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Get a real-valued root?

Kevin Huang
Posted 6 years ago

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As you can see,I want to find the real-valued root in Mathematica, what should i do? I tried so hard to search it up online, but in vein.
POSTED BY: Kevin Huang
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Kevin Huang
Kevin Huang
Posted 6 years ago

Thanks everyone for helping.
POSTED BY: Kevin Huang

The 1/5 power in Mathematica is defined as x^(1/5) == Exp[(1/5) Log[x]]. When x is negative, the principal value of the logarithm has I Pi as imaginary part, so that

In[9]:= (-1)^(1/5) == Exp[(1/5) Log[-1]]

Out[9]= True

and

In[10]:= Exp[(1/5) Log[-1]] // N

Out[10]= 0.809017 + 0.587785 I

This issue is an endless source of confusion and frustration for beginners, which had been raised in a real-centered theory of roots.
POSTED BY: Gianluca Gorni
Kevin Huang
Kevin Huang
Posted 6 years ago

Thanks, it works, but is that really the only way to solve it? What if there are something like 1/5 power and the answer still come up with imaginary number?
POSTED BY: Kevin Huang
Hans Milton
Hans Milton
Posted 6 years ago

Then you can use Surd, as Henrik suggested

Surd[7 + Sqrt[50], 5] + Surd[7 - Sqrt[50], 5] // N

CubeRoot is just a special form of Surd. The following will give the same result as if CubeRoot was used, in case of power 1/3:

Surd[7 + Sqrt[50], 3] + Surd[7 - Sqrt[50], 3] // N
POSTED BY: Hans Milton
Hans Milton
Hans Milton
Posted 6 years ago

Another way, using CubeRoot

CubeRoot[7 + Sqrt[50]] + CubeRoot[7 - Sqrt[50]] // N
POSTED BY: Hans Milton
Kevin Huang
Kevin Huang
Posted 6 years ago

Where do you mean I should try Surd with? I mean I use 1/3 power, isn't that the same?
POSTED BY: Kevin Huang

Maybe try Surd instead of Sqrt.
POSTED BY: Henrik Schachner
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