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 Mar26 asked What does $O\left(\frac{1}{\log\log T}\right)$ mean? Mar25 comment Series with $e^{\frac{1}{n}}$ Shouldn't that be $+O(n^{-4})$ in your equality? Mar25 comment Series with $e^{\frac{1}{n}}$ You could first try expanding $e^{1/n}$ to some order ($O(1/n^k)$). Presumably the value of $k$ will be related to the existence of $1/(2n^2)$ in your summand. Then simplify if possible, and see what happens from there. Haven't tried it so can't be 100% sure, but that's what I'd do first. Mar24 comment Complex integration confusion Try $z=e^{i\theta}$, where $\theta\in[0,2\pi]$. Mar21 comment Why Does $\sum\limits_{k=0}^n \begin{pmatrix} n+1 \\ k+1 \end{pmatrix} p^{k+1} (1-p)^{n-k}$ sum to $(1-(1-p)^{n+1})$? The binomial theorem states that: $$(a+b)^n=\sum_{k=0}^n{n\choose k}a^kb^{n-k}.$$ Mar19 comment Having trouble understanding generalized complex numbers @ Vim yes, that's exactly the problem I was thinking might arise. Mar19 comment Having trouble understanding generalized complex numbers @ Vim yes, as I said I'm not too sure about that part as I haven't looked into it a great deal. I think if the discriminant is negative then you still have a linear equation so you could still solve for $i$. My worry was that the $i$ resulting from the $\sqrt{\cdot}$ was a "different" $i$ from that being defined. But as I say I haven't looked into this in any depth whatsoever! Mar19 comment Having trouble understanding generalized complex numbers Just a thought (and might not be applicable here), regarding (1) could you not use the quadratic formula to obtain a "non-recursive" definition, e.g. $$i=\frac{q\pm\sqrt{q^2-4p}}{2}.$$ You would probably need $q^2-4p\geq 0$, although I'm not too sure about that... Mar18 revised On norms for “more complicated objects” deleted 8 characters in body; edited title Mar17 asked On norms for “more complicated objects” Mar17 comment Any way to simplify integral of Confluent Hypergeometric Function of the First Kind? @NathanMcKenzie out of interest, from what problem does this integral arise? Mar17 comment Any way to simplify integral of Confluent Hypergeometric Function of the First Kind? You may have better luck using the change of variable $y=\frac{\log n}{t}+1$. This gives $dt=\frac{\log(n)dy}{y^2}$ and $t=-\log n\implies y=0$, $t=0\implies y=\infty$. Then the integral becomes: $$-z\log(n)\int_0^\infty \frac{e^{\frac{\log (n)(1-s)}{y-1}}}{(y-1)^2} {}_1F_1\left(1-z,2,\frac{\log n}{y-1}\right)dy.$$ There are plenty of tables of integrals with limits over $\mathbb{R}^+$. See e.g. Gradshteyn and Ryzhik, p.814. Mar11 revised Is there a name for an object with both position and velocity? edited tags Mar11 comment Is there a name for an object with both position and velocity? @Sebastian, so for example a point in phase space might be $p=(x,y,u,v)$, where $(x,y)$ is position and $(u,v)$ is velocity. Thank you. Mar11 asked Is there a name for an object with both position and velocity? Feb26 comment Does the integral $\int_{1}^{\infty} \sin(x\log x) \,\mathrm{d}x$ converge? I understand you're using the Lambert W function which enables you to obtain $x=g(y)$, but could you please explain in a little more detail how you got the substituted integral? Feb23 revised Riemann Zeta Function and Including Complex Numbers added 9 characters in body Feb23 answered Riemann Zeta Function and Including Complex Numbers Feb21 accepted Examples of orthogonal/orthonormal functions which are not finite degree polynomials? Feb19 comment What is known about the complex solutions to $\zeta(s)=-1$? I suppose this is related to "level curves" of a function...