Taylor Polynomials without series (advantages, drawbacks)

April 28, 2023

Taylor Polynomials without series (advantages, drawbacks)

Filed under: calculus, series, Taylor polynomial., Taylor Series — oldgote @ 12:32 am

I was in a weird situation this semester in my “applied calculus” (aka “business calculus”) class. I had an awkward amount of time left (1 week) and I still wanted to do something with Taylor polynomials, but I had nowhere near enough time to cover infinite series and power series.

So, I just went ahead and introduced it “user’s manual” style, knowing that I could justify this, if I had to (and no, I didn’t), even without series. BUT there are some drawbacks too.

Let’s see how this goes. We’ll work with series centered at $c =0$ (expand about 0) and assume that $f$ has as many continuous derivatives as desired on an interval connecting 0 to $x$ .

Now we calculate: $\int^x_0 f'(t) dt = f(x) -f(0)$ , of course. But we could do the integral another way: let’s use parts and say $u = f'(t), dv = dt \rightarrow du = f''(t), v = (t-x)$ . Note the choice for $v$ and that $x$ is a constant in the integral. We then get $f(x) -f(0)=\int^x_0 f'(t) dt = (f'(t)(t-x)|^x_0 -\int^x_0f''(t)(t-x) dx$ . Evaluation:

$f(x) =f(0)+f'(0)x -\int^x_0f''(t)(t-x) dx$ and we’ve completed the first step.

Though we *could* do the inductive step now, it is useful to grind through a second iteration to see the pattern.

We take our expression and compute $\int^x_0f''(t)(t-x) dx$ by parts again, with $u = f''(t), dv =t-x \rightarrow du =f'''(t), v = {(t-x)^2 \over 2!}$ and insert into our previous expression:

$f(x) =f(0)+f'(0)x - (f''(t){(t-x)^2 \over 2!}|^x_0 + \int^x_0 f'''(t){(t-x)^2 \over 2!} dt$ which works out to:

$f(x) = f(0)+f'(0)x +f''(0){x^2 \over 2} + \int^x_0 f'''(t){(t-x)^2 \over 2!} dt$ and note the alternating sign of the integral.

Now to use induction: assume that:

$f(x) = f(0)+f'(0)x +f''(0){x^2 \over 2} + ....f^{(k)}(0){x^k \over k!} + (-1)^k \int^x_0 f^{(k+1)}(t) {(t-x)^k \over k!} dt$

Now let’s look at the integral: as usual, use parts as before and we obtain:

$(-1)^k (f^{(k+2)}(t) {(t-x)^{k+1} \over (k+1)!}|^x_0 - \int^x_0 f^{(k+2)}(t) {(t-x)^{k+1} \over (k+1)!} dt )$ . Taking some care with the signs we end up with

$(-1)^k (-f^{(k+1)}(0){(-x)^{k+1} \over (k+1)! } )+ (-1)^{k+1}\int^x_0 f^{(k+2)}(t) {(t-x)^{k+1} \over (k+1)!} dt$ which works out to $(-1)^{2k+2} (f^{(k+1)}(0) {x^{k+1} \over (k+1)!} )+ (-1)^{k+1}\int^x_0 f^{(k+2)}(t) {(t-x)^{k+1} \over (k+1)!} dt$ .

Substituting this evaluation into our inductive step equation gives the desired result.

And note: NOTHING was a assumed except for $f$ having the required number of continuous derivatives!

BUT…yes, there is a catch. The integral is often regarded as a “correction term.” But the Taylor polynomial is really only useful so long as the integral can be made small. And that is the issue with this approach: there are times when the integral cannot be made small; it is possible that $x$ can be far enough out that the associated power series does NOT converge on $(-x, x)$ and the integral picks that up, but it may well be hidden, or at least non-obvious.

And that is why, in my opinion, it is better to do series first.

Let’s show an example.

Consider $f(x) = {1 \over 1+x }$ . We know from work with the geometric series that its series expansion is $1 -x +x^2-x^3....+(-1)^k x^k + ....$ and that the interval of convergence is $(-1,1)$ But note that $f$ is smooth over $[0, \infty)$ and so our Taylor polynomial, with integral correction, should work for $x > 0$ .

So, nothing that $f^{(k)} = (-1)^k(k!)(1+x)^{-(k+1)}$ our k-th Taylor polynomial relation is:

$f(x) =1-x+x^2-x^3 .....+(-1)^kx^k +(-1)^k \int^x_0 (-1)^{k+1}(k+1)!{1 \over (1+t)^{k+2} } {(t-x)^k \over k!} dt$

Let’s focus on the integral; the “remainder”, if you will.

Rewrite it as: $(-1)^{2k+1} (k+1) \int^x_0 ({(t -x) \over (t+1) })^k {1 \over (t+1)} dt$ .

Now this integral really isn’t that hard to do, if we use an algebraic trick:

Rewrite $({(t -x) \over (t+1) })^k = ({(t+1 -x-1) \over (t+1) })^k = (1-{(x+1) \over (t+1) })^k$

Now the integral is a simple substitutions integral: let $u = 1-{(x+1) \over (t+1) } \rightarrow du = (x+1)( {1 \over (t+1)})^2 dt$ so our integral is transformed into:

$(-1) ({k+1 \over x+1}) \int^0_{-x} u^{k} du = (-1) {k \over (k+1)(x+1)} (-(-x)^{k+1}) = (-1)^{k+1} {k+1 \over (k+1)(x+1)} x^{k+1} =(-1)^{k+1}{1 \over (x+1)}x^{k+1}$

This remainder cannot be made small if $x \geq 1$ no matter how big we make $k$

But, in all honesty, this remainder could have been computed with simple algebra.

${1 \over x+1} =1-x+x^2....+(-1)^k x^k + R$ and now solve for $R$ algebraically .

The larger point is that the “error” is hidden in the integral remainder term, and this can be tough to see in the case where the associated Taylor series has a finite radius of convergence but is continuous on the whole real line, or a half line.

Leave a Comment »

No comments yet.

RSS feed for comments on this post. TrackBack URI

College Math Teaching

April 28, 2023