So, we saw in the last section that we could work out the polynomial expression for $(1+x)^5$ both using combinatorics as well as using calculus. We had also found previously that for small $x$ we could just take the first couple of terms of the polynomial and it was a good approximation of the function itself, depending on how accurate we needed the answer. For instance, for small $x$, $1+5x$ is a reasonable approximation. One thing to note is that the value of these two functions is exactly the same at $x=0$ and the derivative of both functions is the same at $x=0$. The second derivatives are not the same, but had we taken the next term in the polynomial, the second derivatives would have matched as well.

OK, let’s take this one step further and repeat everything exactly as before, but now with the power, not as 5 but as 5.2 and see what happens. This is going to be a copy-paste with edits. We will start by assuming that we can approximate this function with a polynomial expression s let’s presume that we have:

$(1+x)^{5.2}\approx a_0+a_1 x+a_2 x^2+a_3 x^3+a_4 x^4+a_5 x^5+a_6 x^6+...$

where the $a_i$ are the coefficients that we’re going to try and calculate. Well, the first thing we can do is to say what happens when we fix $x=0$ on both sides of the above equation. We are going to get:

$1=a_0$

So $a_0=1$ just as before. Now we have

$(1+x)^{5.2}\approx 1+a_1 x+a_2 x^2+a_3 x^3+a_4 x^4+a_5 x^5+a_6 x^6+...$

What if we now take a derivative of both sides of this with respect to $x$?

$5.2(1+x)^{4.2}\approx a_1+2 a_2 x+3a_3 x^2+4 a_4 x^3+5a_5 x^4+6a_6 x^5+...$

Now let’s set $x=0$ again:

$5.2(1+0)^{4.2}\approx a_1$

So we know what $a_1$ is, ie 5.2. Now let’s repeat. Take another derivative:

$5.2\times 4.2(1+x)^{3.2}\approx 2 a_2+3\times 2 a_3 x+4\times 3 a_4 x^2+5\times 4a_5 x^3+6\times 5a_6 x^4+...$

and setting $x=0$ will give us: $5.2\times 4.2=2 a_2$. Do this again and we get:

$5.2\times 4.2\times 3.2(1+x)^{2.2}\approx 3\times 2 a_3+4\times 3\times 2 a_4 x+5\times 4\times 3 a_5 x^2+6\times 5\times 4a_6 x^3+...$

which gives us $5.2\times 4.2\times 3.2=6 a_3$. Again, take another derivative. Everything is feeling pretty similar to the case we looked at previously, other than the fact that now we don’t have integers:

$5.2\times 4.2\times 3.2\times 2.2(1+x)^{1.2}\approx 4\times 3\times 2 a_4+5\times 4\times 3\times 2 a_5 x+6\times 5\times 4\times 3 a_6 x^2+...$

which gives $5.2\times 4.2\times 3.2 \times 2.2=24 a_4$…and again…

$5.2\times 4.2\times 3.2\times 2.2 \times 1.2(1+x)^{0.2}\approx 5\times 4\times 3\times 2 a_5+6\times 5\times 4\times 3 \times 2 a_6 x+...$

which gives $5.2\times 4.2\times 3.2 \times 2.2\times 1.2=120 a_5$. Now this is interesting. Previously when we looked at this we had $(1+x)^0$ which is a constant, and when we took the next derivative, the term on the left vanished, which gave us $a_6=0$. This isn’t going to happen here. Let’s see. Take another derivative:

$5.2\times 4.2\times 3.2\times 2.2\times 1.2\times 0.2(1+x)^{-0.8}\approx 6\times 5\times 4\times 3 \times 2 a_6+...$

Now, setting $x=0$ we definitely don’t have $a_6=0$ but we have $5.2\times 4.2\times 3.2\times 2.2\times 1.2\times 0.2=6\times 5\times 4\times 3 \times 2 a_6$

Now we see that we could include more terms on the right hand side in the original polynomial expression and we could continue this process indefinitely. The next term (show this for yourself) would give:

$5.2\times 4.2\times 3.2\times 2.2\times 1.2\times 0.2\times (-0.8)(1+x)^{-1.8}\approx 7\times 6\times 5\times 4\times 3 \times 2 a_7+...$

etc. In fact it seems that we can carry on with this indefinitely. The point was that when the power was an integer, as in the first example, we reached a point where we had a constant on the right hand side, and then taking further derivatives of this gave us zero for all higher order terms in the polynomial. In the case of a non-integer we seemed to have skipped the constant completely and gone into negative powers of $(1+x)$ and thus this polynomial goes on for ever (in fact it’s known as an infinite series expansion). If we were to take all of the information we’ve just worked out for the coefficients and put it all together again we would find:

$(1+x)^{5.2}\approx 1. + 5.2 x + 10.92 x^2 + 11.648 x^3 + 6.4064 x^4 + 1.53754 x^5 + 0.0512512 x^6 - 0.00585728 x^7+...$

It was very important that in order to find this series we kept setting $x=0$ to find out what the constants were. What we are really doing is asking that the function on the left, and the polynomial on the right have the same $n^{th}$ derivative at $x=0$. Our polynomial function is going to be a good approximation to the function itself, but only around $x=0$. Could we have chosen another value to “expand around” (this means to approximate a function around a particular value of $x$)? Absolutely. But we would have set the problem up slightly differently. Let’s say we wanted to approximate our function, not around $x=0$ but around $x=0.5$. We would have set up the polynomial as:

$(1+x)^{5.2}\approx a_0+a_1 (x-0.5)+a_2 (x-0.5)^2+a_3 (x-0.5)^3+a_4 (x-0.5)^4+a_5 (x-0.5)^5+a_6 (x-0.5)^6+...$

Such that when we set $x=0.5$ all terms on the right but the constant terms left over will vanish. We can do better than this and improve our notation and right:

$(1+x)^{5.2}\approx\sum_{i=0}^na_i(x-0.5)^i$

where $n$ is the highest power we want in our polynomial expansion. This is up to us as to how accurate we want our approximation. The more terms, the more accurate it will be.

Let’s do this. We will alternately be setting $x=0.5$, calculating what the only $a_i$ remaining on the right hand side is, and then taking another derivative in the expression without having set $x=0.5$. Exactly as above, but instead of setting $x=0$, we set $x=0.5$.

Show for yourself that the answer for the first 3 terms is:

$(1+x)^{5.2}\approx 8.23521 + 28.5487 (-0.5 + x) + 39.9682 (-0.5 + x)^2$

In the next section we’re going to do exactly the same thing, but now for a very different function which looks pretty complicated and we’re going to see how we can approximate it by a polynomial expression. This is going to lead to a general formula for these methods.

What we are doing here is the method of Taylor Polynomials. That is approximating a function near to a certain value of $x$ by a polynomial expression. If we are approximating it near $x=0$ then this is a special case of the Taylor Polynomial, and is called a Maclaurin Polynomial.

 How clear is this post?