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

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.

Teachers' Resources

Why do this problem?

This problem gives a method for building up the Maclaurin series for sine and cosine by creating a sequence of polynomial expansions which are alternately larger and smaller than the actual function. It is an alternative method to the standard formula for Maclaurin expansions and may help students to get more of a sense of these two important standard series.

Possible approach

The problem provides suitable guidance to be completed independently by students who are willing and able to follow the instructions very carefully. However, it might be helpful to go through the first step together, showing that $x- \sin x = 0$ when $x=0$ and that the derivative of $x- \sin x \geq 0$ for all $x \geq 0$. This will allow teachers to check that the students understand how to go about showing each inequality in the problem. They may need to check in again at step (3) when the derivative is $\leq 0$ to ensure the correct conclusion is drawn.

Possible extension

Students are asked in part (5) about generalising this method. What formulae would you expect to derive this way? This method shows a sequence of polynomial approximations to the trig. functions. You are not being asked to give a rigorous proof that the formulae hold in general; that requires a little more work.

Here is another route to the same result. It is not a solution to the problem because the problem specified another method. This should help in seeing the bigger picture more clearly.

We know $\cos x \leq 1$ for all $x$.

The function $f_1(x) = 1 - \cos x$ is positive for all x and so the integral of this function from 0 to $x$ is positive for all $x$. Hence $$\int_0^x f_1 (x) dx =x -\sin x $$ is positive so $\sin x \leq x$.

Integrating again, where $f_2(x) = x - \sin x$ $$\int_0^x f_ 2(x) dx ={x^2\over 2} + \cos x - 1$$ is positive so $\cos x \geq 1 - {x^2\over 2}$.

Integrating again, where $f_3(x)= \cos x - \left(1 - {x^2\over 2}\right)$ $$\int_0^x f_ 3(x) dx =\sin x - \left(x - {x^3\over 3!}\right)$$ is positive so $\sin x \geq x- {x^3\over 3!}$.

Integrating again, where $f_4(x)= \sin x - \left(x - {x^3\over 3!}\right)$ $$\int_0^x f_ 4(x) dx =-\cos x - \left({x^2\over 2!} - {x^4\over 4!}\right)+ 1$$ is positive so $\cos x \leq 1 - {x^2\over 2!} + {x^4\over 4!}$

and so on ...

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