dharr

Dr. David Harrington

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20 years, 148 days
University of Victoria
Professor or university staff
Victoria, British Columbia, Canada

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I am a retired professor of chemistry at the University of Victoria, BC, Canada. My research areas are electrochemistry and surface science. I have been a user of Maple since about 1990.

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These are replies submitted by dharr

@nm There are many Computer Modern fonts on my system (I use Scientific Workplace) and I don't know which are which - I just chose cmr12 and cmr16 at random, but cmbxti10 looks more like what you want. The above x may just be the standard one. 

P1 := plot(x^2):
P2 := plot(x^2,labels=["x","y"],axesfont=[cmr12,roman,16],
           labelfont=[cmbxti10,roman,24]):
plots:-display(Array([ P1, P2 ]));

gives (snapshot from export to .pdf

Latexfonts.mw

@Earl  You/re welcome. As you probably realize, if the control point just goes  x[1] = pts[2][1]..pts[1][1] you will just get the centre part with ellipses in the quadrilateral, which you asked about in a separate question..

@mmcdara  Not at all - I expected someone could help out if that wasn't enough.  It seemed a question about why dsolve,series didn't work, and as is often the case the OP's Maple background wasn't very clear..

I'm out of time these days, but had a quick look. I didn't check the eqns in detail, so can't comment on the method, but didn't see anything obviously wrong, except that "line" is in the plottools package, which wasn't loaded. I just tried regular solve, without the parametric option, but still lost connection with the kernel, which shouldn't happen. So my two tips would be:

Change the floating point numbers like 0.5 to exact 1/2 etc, since solve is for symbolic solutions (maybe solve would do this internally, but it is better to be sure and it is anyways good practice).

Rearrange the tangent equations to remove the denominators so a/b=c changes to a=b*c. Maple's solve is very good for polynomial systems, so that may work better.

@Axel Vogt  Some would say it is zero at x = infinity (complex infinity).  I think this is fairly standard when talking about poles and zeros.

@ecterrab I agree with @nm . The journals I use request \mathrm inside math mode when typesetting equations. I guess it would be \textrm outside.

Here's a variation where one of the coefficients becomes the parameter to vary. Conceptually simpler perhaps, but what the variation leads to and the flexibility is a bit more obscure.

Pencil2.mw

I just copied the output of DirectSearch to bypass that - then it seemed Pen1 should have been Pencil at the point highlighted in red. Then your ellipses work as you describe. But I don't understand what else you want to do, so I'll leave that to someone else.

Pencil.mw

Note that LegndreP and JacobiP are not in the orthopoly package (but P in that package is related).

@max125  If you want the nearest integer then ilog2(8) works, so it is interesting that log2(8) doesn't.

@max125 log10() works as a short form.

@dondomingo Select View/Markers, then open a new Document with File/New/Document mode. Type plot(x^2) and ctrl-= or alt-enter, which produces the plot on the same line. Then edit/document blocks/toggle input/output display.

(If you use enter, the plot is on the next line and you will have to put the cursor back into the document block before toggling the display)

In general, you get more control over layout in document mode, but if you are more worried about the math, use worksheet mode. You can also hide input in Tables.

At least in Maple 2017, only real boundary-value problems can be solved. But since your problem seems to be with fsolve, note you can use the complex option there. Suggest you upload your worksheet with the big green up-arrow, and we can see more clearly what you are trying to do and what the problem is.

@isabelmacpherson  I agree with you - you need the r^2*sin(varphi) as it is part of the volume element. As @nm points out, there are two conventions: you have used the Math convention, with varphi=0..Pi and theta=0..2*Pi.

v(0,0)=1is not acceptable. You need v(zeta,0)=1 for an initial condition (everywhere 1 at time zero) or v(0,t)=1 for a boundary condition (1 at zeta=0 for all time). Similarly for the other conditions. So you need to sort out what you want - I think 6 initial conditions and 2 boundary conditions,

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