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Feb
16
accepted Definition of $L^p$ as a set
Feb
16
revised Definition of $L^p$ as a set
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Feb
16
revised Definition of $L^p$ as a set
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Feb
16
asked Definition of $L^p$ as a set
Feb
15
accepted A class of pairs $(f,g)$ of functions.
Feb
14
comment A class of pairs $(f,g)$ of functions.
@ BaronVT in your first sentence you say "...implies the transforms are both zero," but could they not both be non-zero constant ?
Feb
14
comment A class of pairs $(f,g)$ of functions.
Are you saying that equation (1) implies that $f=g=0$. However, if the limits were instead $-\infty$ and $+\infty$ then equation (1) with the new limits would imply that whatever $f$ and $g$ are, $f$ must be odd and $g$ even ?
Feb
14
revised A class of pairs $(f,g)$ of functions.
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Feb
14
revised A class of pairs $(f,g)$ of functions.
added 184 characters in body
Feb
14
revised A class of pairs $(f,g)$ of functions.
added 184 characters in body
Feb
14
asked A class of pairs $(f,g)$ of functions.
Feb
13
comment Specific form of integral representation of the Riemann zeta function
@Greg that's interesting. I'd not thought about other types of integral. That's more maths to learn :-)
Feb
13
comment Specific form of integral representation of the Riemann zeta function
I've been playing around with other integral representations and noticed there weren't any of the stated form; however, those that do "come close" appear to have an integrand of the form $g(x)/x^{p(s)}$.
Feb
13
asked Specific form of integral representation of the Riemann zeta function
Feb
12
comment Integral $\int_{-1}^1\frac1x\sqrt{\frac{1+x}{1-x}}\ln\left(\frac{2\,x^2+2\,x+1}{2\,x^2-2\,x+1}\right) \ \mathrm dx$
@RonGordon True, and no it certainly was not expected! Then again, I don't expect any particular algebraic number to appear in a result until it does appear, so I guess the surprise here is that $\phi$ has some special significance. I like your expectations idea.
Feb
12
comment Integral $\int_{-1}^1\frac1x\sqrt{\frac{1+x}{1-x}}\ln\left(\frac{2\,x^2+2\,x+1}{2\,x^2-2\,x+1}\right) \ \mathrm dx$
I know that $\phi$ has some nice properties, but at the end of the day it's just another algebraic number: $$\phi = \frac{1+\sqrt{5}}{2},$$ and all we've done is taken the arc cotangent of the square root of it and multiplied by $4\pi$. - just wondering why this result would be seen as "odd". It most likely appears because it happens to be a root of some polynomial which has $(1+\sqrt{5})/2$ as one of its roots.
Feb
11
answered Integers seen as a continuous shape
Feb
11
revised Find $(a+ib)^{492}$ given that $(a+ib)^{493}=1$
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Feb
11
answered Find $(a+ib)^{492}$ given that $(a+ib)^{493}=1$
Feb
11
comment Related paradigms in Computer Science and Mathematics
@user72694 I've added a bit more information with some resources.