# x^(6/4) != (x^6)^(1/4)? x < 0

Discussion in 'Physics & Math' started by genghis37, Jul 2, 2002.

1. ### genghis37Registered Member

Messages:
8
Technically x^(6/4) should equal (x^6)^(1/4).
But for negative value it seems this is not true.

in the first expression it is mandatory to reduce the exponent to a real number and then raise the quantity x to the real number. According to the calculator any negative quantity raised to 1.5 is undef.

in the second expression it is again mandatory to unravel the expression in the proper order. In which case all real will produce a defined result.

so, then it seems that I can conclude that for x<0, x^(6/4) != (x^6)^0.25
Correct? or is there some sort of math law the defines otherwise.

In any case, if this is correct then it seems to me we cannot conclude the general equation is true: x^(y*z) = (x^y)^z where x < 0

3. ### James RJust this guy, you know?Staff Member

Messages:
30,732
Just because your calculator can't do it, doesn't mean it can't be done. Most calculators, for example, struggle with things like sqrt(-1).

If x<0 then x<sup>1/a</sup> can have some real and some complex roots, or no real roots and some complex roots. Your calculator is only happy if it has real roots.

5. ### genghis37Registered Member

Messages:
8
I think you missed the point. Something only has roots if you are solving an equation. There is no equation to solve here! There are to independent expressions that when evaluated do not equal each other!

To rephrase my argument:
x > 0
f(x) = ((-x)^6)^(1/4)
g(x) = (-x)^(6/4)

evaluate f(2) and g(2):
f(2) = 2.828427125
g(2) = -2.828427125* i

then f(x) != g(x)

Try this out in Maple7:
((-2)^6)^0.25 = (-2)^1.5;
2.828427125 = -2.828427125 I

Try this out in Mathematica:
In[2]:=(-2)^1.5 == ((-2)^6)^(1/4)
Out[2]=False

Mathematics software is not a limitation here!

7. ### James RJust this guy, you know?Staff Member

Messages:
30,732
<i>To rephrase my argument:
x > 0
f(x) = ((-x)^6)^(1/4)
g(x) = (-x)^(6/4)</i>

These expressions are equivalent mathematically. When you evaluate them, the answer you get will depend on the order in which you take the sixth power and the fourth root. In the case of f(x), you specify that the sixth power is taken first, meaning that the expression is equivalent to:

f(x) = (x<sup>6</sup>)<sup>1/4</sup>

Do this in mathematica and if you have a real input, it will give you a real output.

In the case of g(x), I suspect that the fourth root is calculated first. There are 4 separate fourth roots of any negative number, but when you evaluate it with something like Mathematica, chances are it only gives you one of them. That accounts for the apparent discrepancy between the results.

I suggest you try this: Plug in (-2)^(1/4) and see what result you get. My bet is that you'll get the simplest root, which is a multiple of i. The other three roots won't be given.

8. ### genghis37Registered Member

Messages:
8
You are very correct in that the software computes the expression in a different order and that is why it comes out with different results. However, the order in which a expression is written is very important and even after you find a final expression that should be equivalent to the first one, what counts is evaluting the solution in the original function in the order the function was meant to be evaluated.

For example, (5+x)+7 could be changed to 5+(x+7) but it is what the original expression generates when evaluated in the order it gave gave that counts even if 5+(x+7) should always equal the former.

The fact that when the two functions evaluate to a different number based on the order the operations were evaluated makes the functions as different as the following two:
f(x) = (x*2)^2
g(x) = x*(2^2)

Okay, so pretty retarted huh? My point is that in my ap calc class we perform this simplification all the time:
(a^x)^y => a^(x*y)
Which can only be garunteed if a is greater than 0. Unfortunantly we rarely know if a is positive or negative. Which would prohibit us from continuing with the simplification. Unfortunantly, this would probably hinder our effort in finding the solution more than help. So what are all the possibilites that would make the given rule safe to apply (other than a > 0)?

9. ### James RJust this guy, you know?Staff Member

Messages:
30,732
<i>My point is that in my ap calc class we perform this simplification all the time:
(a^x)^y => a^(x*y)
Which can only be garunteed if a is greater than 0.</i>

Not true. That statement is guaranteed in mathematics. I am saying that the reason your computer gives you different answers is due to the failings of the computer software, not the mathematics.

10. ### genghis37Registered Member

Messages:
8
Okay, thanks for your helps JamesR. I went ahead and developed a program that would help me visualize all the different roots of a potential base by solving this equation: a = x^n for arbitrary n and complex number a using demoive's (forgive me if i just hacked that spelling) thereom. It seems like it works and I could really see all the different possible values that could come out of... oh say (3 +4i)^(0.123)

I'm not sure if its entirely correct but I did some cross-checking with mathematica and it seems pretty good. Anyone can get it from genghis.remoteserver.org/genghis/home.htm and try it out if they're interested.