Starting from the raising power operator, defined between two numbers x and y over a number field, xy is not the same as yx, due to the missing commutativity property respect than the other sum and multiplication rank binary operators.
So this post is questioning for wich value of x in natural, integers, reals (or complex) number fields this identity is verified:
x2 = 2x .
This is what we call an equation to solve.
This problem could also be interpreted geometrically asking in which points the parabola y=x2 intersect the esponential function y=2x.
Before we start exploring over the natural number (positive integers), we know that 2x > 0 and x2 ≥ 0 for every x Reals.
So, we can exclude immediatly the case for the integer x=0 wich implies: 02 < 20,, that is 0 < 1.
Furthermore, for the same reason and due to negative power of 2, for every negative integer x ≤ 1, we have 2x < x2 .
So, for positive integers x>0, both 2xand x2 are crescent functions : 2x+1 ≥ 2x and (x+1)2 ≥ x2,
and by the fact that:
for x=1, 12 <21 (that is 1 < 2), and
for x=3, 32 >23 (that is 9>8),
this implies that an integer solution is between 1 and 3.
Hence, our 1st trivial solution is for x=2: 22 = 22 = 4 .
In a similar way, from the fact that at x=5 , we have 25 = 32 > 52 = 25, we candidate next integer solution to be between 3 and 5, that is x=4.
Analizing the number 4 as result of this 1st trivial solution, It is interesting to observe that the number 2 let the 4 = 2+2 = 2*2 = 2^2 = ... , for every rank of the binary operator used.
4^2 = (2^2)^2 = 2^(2*2) = 2^(2+2) = 2^4 = 16. (Notice that just for the positive integer 2 you can make (2^2)^2 = 2^(2^2) )
From this point on, infact, 2^x will exceed x^2 for every real x>=4, so no more positive real solution after 4.
Infact for every integer x>0
(x+4)2 ≤ 2(x+4) , applying log in base 2 to both members, we can compare
2 log2(x+4) ≤ x+4 ≤ x+4 + x+4 ≤ 2(x+4)
log2(x+4) ≤ x+4 , for every real x>0.
So for every x>4 we proved that x2 ≤ 2x.
But it's not the only interval where 2x exceeds x2.
infact,
x2 = 2x , can be rewritten as 2nd grade equation giving the:
x = 2x , and x = 2-x solutions.
Using the fact that
(-x)2 = 2x , or
(x)2 = 2-x
x = 2-x/2
From the theorem of the fixed point applied to our exponential,
starting from x0=1,
and proceeding recursively
x1 = -2x0/2 = -√2
x2 = -2x1/2 = -√2-1/√2
x3 = -2x2/2 = -√2-√2/2
...
xn = -2xn-1 = -√2-√2^(-√2^....n-1Times....^-√2))
The greater will be n and the smaller will be the difference xn+1 -xn.
The limit for n that tends to infinity of xn is equal to value x:
x1 = 1
x1 = -1.41421356237310...
x2 = -0.612547326536066...
x3 = -0.808727927081827...
x4 = -0.755569324854238...
x5 = -0.769618475247432...
x6 = -0.765880261444542...
x7 = -0.766873153047868...
x8 = -0.766609309719418...
x9 = -0.766679412545762...
x10 = -0.766660785691404...
x11 = -0.766665734944187...
x12 = -0.766664419898679...
x13 = -0.766664769313767...
x14 = -0.766664676472171...
x15 = -0.766664701140715...
x16 = -0.766664694586141...
x17 = -0.766664696327729...
x18 = -0.766664695864979...
x19 = -0.766664695987935...
x20 = -0.766664695955265...
x21 = -0.766664695963945...
x22 = -0.766664695961639...
x23 = -0.766664695962252...
x24 = -0.766664695962089...
x25 = -0.-66664695962132...
x26 = -0.766664695962121...
x27 = -0.766664695962124...
x28 = -0.766664695962123...
x29 = -0.766664695962123...
The value asimptotically converge quickly according the desired precision.
Another method to find a closed form using the Lambert function W(z) = z ez,
which gives the solution in function of
F. Prowedi
