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
(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?
(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?
If you can show that the derivative of a function is always positive then the function is increasing and if you can show that the derivative is always negative then the function is decreasing.
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.
You 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 ...