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How to reverse square rooting ?

Posted 3 months ago

I am new to Alpha and Mathematica.
I would like to go from sqrt(x^9/2) to get x^9/4. Is it possible to do it in Alpha or Mathematica?
Basically I want go from expression with square root and fractional power exponent to a number with fractional power exponent but without square root and the other way around.
How do I do it?

POSTED BY: Jacek Jacek
7 Replies

It is worth mentioning that PowerExpand assumes the variables are positive (real numbers). If the variable are not always positive, then the transformations made by PowerExpand will be invalid.

PowerExpand is more or less equivalent in this case:

Simplify[Sqrt[x^(9/2)], x > 0]
(* x^(9/4) *)

The second argument specified the assumption that x is positive (and a real number). In this way, we specify exactly which variables are positive.

Another difference is that PowerExpand expands powers and logarithms. Simplify tries to contract the expression to an equivalent expression with the smallest expression tree. Expansion and contraction are sometimes in opposition to each other.

POSTED BY: Michael Rogers
Posted 3 months ago

Gianluca has shown that PowerExpand does what you want. Another way to do it is with Simplify, with its second argument restricting the base to be a positive real number:

Simplify[Sqrt[x^(9/2)], x > 0]

(* x^(9/4) *)
POSTED BY: Hans Milton
Posted 3 months ago

Thank you very much to you ALL for kind repies in this topic, it was helpful. best wishes, Jacek

POSTED BY: Jacek Jacek

To reverse square rooting, you need to square the expression. In Mathematica, you can do this by using the Power function. Here's how you can go from Sqrt[x^9/2] to x^9/4:

expr1 = Sqrt[x^9/2]

This creates the expression Sqrt[x^9/2]. To reverse the square root, you square the expression:

expr2 = expr1^2

This will give you the expression x^9/2. Finally, if you want to change the exponent to 1/4, you can raise the expression to the power of 1/4:

expr3 = expr2^(1/4)

This will yield x^9/4, which is the result you desire.

POSTED BY: Ankit Kantheti
Posted 3 months ago

Thank you very much indeed, I have tried:


but it did not get me what I wanted. Can you please elaborate a bit on this:

Beware that in the complex plane Sqrt[x^(9/2)] is not equivalent to x^(9/4)

Why is that ?

Cheers, Jacek

POSTED BY: Jacek Jacek

By definition

Sqrt[x] == Exp[1/2 Log[x]]

where Log[x] is the principal value of the multivalued complex logarithm. More generally,

x^p == Exp[p*Log[x]]

again with the principal logarithm. Then

Sqrt[x^(9/2)] == Exp[1/2 Log[Exp[9/2 Log[x]]]]


x^(9/4) == Exp[9/4 Log[x]]

Make the calculation with x == I and you will see that the results are different.

POSTED BY: Gianluca Gorni

For that you can use PowerExpand:


Beware that in the complex plane Sqrt[x^(9/2)] is not equivalent to x^(9/4):

FindInstance[Sqrt[x^(9/2)] != x^(9/4), x]

This is the reason why Mathematica does not "simplify" Sqrt[x^(9/2)] automatically.

POSTED BY: Gianluca Gorni
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