Replacing an arbitrary function of a variable

Question

If I do something like

1/Sqrt[x] /. {Sqrt[x] -> a}

it does not lead to an expected result, but can be fixed with

1/Sqrt[x] /. {Sqrt[x] -> a, 1/Sqrt[x] -> 1/a}

I am wondering if, for given g[x] and h[x], there exists a general way of replacing any f[g[x]] with f[h[x]]?

You have to see FullForm of your expression to write a working replacement rule. — Daniel Huber, Apr 11 '22 at 20:30
Sometimes results are better replacing a single variable rather than an expression. 1/Sqrt[x] /. {x -> a^2} // PowerExpand. — Bill Watts, Apr 11 '22 at 20:45
Possible duplicate: https://mathematica.stackexchange.com/a/29219/4999 -- the question at the end seems to have no relation to the implicit question in the beginning examples. — Michael E2, Apr 12 '22 at 03:54

score 5 · Accepted Answer · answered Apr 11 '22 at 20:35

5

{Sqrt[x], 1/Sqrt[x]} // FullForm

List[Power[x,Rational[1,2]],Power[x,Rational[-1,2]]]

Try:

rule = Power[x, Rational[g_, 2]] -> a^Sign[g]
{Sqrt[x], 1/Sqrt[x]} /. rule

{a, 1/a}

For your other query:

f[g[x]] /. g -> h

f[h[x]]

answered Apr 11 '22 at 20:35

Syed

Thank you. Note that, strictly speaking, my second question is the generalization of the first one, yet I cannot do smth like Sqrt -> ... (say, for replacing Sqrt[x] with some constant a) – mavzolej Apr 11 '22 at 20:38
2

Sqrt gets translated as above using Fullform. Replacements work mechanically and don't interpret user's intention so a literal rule is needed. – Syed Apr 11 '22 at 20:39

bmf · Answer 2 · 2022-04-11T20:38:09.280

2

Extended comment

1/Sqrt[x] // FullForm

The following works

expr = 1/Sqrt[x];
Numerator[expr]/(Denominator[expr] /. Sqrt[x] -> a)

The general replacement an be done as follows:

fun = f[g[x]] + D[f[g[x]], x]
fun /. g -> Function[{x}, a^2 h[x]]

edited Apr 11 '22 at 20:38

answered Apr 11 '22 at 20:32

bmf

2 Answers2