dharr

Dr. David Harrington

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21 years, 10 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

@JAMET Well, it's difficult to be sure because you didn't upload your worksheet, but the * after plots:-pointplot suggests you are using 2D input and have a space there that should not be there.

@Joe Riel Thanks; I'd never noticed that. I see now the help page has comptype with the key option more prominently explained than in my version.

@zenterix As I pointed out the problem is with the range. The function goes to r=infinity, so if you do not specify a plot range, Maple tries to choose one, thinking you will want to see all those really big numbers. If you right-click on your original plot (cmd-click on Mac), choose probe info and then cursor position, and run your cursor over the plot, you will see that the x axis runs in a very narrow range around 2, and the y axis runs from about -60000 to 8000. Your polarplot was also a vertical line at 2, it is just that the range was very large, so it looked like at vertical line at zero. In general, if you see a weird plot, you should modify the range to be where you think the most information is.

@Hullzie16 Although these have "I" in they are actually real, as you have seen. You can remove the I with evalc and then simplify - but now you have trig functions

[-10/3-(10/3)*3^(1/2)*sin(-(1/3)*arctan((3/173)*3^(1/2)*373^(1/2))+(1/3)*Pi)-(10/3)*sin((1/3)*arctan((3/173)*3^(1/2)*373^(1/2))+(1/6)*Pi), -10/3+(10/3)*3^(1/2)*sin(-(1/3)*arctan((3/173)*3^(1/2)*373^(1/2))+(1/3)*Pi)-(10/3)*sin((1/3)*arctan((3/173)*3^(1/2)*373^(1/2))+(1/6)*Pi), -10/3+(20/3)*sin((1/3)*arctan((3/173)*1119^(1/2))+(1/6)*Pi)]

 

Download CubicSolve.mw

@jud In general it's poor etiquette to delete a post after it has been answered. Perhaps it will be useful for someone else (the indexing ds[2] is not obvious for example but you had figured that out.) Happy to help in your transition to Maple.

@salauayobami Click the green up-arrow, then Choose the file. Then click "upload". Then perhaps after a delay depending on how complicated your worksheet is you should see the two options "Insert link" and "insert contents". Sometimes "Insert contents" does not work well, so just do "insert link". If that doesn't work, suggest you remove the output from the worksheet by Edit->Remove Output -> From worksheet, and try again.

I deleted the many pages of cut-and-paste, which was not readable. Instead please upload your worksheet, using the green up-arrow.

It would really help if you gave a very specific example of what you want to start with and what the output would be.

@Carl Love "for k from k" - very efficient; I like it. But "is" seems redundant here.

Not sure entirely about your purpose, but from your title you may want to look at de2diffop.

@Carl Love Thanks. Edge {3,6} is missing on the 2 ohm one - still planar after adding it, 2 ohms between vertices 1 and 15.

@Tokoro No-one solved your first one-ohm problem. As I said before, it seems hard to search for these solutions, so if you have a method, you should share it.

@Tokoro The is a nice answer, and quite unexpected. Across which vertices is the equivalent resistance equal to one? Please post your code creating the graph so we can play with it. All vertex degrees are three except two...

I think that the rules should have: "all currents are different and none of them are zero", otherwise my second answer (and adding a pendent edge to your answer) is stlll valid.

I didn't want to do much work here since it wasn't really clear to me that it wasn't a trick question. But after seeing it I wondered how easy it would be to search for a solution by generating connected graphs with iteratively more vertices, with all combinations of edges to the previous vertices (this generates duplicates and might not be the most efficient). However, after 8 vertices it seemed like it would never get to 9. So the real question is - did you find this with Maple?

@evanhowington Hmm - very interesting, I thought the strict inequality case might be clearer. At least in Maple 15:

is(1+I < 2+I);
FAIL

This is what I would expect:  "<" not defined for complex. Now

is(1+I < 1+I);
false

I would still expect FAIL here - if you can't define it, then you can't answer the question. But it seems that Maple decides that if they are equal, then a decision can be made. Then the same for the symbolic case:

is(x<x) assuming x::complex
false

 

@Earl You're welcome. I think plotting is always useful. But also, as @tomleslie pointed out, letting Maple do the definite integral gives immediately the correct answer - Maple is smart about understanding discontinuities and dealing with them. You can also use value() if you have an unevaluated integral (gray integral sign). For example:. 

int1 := ArcLength(Ellipse, theta = 0 .. 2*Pi, output = integral);
value(int1);

 

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