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Home/ Questions/Q 675429
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Editorial Team
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Editorial Team
Asked: May 14, 20262026-05-14T00:49:01+00:00 2026-05-14T00:49:01+00:00

My question is: why doesn’t the following work, and how do I fix it?

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My question is: why doesn’t the following work, and how do I fix it?

Plot[f[t], {t, 0, 2*Pi}] /. {{f -> Sin}, {f -> Cos}}

The result is two blank graphs. By comparison,

DummyFunction[f[t], {t, 0, 2*Pi}] /. {{f -> Sin}, {f -> Cos}}

gives

{DummyFunction[Sin[t], {t, 0, 2 *Pi}],  DummyFunction[Cos[t], {t, 0, 2 * Pi}]}

as desired.

This is a simplified version of what I was actually doing. I was very annoyed that, even after figuring out the annoying “right way” of putting the curly brackets nothing works.

In the end, I did the following, which works:

p[f_] := Plot[f[t], {t, 0, 2*Pi}]
p[Sin]
p[Cos]
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  1. Editorial Team
    Editorial Team
    2026-05-14T00:49:02+00:00Added an answer on May 14, 2026 at 12:49 am

    Mathematica is trying to evaluate Plot before the substitution. You can prevent that with the Hold and ReleaseHold functions:

    ReleaseHold[Hold[Plot[f[t],{t,0,2*Pi}]] /. {{f -> Sin},{f -> Cos}}]
    

    Hold[] will force the entire Plot subexpression to remain unsimplified while the substitution is performed, then ReleaseHold[] will let it proceed with the actual plotting.

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