Re: [math-fun] calculus question
Jim writes: << [W]hy would it make sense pedagogically to use sub-intervals of unequal widths as part of the _definition_ of the integral, and then state a _theorem_ that says that if a function is integrable then you might as well use equal-width sub-intervals (which is what Stewart does), rather than _define_ the integral using equal-width sub-intervals, and then state a _theorem_ that says that you can use more general partitions and still get Riemann sums that converge on the correct value, as long as the mesh goes to zero?
Seems to me that to define the integral in terms of the same limit for a wide variety of partitions would remove any suspicion that the nice way it turns out is solely an artifice of using equal-length intervals. Then to specialize to equal-length partitions when *calculating* the integral makes sense, just for the sake of convenience. This strategy, however, would certainly benefit from a theorem in between the above definition of integrals and the calculation thereof, stating that a large class of functions (e.g., piecewise continuous ones) actually *are* integrable in the sense mentioned. --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele
On Feb 6, 2008 2:52 PM, Dan Asimov <dasimov@earthlink.net> wrote:
Jim writes:
<< [W]hy would it make sense pedagogically to use sub-intervals of unequal widths as part of the _definition_ of the integral, and then state a _theorem_ that says that if a function is integrable then you might as well use equal-width sub-intervals (which is what Stewart does), rather than _define_ the integral using equal-width sub-intervals, and then state a _theorem_ that says that you can use more general partitions and still get Riemann sums that converge on the correct value, as long as the mesh goes to zero?
Seems to me that to define the integral in terms of the same limit for a wide variety of partitions would remove any suspicion that the nice way it turns out is solely an artifice of using equal-length intervals.
Then to specialize to equal-length partitions when *calculating* the integral makes sense, just for the sake of convenience.
Pedagogically, I think it makes sense to do things the other way around. This is a class for people who do not have a lot of experience with mathematics, and the problem people have understanding mathematics is the high level of abstraction involved. The way to combat this is by being as concrete as possible at first, and only introducing abstraction as needed. If you start out with the "any partition" definition, then you have a very complex definition, one that you cannot use to calculate, and you have no examples. You do not have any functions that you can prove have an integral with the given definition. You then proceed to prove a difficult and complex theorem about these integrals (that says that if the calculation with equal-length partitions converges to a limit, then this value is the integral). Only then can you begin to exhibit examples. If you go the other direction, first you have a simple, computationally effective, definition. You can give examples, and calculate integrals. The next step is to prove a theorem to show that your definition is a useful and interesting one, because the integral you have calculated isn't an artifact of the particular way you have done the subdivision. This theorem is what "removes any suspicion that the nice way it turns out is solely an artifice of using equal-length intervals." But now you are proving a theorem about a concept that is simply defined and easy to understand, for which you have concrete examples, and that you therefore already understand. Moving back from pedagogy to mathematics, it would be even better to make the simple version even simpler, if that doesn't change the definition. Are there any functions where the Riemann integral (with either of the above definitions) does not exist, but Lim{n->infty}(Sum{i = 1 to n} (f(a + ((b-a)/n)))/n) does exist? I don't think so, but this may be harder to prove. If the limit in the above formula (that is, always use equal subdivisions, and evaluate the function at the left endpoint of each subdivision) is defined exactly when the conventional Riemann interval is defined, why not use the above formula as the definition, and then prove theorems that the result does not depend on either the point chosen within each subdivision, nor on the subdivision scheme, as long as the largest subdivision length approaches zero? -- Andy.Latto@pobox.com
When you go from equal intervals to unequal ones, it's important to specify exactly how all the intervals approach zero width. I imagine that the limit is affected by how the limit is reached. Steve Gray -----Original Message----- From: math-fun-bounces@mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Andy Latto Sent: Wednesday, February 06, 2008 12:18 PM To: math-fun Subject: Re: [math-fun] calculus question On Feb 6, 2008 2:52 PM, Dan Asimov <dasimov@earthlink.net> wrote:
Jim writes:
<< [W]hy would it make sense pedagogically to use sub-intervals of unequal widths as part of the _definition_ of the integral, and then state a _theorem_ that says that if a function is integrable then you might as well use equal-width sub-intervals (which is what Stewart does), rather than _define_ the integral using equal-width sub-intervals, and then state a _theorem_ that says that you can use more general partitions and still get Riemann sums that converge on the correct value, as long as the mesh goes to zero?
Seems to me that to define the integral in terms of the same limit for a wide variety of partitions would remove any suspicion that the nice way it turns out is solely an artifice of using equal-length intervals.
Then to specialize to equal-length partitions when *calculating* the integral makes sense, just for the sake of convenience.
Pedagogically, I think it makes sense to do things the other way around. This is a class for people who do not have a lot of experience with mathematics, and the problem people have understanding mathematics is the high level of abstraction involved. The way to combat this is by being as concrete as possible at first, and only introducing abstraction as needed. If you start out with the "any partition" definition, then you have a very complex definition, one that you cannot use to calculate, and you have no examples. You do not have any functions that you can prove have an integral with the given definition. You then proceed to prove a difficult and complex theorem about these integrals (that says that if the calculation with equal-length partitions converges to a limit, then this value is the integral). Only then can you begin to exhibit examples. If you go the other direction, first you have a simple, computationally effective, definition. You can give examples, and calculate integrals. The next step is to prove a theorem to show that your definition is a useful and interesting one, because the integral you have calculated isn't an artifact of the particular way you have done the subdivision. This theorem is what "removes any suspicion that the nice way it turns out is solely an artifice of using equal-length intervals." But now you are proving a theorem about a concept that is simply defined and easy to understand, for which you have concrete examples, and that you therefore already understand. Moving back from pedagogy to mathematics, it would be even better to make the simple version even simpler, if that doesn't change the definition. Are there any functions where the Riemann integral (with either of the above definitions) does not exist, but Lim{n->infty}(Sum{i = 1 to n} (f(a + ((b-a)/n)))/n) does exist? I don't think so, but this may be harder to prove. If the limit in the above formula (that is, always use equal subdivisions, and evaluate the function at the left endpoint of each subdivision) is defined exactly when the conventional Riemann interval is defined, why not use the above formula as the definition, and then prove theorems that the result does not depend on either the point chosen within each subdivision, nor on the subdivision scheme, as long as the largest subdivision length approaches zero? -- Andy.Latto@pobox.com _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
The wikipedia article on Riemann Integrals has a nice discussion of this, and points out that if you subdivide the interval [0,1] into n equal sized pieces then the indicator function of the rationals appears to be integrable (and it is not). See http://en.wikipedia.org/wiki/Riemann_integral Victor On Feb 6, 2008 8:00 PM, Stephen Gray <stevebg@roadrunner.com> wrote:
When you go from equal intervals to unequal ones, it's important to specify exactly how all the intervals approach zero width. I imagine that the limit is affected by how the limit is reached.
Steve Gray
-----Original Message----- From: math-fun-bounces@mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Andy Latto Sent: Wednesday, February 06, 2008 12:18 PM To: math-fun Subject: Re: [math-fun] calculus question
On Feb 6, 2008 2:52 PM, Dan Asimov <dasimov@earthlink.net> wrote:
Jim writes:
<< [W]hy would it make sense pedagogically to use sub-intervals of unequal widths as part of the _definition_ of the integral, and then state a _theorem_ that says that if a function is integrable then you might as well use equal-width sub-intervals (which is what Stewart does), rather than _define_ the integral using equal-width sub-intervals, and then state a _theorem_ that says that you can use more general partitions and still get Riemann sums that converge on the correct value, as long as the mesh goes to zero?
Seems to me that to define the integral in terms of the same limit for a wide variety of partitions would remove any suspicion that the nice way it turns out is solely an artifice of using equal-length intervals.
Then to specialize to equal-length partitions when *calculating* the integral makes sense, just for the sake of convenience.
Pedagogically, I think it makes sense to do things the other way around. This is a class for people who do not have a lot of experience with mathematics, and the problem people have understanding mathematics is the high level of abstraction involved. The way to combat this is by being as concrete as possible at first, and only introducing abstraction as needed.
If you start out with the "any partition" definition, then you have a very complex definition, one that you cannot use to calculate, and you have no examples. You do not have any functions that you can prove have an integral with the given definition. You then proceed to prove a difficult and complex theorem about these integrals (that says that if the calculation with equal-length partitions converges to a limit, then this value is the integral). Only then can you begin to exhibit examples.
If you go the other direction, first you have a simple, computationally effective, definition. You can give examples, and calculate integrals. The next step is to prove a theorem to show that your definition is a useful and interesting one, because the integral you have calculated isn't an artifact of the particular way you have done the subdivision. This theorem is what "removes any suspicion that the nice way it turns out is solely an artifice of using equal-length intervals." But now you are proving a theorem about a concept that is simply defined and easy to understand, for which you have concrete examples, and that you therefore already understand.
Moving back from pedagogy to mathematics, it would be even better to make the simple version even simpler, if that doesn't change the definition. Are there any functions where the Riemann integral (with either of the above definitions) does not exist, but Lim{n->infty}(Sum{i = 1 to n} (f(a + ((b-a)/n)))/n) does exist? I don't think so, but this may be harder to prove.
If the limit in the above formula (that is, always use equal subdivisions, and evaluate the function at the left endpoint of each subdivision) is defined exactly when the conventional Riemann interval is defined, why not use the above formula as the definition, and then prove theorems that the result does not depend on either the point chosen within each subdivision, nor on the subdivision scheme, as long as the largest subdivision length approaches zero?
-- Andy.Latto@pobox.com
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participants (4)
-
Andy Latto -
Dan Asimov -
Stephen Gray -
victor miller