Prove that $f(x) = \varphi (x) + P_{n+1}(x)$ given $f(x+T) = f(x) + Q_n(x)$ where $\varphi (x)$ is periodic with period $T$.

Question

The question below is an attempt to generalize this idea.

Given a function $f(x)$ there exist $T \ne 0$ and some polynomial $Q_n(x)$ of $n$-th power such that $\forall x \in D(f)$: $$ x+T \in D(f) \\ x-T \in D(f) $$

And $f(x+T) = f(x) + Q_n(x)$.

Prove that: $$ f(x) = \varphi (x) + P_{n+1}(x) $$ Where $\varphi (x)$ is periodic with period $T$ and $P_{n+1}(x)$ is a polynomial of $(n+1)$-th power.

As shown in the linked question we may use the fact that $\varphi (x)$ is periodic. So it has been shown that:

$$ \varphi (x) = \varphi (x+T) = f(x) - {a\over T}x $$ in case $f(x+T) = f(x) + a$.

Using the same idea I have shown that:

$$ f(x) = \varphi (x) + {a\over {2T}}\left(x^2 - Tx\right) $$

in case $f(x+T) = f(x) + ax$.

And now i'm struggling to show that this idea may be generalized. I've tried to use the same technique suggested in the linked question.

I started with presenting $Q_n(x)$ and $P_{n+1}(x)$ in the form of a sum with some unknown coefficients, but that got ugly very quickly, since $(x+T)^k$ appears in the sum when applying $\varphi (x)$ to $x+T$.

How can i prove the above?

upd 1:

Here are some further thoughts. So i want $\varphi (x) = \varphi (x+T)$. Let's try to find coefficients of $Q_0(x)$ and $P_1(x)$ to see whether we could deduce the polynomials ($a_k$ is coefficient of $Q_n(x)$ and $b_k$ is for $P_n(x)$):

$$ f(x+T) = f(x) + a_1 \\ f(x) = \varphi (x) + b_1x + b_2 \\ \varphi (x) = f(x) - b_1x - b_2 \\ \varphi (x + T) = f(x+T) - b_1(x+T) - b_2 \\ \varphi (x+T) = \varphi (x) \\ f(x) + a_1 - b_1(x+T) - b_2 = f(x) - b_1x - b_2 \\ a_1 - b_1x -b_1T-b_2 = -b_1x -b_2 $$

This gives $b_2 = 0$ and $b_1 = {a_1\over T}$. So now if $a_1 = a$ we get $f(x+T) = f(x) + a$ and $f(x) = \varphi(x) + {a\over T}x$. The exact situation as in the linked post.

For $Q_1(x)$ and $P_2(x)$ i get that:

$$ 2b_1Tx + b_1T^2 + b_2T = a_1x+a_2 $$

So $b_1 = {a\over 2T}$ and $b_2$ may be obtained only by the assumption that $a_2 = 0$. If $a_2 = 0$ then $b_2 = {a\over 2}$ and we get the case for $Q_1$ and $P_2$. But the assumption is taken from nowhere.

upd 2:

I've just tried to find $Q_2(x)$ and $P_3(x)$ and finally arrived at a system of equations for the coefficients:

$$ \cases { b_1 = {a_1 \over 3T} \\ b_2 = {{a_2 - 3b_1T^2}\over {2T}} \\ b_3 = {{a_3 - b_1T^3 - b_2T^2} \over {T}} } $$

In this case I've tried to assume that $a_2 = 0$ and $a_3 = 0$ and therefore the following is true:

$$ f(x) = \varphi (x) + \frac{a}{3T} (x^3 - \frac{3x^2T}{2} + 2T^2x) $$

in case $f(x+T) = f(x) + ax^2$. It seems like $Q_n(x)$ is in the form of $ax^n$. But $P_n(x)$ has somewhat more complicated form.

Answer 1

Generalizing your computations, we can find (without the theory of vector spaces) a polynomial $P(X)=\sum_{k=0}^{n+1}b_kX^k$ such that $P(X+T)-P(X)=Q(X)=\sum_{k=0}^na_kX^k$ (and then we finish as in Hagen von Eitzen's answer).

Compute $$ P(X+T)-P(X)=\sum_{k=0}^{n+1}\sum_{i=0}^kb_k\binom{k}{i}X^iT^{k-i} - \sum_{k=0}^{n+1}b_kX^k = \sum_{i=0}^{n+1}\left(\sum_{k=i+1}^{n+1}b_k\binom{k}{i}T^{k-i}\right)X^i $$ so that the condition is equivalent to having, for all $i$ from $0$ to $n$ : $$ \sum_{k=i+1}^{n+1}b_k\binom{k}{i}T^{k-i} = a_i $$ which is, taking $j=i+1$ ($j$ goes from 1 to $n+1$) and taking out $k=j$ from the sum : $$ b_j = \frac{1}{jT}\left(a_{j-1}-\sum_{k=j+1}^{n+1}b_k\binom{k}{j-1}T^{k-j+1}\right) = \frac{1}{j}\left(\frac{a_{j-1}}{T}-\sum_{k=j+1}^{n+1}b_k\binom{k}{j-1}T^{k-j}\right) $$ so the coefficients $b_j$ can be computed recursively by starting from $j=n+1$ and going down :

• first $b_{n+1} = \frac{a_n}{(n+1)T}$
• then $b_n = \frac{1}{n}\left(\frac{a_{n-1}}{T}-b_{n+1}\binom{n+1}{n-1}T\right) = \frac{1}{n}\left(\frac{a_n}{T}-\frac{(n+1)nb_{n+1}}{2}\right)$ which we can compute because we already computed $b_{n+1}$
• then we can compute $b_{n-1}$ using $b_{n+1}$ and $b_n$
• and so on until $b_1$ (as for $b_0$, from our analysis there is no condition on $b_0$, so any choice works)

This shows that given $Q_n$ we can always find a $P_{n+1}$ satisfying our requirements (and this $P_{n+1}$ is unique up to its constant coefficient).

Answer 2

Let $\mathcal P_n$ be the vector space of polynomials of degree $\le n$, and $T>0$ fixed. Then $p(x)\mapsto p(x+T)-p(x)$ is a linear map $\Delta_T\colon\mathcal P_n\to \mathcal P_{n-1}$ because the highest powers cancel. If $p\in\ker\Delta_T$, then $p$ is periodic and being, being a polynomail, $p$ is constant (and indeed, if $p$ is constant, then $\Delta_Tp=0$). We conclude that $\dim\ker\Delta_T=1$, hence from $\dim \mathcal P_n=n+1$, $\dim \mathcal P_{n-1}=n$, we conclude that $\Delta_T$ is onto.

Now let $f,Q$ as in the problem statement. From the surjectivity of $\Delta_T$, we obtain $P\in \mathcal P_{n+1}$ with $\Delta_TP=Q$. Let $\phi(x)=f(x)-P(x)$. Then $$\phi(x+T)-\phi(x)=f(x+T)-f(x)-P(x+T)+P(x)=Q(x)-Q(x)=0,$$ henec $\phi$ is periodic as desired.