increasingly larger than sin x and therefore, when $x \geq 0$, $\sin x \leq x.$

(2)\begin{eqnarray}f(x) &=& \cos x - \left(1 - {x^2\over 2}\right)\\ f(0) &=& 0 \\ f'(x) &=& -\sin x + x \geq 0 \quad \rm {by \ (1)}\end{eqnarray} We have already shown that $\sin x \leq x$, therefore $x - \sin x \geq 0$. So the function $f(x) = \cos x - \left(1 - {x^2\over 2}\right)$ is always increasing when $x\geq 0$ and $f(0)=0$. For this to be true $\cos x \geq 1 - {x^2\over 2}$ when $x \geq 0$.

(3) \begin{eqnarray} f(x) &=& \left(x - {x^3 \over 3!}\right) - \sin x \\ f(0) &=& 0 \\ f'(x) &=& 1 - {x^2\over 2} - \cos x \leq 0 \quad \rm{by\ (2)} \end{eqnarray} As $\cos x \geq 1 - {x^2\over 2}$ then $f'(x)$ here must always be negative. Therefore $f(x)=\left(x - {x^3 \over 3!}\right) - \sin x $ is always decreasing for $x\geq 0$. As $f(0)= 0$, and the function decreases as $x$ increases, then $f(x)\leq 0$ when $x \geq 0$. Hence $\sin x \geq \left(x - {x^3 \over 3!}\right)$ for $x \geq 0$.

Note that the derivative of ${x^n\over n!}$ is ${nx^{n-1}\over n!}= {x^{n-1}\over (n-1)!}$ and we can write $2=2!$.

(4) \begin{eqnarray}f(x) &=& \cos x - \left(1 - {x^2 \over 2!} + {x^4\over 4!}\right) \\ f(0)&=&0 \\ f'(x)&=& - \sin x + x - {x^3\over 3!}\leq 0 \quad \rm{by\ (3)} \end{eqnarray} As $\sin x \geq \left(x - {x^3 \over 3!}\right)$ then the function $f(x) = \cos x - \left(1 - {x^2 \over 2!} + {x^4\over 4!}\right)$ is always decreasing, and again $f(0) = 0$ so $f(x)\leq 0$ for $x\geq 0$. If this is true then $\cos x \leq \left(1 - {x^2\over 2!} + {x^4 \over 4!}\right)$ for $x \geq 0$.

(5) If we continue this process, we see that $$1 - {x^2\over 2!} + {x^4\over 4!} - {x^6\over 6!} \leq \cos x \leq 1 - {x^2\over 2!} + {x^4\over 4!}$$ As we introduce more terms this series gets closer to $\cos x$.

This process can be repeated indefinitely to give the infinite Maclaurin series, valid for all $x$ (in radians) $$\cos x = 1 - {x^2\over 2!}+ {x^4\over 4!} -{x^6\over 6!} + ...$$ By a similar look at the terms for $\sin x$, we get the infinite Maclaurin series for $\sin x$, again valid for all $x$ (in radians) $$\sin x = x - {x^3\over 3!}+ {x^5\over 5!} -{x^7\over 7!} + ...$$

A Maclaurin series is a Taylor series expansion of a function about $0$.

The mathematicians Brook Taylor and Colin Maclaurin were contemporaries who both developed Newton's work on calculus.

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