Thanks!
El martes, 23 de noviembre de 2021 a las 21:06:15 UTC+1, slelievre escribió:
> 2021-11-23 19:42:12 UTC, Juan Luis Varona:
> >
> > In the expression (5^x)^2-7*5^x+4, I want to substitute x^5 by t.
> >
> > With sagemath 7.2 (or another old versions), I can do
> > ((5^x)^2-7*5^x+4).subs(5^x==t)
> > and I get t^2 - 5*t + 4
> >
> > But sagemath 9.4 does not change the first 5^x and he gives
> > 5^(2*x) - 7*t + 4
> >
> > Why?
> >
> > (In both cases, var("t") has been previously used)
> >
> > Yours,
> > Juan Luis Varona
>
> In recent versions of Sage, defining:
> ```
> sage: t, x = SR.var('t, x')
> sage: a = (5^x)^2-7*5^x+4
> ```
> automatically groups exponents and gives:
> ```
> sage: a
> 5^(2*x) - 7*5^x + 4
> ```
> in which `5^x` is only seen once as such (old versions
> of Sage possibly did not group exponents, thus keeping
> two visible occurrences of `5^x` in the resulting `a`).
>
> This means that only one `5^x` gets replaced
> by `t` when we do the following substitution:
> ```
> sage: aa = a.subs(5^x == t)
> sage: aa
> 5^(2*x) - 7*t + 4
> ```
>
> To work around this, we can instead think of
> rewriting `x` as `log(t, 5)` as in the following
> substitution, which gives the expected result:
> ```
> sage: ab = a.subs(x == log(t, 5))
> sage: ab
> t^2 - 7*t + 4
> ```
>
> Now we have a polynomial expression in t and
> we can use corresponding tools. --Samuel
>
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