With f defined as , show the fixed point iteration
,
must converge to with
Had a go at this by using contraction mapping theorem and putting it in the form , and as f(x)=0 then can work out what g(x) is but I'm not sure it is a contraction so not sure how can know it converges?
Hello again at this post Thank you very much, I think I have found another way to solve this exercise if you can review it and comment on how rigorous it is please! (and if it is correct)
abs(g(x)-g(y))=abs((e^(y)-e^(x)) / 6) = 1/6 * abs ( e^(x) * (e^(y-x)-1))=e^(x)/6 * abs (e^(y-x)-1), since e^(x)=abs(e^(x))
Then e^(x)/6 * abs (e^(y-x)-1) <= e/6 * abs (e^(y-x)-1) (since x=1 gives e^(x)=1 and x=1 is the maximum of the interval [0,1]
Now e/6 * abs (e^(y-x)-1) <= e/6 * abs( (e-1) (y-x))= e/6 * (e-1) * abs(x-y) [using that abs(e-1) = e-1 and that abs(y-x)= abs (x-y)
So finally abs(g(x)-g(y)) <= k abs(x-y) where k= e/6 * (e-1) and 0<k<1
Hope I found a rigorous, correct proof for contraction of our function, I will really appreciate if you review it!! Thank you very much again!!
Not sure where that's coming from. How do you get that
= e/6 * (e-1) * abs(x-y) [using that abs(e-1) = e-1 and that abs(y-x)= abs (x-y)
So finally abs(g(x)-g(y)) <= k abs(x-y) where k= e/6 * (e-1) and 0<k<1
Hope I found a rigorous, correct proof for contraction of our function, I will really appreciate if you review it!! Thank you very much again!!
To use the derivative approach, all you need do is show that the derivative exists everywhere and is less than 1 in magnitude on the interval. That implies the function is Lipschitz with Lipschitz constant less than one, which implies that the function has a unique fixed point on that interval.
Thank you again, sorry I edited my post, I just found that that inequality can be justified by a sketch as I wrote above. Do you think now is a good proof?
I think if I can justify the inequality even by a sketch is still justified and I am obtaining an exact L and at the same moment proving directly that it is a contraction!
But you may have a different opinion (your opinion is obviously "heavier" than mine )