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I want to evaluate a function of "two" variables using some data points (x,y(x)) (but one of them is related to the other) that I have made with the function Table.

My table of data (x,y(x)) that I have build in Mathematica using Table with a given complicated implicit equation is something like:

data = {{1.35, 3.56}, {1.55, 4.78}, {1.76, 8.89},...}

And imagine that I have the following function:

f[x_, y_] := x*y / (x^2 + y^2)

Now I want to evaluate my list of data points $(x,y(x))$ using my function, and make a plot of that, so plotting actually $f(x)$, because $y(x)$ is not an independent variable. How can that be done?

EXAMPLE:

If I evaluate the list data under the previous function, I will obtain something like

evaluation = {{1.35, 0.331}, {1.55, 0.293}, {1.76, 0.191}}

So precisely I want to do a plot of these points, but in a systematic way of course.

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closed as off-topic by MarcoB, Coolwater, Henrik Schumacher, LLlAMnYP, Karsten 7. May 18 '18 at 3:05

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question arises due to a simple mistake such as a trivial syntax error, incorrect capitalization, spelling mistake, or other typographical error and is unlikely to help any future visitors, or else it is easily found in the documentation." – MarcoB, Coolwater, Henrik Schumacher, Karsten 7.
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ very closely related: 2688. Is ListPlot3D[ {##, f[##]}& @@@ data] what you want? $\endgroup$ – Kuba May 9 '18 at 14:02
  • $\begingroup$ okay, so I have eddited my post. Actually it's not a function f(x,y), it's given by y(x), so at the end the function is only for a given variable x, because the other is related like y(x) by the table of points $\endgroup$ – Joe May 9 '18 at 14:13
  • $\begingroup$ Then isn't data already the table of {x,y}? Use ListLinePlot[data]? $\endgroup$ – eyorble May 9 '18 at 14:17
  • $\begingroup$ no, I have edited the post again for clarify $\endgroup$ – Joe May 9 '18 at 14:28
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    $\begingroup$ Generate the processed data using {#1, f[#1, #2]} & @@@ data then use ListPlot. $\endgroup$ – MarcoB May 9 '18 at 15:15
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A simple solution is define f a little different than the way you show.

f[{x_, y_}] := {x, x y/(x^2 + y^2)}

With this definition,

data = {{1.35, 3.56}, {1.55, 4.78}, {1.76, 8.89}};
f /@ data

gives

{{1.35, 0.331537}, {1.55, 0.293415}, {1.76, 0.190508}}

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