215 reputation
18
bio website tudelft.nl
location Delft, Netherlands
age 27
visits member for 1 year, 10 months
seen Nov 11 at 17:29

Jul
28
comment Zero-order interpolation not going fully around
Terrific summary of the problem and our discussion in the comments. Thanks again
Jul
26
comment Extrude 2D cross-section to 3D shape with shrinkfactor
This is indeed the route that I explored at first, but it is not entirely what I was looking for, because this remains a body of slices instead of a complete surface.
Jul
26
comment Extrude 2D cross-section to 3D shape with shrinkfactor
Thanks a bunch for the thorough answer, it really makes for beautiful shapes now!!
Jul
26
comment Extrude 2D cross-section to 3D shape with shrinkfactor
Thanks, but never mind. I already figured out the issue (see my edit in the question). Of course, the error was due to an adaptation that I made to your code, so I've only got myself to blame.
Jul
26
comment Extrude 2D cross-section to 3D shape with shrinkfactor
This works brilliantly, except for one thing that I cannot figure out: the cross-section for my own data is scaled to half the size in the extruded shape?! If I put a factor 2 inside the append (like Append[2*thicknessFunc[u,2]*fdata[t], u]) then it is fixed, but I really don't understand. I will add this to my question in an edit
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"?
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
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
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.
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
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 $\pi/2$
Feb
27
comment Stop Mathematica from giving imaginary solutions
@DanielLichtblau thanks for the explanation, I was not aware of that phenomenon. The Cubics->False option doesn't solve it however
Feb
26
comment Stop Mathematica from giving imaginary solutions
I have done what you mentioned, except for the //First//First part, because that will give me the wrong 1 of the 3 answers.
Feb
26
comment Stop Mathematica from giving imaginary solutions
Never mind this comment..... Too quick to respond while you where still explaining
Feb
26
comment Stop Mathematica from giving imaginary solutions
That is actually a good point. I didn't know that Solve had that option. I tried using $Assumptions, but that didn't help. Thanks!
Feb
26
comment Stop Mathematica from giving imaginary solutions
Not exactly. I found this post but I would like Solve to work with the assumptions already 'in mind'. Not `change' the answer afterwards