The beginnings of Taylor polynomials and approximations

We’ll start off with some of the discussion we had in class today about treating the binomial expansion as a way to approximate a function.

We know that $(x+y)^5=\sum_{r=0}^5{_5C_r} x^r y^{5-r}$. We can then see that if we have a function like $(1+x)^5$ we can simply replace the $y$ in the above expression by 1 and we get (writing out the sum explicitly): $(1+x)^5=1+5x+10 x^2+10x^3+5 x^4+x^5$

If you can’t remember these binomial coefficients offhand, just write out the first 6 lines of Pascal’s triangle and read them off.

ok, so what if I asked you to give me an approximate value for $(1+0.01)^5$. This isn’t so easy because you’ll have to multiply 1.01 by itself five times. How about if we use the polynomial form from the binomial expansion? Well, we can see that this is: $(1+0.01)^5=1+5(0.01)+10 (0.01)^2+10(0.01)^3+5 (0.01)^4+(0.01)^5$

But we can also see that each term gets smaller and smaller. The last term is really tiny compared to the first. How about we just take the first two terms? Indeed $(1+0.01)^5=1.05101$, and taking the first two terms we get 1.05. This is then a pretty good approximation to the question. If you included the next term then you’d get something even more accurate, but maybe we are happy with just 2 decimal places of accuracy.

ok, so we can see that we can approximate the function $(1+x)^5$ by a few terms in the polynomial as long as $x$ is small. In order to do this though we had to know the binomial expansion. If we had $(1+x)^{5.2}$ we wouldn’t be in luck because the binomial coefficients are only defined for non-negative integers. $5.2!$ doesn’t make sense.

Can we derive the expression for the binomial expansion in a way which can generalise? We’re actually going to use calculus for this.

Let’s say that we know that the function can be written as a polynomial, but we don’t know what the coefficients of the polynomial are. Can we figure them out in any way other than using combinatorics (which will not work when we have non-integer powers)?

Well, let’s presume that we have: $(1+x)^5=a_0+a_1 x+a_2 x^2+a_3 x^3+a_4 x^4+a_5 x^5+a_6 x^6+...$

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: $1=a_0$

Well, that’s great! We’ve just found out what $a_0$ must be! So we have: $(1+x)^5=1+a_1 x+a_2 x^2+a_3 x^3+a_4 x^4+a_5 x^5+a_6 x^6+...$

What if we now take a derivative of both sides of this with respect to $x$? $5(1+x)^4=a_1+2 a_2 x+3a_3 x^2+4 a_4 x^3+5a_5 x^4+6a_6 x^5+...$

Now let’s set $x=0$ again: $5(1+0)^4=a_1$

So we know what $a_1$ is, ie 5. Now let’s repeat. Take another derivative: $5\times 4(1+x)^3=2 a_2+3\times 2 a_3 x+4\times 3 a_4 x^2+5\times 4a_5 x^3+6\times 5a_6 x^4+...$

and setting $x=0$ will give us: $20=2 a_2$, so $a_2=10$. Do this again and we get: $5\times 4\times 3(1+x)^2=3\times 2 a_3+4\times 3\times 2 a_4 x+5\times 4\times 3 a_5 x^2+6\times 5\times 4a_6 x^3+...$

which gives us $60=6 a_3$. Note that this is $\frac{5!}{2!}=3! a_3$. Again, take another derivative: $5\times 4\times 3\times 2(1+x)=4\times 3\times 2 a_4+5\times 4\times 3\times 2 a_5 x+6\times 5\times 4\times 3 a_6 x^2+...$

which gives $120=24 a_4$. Note that this is $\frac{5!}{1!}=4! a_4$…and again… $5\times 4\times 3\times 2=5\times 4\times 3\times 2 a_5+6\times 5\times 4\times 3 \times 2 a_6 x+...$

which gives $120=120 a_5$…and again: $0=6\times 5\times 4\times 3 \times 2 a_6+...$

which gives $0=a_6$. And if we carried on with higher $a_i$ we would find that they were all zero.

So this is interesting. We now have that $(a_0,a_1,a_2,a_3,a_4,a_5,a_6,...)=(1,5,10,10,5,1,0,...)$ which were the binomial coefficients we came up with before! In fact, looking above, we can see that we had expressions like: $\frac{5!}{2!}=3! a_3$

which is the same as saying: $a_3=\frac{5!}{2!3!}$ which is precisely the form of the binomial coefficient we know and love…ok, it’s the form of the binomial coefficient we know.

So, we’ve derived the expansion for $(1+x)^5$ using calculus, and not combinatorics. We happened to note that the coefficients could be written as factorials, but we didn’t use it in the derivation of the coefficients. Can we somehow generalise this to cases which are not simply binomial expressions. In the next section we are going to see that we can do precisely this.

 How clear is this post?