What Do Functions Do for Tiny X?

Looking at small values of functions. Motivating the existence of the Taylor expansion.

Building Approximations for Sin(x)

Build up the concept of the Taylor series

Taking Trigonometry Series-ly

Look at the advanced way of viewing sin and cos through their power series.

Towards Maclaurin

Age 16 to 18 Challenge Level:

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 ...