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 \(x_n\). Such an equation is called \(functional \space equation\). A function equation of the form
\[x_n=p.x_{n-1}+q.x_{n-2} , (q \neq 0) .................(1) \]
is a **(homogeneous)** \(linear \ difference \ equation \ of \ order \ 2\) **(with constant coefficients)**.

To find the general solution of \((1)\), first we try to find a solution of the form \(x_n=\lambda^n\) for a suitable number \(\lambda\). To find \(\lambda\), we plug \(\lambda^n\) into \((1)\) and get \[\lambda^n=p.\lambda^{n-1}+q.\lambda^{n-2}\] \[\implies \lambda^2-p.\lambda-q=0 .................(2) \] And \((2)\) is called the \(characteristic \space equation\) of \((1)\). For distinct roots \(\lambda_1, \space and \space \lambda_2\), \[x_n=a.\lambda_1^n+b.\lambda_2^n\] is the general solution. \( a \ and \ b\) can be found from the initial values \(x_0\) and \(x_1\).

\(\bullet\) If \(\lambda_1=\lambda_2=\lambda\), the general solution has the form \[x_n=(a+b.n)\lambda^n\]

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

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

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

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

\(\bullet\) Yet it is unlikely that this irrational number gives a rational multiple of \(\pi\) for the angle \(\phi\), 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 ?

## Comments

Sort by:

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

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

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 "\(f\) is non-zero" is an required assumption. You need that one of the \( a_i \) is non-zero.

4) Several times, you used \(n\) as an integer. Other times, you get away with the "assumption" that \(n\) is a real number, ie \( \frac{ 2 \pi } { n } \) is unlikely to be a rational multiple of \( \pi \). 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 · 2 years, 2 months ago

Log in to reply

\(5\). We have worked here on trying to find the solution by taking \[f(x)=\lambda^x\] and In this case \(\lambda\) comes out to be \(\large e^{i.\frac{2.\pi}{n}}\). From here I can say that \(\lambda\) is one of the \(nth\) roots of unity. Hence it lies on the circle \(|\lambda|=1\). 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 \[f(x)=i^x\] where \(i=\sqrt{-1}\) which is one of the fourth roots of unity. So it's fundamental period will be \(4\). There exists no positive real number \(<4\) for which \(f(x)\) will keep repeating its value after that interval. On the basis of this, \(4\) will be the fundamental period here.

Similarly, \(\lambda\) is one of the \(nth\) roots of unity. Hence its fundamental period will be \(n\). @Calvin Lin@Sanjeet Raria – Sandeep Bhardwaj · 2 years, 2 months ago

Log in to reply

For example, a non-continuous function that satisfies your equation is

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

5) How do you know that the solution must be of the form \( f(x) = \lambda ^ x \)? Why can't it be of the form \( f(x) = A \lambda ^ x \) or \( f(x) = A \lambda ^ x + B \gamma ^ x \)? – Calvin Lin Staff · 2 years, 1 month ago

Log in to reply

5) I didn't imposed it that you have to take \(f(x)=\lambda^x\). You can even take function \(f\) of the form \(f(x)=A.\lambda^x \quad or \quad f(x)=A.\lambda^x+B.\gamma^x\). But if you take the later ones, then finally It will reduce to the form \(f(x)=\lambda^x\). 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

fundamentalone. 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 · 2 years, 1 month agoLog in to reply

musttalk 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 \( f(x) = \lambda ^ x \).

All that the recurrence relations shows, is that for each \( 0 \leq r < 1 \), there exists constants \(A_r , B_r \) such that \( f(n + r) = A_r \lambda^n + B_r \gamma^n \).

It is at this point that we

need a continuity argument, in order to show that \( f(x) = A \lambda^x + B \gamma ^ x \) for some fixed constants \(A\) and \(B\).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 \( \lambda \neq \gamma \). So, for certain values of \(t\), the functions need not be periodic. What are the value(s) of \(t\)?

As to whether or not the period is fundamental, that would depend on the values of \(t, A, B \). As stated, if \(A = B = 0 \), 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 · 2 years, 1 month ago

Log in to reply

Nice note,@Sandeep Bhardwaj – Anuj Shikarkhane · 2 years, 2 months ago

Log in to reply

– Sandeep Bhardwaj · 2 years, 2 months ago

thank you very much. :)Log in to reply

– Anuj Shikarkhane · 2 years, 2 months ago

You're welcome😄👍👍👍Log in to reply

good work – Akash Gupta · 1 year, 10 months ago

Log in to reply

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 · 2 years ago

Log in to reply