Timeline for The role of the mean value theorem (MVT) in first-year calculus
Current License: CC BY-SA 2.5
12 events
| when toggle format | what | by | license | comment | |
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| Jul 12, 2010 at 4:53 | history | made wiki | Post Made Community Wiki by S. Carnahan♦ | ||
| Jul 11, 2010 at 21:05 | comment | added | The Mathemagician | @Pete Rolle's theorum ----> MVT is the way I learned it twice-once in calculus and again in baby real analysis and it seems to work fine. The geometric content this approach gives is very helpful because in many ways,the MVT is really a result about approximation by linear functions. | |
| Jul 11, 2010 at 19:28 | comment | added | Pete L. Clark | @Robin: That is the version I presented in an undergraduate course "Real Analysis II", at McGill University. My statement and proof is the same as that of, e.g., Rudin's Principles of Mathematical Analysis. | |
| Jul 11, 2010 at 19:10 | comment | added | Robin Chapman | Pete, is that the version of the FTC you present in beginning analysis courses? The vanilla version involving a continuous integrand is more standard surely? | |
| Jul 11, 2010 at 19:03 | comment | added | O.R. | All the first year student need to be able to do (seen you don't know whether he/she is going to use calculus for) is to be able to use it. Compute, determine, resolve, expand, ... | |
| Jul 11, 2010 at 19:02 | comment | added | O.R. | One annotation to this answer. You are assuming (and is not your fault, it is done that way in most of the books, schools,...) that you need to prove it. What better intuition builder than to expend the time instead to use it. The real problem of freshman courses is not expending enough time in developing skills and too much in a professor repeating centuries old proves. Actually most of the proves in freshman calculus are just exercises. Imagine a problem-solving oriented course in which you just present a few fact intuitive geometrical ideas and right to the list of exercises. | |
| Jul 11, 2010 at 18:46 | comment | added | Pete L. Clark | Ahem -- TILTING my head. Although what I wrote sounds like a good way to liven up a monotonous calculus lecture. | |
| Jul 11, 2010 at 18:44 | comment | added | Pete L. Clark | @RC: There are several different versions of FTC. For the following version of "part b)" -- Let $f: [a,b] \rightarrow \mathbb{R}$ be a differentiable function such that $f'$ is Riemann integrable (but not necessarily continuous). Then $\int_a^b f' = f(b) - f(a)$ -- the proof I know of this uses MVT: see e.g. p.8 of math.uga.edu/~pete/243integrals1.pdf. [And yes, the concept of a discontinuous but Riemann integrable derivative is too subtle for a freshman calculus class.] I would be interested to know what you have in mind that avoids MVT. | |
| Jul 11, 2010 at 18:36 | comment | added | Pete L. Clark | ...There is an issue here that this tilting process might result in a graph which is not the graph of a function. For a very strong class, I would mention this point, since it leads to the interesting conclusion that MVT is actually true for graphs of parameterized curves ("Cauchy's MVT"). In recent years, I have not mentioned it. I used to also include the textbook proof of subtracting off a linear function to reduce to Rolle's Theorem, but I now feel that my clientele do not get much out of this. | |
| Jul 11, 2010 at 18:35 | comment | added | Robin Chapman | Tom Korner makes some interesting points about geometric intuition with respect to the MVT in his Companion to Analysis books.google.co.uk/… . Also I cannot see how the MVT is supposedly useful in proving the fundamental theorem of calculus. (The (to me) obvious proof of FTC only uses the definitions of continuity and derivative and the fact that $f\le g$ implies $\int f\le \int g$ on any interval.) | |
| Jul 11, 2010 at 18:33 | comment | added | Pete L. Clark | I disagree that it is a long side track to prove. The way I do it in freshman calculus takes about 10-12 minutes. First, prove "Rolle's Theorem". This proof I stand by, since it's visual, relatively easy, and reinforces the important concept that a function can't have a local max or local min at a point where it is differentiable with nonzero derivative. Then to prove MVT from Rolle's Theorem, I draw the line between $f(a)$ and $f(b)$ and make a big point of tiling my head until that line looks like the $x$-axis.... | |
| Jul 11, 2010 at 18:18 | history | answered | Matt | CC BY-SA 2.5 |