Reputation
3,190
Top tag
Next privilege 5,000 Rep.
Approve tag wiki edits
Badges
9 30
Newest
 Nice Answer
Impact
~105k people reached

Mar
26
asked What does $O\left(\frac{1}{\log\log T}\right)$ mean?
Mar
25
comment Series with $e^{\frac{1}{n}}$
Shouldn't that be $+O(n^{-4})$ in your equality?
Mar
25
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.
Mar
24
comment Complex integration confusion
Try $z=e^{i\theta}$, where $\theta\in[0,2\pi]$.
Mar
21
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: $$(a+b)^n=\sum_{k=0}^n{n\choose k}a^kb^{n-k}.$$
Mar
19
comment Having trouble understanding generalized complex numbers
@ Vim yes, that's exactly the problem I was thinking might arise.
Mar
19
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!
Mar
19
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...
Mar
18
revised On norms for “more complicated objects”
deleted 8 characters in body; edited title
Mar
17
asked On norms for “more complicated objects”
Mar
17
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?
Mar
17
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.
Mar
11
revised Is there a name for an object with both position and velocity?
edited tags
Mar
11
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.
Mar
11
asked Is there a name for an object with both position and velocity?
Feb
26
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?
Feb
23
revised Riemann Zeta Function and Including Complex Numbers
added 9 characters in body
Feb
23
answered Riemann Zeta Function and Including Complex Numbers
Feb
21
accepted Examples of orthogonal/orthonormal functions which are not finite degree polynomials?
Feb
19
comment What is known about the complex solutions to $\zeta(s)=-1$?
I suppose this is related to "level curves" of a function...