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| bio | website | |
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| location | ||
| age | ||
| visits | member for | 1 year, 3 months |
| seen | Apr 28 at 19:30 | |
| stats | profile views | 54 |
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Jan 28 |
accepted | Why doesn't FullSimplify get rid of the common factor in this expression? |
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Jan 28 |
comment |
Why doesn't FullSimplify get rid of the common factor in this expression? All these suggestions work. I'm not sure how to judge this, but the one involving Cancel seems like the most direct. As for an explanation for this strange behavior, I guess I shouldn't hold my breath. @PinguinDirk - do you want to turn your comment into an answer so I can accept it? |
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Jan 27 |
asked | Why doesn't FullSimplify get rid of the common factor in this expression? |
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Jan 26 |
awarded | Yearling |
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Dec 7 |
awarded | Nice Question |
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Nov 15 |
comment |
An elegant way to plot a numeric function that returns a list, and have each element in a different color By using ListPlot you're not taking advantage of the adaptive sampling algorithm of Plot. |
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Nov 15 |
awarded | Critic |
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Nov 15 |
accepted | An elegant way to plot a numeric function that returns a list, and have each element in a different color |
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Nov 15 |
comment |
Plotting the components of a function that returns a list in different colors without redundant evaluations of the function Another related SO question: "Telling Plot to style vector-valued black-box functions in Mathematica" |
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Nov 15 |
comment |
An elegant way to plot a numeric function that returns a list, and have each element in a different color @Sasha - in the question you linked to the problem is to avoid redundant evaluations of the function, since it is expensive to evaluate (just like in this question, which is already mentioned above). In my question I don't mind redundant evaluations, my problem is coding style. These are two different issues. |
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Nov 14 |
asked | An elegant way to plot a numeric function that returns a list, and have each element in a different color |
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Oct 31 |
accepted | Why doesn't Mathematica simplify a square root of an expression that equals a square of a positive real? |
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Oct 31 |
comment |
Why doesn't Mathematica simplify a square root of an expression that equals a square of a positive real? Great idea! It can be a bit simplified by using just one replacement rule: x_^2 + 2 y_^2 + 2 y_ Sqrt[x_^2 + y_^2] -> (Sqrt[x^2 + y^2] + y)^2. |
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Oct 31 |
comment |
Why doesn't Mathematica simplify a square root of an expression that equals a square of a positive real? Nice! Then I just add /.{r -> Sqrt[x^2 + y^2], t -> ArcTan[x, y]} and simplify again and I have the simplified expression in my original variables. On the other hand, I have this expression appearing multiple times with different variables instead of x and y, and I'm not sure if your approach can be generalized to such a case. |
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Oct 25 |
accepted | Is Abs[z]^2 a bad way to calculate the square modulus of z? |
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Oct 25 |
comment |
Is Abs[z]^2 a bad way to calculate the square modulus of z? Awesome analysis! By slightly adjusting your code, I plotted the difference in the error between the functions Abs[z]^2 and Re[z]^2+Im[z]^2 when they are applied to the same argument, and indeed for most of the points you used Abs[z]^2 has a smaller error! This is really weird and counterintuitive to me, but at least now I know. |
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Oct 25 |
comment |
Why doesn't Mathematica simplify a square root of an expression that equals a square of a positive real? Since I am trying to simplify a very long expression that contains multiple versions of the expression I'm asking about, and I don't want to go through my long expression manually. |
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Oct 25 |
comment |
Is Abs[z]^2 a bad way to calculate the square modulus of z? Please do feel free to edit my posts, this was never an issue for me! (I corrected the mistakes, thank you :). |
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Oct 25 |
revised |
Is Abs[z]^2 a bad way to calculate the square modulus of z? corrected spelling |
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Oct 25 |
asked | Is Abs[z]^2 a bad way to calculate the square modulus of z? |
