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Towards Maclaurin

Stage: 5 Challenge Level: Challenge Level:3 Challenge Level:3 Challenge Level:3

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.