Question

Trying to solve x=cosx algebraically, any ideas?

Answer 1

nope. Can not be done

Answer 2

Yeah, how do you know?

Answer 3

how would you isolate x?

Answer 4

Plug it into itself an infinite number of times is one way I guess.

x=cos(cos(cos(cos(cos(cos(cos(cos(cos(cos(cos(cos(cos(cos(cos(cos(cos(cos(cos(cos(cos(cos(cos(cos(cos(cos(cos(cos(cos(cos(cos(cos(cos(cos(cos(cos(cos(cos(cos(cos(cos(cos(cos(cos(cos(cos(cos(cos(cos(cos(cos(cos(cos(cos(cos(cos(cos(cos(cos(cos(cos(cos(cos(cos(cos(cos(cos(cos(cos(cos(cos(cos(cos(cos(cos(cos(cos(cos(cos(cos(cos(cos(cos(cos(...))))))))))))))))))))))))))))))))))))))))))))))))))))

Answer 5

depends on the x

Answer 6

omg youre so hot.

Answer 7

\[x=1-\frac{ x^2 }{ 2! }+\frac{ x^4 }{ 4! }-\frac{ x^6 }{ 6! }+...\] I suppose there's another infinite thing going on there idk, just looking for stuff.

How about we invent new algebra that can solve this?

Or is there some way that we can solve this like an eigenvalue problem?

@FibonacciChick666 there is only one x.

Answer 8

\[x=\cos x \implies x = 1-\frac{ x^{2} }{ 2 } + \frac{ x^{4} }{ 24 } \]

solve it

Answer 9

no I mean, you do 0=cosx-x then you find x values where the cosx =x

Answer 10

do they exist?