Integration from first principles
Nick Grey
Is it possible to do integration from first principles, in a similar way
to which differentiation is done from first principles? All the
references I can find do differentiation from first principles to find the
"shortcuts" for differentiation, then basically say that to do
integrations you reverse the shortcuts.
Regards,
Nick
A N Neil
<nick...@farrand.net> wrote:
???
Integration done with Riemann sums. No "reverse the shortcuts"
involved.
Virgil
It is possible, at least for simple enough integrands, to do
definite (Riemann) integrals by finding limits of Darboux sums.
But it is not very useful except in proving theorems about
properties of Riemann integrals, or as a learning device.
The whole point of the Fundamental Theorems of Calculus is that
integration and differentiation are closely related operations, and
that knowledge of one may help with the other.
Lovecraftesque
Maybe he mean antidifferentiation, rather than
integration.
theodore hwa
: Hi,
For differentiation, the product, quotient, and chain rules allow you to
differentiate any elementary function in a straightforward manner. There
are no such general rules for integration. The methods are necessarily
heuristic, based on reversing the rules for differentiation. (Yes, there
are algorithms by Risch and others for symbolic integration, but no one
does those by hand, and even computer programs, as far as I know, use a
mixture of Risch algorithms with heuristics.)
Ted
Darren G. Lorent
Actually, indefinite integrals are usually evaluated using the
Fundamental Theorem of Calculus. The FTC tells us that to evaluate the
integral of f(x), find a function whose derivative is f(x). So I would
say that you usually DO "reverse the (differentiation) shortcuts". All
of the tricks we know for doing this are another set of shortcuts. I
guess it's really all in how you look at it.
Darren
Charles Matthews
"Nick Grey" wrote
> Is it possible to do integration from first principles, in a similar way
> to which differentiation is done from first principles?
Go back to Archimedes, and I think that's what was done, really. Not easy
except for rectangles, triangles. The area under a parabola?
Charles
Herman Rubin
I believe that what you are calling "integration" is really
"antidifferentiation". Also, you seem to be interested in
the tricks for calculation, not the concepts of what is behind
the mechanics. This is computationally useful, and should be
considered as such.
So all of the procedures are the inverse of the procedures
for differentiation, even if some do not look like it.
--
This address is for information only. I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Deptartment of Statistics, Purdue University
hru...@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558
John Mitchell
The answer is probably "no" no matter how your question is
interpreted. For example, it is known that certain antiderivatives
cannot be expressed in elementary form (the best-known being
Int{exp(-x^2)dx}).
Or you might ask: "Can we derive the formula for Int{x^2 dx} without
using some variation of guessing an antiderivative?" (in contrast to
the computation of the derivative of x^3, say, which is a direct,
finite computation requiring no guesswork). Not that I know of, but
I'm not even sure how to make that question precise.
Or you might ask: "Is there a product rule for integrals?" One precise
version of this question is: "Is there a function A of four real
arguments such that
Int{f(x)g(x)dx} = A(f(x), g(x), Int{f(x)dx}, Int{g(x)dx})
for all (integrable) functions f, g?" Here Int can represent either
antidifferentiation or a definite integral from x0 to x with x0 fixed.
It's not hard to see that there is no such product rule: f and g may
be varied in a way that keeps the left side of the equation and any
three of the arguments to A fixed at arbitrary pre-assigned values,
while the remaining argument to A varies over arbitrary values. This
implies that A doesn't depend on any of its arguments, which is
impossible. Of course, that's not the most general imaginable form for
a product rule for integrals, but I suspect that there answer is "no"
even in the most general case. The theory of Differential Algebra
might answer the more general question.
Regards,
John Mitchell
Virgil
john.m...@autodesk.com (John Mitchell) wrote:
> Or you might ask: "Can we derive the formula for Int{x^2 dx} without
> using some variation of guessing an antiderivative?" (in contrast to
> the computation of the derivative of x^3, say, which is a direct,
> finite computation requiring no guesswork). Not that I know of, but
> I'm not even sure how to make that question precise.
There is a general ->direct<- method for (Riemann) definite
integrals of form Integral[x=a..b, x^n] = ( b^(n+1)-a^(n+1) )/(n+1),
for positive integers n, as the limit of a Darboux sum.
The antiderivative formula for x^n is an immediate consequence.
John Mitchell
The problem certainly can be reduced to computing the sum Sum{k^n},
k=1...n, which is presumably what you mean by "Darboux sum", but
that's just a discrete version of the same problem. Can the sum be
computed without resorting to "guessing the antiderivative" (now in
the context of finite differences)?
John Mitchell
Virgil
> > The antiderivative formula for x^n is an immediate consequence.
>
> The problem certainly can be reduced to computing the sum Sum{k^n},
> k=1...n, which is presumably what you mean by "Darboux sum", but
> that's just a discrete version of the same problem. Can the sum be
> computed without resorting to "guessing the antiderivative" (now in
> the context of finite differences)?
>
> John Mitchell
I don't see that Sum[k^n,k=1..n] will get you much of anywhere.
Try Sum[k^m,k=1..n] = n^(m+1)/(m+1) + o(n^(m+1)) as n -> +oo.
Herman Rubin
John Mitchell <john.m...@autodesk.com> wrote:
>"Nick Grey" <nick...@farrand.net> wrote in message news:<pan.2002.12.04....@farrand.net>...
>> Hi,
>> Is it possible to do integration from first principles, in a similar way
>> to which differentiation is done from first principles? All the
>> references I can find do differentiation from first principles to find the
>> "shortcuts" for differentiation, then basically say that to do
>> integrations you reverse the shortcuts.
>The answer is probably "no" no matter how your question is
>interpreted. For example, it is known that certain antiderivatives
>cannot be expressed in elementary form (the best-known being
>Int{exp(-x^2)dx}).
>Or you might ask: "Can we derive the formula for Int{x^2 dx} without
>using some variation of guessing an antiderivative?" (in contrast to
>the computation of the derivative of x^3, say, which is a direct,
>finite computation requiring no guesswork). Not that I know of, but
>I'm not even sure how to make that question precise.
This, and many other simple situations, such as sin(x),
can be done directly. Fermat computed Int(x^r dx), r
rational, without guessing, by using a geometric series
for the partition points.
The Fundamental Theorem of Calculus is a tautology to
most students because they do not understand that the
integral can be directly studied. Antidifferentiation
is a useful method for evaluating some integrals, not
part of the definition.
John Mitchell
That's what I meant (I'll have to reprimand my editor :-)
My question was whether the summation formula can be derived directly.
It's pretty easy to conclude that the solution is a polynomial in n of
degree m+1, and then to find that polynomial, but I consider that to
involve "guessing" (at least in contrast to computing the finite
difference of n^m).
My general point was that there is no formal recipe for computing
integrals of of the types of functions typically encountered in
applications, in constrast to the problem of computing derivatives,
where linearity, the product rule, and the chain rule provide such a
recipe.
John Mitchell
> In article <dc44dd30.02120...@posting.google.com>,
> John Mitchell <john.m...@autodesk.com> wrote:
> >"Nick Grey" <nick...@farrand.net> wrote in message news:<pan.2002.12.04....@farrand.net>...
> >> Hi,
>
> >> Is it possible to do integration from first principles, in a similar way
> >> to which differentiation is done from first principles? All the
> >> references I can find do differentiation from first principles to find the
> >> "shortcuts" for differentiation, then basically say that to do
> >> integrations you reverse the shortcuts.
>
>
> >The answer is probably "no" no matter how your question is
> >interpreted.
> >Or you might ask: "Can we derive the formula for Int{x^2 dx} without
> >using some variation of guessing an antiderivative?" (in contrast to
> >the computation of the derivative of x^3, say, which is a direct,
> >finite computation requiring no guesswork). Not that I know of, but
> >I'm not even sure how to make that question precise.
>
> This, and many other simple situations, such as sin(x),
> can be done directly. Fermat computed Int(x^r dx), r
> rational, without guessing, by using a geometric series
> for the partition points.
That's interesting, I wasn't aware of Fermat's method. Do you know of
a reference? My main point, though, was that there is no formal recipe
for computing arbitrary integrals (or antiderivatives) of the kinds of
functions one typically encounters in applications, in contrast to the
situation for derivatives.
> The Fundamental Theorem of Calculus is a tautology to
> most students because they do not understand that the
> integral can be directly studied. Antidifferentiation
> is a useful method for evaluating some integrals, not
> part of the definition.
I once tried to drive home the importance of the Fundamental Theorem
of Calculus to a class I was teaching by spending a few days exploring
the direct computation of (definite) integrals. For example, Int{x^2
dx} would be reduced to finding a formula for the summation Sum{k^2}
k=1...n and then taking a limit. My hope was that when the students
later learned how to calculate integrals using antiderivatives, they
would better appreciate the power of that method and the reason it's
called fundamental. I don't recall any audible gasps of astonishment,
so maybe it didn't sink in.
John Mitchell
Virgil
john.m...@autodesk.com (John Mitchell) wrote:
You don't need the whole polynomial, you only need its highest
degree term, as everything else drops out in the limit of the
summation.
And that term requires only a minimum of guessing, as indicated in
my litle-oh expression above.
On the other hand, it is quite clear that integration of whatever
sort is not as mechanical as differentiation.
Herman Rubin
John Mitchell <john.m...@autodesk.com> wrote:
>hru...@odds.stat.purdue.edu (Herman Rubin) wrote in message news:<astga9$51...@odds.stat.purdue.edu>...
>> In article <dc44dd30.02120...@posting.google.com>,
>> John Mitchell <john.m...@autodesk.com> wrote:
>> >"Nick Grey" <nick...@farrand.net> wrote in message news:<pan.2002.12.04....@farrand.net>...
>> >> Hi,
.........................
>That's interesting, I wasn't aware of Fermat's method. Do you know of
>a reference? My main point, though, was that there is no formal recipe
>for computing arbitrary integrals (or antiderivatives) of the kinds of
>functions one typically encounters in applications, in contrast to the
>situation for derivatives.
I heard it mentioned in a talk. Fermat preceded differentiation.
>> The Fundamental Theorem of Calculus is a tautology to
>> most students because they do not understand that the
>> integral can be directly studied. Antidifferentiation
>> is a useful method for evaluating some integrals, not
>> part of the definition.
>I once tried to drive home the importance of the Fundamental Theorem
>of Calculus to a class I was teaching by spending a few days exploring
>the direct computation of (definite) integrals. For example, Int{x^2
>dx} would be reduced to finding a formula for the summation Sum{k^2}
>k=1...n and then taking a limit. My hope was that when the students
>later learned how to calculate integrals using antiderivatives, they
>would better appreciate the power of that method and the reason it's
>called fundamental. I don't recall any audible gasps of astonishment,
>so maybe it didn't sink in.
The only way to avoid this in the typical program is to
teach general simple integration. Euclid's students
understood the "Riemann integral", at least for nice
functions including discontinuous ones with a finite
number of pieces; they just did not know how to calculate
many of them. One does not learn concepts by knowing how
to compute answers.
One should start with the idea of a measure early. While
the general concept did not appear until the 1890's, it
is simple enough. In fact, every measure is discrete, or
a limit of discrete. Integral is a sum of products, or a
limit of such, and yes, Euclid's students could use limits,
even though they could not use variables, as they had not
yet been invented. A good example for college students is
the total number of grade points. The oldest example of
computing an integral is determining the amount of the
bill presented by a merchant.
Bob Pease
"John Mitchell" <john.m...@autodesk.com> wrote in message
news:dc44dd30.02120...@posting.google.com...
It probably sunk in for about a third of them.
The lesson that I think that most of them would get is
"how come is he doing it the HARD way when you can do it by the"Rules"?"
The FTC is really amazing when you think about it. it's not really
fundamentally obvious, and the discovery of the Proof must rank very high in
the list of achievements of mankind.!!
RJ Pease
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Dave L. Renfro
[sci.math Dec 8 2002 11:24:36:000PM]
http://mathforum.org/discuss/sci.math/m/463344/464148
wrote (in part, in response to Herman Rubin):
>> This, and many other simple situations, such as sin(x),
>> can be done directly. Fermat computed Int(x^r dx), r
>> rational, without guessing, by using a geometric series
>> for the partition points.
>
> That's interesting, I wasn't aware of Fermat's method. Do you
> know of a reference? My main point, though, was that there is
> no formal recipe for computing arbitrary integrals (or
> antiderivatives) of the kinds of functions one typically
> encounters in applications, in contrast to the situation for
> derivatives.
You can find Fermat's calculation of Int(x^r) for r rational
(and not equal to -1) in this book --->
Ernst Hairer and Gerhard Wanner, "Analysis by Its History",
Undergraduate Texts in Mathematics, Springer-Verlag, 1995.
http://makeashorterlink.com/?Z522139B2 [amazon.com page]
http://www.unige.ch/math/folks/wanner/analysis.html
http://www.netstoreusa.com/mabooks/038/0387945512.shtml
The only copy I have is in my office and I'm at home right
now, otherwise I'd give you the page reference. I don't think
it's hard to find in this book (index: "Fermat" will probably
do it), but let me know if you need the page(s).
I'm sure Fermat's "trick" is also discussed in several
well-known history of calculus/analysis books, such as
the books you can find at the following web page, but
again I'm at home and everything like this is in my office.
http://aleph0.clarku.edu/~djoyce/mathhist/analysis.html
Does anyone happen to know whether calculations using
subintervals of varying relative lengths (such as what
Fermat did) played any role in Riemann's decision to
define his integral in this way (i.e. using Riemann sums
in which the subintervals don't have to have the same
length)?
My guess is that Riemann was motivated by calculations
that involved decomposing an interval into subintervals
over which a (sufficiently nice) function was monotone.
Incidentally, on a related note that others might be interested
in, there is a notion that is sometimes called the "Cauchy
integral". No, I'm not talking about a type of improper integral
(the Cauchy-Principle Value), but rather what appears to be a
less general version of the Riemann integral. The Riemann integral
uses limits involving arbitrary finite partitions (i.e. the
subintervals don't have the same lengths) and arbitrary choices
of a point in each of these subintervals. The Cauchy integral is
when the partitions can still be arbitrary, but the point picked
from each subinterval is the left endpoint of that subinterval.
I came across a paper a few months ago in one of the U.S. math
journals (it was probably Trans. Amer. J. Math., Amer. J. Math.,
or Annals of Math.) that was published somewhere between 1910
and 1930 which proved that the Riemann integral is no more
general than the Cauchy integral. That is, the class of functions
integrable in the "Cauchy sense" is the same as the class of
functions integrable in the Riemann sense. I don't remember
anything more specific about it, but if anyone is interested I
can give the precise reference tomorrow. [I have a copy of this
paper at my office.]
This leads me to wonder about two other variations on Riemann
integration --->
(A) Partitions using subintervals of the same length,
but the point in each subinterval can be chosen
arbitrarily.
(B) Partitions using subintervals of the same length
AND the point in each subinterval is chosen to be
the left endpoint of that subinterval.
I haven't thought about this much, but it seems to me
that an investigation of (A) and (B), in conjunction
with reading the paper I mentioned above, might be
suitable for an undergraduate honors project. My guess
is that it wouldn't be very difficult to show that (A)
defines the same class of functions that Cauchy integration
does, which would show that (A) is equivalent to Riemann
integrability). However, I'd be very surprised if (B)
were equivalent to Riemann integrability (I'd also be
surprised if a counterexample was difficult to come by),
and so the part that might be worth looking into would
be characterizing the functions that (B) defines.
Dave L. Renfro
Dave L. Renfro
[sci.math Dec 8 2002 (around 4:00 P.M. EST)]
http://mathforum.org/discuss/sci.math/t/463344
wrote (in part):
> This leads me to wonder about two other variations on Riemann
> integration --->
>
> (A) Partitions using subintervals of the same length,
> but the point in each subinterval can be chosen
> arbitrarily.
>
> (B) Partitions using subintervals of the same length
> AND the point in each subinterval is chosen to be
> the left endpoint of that subinterval.
>
> I haven't thought about this much, but it seems to me
> that an investigation of (A) and (B), in conjunction
> with reading the paper I mentioned above, might be
> suitable for an undergraduate honors project. My guess
> is that it wouldn't be very difficult to show that (A)
> defines the same class of functions that Cauchy integration
> does, which would show that (A) is equivalent to Riemann
> integrability). However, I'd be very surprised if (B)
> were equivalent to Riemann integrability (I'd also be
> surprised if a counterexample was difficult to come by),
> and so the part that might be worth looking into would
> be characterizing the functions that (B) defines.
Naturally, I'd think of a counterexample less than a minute
AFTER I sent my post in!
An example of a function integrable in the sense of (B) on [0,1]
that is not Riemann integrable on [0,1] is the characteristic
function of a Cantor set in [0,1] that has positive measure and
which contains no rational numbers. Every (B)-Riemann sum for
this function is zero, and hence the (B)-integral of this function
is zero. This function isn't Riemann integrable because it's
discontinuous at every point of this Cantor set, and hence it
has a non-measure zero set of discontinuities.
Dave L. Renfro
Robert Israel
John Mitchell <john.m...@autodesk.com> wrote:
>hru...@odds.stat.purdue.edu (Herman Rubin) wrote in message
>news:<astga9$51...@odds.stat.purdue.edu>...
>> This, and many other simple situations, such as sin(x),
>> can be done directly. Fermat computed Int(x^r dx), r
>> rational, without guessing, by using a geometric series
>> for the partition points.
>That's interesting, I wasn't aware of Fermat's method. Do you know of
>a reference?
I don't know of a reference, but it's not too hard. For
int_1^b x^r dx with b > 1 and r <> -1, take x_n = b^(n/N) for n=0..N so
x_{n+1}-x_n = b^(n/N) (b^(1/N)-1). The left Riemann sum is a geometric
series
sum_{n=0}^{N-1} b^(nr/N) b^(n/N) (b^(1/N)-1)
= (b^(1/N) - 1) (b^(1+r) - 1)/(b^((1+r)/N)-1)
and use the fact that the limit as N -> infinity of
(b^(1/N)-1)/(b^((1+r)/N)-1) is 1/(1+r). Similarly for the right sum.
Robert Israel isr...@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2