Differential and Integral Calculus

\(n=1\).

When \(n=1\text{,}\) we want a degree-1 polynomial approximation. That is, we want the tangent line. So we use

\(n=2\).

A degree-2 polynomial looks like \(q(x)=Ax^2+Bx+C\text{.}\) Since we're interested in what happens when \(x\to c\text{,}\) let's rewrite this form as \(q(x)=a_2(x-c)^2+a_1(x-c)+a_0\text{.}\) (The exact relationship between \(A,B,C\) and \(a_0,a_1,a_2\) isn't important -- we just want to make sure that we're writing everything in terms of \(c\text{.}\))

Let's again write down our demands. We want \(f(x)\sim q(x)\text{ as }x\to c\text{,}\) i.e.,

As \(x\to c\text{,}\) \(x-c\to 0\text{.}\) So \(\lim_{x\to c} q(x)=a_0\text{.}\) As long as \(f\) is continuous, \(\lim_{x\to c}f(x)=f(c)\text{.}\) So we need \(a_0=f(c)\text{.}\)

As we did in the linear case, let's imagine a competition among all quadratic polynomials of the form \(q(x)=a_2(x-c)^2+a_1(x-c)+f(c)\text{.}\) We demand:

But \(f(x)-f(c)\sim a_2(x-c)^2+a_1(x-c)\text{ as }x\to c\) just means

We can rearrange the fraction to:

As \(x\to c\text{,}\) \(\frac{1}{a_2(x-c)+a_1}\to\frac{1}{a_1}\text{.}\) We end up with \(a_1=f'(c)\text{.}\)

So now we know that \(q(x)=a_2(x-c)^2+f'(c)(x-c)+f(c)\text{.}\) Let's rewrite our demands again:

This means we need

The form of this limit is \(\frac{0}{0}\text{,}\) so we can apply L'Hôpital's Rule. We need to be a bit careful to differentate with respect to \(x\). The numerator's derivative is

and the denomator's is

Applying L'Hôpital's Rule, we get

So \(a_2=\frac{1}{2}f''(c)\text{.}\)

That is, the quadratic which best approximates \(f(x)\) near \(x=c\) is

\(n=3\).

Now let's approximate \(f(x)\) by a cubic of the form \(q(x)=a_3(x-c)^3+a_2(x-c)^2+a_1(x-c)+a_0\text{.}\) It probably won't surprise you that \(a_0=f(c)\text{.}\) Let's see about \(a_1\text{:}\)

which just means

or in other words

Because \(\frac{1}{a_3(x-c)^2+a_2(x-c)^+a_1}\to\frac{1}{a_1}\) as \(x\to c\text{,}\) we see that \(a_1=f'(c)\text{.}\)

So now we have

In order words, we need

L'Hôpital tells us that

So \(a_2=\frac{1}{2}f''(c)\text{.}\)

Let's update our demand:

So we consider the limit

Again L'Hôpital's Rule applies here, so we differentiate the numerator and the denominator to get:

This form is still indeterminate, so we apply L'Hôpital's Rule again:

Thus, we need to use \(a_3=\frac{1}{6}f'''(c)\text{.}\) Our approximating cubic is therefore:

Simple Compositions.

Let's say we want to compute the degree-8 Maclaurin polynomial for \(f(x)=\sin(x^2)\text{.}\) We could go ahead and differentiate 8 times. But to save ourselves some work, let's switch variables. Set \(t=x^2\text{.}\) Table 15.10 tells us that

and that this is the best approximation by a polynomial which is degree 4 in \(t\text{.}\) Notice that being degree 4 in \(t\) is the same as being degree 8 in \(x\text{.}\) If we substitute \(t=x^2\) back into our polynomial, we get:

which is our answer.

This trick works with powers of \(x\) and multiples of \(x\text{.}\) For example, the degree-5 Maclaurin polynomial for \(f(x)=e^{3x}\) is

We can also do this with shifts: the degree-3 Maclaurin polynomial for \(\ln(2+x)=\ln(1+(x+1))\) is

Products.

Let's say we want to compute the degree-6 Maclaurin polynomial for the function \(h(x)=e^x\sin(x)\text{.}\) To use the formula in Definition 15.2, we would have to take sixth derivatives. Because of the product rule, \(h'(x)\) will have two terms; \(h''(x)\) will have four terms; etc. So that seems like kind of a mess. On the other hand, Table 15.10 tells us that

The Taylor polynomials are polynomials, so let's treat them like polynomials and multiply!

Let's go ahead and do the multiplication. I've color-coded the terms coming from \(e^x\) to help with the bookkeeping.

I've left off the remaining rows; you are an expert at manipulating polynomials so you can complete them yourself. But I want to point something out here: there are lots of terms with degrees greater than 6. Because we want a polynomial of degree 6, we can just omit these.

More Complicated Compositions.

We can also use it with more complicated compositions such as \(f(x)=\cos(e^x-1)\text{.}\) Let's find the degree-3 Maclaurin polynomial for \(\cos(e^x-1)\text{.}\) We start with the degree-3 Maclaurin polynomial for \(\cos(t)\text{,}\) applied with \(t=e^x-1\text{:}\)

Now, we remind ourselves that \(e^x\approx 1+x+\frac{1}{2}x^2+\frac{1}{6}x^3\text{,}\) so that \(t=e^x-1\sim x+\frac{1}{2}x^2+\frac{1}{6}x^3\text{,}\) and make that substitution:

Like we did with products, now we bite the bullet and do the algebra:

Notice that we actually got a degree 6 polynomial. Since we only needed to approximate \(f\) to order 3, we take the cubic part and our answer is:

Plotting \(T_3\) together with \(f\) confirms that this is a good approximation, at least near \(x=0\text{:}\)