176 reputation
7
bio website tudelft.nl
location Delft, Netherlands
age 27
visits member for 1 year, 5 months
seen Jul 5 at 10:36

Jul
5
accepted Forcing an integral to be solved in separate terms
Jul
4
revised Forcing an integral to be solved in separate terms
made the result in latex
Jul
4
comment Forcing an integral to be solved in separate terms
Great, thanks! Quick comment: did you know that you can copy the output of mathemetica directly as latex by right-clicking on the equation and selecting "Copy as -> Latex"?
Jul
4
suggested suggested edit on Forcing an integral to be solved in separate terms
Jul
4
asked Forcing an integral to be solved in separate terms
Jan
20
accepted Count the number of sqrt and 4th powers in a function
Jan
20
comment Count the number of sqrt and 4th powers in a function
Nice, thanks! A small follow-up: how does {0, Infinity} differ from just Infinity? Because if I use for example Count[Sec[x], Sec[__],..] mathematica returns 0 if I only use Infinity, but 1 if I use the `{0,Infinity}'
Jan
20
asked Count the number of sqrt and 4th powers in a function
Jan
18
accepted How to find the most compact form of an equation
Jan
15
comment How to find the most compact form of an equation
So if I understand correctly I can in theory give any ComplexityFunction I like?! Do you know what happens when I give a very simple ComplexityFunction e.g. the number of times a certain variable appears in the equation, and there are multiple equal solutions. Does it then use its default ComplexityFunction to distinguish between those or is one of the options 'randomly' given as best solution?
Jan
15
awarded  Commentator
Jan
15
comment How to find the most compact form of an equation
@SjoerdC.deVries I did indeed specify assumptions on variables being both real and positive. Without that the original equation would have been several lines longer. The ComplexityFunction seems to be what I am looking for.
Jan
15
asked How to find the most compact form of an equation
Apr
11
awarded  Teacher
Mar
26
awarded  Supporter
Mar
23
revised Evaluating the integral of $\cos (\theta )\cos\left[\tan ^{-1}\left(\frac{a^2 \tan (\phi )}{b^2}\right)\right]$
edited body
Mar
22
comment Evaluating the integral of $\cos (\theta )\cos\left[\tan ^{-1}\left(\frac{a^2 \tan (\phi )}{b^2}\right)\right]$
I had tried it, but it gave a very different result. However, looking at the documentation I found that I should use ArcTan[b^2 Cos[phi]],a^2 Sin[phi]] to get the correct result. Thanks! If you could write it as an answer I will definitely accept it.
Mar
22
revised Evaluating the integral of $\cos (\theta )\cos\left[\tan ^{-1}\left(\frac{a^2 \tan (\phi )}{b^2}\right)\right]$
added 258 characters in body
Mar
22
comment Evaluating the integral of $\cos (\theta )\cos\left[\tan ^{-1}\left(\frac{a^2 \tan (\phi )}{b^2}\right)\right]$
@MichaelE2 no I am sure the phase-shift is the correct thing to do. The reason is that ArcTan[(a/b)^2 Tan[[Phi]]] represents the angle the interface normal of an ellipse makes with the x-axis as determined from the angle the local x,y coordinates make with the x-axis (this angle is $\phi$). So I know that the whole thing should run from 0 to 360 deg and that the minima and maxima should coincide with those found when a=b. ---- So maybe I didn't explain enough of the background of the problem in my original question, but this is what I was looking for
Mar
21
comment Evaluating the integral of $\cos (\theta )\cos\left[\tan ^{-1}\left(\frac{a^2 \tan (\phi )}{b^2}\right)\right]$
Thanks for the explanation of how to get the answer to the integral, but I found out that it was in fact not the answer to the integral that was incorrect, but rather what I fed it: the Cos[ArcTan[ Tan[[Phi]]]] which has a modulus at $\pi/2$