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Just out of curiosity I was trying to solve the equation $x^2=2^x$, initially I thought there would be just the two solutions $x=2$ and $x=4$, but wolfram shows that the two equations intersect at not 2 but 3 locations, the third being a negative value of $x$. The third solution isn't obvious like the other two, so I just have a few questions about the negative solution. Is it rational? is it commonly represented with a greek letter? If it is irrational is there a way to approximate it?

user217339
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  • It might help to note that the (negative of the) solution you seek is the positive solution to $x=2^{-x/2}$. – Antonio Vargas Nov 24 '15 at 18:06
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    $-\frac{23}{30}$ is a very good approximation. – Peter Nov 24 '15 at 18:07
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    Well $x^2$ is always positive so if there is a negative solution it would be x = -y where y is positive and $y^2 = 1/2^y$. So y < 1. If g(x) = $y^2 - 1/2^y$ then g(x) is continuous. g(0) = -1. g(1) = 1/2. So there is a solution in there. – fleablood Nov 24 '15 at 18:11
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    Definitely not rational as it that would imply 2 (or 1/2) would have rational roots. – fleablood Nov 24 '15 at 18:13
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    @fleablood by a negative solution, I meant the negative value of x for which $x^2=2^x$ – user217339 Nov 24 '15 at 18:16
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    Yes, I know. Why do you point that out? – fleablood Nov 24 '15 at 18:18
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    A negative solution to $x^2 = 2^x$ would be the negative equivalent to a positive solution to $y^2 = 2^{-y}$. – fleablood Nov 24 '15 at 18:20
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    $2^2 = 4$ $(1/2)^2 = 1/4$. That is not a solution. At least not to my equation. It is to $x^2 = 2^x$ but we already knew that. We are looking for a negative solution to that. – fleablood Nov 24 '15 at 18:23
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    @Peter what argument can't be valid without restrictions? – fleablood Nov 24 '15 at 18:26
  • @user217339 It shows once again that making a proper graph to start out with, can do miracles. It clearly show 3 solutions... – imranfat Nov 24 '15 at 18:26
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    @peter The negative solution can not be rational. If x = -y is the negative solution then $y^2 = (1/2)^y$ where y is positive. y can not be 1 or greater as that'd make $y^2 \ge 1$ while $(1/2)^y <\le 1/2$. So if y is less than 1 and rational it mean a root of 2 is rational, which is impossible. (All solutions must be irrational or integers, by the way. The positive solutions are greater than 1 and happen to be integers. The negative solution is between 0 and -1 so can not be an integer.) – fleablood Nov 24 '15 at 18:33
  • OK, so you only looked at the negative solution. – Peter Nov 24 '15 at 18:36
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    @peter. NO!!! I looked at both positive solutions. They were in the OP for !!!!'s sake! So of course I looked at them! The OP was asking about the negative solutions! so I was hypothesizing about the strictly negative solution. I wasn't "not looking at" the positive solutions. The positive solutions are known so there is nothing to say about them. – fleablood Nov 24 '15 at 18:42
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    If you look for a "standard notation" for the negative solution, you might want to check out the Lambert W function. Using it, the negative solution is something like $x=-2W((\ln 2)/2)/\ln 2$. – mickep Nov 24 '15 at 18:45
  • Is there a name for this number or it part of a certain set of numbers, e.g. negative solutions to $a^x=x^a$, for $a\beq 2$ – Andrew Brick Nov 24 '15 at 19:30
  • @peter was that approx calculated by simply narrowing the range (-1, 0) to get closer to the solution – Andrew Brick Nov 24 '15 at 19:36

2 Answers2

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Suppose , $gcd(a,b)=1$ , $x=\frac{a}{b}$ and $x^2=2^x$

We have $$a^2=b^2\times 2^{a/b}$$

implying

$$a^{2b}=b^{2b}\times 2^a$$

This is impossible, if $gcd(a,b)=1$ and $b>1$. The negative solution is obviously not an integer.

If $x$ is irrational algebraic, then $2^x$ is transcendental, but $x^2$ is not.

So, $x$ , the negative solution, must be a transcendental number.

Peter
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So we have $x^2 = 2^x$. Taking the square root of both sides and assume the solution is negative gives $x=-\sqrt{2}^x$. We can then establish a recursive sequence, $x_n = -\sqrt2^{x_{n-1}}$. Assuming that this converges gives us the answer, $x=-\sqrt2 ^{-\sqrt2 ^ {-\sqrt2 ^\cdots}}$. After five iterations, we get $$x\approx-0.76961847524.$$ Substituting the answer back in we get, $$2^{-0.76961847524} = 0.58657257487 \approx 0.59231259743 =(-0.76961847524)^2.$$

Calvin Khor
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Trevor Norton
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