A special functional equation Analysis.

A \(sequence\) is a function \(f\) defined for every non-negative integer n. For sequences, generally it is set that \(x_n=f(n)\). Usually we are given equation of the form \[x_n=F(x_{n-1},x_{n-2},x_{n-3},......)\]

Sometimes we are expected to find a "closed expression" for xnx_n. Such an equation is called functional equationfunctional \space equation. A function equation of the form xn=p.xn1+q.xn2,(q0).................(1)x_n=p.x_{n-1}+q.x_{n-2} , (q \neq 0) .................(1) is a (homogeneous) linear difference equation of order 2linear \ difference \ equation \ of \ order \ 2 (with constant coefficients).

To find the general solution of (1)(1), first we try to find a solution of the form xn=λnx_n=\lambda^n for a suitable number λ\lambda. To find λ\lambda, we plug λn\lambda^n into (1)(1) and get λn=p.λn1+q.λn2\lambda^n=p.\lambda^{n-1}+q.\lambda^{n-2}     λ2p.λq=0.................(2)\implies \lambda^2-p.\lambda-q=0 .................(2) And (2)(2) is called the characteristic equationcharacteristic \space equation of (1)(1). For distinct roots λ1, and λ2\lambda_1, \space and \space \lambda_2, xn=a.λ1n+b.λ2nx_n=a.\lambda_1^n+b.\lambda_2^n is the general solution. a and b a \ and \ b can be found from the initial values x0x_0 and x1x_1.

\bullet If λ1=λ2=λ\lambda_1=\lambda_2=\lambda, the general solution has the form xn=(a+b.n)λnx_n=(a+b.n)\lambda^n

Now we will talk about the functional equation of function ff (non-zero) of the form f(x+1)+f(x1)=t.f(x)f(x+1)+f(x-1)=t.f(x) No it looks like a linear difference equation of order 2. But the discrete variable nn is replaced by the continuous variable xx. So we try to find the solutions f(x)=λxf(x)=\lambda^x. For the value of λ\lambda , we get λ2t.λ+1=0\lambda^2-t.\lambda+1=0 with the solutions λ=t2±t241\lambda=\dfrac{t}{2} \pm \sqrt{\dfrac{t^2}{4}-1}. For t<2t <2, we have the solutions λ=t2+i.1t24\lambda=\dfrac{t}{2}+i. \sqrt{1-\dfrac{t^2}{4}} λˉ=t2i.1t24\bar{\lambda}=\dfrac{t}{2}-i. \sqrt{1-\dfrac{t^2}{4}}     λ=λˉ=1\implies |\lambda|=|\bar{\lambda}|=1

So λ\lambda and its conjugate λˉ\bar{\lambda} are unit vectors in the complex plane, that is , λ=cosϕ+i.sinϕ\lambda=cos\phi+i.sin\phi λˉ=cosϕi.sinϕ\bar{\lambda}=cos\phi-i.sin\phi Thus λ\lambda has a period nn, if λn=1\lambda^n=1     λ=cos(2.πn+i.sin(2.πn)\implies \lambda=cos(\frac{2.\pi}{n}+i. sin(\frac{2.\pi}{n})

Putting the value λ=t2+i.1t24\lambda=\dfrac{t}{2}+i. \sqrt{1-\dfrac{t^2}{4}} above, we get t2=cos(2.πn)\dfrac{t}{2}=cos(\frac{2.\pi}{n})

Therefore Fundamental period n=2.πcos1(t2)n=\dfrac{2.\pi}{cos^{-1}(\frac{t}{2})} for non-zero functions satisfying f(x+1)+f(x1)=t.f(x)f(x+1)+f(x-1)=t.f(x).

\bullet Yet it is unlikely that this irrational number gives a rational multiple of π\pi for the angle ϕ\phi, the only way to secure periodicity.

You can try the problems based on it problem 1, and problem 2

You can try more such problems of the set Do you know its property ?

Note by Sandeep Bhardwaj
6 years, 7 months ago

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Good note to start the investigation. Here are some pointers for improvement.

1) As opposed to saying "We try to find solutions of the form f(x)=λx f(x) = \lambda ^ x , you should say that "from the theory of linear recurrences, all continuous functions which satisfy the functional equation have the form f(x)=aiλx f(x) = \sum a_i \lambda^x .

2) Notice that I made mention of the fact "all continuous functions". This is an extremely important part that is left out of your analysis. Without this assumption, your conclusion is not true. Please figure out where to add this in, and why it matters.

3) Please explain why "ff is non-zero" is an required assumption. You need that one of the ai a_i is non-zero.

4) Several times, you used nn as an integer. Other times, you get away with the "assumption" that nn is a real number, ie 2πn \frac{ 2 \pi } { n } is unlikely to be a rational multiple of π \pi . You should make that distinction extremely clear.

5) You have only shown that this is a period. Why must it be a fundamental period?

Calvin Lin Staff - 6 years, 7 months ago

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11. Yeah, off course f(x)f(x) must be a continuous non-zero function here. And it is clearly mentioned in the note "x is a continuous variable" which implies that f(x)f(x) is a continuous function. Otherwise there can be many discrete-valued functions following the functional equation.

55. We have worked here on trying to find the solution by taking f(x)=λxf(x)=\lambda^x and In this case λ\lambda comes out to be ei.2.πn\large e^{i.\frac{2.\pi}{n}}. From here I can say that λ\lambda is one of the nthnth roots of unity. Hence it lies on the circle λ=1|\lambda|=1. So it will repeat its value once, when it covers the circle one time. So that will be the minimum value of the Period, that is called the Fundamental Period.

For eg. : let f(x)=ixf(x)=i^x where i=1i=\sqrt{-1} which is one of the fourth roots of unity. So it's fundamental period will be 44. There exists no positive real number <4<4 for which f(x)f(x) will keep repeating its value after that interval. On the basis of this, 44 will be the fundamental period here.

Similarly, λ\lambda is one of the nthnth roots of unity. Hence its fundamental period will be nn. @Calvin Lin@Sanjeet Raria

Sandeep Bhardwaj - 6 years, 7 months ago

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1) xx being continuous does not imply that f(x) f(x) is continuous. E.g. f(x)=x f(x) = \lfloor x \rfloor .
For example, a non-continuous function that satisfies your equation is

f(x)={λxxN0x∉N f(x) = \begin{cases} \lambda ^ x & x \in \mathbb{N} \\ 0 & x \not \in \mathbb{N} \end{cases}

5) How do you know that the solution must be of the form f(x)=λx f(x) = \lambda ^ x ? Why can't it be of the form f(x)=Aλx f(x) = A \lambda ^ x or f(x)=Aλx+Bγx f(x) = A \lambda ^ x + B \gamma ^ x ?

Calvin Lin Staff - 6 years, 6 months ago

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@Calvin Lin 1) If I take f(x)=λx,xRf(x)=\lambda^x , \forall x \in R, in this particular case, if xx is continuous, then f(x)f(x) will also be continuous function. And also there exist many function of the type f(n)f(n) for discrete values of nn (let's say Natural no.) which are not continuous over the whole real No. line.

5) I didn't imposed it that you have to take f(x)=λxf(x)=\lambda^x. You can even take function ff of the form f(x)=A.λxorf(x)=A.λx+B.γxf(x)=A.\lambda^x \quad or \quad f(x)=A.\lambda^x+B.\gamma^x. But if you take the later ones, then finally It will reduce to the form f(x)=λxf(x)=\lambda^x. And this form is taken for the simplicity, to avoid too much calculations.

And as far as, it is concerned that the period here will be fundamental period or not. According to me, the period here is the fundamental one. Without going into deep proofs, first I would like to know that whether it is fundamental or not ??? After that, I will look for the proofs in favour of this. Waiting for your reply sir. @Calvin Lin

Sandeep Bhardwaj - 6 years, 6 months ago

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@Sandeep Bhardwaj 1) Right. My point is that you must talk about continuous functions, and explain why/how this is required. You made no mention of this in the writeup, and instead merely assumed that the function must have a certain form. I have demonstrated the existence of a discontinuous function which satisfies the initial conditions that you listed, which contradicts the conclusion of your proof. The main difference is that you need the condition of continuous functions.

5) (and, as an explanation of the above) No. You did impose the form that f(x)=λx f(x) = \lambda ^ x .
All that the recurrence relations shows, is that for each 0r<1 0 \leq r < 1 , there exists constants Ar,BrA_r , B_r such that f(n+r)=Arλn+Brγn f(n + r) = A_r \lambda^n + B_r \gamma^n .
It is at this point that we need a continuity argument, in order to show that f(x)=Aλx+Bγx f(x) = A \lambda^x + B \gamma ^ x for some fixed constants AA and BB.
Note: I am unclear what you mean exactly by "and this form is taken for the simplicity, to avoid too much calculations. I can guess at what you mean, but because it is not expressed clearly above, it currently represents a leap of logic to other readers.

6) The periodicity of the function is dependent on being in case 2, IE λγ \lambda \neq \gamma . So, for certain values of tt, the functions need not be periodic. What are the value(s) of tt?
As to whether or not the period is fundamental, that would depend on the values of t,A,Bt, A, B . As stated, if A=B=0A = B = 0 , then we get the zero function, which has no fundamental period. When you continue your analysis, you will have to take that into account.

Calvin Lin Staff - 6 years, 6 months ago

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Nice note,@Sandeep Bhardwaj

Anuj Shikarkhane - 6 years, 7 months ago

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thank you very much. :)

Sandeep Bhardwaj - 6 years, 7 months ago

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You're welcome😄👍👍👍

Anuj Shikarkhane - 6 years, 7 months ago

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Why we always take x=r cos theta & y=r sin theta in Cartesian geometry w.r.t. X axis and Y axis respectively?

Kavita Thakur - 6 years, 6 months ago

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The angle is by convention measured in anticlockwise direction from the positive x-axis.

Tapas Mazumdar - 3 years, 2 months ago

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good work

Akash Gupta - 6 years, 3 months ago

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