x^(2/3) (1 Viewer)

[Estel]

When x is negative, the calculator says error...
I do it manually, and take it as cube root of x squared which is fine, throw it in a graphing program and it's fine...

Quick q... which is right?

Edit: Graphing y=x^x x must be an integer when x<0... but how is it different from the above situation?

 

[CM_Tutor]

y = x^2/3 is defined for all x.

 

[Xayma]

Most calculators have problems with some powers.

Eg if you change a power of 1/2 to a power of 0.5 even though they are the same thing it wont compute it if its negative (even if you have it in complex mode where it normally does the sqrt of it). It will come up with an error for all calculators where it cant get the numerator to 1 for negative numbers.

 

[CrashOveride]

Originally posted by Estel
When x is negative, the calculator says error...
I do it manually, and take it as cube root of x squared which is fine, throw it in a graphing program and it's fine...

Quick q... which is right?

Edit: Graphing y=x^x x must be an integer when x<0... but how is it different from the above situation?

My calc encounters no errors. Also, What about if you had x=-4 in your y=x^xx equation. That wouldn't yield an integer, you get the fraction -1/64.

 

[Estel]

You misunderstand:
y=x^x, x must be an integer when x<0.

 

[CM_Tutor]

y = x^x is defined for penty of x values, x < 0, where x is not an integer. Consider x = -1 / 3, for example.

 

[CrashOveride]

Originally posted by Estel
You misunderstand:
y=x^x, x must be an integer when x<0.

Why do you say it must? As CM_Tutor just showed, if u put in -1/3 u dont get an integer. Do you mean to say perhaps when x<0 and is an integer. Well an integer to the power of an integer will always be an integer

 

[Estel]

Alright I'm the confused one.

I put -0.2 does work.
I put -0.4, it doesn't work.

I put in graph calc, dots aren't joined.

hmmm

 

[CrashOveride]

ok i tried -0.4
didnt work for me either

 

[CrashOveride]

You can do it maually.....for -0.4 i get 5th root of -0.4 all squared.

MA Error with the calculator when u plug it str8 in

 

[Estel]

hmm

suppose x = a/b where a/b is negative
y = x^x
= (a/b)^(a/b)
= 1/[(a/b)^(|a/b|)]
= 1/[(|b|rt)(a/b)^|a|]
hence if a is odd b must be odd
and if a is even b can be either.

so x = -0.4 = -2/5 should work then.
Where's the fault in my logic?

Edit: I see. Cursed calculator.

 

[Xayma]

Basically most calcs wont compute negative numbers to negative fractional powers unless the numerator will simplify to 1.

 

[CrashOveride]

Originally posted by Estel
hmm

suppose x = a/b where a/b is negative
y = x^x
= (a/b)^(a/b)
= 1/[(a/b)^(|a/b|)]
= 1/[(|b|rt)(a/b)^|a|]
hence if a is odd b must be odd
and if a is even b can be either.

so x = -0.4 = -2/5 should work then.
Where's the fault in my logic?

Edit: I see. Cursed calculator.

the a/b in the final square root is negative as you defined it. Therefore it doesnt matter what a or b is, the bth root must be odd because you cant have an even root of a negative number.

So if a is even, how can B be either ?

 

[Estel]

If a is even, then if b is even you can simplify, and if b is odd it works either way.
If a is even, (a/b)^a i positive, and you can take the bth root regardles...

I *think*...
i'm totally lost here.

 

[CrashOveride]

Hold on

But uve got to take the bth root first, then everything goes to power A

And since the root is of a negative, b must be odd, always.

 

[Estel]

hmmm
I'm just utterly confused.
Is it a simulatenous process or does one go first?

 

[CrashOveride]

Heh

Dude the power of a applies to that entire bottom expression.

Therefore, if the bth root is even, the expression in the brackets wont work so dont worry about raiisng it to the power of a =p

Is it clear?

 

[Estel]

Ok.
I'll probably get a chance to go over it when I do exponentials or logarithms or something.

 

[CrashOveride]

Mind you, this wasn't either

 

[Estel]

Yeh but we get to review indices as a syllabus dot point in one of those topics.

Quite sad as I've done indices three times in the space of a year and I still don't get things like that.