**Please don't make me think!**

You can look in almost any calculus book and find the following statement in the section on power series and polynomial approximations of functions:

If a functionThat seems straightforward enough, right?fis differentiable through ordernin an open interval containingc, thenth-degreeTaylor polynomialoffatx=cis given byf(c) +f′(c)(x−c) + (1/2)f″(c)(x−c)^{2}+ ... + (1/n!)f^{(n)}(c)(x−c)^{n}.

But you are asking for

*major*trouble if you ask a student to find the third-degree Taylor polynomial for

*f*(

*x*) = sin

*x*at

*x*= 0. You know why? The student will say, “I can't do this problem. You didn't tell me the value of

*c*.”

Look again at the definition of a Taylor polynomial. It contains the statement “

*x*=

*c*.” In stating my problem, I asked for the Taylor polynomial at “

*x*= 0.” So why can't the student discern the exceedingly subtle and mysterious fact that

*c*has been replaced with 0 (and that they should do the same thing in computing the polynomial)?

You'll find the answer if you go to the exercise set in your textbook (doesn't matter which one; they're practically all the same). The Taylor problems are all written in a special way. Let me restate my problem in book-speak:

Find the third-degree Taylor polynomial for

*f*(

*x*) = sin

*x*at

*c*= 0.

See the subtle difference? You're supposed to rub their noses in the value of

*c*. Saying that “

*x*= 0” in lieu of expressly stating that “

*c*= 0” is not enough. If I daresay that I want the polynomial at “

*x*= 0,” I will get travesties that start out with

*f*(

*c*) +

*f*′(

*c*)(0 −

*c*) + (1/2)

*f*″(

*c*)(0 −

*c*)

^{2}+ ... + (1/

*n*!)

*f*

^{(n)}(

*c*)(0 −

*c*)

^{n},

followed by a plaintive request for the secret value of

*c*. (And indeed I have.)

I've taken several runs at this with different groups of calculus students, but it's no good. Their pattern-matching is extremely rudimentary and not equal to the task of seeing that the

*c*in the statement of the definition of Taylor polynomial corresponds to the zero in my request for the Taylor polynomial for the sine function (or whatever other function and initial value I choose).

As far as I can tell, the calculus books of today differ only in whether they choose to define Taylor polynomials and series in terms of

*x*=

*c*or

*x*=

*a*. Their problem sets are uniformly explicit about the value of

*c*(or

*a*). The same thing is true, for that matter, in the calculus books of yesterday, including early editions of Thomas that go back into the sixties. Since I don't have a copy of one of the Thomas editions from the fifties, I dug out my 1954 third edition of Sherwood & Taylor (who, regrettably, is not the Taylor of Taylor polynomials). They finesse the entire matter by asking for Taylor expansions “in powers of

*x*+ 1,” which implies that

*c*= −1, or “in powers of

*x*− 2,” which implies that

*c*= 2.

Hmm. There's an idea.

## 8 comments:

I don't think I get it.

Sorry. The post got published before it was finished. You saw a stub.

I'm guilty of this sort of thing myself. Various second-year physics courses have scrubbed it out of me, but I've gotten the impression at various times that what's needed is for instructors to explicitly introduce students to various permutations of ways to ask the question (by way of examples), then throw them something

reallyconvoluted on the exam. Repeated application should improve their pattern-matching.This is assuming that the students' little brains can handle that information in a reasonably time period. This may be a flaw in my suggestion…

I think the problem is that x is a variable. Thus the statement "at x=c" is so counterintuitive that the

students become confused (and rightly so since x is not c). If you focus on "around x=c" instead and hit them over the head with it a few

times they will probably do better.

It sounds like another case of favoring textual substitution in lieu of understanding the concepts.

I honestly can't remember how the problems were set twelve years ago when I took calculus.

But I'm shaking my head in disbelief now. (Prolly doesn't help that I was directed to http://notalwaysright.com/ earlier today.)

If I'm being honest though, I wouldn't be surprised if I couldn't do it, myself, without prompthing now. I'd very likely forget the coefficients and even at my best I often 'diffegrated' when doing derivatives. And I have

norecollection of how to estimate the bounds on the error term.Oh - I forgot the anecdote.

Supposedly the calculus exam at my uni years ago broke a few heads by asking people to find the derivative of a function given in terms of x(f).

Physics students have no problem with this expansion. They all know c is the speed of light.

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