can Sup Forums solve for x?
x^x=x+1
Can Sup Forums solve for x?
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/-1
x^x-1=x
It's about 0.35
X=√π
about three fidy
nope
no, but kinda close
x=rape
It's 0
>0^0 = 0+1
>1 = 1
>nope
>.35^.35 = 0.6925
>0.6925/2 = 0.3462
So really the answer is 0.3463
0 works as well I suppose
You just take an x root of both sides. It will be something like x = x root (X+1)
x^x = x + 1
x^x - x - 1 = 0
x1 = (1 + (1 - 4 × 1 × -1)^1/2) / 2 = 1 + 5^1/2
x2 = (1 - (1 - 4 × 1 × -1)^1/2) / 2 = 1 - 5^1/2
Might have made a mistake somewhere
Fuck this man is a genius
>0^0=1
you are wrong
Are you retarded or what
its x^x=x+1
not x^x=x+x
Two solutions, 0 and 1.78
its not a cuadratic equation, you cant do that
0^0=1=0+1
technically the two solutions are y (lim y->0) and 1.78
Lambert W functions
no, 0^0 is undefined, if you used limits instead of 0, yes
fuk
>not x^x=x+x
shit you're right.
1.11
more digits
1.77677504...
x^x=x+1
(x^x)-x=1
((x^x)-x)^1/2 = 1^1/2
sub x = 0
((0^0)-0)=1
=1-0=1
.*. x =0
0^0 isnt 1
It is, dude. Anything with an exponent of 0 = 1.
0^n=0
n^0=1
0^0 is undefined
Can you, summer?
yes it's only a decimal approximation though
No such number.
X doesn't exist as real number.
see
True. The answer is about x=1,777 (not an infinite seq of 7s, just rounded).
You can get the result if you plot both sides of the equation and see where they intersect.
Are you retarded? It absolutely is a quadratic equation
i calculated it with newtons method
a quadratic would be x^2 = x+1, this is x^x=x+1
So you solved for equation x^x - x - 1? I think that should be doable, the equation seems differentiable. Not a 100% sure about x^x but I think so. If you got the same result, cool.
The answer isn’t less than 2 and greater than 1. Simply plot the linear function x+1 and plot x^x and find the intersect on a graphing calculator. However, one can easily determine that 1
Here's a graph of it. Note that the image is not entirely accurate as both sides of the equation are plotted with x intervals of 0.01. I didn't include the negative x part of the plot where there's another intersection point because things get iffy at that point (basically, you get negative values to the power of all kinds of values from negative to positive and it gets complex).
*Is less than 2 greater than 1
it's so close to 0, could it be .000000000001?
The limes of x^x is 1, as is the limes of x+1. However, they will never intersect at zero as x^x is not defined at x=0.
See
its the limit as it goes to 0
yup
x^x=x+1
∴x^x-x-1=0
(f(x)=x^x-x-1)
∴x=root of f(x)
f'(x)=d/dx(x^x-x-1)
=d/dx(x^x)+d/dx(-x)+d/dx(-1)
=d/dx(x^x)-1-0
=(x^x)(1+ln(x))-1
Apply newton
result=1.77677504...
its either 0 or 1.777 recurring
its not 1.7..., its irrational
also it isnt 0 because 0^0 is undefined