# Polynomials - Examples

Does the algorithm work for actual polynomials? Any such exercise is going to seem like a tautology if it works, but lets do it one time, just to show that it is consistent. Consider the quadratic given by $$p(x)=3-2x+x^2.$$ If the algorithm works, it should just give the same polynomial back.

Thus, we suppose that $$p(x)=a_0+a_1x+a_2x^2+R(x)x^3$$. Note that $$R(x)$$ should be zero, but we'll just treat it formally for the moment. Steps one and two of the algorithm say that $$a_0=p(0)$$. We can see that $$p(0)=3-2(0)+(0)^2=3,$$ so $$a_0 = 3.$$. The next step of the algorithm says we should solve for $$a_1$$ in terms of $$x$$ and take its limit as $$x\to0$$. This looks like \begin{align*} a_1 &= \lim_{x\to0}\frac{p(x)-3-a_2x^2-R(x)x^3}{x}\\ &=\lim_{x\to0}\left(\frac{(3-2x+x^2)-3}{x}-a_2x-R(x)x^2\right)\\ &=\lim_{x\to0}(-2+x)\\ &=-2. \end{align*}

Ok, so now our polynomial approximation to $$p$$ looks like $$3-2x+a_2x^2+R(x)x^3$$. We repeat step 3 of the algorithm for $$a_2$$: \begin{align*} a_2 &= \lim_{x\to0}\frac{p(x)-(3-2x+R(x)x^3)}{x^2}\\ &=\lim_{x\to0}\left(\frac{(3-2x+x^2)-(3-2x)}{x^2}-R(x)x\right)\\ &=\lim_{x\to0}\left(\frac{x^2}{x^2}\right)\\ &=1. \end{align*} This leaves our polynomial approximation looking like $$3-2x+x^2+R(x)x^3$$. At this point we can see that $$p(x)-(3-2x+x^2)=0$$ only if $$R(x)=0$$.

Example: [ExponentialPolynomial] Find a polynomial approximate near $$x_0=0$$ to the function given by $$\exp(x)=e^x.$$

We suppose that we know the value of $$e$$, it is approximately $$2.718281828459045235360287$$. We also know $$e^0=1$$. We can do some numerical calculations using those numbers - we have no other way of attacking the problem at this stage. We suppose, as usual, that $$\exp(x)=a_0+a_1x+a_2x^2+R(x)x^3.$$ Also as usual we find that $$a_0 = \lim_{x\to0}\exp(x)=\exp(0)=1.$$ The next step is to evaluate $\lim_{x\to0}a_1=\lim_{x\to0} \left(\frac{\exp(x)-1}{x}-a_2x-R(x)x^2\right).$ We don't really have any means of doing that analytically, but we can run some numerical experiments. Table 1 shows values of $$\frac{\exp(x)-1}{x}$$ as $$x$$ approaches zero, using 25-digit arithmetic. We see that the numbers appear to be approaching 1. We conclude that $a_1=\lim_{x\to0}\frac{\exp(x)-1}{x}=1.$ It is no proof, but convincing nonetheless.

$$x$$$\frac{\exp(x)-1}{x}$
$$10^{-3}$$1.000500166708341668055755
$$-10^{-3}$$0.9995001666250083319446481
$$10^{-6}$$1.000000500000166666717442
$$-10^{-6}$$0.9999995000001666666304228
$$10^{-9}$$1.000000000500000007640262
$$-10^{-9}$$0.9999999995000000006006477
$$10^{-12}$$1.000000000000487685252691
$$-10^{-12}$$0.9999999999995054082752847
Table 1: Numbers illustrating $$\lim_{x\to0}\frac{\exp(x)-1}{x}=1$$.

Now we do this again to find $$a_2.$$ We have $a_2 = \left(\frac{\exp(x)-1-x}{x^2}-R(x)x\right).$

$$x$$$\frac{\exp(x)-1-x}{x^2}$
$$10^{-3}$$0.5001667083416680557545491
$$-10^{-3}$$0.4998333749916680553518673
$$10^{-6}$$0.5000001666667174424940336
$$-10^{-6}$$0.4999998333333695771623772
$$10^{-9}$$0.5000000076402615138050785
$$-10^{-9}$$0.4999999993993522514395795
Table 2: Numbers illustrating $$\lim_{x\to0}\frac{\exp(x)-1-x}{x^2}=\frac12$$.

It looks like this limit goes to $$\frac12$$. We conclude that our numerical experiments show that $\exp(x)\approx 1+x+\frac{x^2}2.$ Figure 1 illustrates this and shows the error. The exponential function is shown in blue; the quadratic is green.

Choose $$x_0$$:

Figure 1: Polynomial approximation to $$\exp(x)$$