X^x = -2

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  • #1
Alkatran
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What's the answer to this? Obviously x is not a real number, since if x is within R, the range is from -1 to positive infinity. Is it imaginary?

This is one of those questions that's been bugging me for a long time... as well as things like

x^^x = -2 (x^x^x^...x times...^x^x)
 

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  • #2
Tide
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x is approximately equal to: -0.1222025964 + 1.182348981 i

:-)
 
  • #3
Alkatran
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Tide said:
x is approximately equal to: -0.1222025964 + 1.182348981 i

:-)

Awesome, how'd you solve it? ... I get the feeling I'm going to hear 'Euler'.

Nothing on x^^x = y?
 
  • #4
shmoe
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Alkatran said:
x^^x = -2 (x^x^x^...x times...^x^x)

If x is not a positive integer, this makes no sense. What's i^^i? 1.23^^1.23? (-2)^^(-2)?

If x is a positive integer, then you'll never get -2.
 
  • #5
Tide
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Oops! I discovered an error!

This is how I approached it: take the natural log of both sides so z ln z = ln(-2) then substitute [itex]z = r e^{i \theta}[/itex], equate real and imaginary parts then solve numerically. On my first go around I mistakenly wrote -2 on the RHS instead of the log. Of course there will be an infinite set of solutions depending on which branch you choose for the logs.

Regarding your other problem - hey, I haven't had breakfast yet! ;-)
 
  • #6
Alkatran
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shmoe said:
If x is not a positive integer, this makes no sense. What's i^^i? 1.23^^1.23? (-2)^^(-2)?

If x is a positive integer, then you'll never get -2.

I would say it's probably similar to 1.23^1.23: You convert it to a fraction:

1.23^1.23 = 1.23^(123/100)
That's 1.23 to the power of 123, then taking the hundreth root. As for i, x^i is a bit hard to get your head around, too...
 
  • #7
shmoe
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Alkatran said:
I would say it's probably similar to 1.23^1.23: You convert it to a fraction:

1.23^1.23 = 1.23^(123/100)
That's 1.23 to the power of 123, then taking the hundreth root. As for i, x^i is a bit hard to get your head around, too...

Maybe I'm missing the obvious, but I fail to see how that will help. Please, find x^^x for any non-positive integer for me. Try something rational first, like 123/100 if you like. I'd like to know what "123/100 times" translates to in your definition of x^^x.

For that matter, find x^^x for the first few positive integers. It's unclear to me the order you are exponentiating. is 3^^3=3^(3^3) or is 3^^3=(3^3)^3?
 
  • #8
Alkatran
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shmoe said:
Maybe I'm missing the obvious, but I fail to see how that will help. Please, find x^^x for any non-positive integer for me. Try something rational first, like 123/100 if you like. I'd like to know what "123/100 times" translates to in your definition of x^^x.

For that matter, find x^^x for the first few positive integers. It's unclear to me the order you are exponentiating. is 3^^3=3^(3^3) or is 3^^3=(3^3)^3?

I think the standard is 3^(3^3)

The question I was asking was about the negative numbers, how can I answer what I don't know (without looking for a long time...)?
 
  • #9
shmoe
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Alkatran said:
I think the standard is 3^(3^3)

The question I was asking was about the negative numbers, how can I answer what I don't know (without looking for a long time...)?

Yes, I saw your question. I maintain that your function is not even definied when x is not a positive integer.

If the domain of x^^x is only the positive integers then there are no solutions to x^^x=-2

If you want to insist that it's defined for non-positive integers, then I ask again for you to show me how to find any of i^^i, 1.23^^1.23, (-2)^^(-2). If you can't even begin to do this, you need to go back and re-examine your question.
 
  • #10
Alkatran
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shmoe said:
Yes, I saw your question. I maintain that your function is not even definied when x is not a positive integer.

If the domain of x^^x is only the positive integers then there are no solutions to x^^x=-2

If you want to insist that it's defined for non-positive integers, then I ask again for you to show me how to find any of i^^i, 1.23^^1.23, (-2)^^(-2). If you can't even begin to do this, you need to go back and re-examine your question.

People used to say that the square root was undefined for non positive numbers.

assuming similar rules to exponentials:
1.5^^1.5 = 1.5^^(3/2) = (1.5^^3)^^(1/2) = (1.5^1.5^1.5)^^(1/2)
=~ 2.106^^(1/2) =~~ 1.6

I'm sorry I didn't use 1.23, I just didn't want to calculate 1.23^1.23^1.23 ... 100 times. The number would be massive.
 
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  • #11
shmoe
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Alkatran said:
assuming similar rules to exponentials:
1.5^^1.5 = 1.5^^(3/2) = (1.5^^3)^^(1/2) = (1.5^1.5^1.5)^^(1/2)
=~ 2.106^^(1/2) =~~ 1.6

How did you get 2.106^^(1/2)=~~1.6?

Also, is the two "~" you use signifigant? If so, what' does it mean?
 
  • #12
Alkatran
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shmoe said:
How did you get 2.106^^(1/2)=~~1.6?

Also, is the two "~" you use signifigant? If so, what' does it mean?

I got that one via trial and error on the windows calculator. The double ~~ just means approximately approximately... it's only accurate to maybe one decimal place.
 
  • #13
shmoe
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Alkatran said:
I got that one via trial and error on the windows calculator. The double ~~ just means approximately approximately... it's only accurate to maybe one decimal place.

I'm going to have to ask you to elaborate please. What equation are you putting into the calculator that justifies your conclusion? In other words, what functions of the calculator are you using in your trial and error? I really don't see how you are relating ^^(1/2) back to the usual operations.
 
  • #14
Alkatran
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shmoe said:
I'm going to have to ask you to elaborate please. What equation are you putting into the calculator that justifies your conclusion? In other words, what functions of the calculator are you using in your trial and error? I really don't see how you are relating ^^(1/2) back to the usual operations.

Note the: "Assuming similar to exponentials" at the top.

if x^2 = y then y^(1/2) = x
so I assumed if x^^2 = y then y^^(1/2) = x, or, if x^x = y then y^^(1/2) = x

So I just tried values for x in the equation x^x. 1.6 came close to the answer I wanted (2.106) so I used it.
 
  • #15
shmoe
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To be totally clear, since

1.5^^3=1.5^(1.5^15)=~2.106..=~~2.12..=~1.6^1.6=1.6^^2

you are asserting that

1.5^^(3/2)=1.5^^1.5=~~1.6

You could be much more accurate here,

1.5^(1.5^15)=~2.10620...=~1.595152^^2

to get 1.5^^(3/2)=~~1.595

and you can maybe agree this is accurate to more than one decimal place? If that's correct, then since

1.5^(1.5^(1.5^(1.5^(1.5^1.5))))==~2.8604..=~1.552561^(1.552561^(1.552561^1.552561)))

I'm going to assert that

1.5^^6=~1.552561^^4

or

1.5^^1.5=1.5^^(6/4)=~1.552561

If you agree this is accurate to more than 1 decimal place, you have some serious problems.

If you don't agree, that I can probably come up with a more convincing argument that your method to extend ^^ to rationals is not well-defined.
 
  • #16
robphy
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shmoe said:
Yes, I saw your question. I maintain that your function is not even definied when x is not a positive integer.

If the domain of x^^x is only the positive integers then there are no solutions to x^^x=-2

If you want to insist that it's defined for non-positive integers, then I ask again for you to show me how to find any of i^^i, 1.23^^1.23, (-2)^^(-2). If you can't even begin to do this, you need to go back and re-examine your question.

[tex]
\begin{align*}
i^i
&=(e^{i\frac{\pi}{2}})^i\\
&=e^{i^2\frac{\pi}{2}}\\
&=e^{-\frac{\pi}{2}}\\
&\approx 0.2078795
\end{align*}

[/tex]
 
  • #17
shmoe
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Here'ss a more convincing argument avoiding approximations:

If you can treat ^^ like normal exponents, then

2^^(3/2)=y if and only if

16=2^(2^2)=2^^3=y^^2=y^y

so clearly y<3 since 3^3=27.


Now 2^^(6/4)=2^^(3/2)=y. Therefore:

2^^6=y^^4

but 2^^6=2^(2^65536)

compare to 3^^4=3^(3^27)

Now log(3)<log(4) so log(3)/log(2)<2

So 65535/27>log(3)/log(2). Rearrage and take exponents to get 2^65535>3^27, rearrange a little more and multiply by 2 to get 2^65536/3^27>2>log(3)/log(2).

Rearrange 2^65536/3^27>log(3)/log(2) and exponentiate to get 2^^6=2^(2^65536)>3^(3^27)=3^^4.

Thus we see y>3, so this is not well defined on the rationals.



robphy-look closely, the double "^^" was not a typo.
 
  • #18
BobG
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Alkatran said:
What's the answer to this? Obviously x is not a real number, since if x is within R, the range is from -1 to positive infinity. Is it imaginary?

This is one of those questions that's been bugging me for a long time... as well as things like

x^^x = -2 (x^x^x^...x times...^x^x)

There is no number, real or imaginary, that satifies your equation, since a negative exponent is equivalent to a positive exponent in the denominator i.e. - -1^(-1) is the same as 1/[(-1)^1].

But, in general, x^x = n, where n < 0 does not necessarily mean x is a complex or imaginary number. It just means the graph of x^x will be discontinous (a few selected values of x have a real solution, a large number have complex solutions).

For example, (-1/3)^(-1/3) has a real solution of around -1.44. By the way, I think this is the peak absolute value for negatives, which is why there is no solution to your problem.

(-2/3)^(-2/3) would also have a real solution of about 1.31.

The real solutions (when they exist) cycle back and forth across zero, depending upon whether the denominator is even or odd.
 
  • #19
Alkatran
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BobG said:
There is no number, real or imaginary, that satifies your equation, since a negative exponent is equivalent to a positive exponent in the denominator i.e. - -1^(-1) is the same as 1/[(-1)^1].

But, in general, x^x = n, where n < 0 does not necessarily mean x is a complex or imaginary number. It just means the graph of x^x will be discontinous (a few selected values of x have a real solution, a large number have complex solutions).

For example, (-1/3)^(-1/3) has a real solution of around -1.44. By the way, I think this is the peak absolute value for negatives, which is why there is no solution to your problem.

(-2/3)^(-2/3) would also have a real solution of about 1.31.

The real solutions (when they exist) cycle back and forth across zero, depending upon whether the denominator is even or odd.

So are we going to need some new constant to solve x^x for negatives, similar to how we needed one to solve x*x for negatives?
 
  • #20
shmoe
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Alkatran said:
So are we going to need some new constant to solve x^x for negatives, similar to how we needed one to solve x*x for negatives?

There are complex answers. You can express them in terms of the Lambert W function, which is a multi-branched function giving infinitely many complex solutions and satisfies W(x)e^W(x)=x

The principle branch in Maple spits out ~1.99729+1.74769i, for x^x=-2
 

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