network profile
| bio | website | facebook.com/paterz1 |
|---|---|---|
| location | Montreal, Canada | |
| age | 22 | |
| visits | member for | 10 months |
| seen | Nov 25 '12 at 10:39 | |
| stats | profile views | 26 |
I completed my second year of studies in mathematics at University of Montreal. My current interests are number theory, analysis, measure theory and abstract algebra.
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Sep 21 |
awarded | Custodian |
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Sep 10 |
accepted | Trying to plot ugly expression, not working |
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Sep 10 |
comment |
Trying to plot ugly expression, not working I just saw that the typo was also in my question. It was really just a transcription mistake though because I've always been able to plot the first equation, it was with the second one I had trouble. And you helped! Thank you again. +1 & check! |
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Sep 10 |
revised |
Trying to plot ugly expression, not working added 12 characters in body |
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Sep 10 |
comment |
Trying to plot ugly expression, not working I found the typo in your code! The curves now perfectly match. The curve in brown is actually the original one, the right line for 'equation' was equation =
ComplexExpand[
Abs[A[z] + (3 z + z^2)/A[z] + z] -
Abs[rho A[z] + rho^2-----> (3 z + z^2)/A[z]<------ + z] /. z -> x + I y,
TargetFunctions -> {Re, Im}]; |
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Sep 10 |
comment |
Trying to plot ugly expression, not working I don't know why in case of equation2 you felt the need to use the command 'TargetFunctions->{Re,Im}'? I must admit I don't know what it does, but I don't see how what you're trying to do is any different from the 'equation' case, where you just wrote 'equation==0' in the end. |
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Sep 10 |
comment |
Trying to plot ugly expression, not working Sure. I'll give it a try and look what happens. But you definitely don't have the right expression; that curve in brown down there is not natural in the problem. Maybe I typed something wrong when entering the question in Mathematica. I'll double check |
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Sep 10 |
awarded | Commentator |
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Sep 10 |
comment |
Trying to plot ugly expression, not working I really don't understand why you added those two lines though! equation /. {x -> 1.2, y -> 3.4} equation2 /. {x -> 1.2, y -> 3.4} |
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Sep 10 |
comment |
Trying to plot ugly expression, not working The picture in purple is quite similar to the one I have, except mine was way smoother. The equation you have in brownish looks like the conjugate of the equation in purple. I expected this to happen, but I didn't know about the ComplexExpand command. |
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Sep 10 |
comment |
Trying to plot ugly expression, not working @belisarius : Sorry yes, I just typed it on MSE a little too fast. But I did put the x + Iy when I tried. I know the expression is a bit ugly but it's just something of the form f(x) = conjugate( f(x) ) rho^2, so if you just remember that f(x) = A[x]...+(3x+x^2)/A[x], I tried something like ContourPlot[F[x+Iy] == Conjugate[F[x+I y]] rho^2, {x,-3,3},{y,-3,3}], and I tried with+without using a function F or the actual expression. |
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Sep 10 |
revised |
Trying to plot ugly expression, not working added 96 characters in body |
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Sep 10 |
comment |
Trying to plot ugly expression, not working I guessed that ContourPlot[A[x] Conjugate[x] + Conjugate[(3x+x^2)/A[x]] x +
Conjugate[A[x]] (3x + x^2)/A[x] == Conjugate[A[x] Conjugate[x] + Conjugate[(3x+x^2)/A[x]] x +
Conjugate[A[x]] (3x + x^2)/A[x]] rho^2
,{x,-3,3},{y,-3,3}] should have worked, but unfortunately it didn't. |
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Sep 10 |
comment |
Trying to plot ugly expression, not workingContourPlot[
Abs[H[x + I y]] == Abs[J[x + I y]], {x, -3, 3}, {y, -3, 3}]
Where
H[x_] := AA[x] + (3 x + x^2)/AA[x] + x
and
J[x_] := rho AA[x] + rho^2 (3 x + x^2)/AA[x] + x.
This is the one that worked (using the initial equations, not the re-worked up ones). All my other attempts with the new equation failed, I tried a few using ContourPlot. |
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Sep 10 |
comment |
Trying to plot ugly expression, not working @R.M : It is utterly simple, but it's really huge, because one needs to expand two products and each product is a product of two factors with three terms, so you end up adding 18 things. But it simplifies really well, you just need to know that (1-rho^2)/(1-rho) = rho+1 = -rho^2 and everything follows naturally. I really considered seriously the answer to your question. If you don't trust me, feel free to check it yourself, but re-transcribing it might end up in transcription mistakes more than anything else. |
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Sep 10 |
awarded | Editor |
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Sep 10 |
revised |
Trying to plot ugly expression, not working added 64 characters in body |
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Sep 10 |
reviewed | Approve suggested edit on Trying to plot ugly expression, not working |
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Sep 10 |
asked | Trying to plot ugly expression, not working |
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Aug 2 |
comment |
Finding real roots of negative numbers (for example, $\sqrt[3]{-8}$) Was very useful to me. +1 |
