Aleksander from Gdynia Bilingual High
School No 3, Poland used the properties of the sine function to
find a polynomial approximation.
As we know, polynomials are one of the most flexible functions
and hence, can have very different shapes. They are probably the
most flexible functions you can find, thus we can state a
hypothesis that every function can be approximated by a polynomial
of an appropriately high degree. Let's now consider sine, the
functions that is very difficult to calculate and thus we usually
need its approximation. In the beginning let's find coefficients in
the following expression: $$\sin x \approx a + bx +cx^2 +dx^3.$$
Let's make a few observations:
(1) Sine is an odd function
(2) $\sin [0]=0$
(3) For sufficiently small $x$, $\sin x \approx x$.
Polynomials are odd only if coefficients of even powers of $x$
are equal to 0.
From the statements above we have:
$\sin (0) = 0 \Rightarrow a=0$
From (3), $b=1$\\ From (1), $c=0$.
After simplification $\sin x \approx x +dx^3$. Taking $x =
{\pi \over 6}$ and $\sin x = 0.5$ a simple calculation gives $d = -
0.1643968$ which is close to $d = -{1\over 6}$ and we have $\sin x
\approx x - {x^3 \over 6}$.
To get approximation of $\sin x$ using polynomials of higher
degrees, we have to remember that coefficients of even powers
should be equal to 0. For that reason the next degree of a
polynomial that can be used here is the 5th.
Andrei from Romania used the Taylor
series and drew graphs to show the polynomial approximations
.
To solve this problem, I use the Taylor series expansion around the
origin. I see that the derivatives of $f(x) = \sin x $ are:
$$\eqalign{ f^{4k}(x) &= \sin x \cr f^{4k+1}(x) &= \cos x
\cr f^{4k+2}(x) &= -\sin x \cr f^{4k+3}(x) &= -\cos x }.$$
The formula for the Taylor expansion is: $$f(x) = f(x_0) + {(x -
x_0)\over 1!}f'(x_0)+ {(x - x_0)^2\over 2!}f''(x_0)+... + {(x -
x_0)^k\over k!}f^{(k)}(x_0)+ ...$$ Evidently, the odd-order
derivatives are 0 for $x_0=0$. So, $\sin x$ could be written around
the origin as: $$\sin x = x - {x^3\over 3!} + {x^5\over 5!} -
{x^7\over 7!} + ...$$ The simplest method to test the accuracy of
the series expansion is to represent on the same graph the function
and its different order expansions.
The function sin(x) is represented in white, the first order
polynomial in red, the third in cyan, the fifth in green and the
seventh in yellow. It can be observed that the accuracy is better
and better. As the order of the polynomial increases, the accuracy
increases.
It is remarkable that, using only up to the seventh order
polynomial, I obtain a very good approximation of the function. Sin
is a periodic function and so it is sufficient to work on the
interval $[-\pi, \pi]$, and, observing that sin is an odd function
the interval $[0, \pi]$ is enough. Around $\pi$, in fact I should
use the expansion of the function around this value. This is
equivalent to moving the y-axis $\pi$ to the right.
I work in a similar manner for $\cos x$ and $\log (1+x).$ For $\cos
x$, the MacLaurin series expansion is: $$\cos x = 1 - {x^2\over 2!}
+ {x^4\over 4!} - {x^6\over 6!} +...$$
The function is in blue, the second order polynomial in violet, the
fourth - in white and the sixth in red. I see that the sixth order
polynomial is a rather good approximation on the whole
interval.
As cos is periodic, the interval $[-\pi, \pi]$ is sufficient, and
more, as cos(x) is even $[0, \pi]$ is enough.
For, $\log (1+x)$, I consider the logarithm to the base e. Here, I
obtain the following MacLaurin series expansion: $$\log (1+x) = x -
{x^2\over 2} + {x^3\over 3} - {x^4\over 4} + {x^5\over 5} ...$$

Here, the colours of the curves are as follows: violet for the
function, white for the first order approximation, red for the
second, cyan for the third, green for the fourth and yellow for the
fifth.
In this case, the approximations are good only on a small interval
around the origin. I think that for approximations of the function
valid on different intervals, I have to use the series expansion
around the chosen point. So, around x=1: $$\log (1 + x) = \log 2 +
{x-1\over 2} - {(x-1)^2\over 2^3} + {(x-1)^3\over 24} -
{(x-1)^4\over 64} + {(x-1)^5\over 160}1 ...$$

The colours are as follows: the function in red, the first order
polynomial in cyan, the second order one in green, the third in
yellow, the fourth in blue, and the fifth in violet.