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  • $\begingroup$ Thank you very much, Dominic. Can I ask you one more thing: How can I convert a function like $x^{1/2} + x^{1/3}$ to its algebraic form? $\endgroup$ Commented Nov 24, 2019 at 3:43
  • $\begingroup$ That's $w^6=x(x^3+x^2)^6$. However there are limits to simple algebraic mainpulation. In those cases, would need to construct analytically-continuous version of each multivalued part of the expression, for example, above, letting $w_1=x^{1/2}$ and $w_2=x^{1/3}$ and solving two ivps and then adding them. $\endgroup$ Commented Nov 24, 2019 at 9:01
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    $\begingroup$ @Hayashi and Dominic: you can use GroebnerBasis[] for that, e.g. Simplify[First[GroebnerBasis[{w == (x + x^(1/3))^(1/3)}, {x, w}]]] yields -w^9 + x + 3 w^6 x - 3 w^3 x^2 + x^3. Due to the nature of the method, it might take a while if the function to be "algebraicized" is fairly complicated, but it mostly works well. On that note, Simplify[First[GroebnerBasis[{w == x^(1/2) + x^(1/3)}, {x, w}]]] gives a result that is quite different from Dominic's comment. $\endgroup$ Commented Nov 29, 2019 at 23:01
  • $\begingroup$ Thanks for pointing that out to me J.M. I made a mistake with exponents above. It's more complicated and I don't see how to convert it to the GroebnerBasis with simple algebraic manipulations by hand. $\endgroup$ Commented Nov 30, 2019 at 13:52