Another fixed-point theorem

Algebra Level 4

Let function $f(x)=\sqrt{e^{x}+x-a}$ for $a \in \mathbb R$ . If there exists $x_{0} \in [-1,1]$ such that $f(f(x_{0}))=x_{0}$ , what is the range of $a$ ?

Note: This problem requires a little-known and trivial fixed-point theorem. Can you find and prove it?

[1,e]

[\dfrac{1}{e}-1,1]

[\dfrac{1}{e}-1,e+1]

[1,e+1]

2 solutions

Shreyas Iyer
Jul 10, 2019

I don’t know what a fixed-point theorem is (my knowledge terminology doesn’t really extend too far beyond A-level), but I’ve attached my solution nevertheless, just please take what you will. I hope an image will suffice, I have no experience of using LATEX.

Yes, that is my expected fixed-point theorem. It is so trivial that it is not worth appearing online, but indeed you need it.

Alice Smith - 1 year, 11 months ago

wait what was the fixed-point theorem?

Razzi Masroor - 1 year, 4 months ago

Satyam Bhardwaj
Jul 10, 2019

% solution using MATLAB
ind=1;
range=zeros(1,2000*10000);
for x=-1:0.001:1
    for a=-5:0.001:5
        if abs(f(f(x,a),a)-x)<1e-2
            range(ind)=a;
            ind=ind+1;
        end
    end
end
function out=f(x,a)
out = sqrt(exp(x)+x-a);
end

then find min(range) and max(range) haha

I admire your good application of programming for this question haha

Shreyas Iyer - 1 year, 11 months ago

Another fixed-point theorem

2 solutions

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