2,991 reputation
11430
bio website asmeurer.github.io
location Austin, TX
age 25
visits member for 4 years, 3 months
seen Nov 17 at 20:16

I am a software developer at Continuum Analytics. I am also the lead developer for the SymPy project.

Unless otherwise noted, all my answers on StackExchange, including code snippits, are released under ​Creative Commons CC0 (public domain).


Dec
18
comment Which one result in mathematics has surprised you the most?
${}{}{}{}{}{}x^3$?
Dec
18
comment Which one result in mathematics has surprised you the most?
Also known as $\sqrt{\pi}$.
Dec
18
accepted Is there a fundamental reason that $\int_b^a = -\int_a^b$
Dec
18
comment Is there a fundamental reason that $\int_b^a = -\int_a^b$
"The fundamental theorem of calculus...is based on this interpretation of the integral." This, I think, is the key insight here.
Dec
18
comment Is there a fundamental reason that $\int_b^a = -\int_a^b$
But what about signed measures?
Dec
18
comment Is there a fundamental reason that $\int_b^a = -\int_a^b$
Your very last sentence also gives a very nice motivation (it's ultimately the same as the accepted answer from math.stackexchange.com/questions/232455/…). I need to think on how this can apply to summations.
Dec
18
comment Is there a fundamental reason that $\int_b^a = -\int_a^b$
"...and $\int_A f+\bar\int_A\,f=0$." As you've defined it, how could this not hold?
Dec
18
awarded  Nice Answer
Dec
18
comment Is there a fundamental reason that $\int_b^a = -\int_a^b$
I'm not sure we're understanding each other. What I'm saying is that a summation is a Lebesgue integral with the counting measure.
Dec
18
comment Is there a fundamental reason that $\int_b^a = -\int_a^b$
Ah, that's a good reason. But I'm especially curious if this can be explained for Lebesgue integrals, since summations are special cases of Lebesgue integrals, not Riemann integrals (or at least not as far as I know).
Dec
18
revised Is there a fundamental reason that $\int_b^a = -\int_a^b$
Fix Karr's convention for summation
Dec
18
comment Is there a fundamental reason that $\int_b^a = -\int_a^b$
@CameronBuie that question doesn't really give the answer I'm looking for. At least to me, integrals are Lebesgue integrals (especially if I want to consider sums as special cases of integrals), and things like the fundamental theorem of calculus don't always hold, at least not without some generalized definitions of derivatives.
Dec
18
asked Is there a fundamental reason that $\int_b^a = -\int_a^b$
Dec
18
answered What exactly is infinity?
Dec
17
comment Purely “algebraic” proof of Young's Inequality
I thought the point here was to find them to prove the inequality. Artin's theorem tells us that our search is a reasonable direction to take (and that purely algebraic proofs are possible, at least modulo choosing rational exponents only).
Dec
17
accepted Purely “algebraic” proof of Young's Inequality
Dec
17
comment Purely “algebraic” proof of Young's Inequality
Young's inequality holds for any $p$ and $q$, so one would need to find a pattern for any exponent. According to Wikipedia, explicit algorithms exist to find these. I wonder if any of them are efficient, and if so, if they are implemented anywhere.
Dec
17
comment Purely “algebraic” proof of Young's Inequality
"Every polynomial inequality is a consequence..." OK, that and the other facts of inequalities, namely the basic poset relations, and also the algebraic properties (like $a < b$, $c < d$ $\implies$ $a + c < b + d$), which are important to assert that a sum of squares is nonnegative.
Dec
17
comment Purely “algebraic” proof of Young's Inequality
Cool. Of course, this doesn't actually prove that a proof exists, because the exponents in this case are symbolic, so there's no guarantee a priori that a general sum of squares formula can be found for arbitrary $p$. But the AM-GM argument gives hope that it does. I would be very interested if it could be determined.
Dec
17
comment Some ways to get a set of primes
Oh I get it. You're guaranteed to find a new prime each time, either as the number or as one of its factors.