functional-equations

continuous linear recurrent relations

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For a function $f:\mathbb{R}\rightarrow \mathbb{R}$ denote $f_0(x)=x$, $f_n(x)=f(f_{n-1}(x))$. Assume that $f$ satisfies a functional equation $$f_n(x)+a_{n-1} f_{n-1}(x)+\dots+a_0 f_0(x)\equiv 0$$ for some constant real coefficients $a_{0},a_1,\dots,a_{n-1}$. Assume also that $f$ is continuous. What are conditions for polynomial $t^n+a_{n-1}t^{n-1}+\dots +a_0$ which allow to conclude that $f$ is linear? Answer AttributionSource : Link , Question Author : Fedor … Read more

modularity of Theta functions attached to Hecke characters

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Let K/Q be a quadratic imaginary field, and let χ be a Hecke character on K. Using Poisson summation, one can show that the theta function θ(z):=∑I⊆OKχ(I)q|I|,q=e2πiz satisfies a functional equation relating θ(z) to θ(−1/Nz), where N is equal to the discriminant of K times the norm of the conductor of χ. But from this … Read more

solution of functional equation f∘k(x)=xf^{\circ k}(x) = x

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The equation f∘k(x)=Id for x∈E is called the Babbage equation and the general solution is given in the following way [M. Kuczma, Functional equations in a single variable]: Let 1=n0<⋯<nr=k be the complete set of divisors of k and let E=∪ri=0∪nij=1Uij be an arbitrary decomposition of E into disjoint sets such that for every i … Read more

approximately linear functions — more

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Suppose f,g are continuous functions from R to R, with the property that f(x)+f(y)=g(x+y) for all x,y. Taking x=y=z/2 implies that g(x)=2f(x/2) so that the above condition becomes f(x)+f(y)=2f((x+y)/2). This is known as Jensen’s functional equation, and it implies that f is linear. There’s also a generalization of Jensen’s equation (I’ve seen it in work … Read more

Techniques to solve equations involving a definite integral [closed]

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Closed. This question needs details or clarity. It is not currently accepting answers. Want to improve this question? Add details and clarify the problem by editing this post. Closed 7 years ago. Improve this question Are there any well known techniques to solve a problem of the following form: ∫baf(x,α)dx=g(α), where a,b∈R are fixed, f … Read more

A Differential Equation with Nested Functions

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This was posted to Math Stackexchange, but got no useful answers, and the more I think about it, the harder it seems. I would like to know whether there exists a differentiable function from the (open or closed) unit interval to itself satisfying 1−x−f(f(x))−f(x)f′(f(x))=0 for all x. Ideally, I’d also like a list of all … Read more

The functional equation f(x)=qx+qxf(x)−f(x2)f(x) = qx + qxf(x) – f(x^2)

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A word (i.e., ordered string of letters) is bifix-free provided it has no proper initial string and terminal string that are identical. For example, the word ingratiating has bifix ing, but the word ingratiate is bifix-free. Let a(q)n be the number of bifix-free words over a fixed q-letter alphabet. The generating function f(x)=f(q)(x)=∑∞n=0anxn satisfies the … Read more

Zappa-Szép products of the monoid of integers with itself

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Question What are all the functions α,β:N→N satisfying the following functional equations? ∙    α(0)=0,β(0)=0∙   αn(m+m′)=αn(m)+αβm(n)(m′)∙   βm(n+n′)=βm(n)+βαn(m)(n′) Here, αn denotes the n-fold composition of α. So far I have only found the following solutions: 1) α(n)=n, β(n)=pn, where p∈N 2) α(n)=pn, β(n)=n, where p∈N 3) α(n)=β(n)=max I am not even sure if there are more solutions. Of course, 1) … Read more

Is this method of “fractional sums” using a Fourier series viable?

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Hi. I have this idea about developing what I call a “continuum sum”, that is, a method to “add up a non-integer number of terms”, i.e. to see if there is a “natural” way to assign a meaning to the expression ∑bn=af(n) where a and b are non-integer fractional, real, or even complex numbers, for … Read more

On a generalization of the classical Cauchy’s functional equation

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I start with some known preliminaries on the problem: Classical result. The one-dimensional Cauchy functional equation ∀x,y∈R,f(x+y)=f(x)+f(y) with f:R→R is only solved by the trivial solutions f(x)=cx, for some c∈R, if f satisfies for some additional conditions, e.g., continuity. Classical result with restricted domain. Now let R+:=(0,∞). It is clear from the proof of the … Read more