Advanced Search

Question: Simplifying Radicals: Part III

Question:Simplifying Radicals: Part III

H-R 155 Maple 2015
radical simplify assuming
February 06 2016
0

In this question, I asked for a way to simplify an expression containing radicals. The discussion led us to that as default field for simplicfication is the Complex number system we should use assume or assuming command to simplify the radicals. However, the mothod suggested there seems to not work in this new case that I have. For details please see the attached file. The terms sqrt{u} and sqrt{u-1} should cancel in denominator.

 What Maple Does

> 

restart
> 

`ϕ` := (1+sqrt(5))*(1/2)

1/2+(1/2)*5^(1/2)

 (1)
> 

f := (1/2)*sqrt(-(u-1)*(u+1)*(u^2-u-1))*u*(4*u-3)/sqrt(u*(u-1))

(1/2)*(-(u-1)*(u+1)*(u^2-u-1))^(1/2)*u*(4*u-3)/(u*(u-1))^(1/2)

 (2)
> 

`assuming`([combine(f)], [1 < u and u < `&varphi;`])

(1/2)*u*(4*u-3)*((u+1)*(-u^2+u+1)/u)^(1/2)

 (3)
> 

`assuming`([simplify(f)], [1 < u and u < `&varphi;`])

(1/2)*(-u^2+u+1)^(1/2)*(u^2-1)^(1/2)*u^(1/2)*(4*u-3)/(u-1)^(1/2)

 (4)
> 

`assuming`([combine(f, radical)], [1 < u and u < `&varphi;`])

(1/2)*u*(4*u-3)*((u+1)*(-u^2+u+1)/u)^(1/2)

 (5)
> 

`assuming`([simplify(f, radical)], [1 < u and u < `&varphi;`])

(1/2)*((u-1)*(u+1)*(-u^2+u+1))^(1/2)*u*(4*u-3)/(u*(u-1))^(1/2)

 (6)
> 

``

Radical.mw

 Remark by Markiyan Hirnyk. The below content is added by the questionner on 08.02.2016 .

What Mathematica Does

 
Please Wait...