What is the third zero of f(x) = x^2 - 2^x?

Question

The positive zeros are obviously 2 and 4. But there is a third root as well, which is approximately equal to -0.7666647. However, I have not been able to obtain a closed-form expression for this number. The closest I have gotten is a proof that it is irrational.

Is there a closed-form expression for this number?

If not, why?

Answer 1

There is no closed form expression.

I don't know which proof you have that x is irrational, but I'll prove it my way ;)

Suppose x = p/q, then the equation becomes

(p/q)^2 = 2^(p/q)
(p/q)^2q = 2^p
p^(2q) = 2^p * q^(2q)

It is easy to see that p and q must have the same prime factors, except for the prime factor 2. We conclude p = 2^n * q with n integer.

The equation now becomes

(2^n * q)^(2q) = 2^(2^n * q) * q^(2q)
(2^n)^(2q) = 2^(2^n * q)
2^(2nq) = 2^(2^n q)
2nq = 2^n q
2n = 2^n

Integer solutions are n = 1 and n = 2, giving the solutions you already found.

Answer 2

There is a expression for this:

x^(2/x) = 2 or x^(1/x) = sqrt(2)

provided x is a positive integer.

If x=2 then we have 2^1 =2
If x=4 then we have sqrt(4) = 2
x cannot equal to -0.7666647

Of course not every value of x will work.

Sherman81:
How can you say that x = -0.7666647 is valid? You cannot compute the logarithm of a negative value. In your equation:

x/ln x = 2/ln 2

ln x is undefined when x is negative or zero.

Answer 3

do a search in answers.yahoo.com for x^2-2^x. you will find that this is not the first time this question has been posted. just slight variations in the wording. The consensus seems to be a numeric approach.

let x (lower case) be the guess; Let X be the next approximation.
If you rewrite your expression you can get this:

X = -sqrt(2^x)

start with x = 0; then X will be -1
then let x = X
and start over again.

it converges very quickly to -0.766664689

Answer 4

f(x) = x^2 - 2^x
0 = x^2 - 2^x
2^x = x^2

xln2 = 2lnx
x/lnx = 2/ln2
as you can tell 2 is the answer for this one

x = 2, 4, -.766665

Answer 5

No