Solving (quadratic) equations of iterated functions, such asf(f(x))=f(x)+x

Question

Cristian Rosales · Accepted Answer

Step 1In fact this belongs to a functional equation of the form{x=u(t)f=u(t+1)Then u(t+2)=u(t+1)+u(t)u(t+2)−u(t+1)−u(t)=0u(t)=C1(t)(1+52)t+C2(t)(1−52)t,where C1(t) and C2(t) are arbitrary periodic functions with unit period∴{x=C1(t)(1+52)t+C2(t)(1−52)tf=C1(t)(1+52)t+1+C2(t)(1−52)t+1and C2(t) are arbitrary periodic functions with unit period

Ann Mathis · Accepted Answer

Step 1Assume f is a linear transformation, then linear algebra lets us give it a matrix representationf(x)=F×xIt is also the case that f∘g=FG, function composition is represented by matrix multiplication.In your derivation you do not make any distinction between f and F, this is fine as long as you know you are doing it. Also the matrix F is 1×1 because R is a 1-dimensional vector space.That is why we have:f(f(x))=f(x)+x⇔F2=F+I⇔φ2=φ+1(Where f(x)=F⋅x=φxThe only problem is that we cannot be sure f is a linear transform just based on what was told. So it is not known whether this gives all solutions.

Solving (quadratic) equations of iterated functions, such as f(f(x))=f(x)+x

Answered question

Answer & Explanation

New Questions in Abstract algebra