Functions f:N→R, g:N→R are such that
f(n+1)=f(n)cos(g(n))−g(n)sin(g(n))
g(n+1)=f(n)sin(g(n))+g(n)cos(g(n))
If f(1)=0.8 and g(1)=0.6, find the limit of f(n) as n tends towards infinity.
If there is a limit (which I have not proven), then we can deduce the following:
Let the respective limits be f and g.
Then, f = f(cos(g)) - g(sin(g))
g = f(sin(g)) + g(cos(g))
Manipulating the equations gives:
f(1-cos(g)) = -g(sin(g))
f(sin(g)) = g(1-cos(g))
Multiplying the lhs and the rhs gives:
f^2*(1-cos(g))sin(g) = -g^2*(1-cos(g))sin(g)
By inspection, solutions are
f = 0 and g=0
or sin(g) = 0
or cos(g) = 1
This can be simplified to just sin(g) = 0, as the other two are just specific cases of sin(g) = 0.
And given the starting point, we can expect that the only g which could be the limit is g = 0.
But this method gives no hint what f is. g = 0 is consistent with any f. Other methods are called for.
Edited on May 20, 2020, 7:23 am
If there is a limit ...
