Re: [math-fun] The difficulty of integration vs. the ease of differentiation
I was hinting at the fact that integrate(A(x)/B(x),x), where A(x), B(x) are polynomials in x, is quite easy when B(x) factors into linear factors; use partial fraction expansion. A lot of trig functions crop up when you are unable or unwilling to move into the complex domain and B(x) has quadratic factors. At 02:27 PM 12/16/2016, Fred Lunnon wrote:
Numerically speaking, integration is in contrast easier than differentiation ... WFL
On 12/16/16, Henry Baker <hbaker1@pipeline.com> wrote:
2. Some closed form integrals may require polynomial root-finding (or root expressing), so you may not like the look of the resulting expression. E.g., you might find floating point approximations to certain numbers instead of a "perfectly precise" answer.
Actually the only trig function that pops up is arctan (which is essentially a logarithm). If you use int(1/(x^2+1)) dx = arctan x, a linear change of variables will reduce the quadratic factors to this. Victor On Fri, Dec 16, 2016 at 5:52 PM, Henry Baker <hbaker1@pipeline.com> wrote:
I was hinting at the fact that integrate(A(x)/B(x),x), where A(x), B(x) are polynomials in x, is quite easy when B(x) factors into linear factors; use partial fraction expansion. A lot of trig functions crop up when you are unable or unwilling to move into the complex domain and B(x) has quadratic factors.
At 02:27 PM 12/16/2016, Fred Lunnon wrote:
Numerically speaking, integration is in contrast easier than differentiation ... WFL
On 12/16/16, Henry Baker <hbaker1@pipeline.com> wrote:
2. Some closed form integrals may require polynomial root-finding (or root expressing), so you may not like the look of the resulting expression. E.g., you might find floating point approximations to certain numbers instead of a "perfectly precise" answer.
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Since a rational functions is analytic, its integral in "the complex domain" has the same expression as in the real domain. So just use partial fractions for every ratio of polynomials, and if the integrand was real, the integral is equal to a real-valued function of real arguments in any case, even if its metamorphosis included a complex stage. (Like the casus irreducibilis of the cubic.) —Dan
On Dec 16, 2016, at 2:52 PM, Henry Baker <hbaker1@pipeline.com> wrote:
I was hinting at the fact that integrate(A(x)/B(x),x), where A(x), B(x) are polynomials in x, is quite easy when B(x) factors into linear factors; use partial fraction expansion. A lot of trig functions crop up when you are unable or unwilling to move into the complex domain and B(x) has quadratic factors.
At 02:27 PM 12/16/2016, Fred Lunnon wrote:
Numerically speaking, integration is in contrast easier than differentiation ... WFL
On 12/16/16, Henry Baker <hbaker1@pipeline.com> wrote:
2. Some closed form integrals may require polynomial root-finding (or root expressing), so you may not like the look of the resulting expression. E.g., you might find floating point approximations to certain numbers instead of a "perfectly precise" answer.
On Friday, December 16, 2016, Dan Asimov <asimov@msri.org <mailto:asimov@msri.org>> wrote:
Why is it usually so easy to differentiate a function defined by an exact formula, but so much more difficult to integrate?
If this question can be made rigorous, how might that be done?
(And if so, what is the rigorous answer, or at least a method of approaching it?)
Maybe we can just as well ask "why is multiplication harder than addition" -- i.e. if we have two nontrivial binary operations that obey the distributive law, is it somehow impossible for them to have equal computational complexity? On Fri, Dec 16, 2016 at 6:27 PM, Dan Asimov <asimov@msri.org> wrote:
Since a rational functions is analytic, its integral in "the complex domain" has the same expression as in the real domain. So just use partial fractions for every ratio of polynomials, and if the integrand was real, the integral is equal to a real-valued function of real arguments in any case, even if its metamorphosis included a complex stage. (Like the casus irreducibilis of the cubic.)
—Dan
On Dec 16, 2016, at 2:52 PM, Henry Baker <hbaker1@pipeline.com> wrote:
I was hinting at the fact that integrate(A(x)/B(x),x), where A(x), B(x) are polynomials in x, is quite easy when B(x) factors into linear factors; use partial fraction expansion. A lot of trig functions crop up when you are unable or unwilling to move into the complex domain and B(x) has quadratic factors.
At 02:27 PM 12/16/2016, Fred Lunnon wrote:
Numerically speaking, integration is in contrast easier than differentiation ... WFL
On 12/16/16, Henry Baker <hbaker1@pipeline.com> wrote:
2. Some closed form integrals may require polynomial root-finding (or root expressing), so you may not like the look of the resulting expression. E.g., you might find floating point approximations to certain numbers instead of a "perfectly precise" answer.
On Friday, December 16, 2016, Dan Asimov <asimov@msri.org <mailto: asimov@msri.org>> wrote:
Why is it usually so easy to differentiate a function defined by an exact formula, but so much more difficult to integrate?
If this question can be made rigorous, how might that be done?
(And if so, what is the rigorous answer, or at least a method of approaching it?)
math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
----- As the animals left the ark, Noah told them to go forth and multiply. After some while, Noah happened upon two snakes sunning themselves. "Why aren't you multiplying?" Noah asked. The snakes replied, "We can't, we're adders." So Noah and his sons went into the nearby forest and felled some trees. They made a platform of logs onto which they placed the snakes. You see, even adders can multiply on a log table. ----- But seriously, your argument *might* explain why integration and differentiation are unequal, but not why they are markedly different. —Dan
On Dec 16, 2016, at 4:25 PM, Michael Collins <mjcollins10@gmail.com> wrote:
Maybe we can just as well ask "why is multiplication harder than addition" -- i.e. if we have two nontrivial binary operations that obey the distributive law, is it somehow impossible for them to have equal computational complexity?
Noah: Why aren't you adders adding? Snakes: We're only half adders, our father was a cobra. His genes didn't carry (over). Noah: Well, you could if you or your associate would just work together... At 05:53 PM 12/16/2016, Dan Asimov wrote:
As the animals left the ark, Noah told them to go forth and multiply. After some while, Noah happened upon two snakes sunning themselves. "Why aren't you multiplying?" Noah asked. The snakes replied, "We can't, we're adders." So Noah and his sons went into the nearby forest and felled some trees. They made a platform of logs onto which they placed the snakes. You see, even adders can multiply on a log table.
Cheap remark: max and min distribute over each other and have equal computational complexity. Jim Propp On Friday, December 16, 2016, Michael Collins <mjcollins10@gmail.com> wrote:
Maybe we can just as well ask "why is multiplication harder than addition" -- i.e. if we have two nontrivial binary operations that obey the distributive law, is it somehow impossible for them to have equal computational complexity?
On Fri, Dec 16, 2016 at 6:27 PM, Dan Asimov <asimov@msri.org <javascript:;>> wrote:
Since a rational functions is analytic, its integral in "the complex domain" has the same expression as in the real domain. So just use partial fractions for every ratio of polynomials, and if the integrand was real, the integral is equal to a real-valued function of real arguments in any case, even if its metamorphosis included a complex stage. (Like the casus irreducibilis of the cubic.)
—Dan
On Dec 16, 2016, at 2:52 PM, Henry Baker <hbaker1@pipeline.com <javascript:;>> wrote:
I was hinting at the fact that integrate(A(x)/B(x),x), where A(x), B(x) are polynomials in x, is quite easy when B(x) factors into linear factors; use partial fraction expansion. A lot of trig functions crop up when you are unable or unwilling to move into the complex domain and B(x) has quadratic factors.
At 02:27 PM 12/16/2016, Fred Lunnon wrote:
Numerically speaking, integration is in contrast easier than differentiation ... WFL
On 12/16/16, Henry Baker <hbaker1@pipeline.com <javascript:;>> wrote:
2. Some closed form integrals may require polynomial root-finding (or root expressing), so you may not like the look of the resulting expression. E.g., you might find floating point approximations to certain numbers instead of a "perfectly precise" answer.
On Friday, December 16, 2016, Dan Asimov <asimov@msri.org <javascript:;> <mailto: asimov@msri.org <javascript:;>>> wrote:
Why is it usually so easy to differentiate a function defined by an exact formula, but so much more difficult to integrate?
If this question can be made rigorous, how might that be done?
(And if so, what is the rigorous answer, or at least a method of approaching it?)
math-fun mailing list math-fun@mailman.xmission.com <javascript:;> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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