Find all real polynomials $f,g$ such that $\forall n\in\Bbb N,\;f(n) = \frac{g(1) + g(2) + \dots + g(n)}{g(n)}$

Question

Question:

Let $f,g: \mathbb{R} \to \mathbb{R}$ be polynomials such that

$$f(n) = \frac{g(1) + g(2) + \dots + g(n)}{g(n)}$$

for all integers $n \ge 1$.

We want to find all such pairs of polynomials $(f,g)$.

Here is my attempt

WLOG $g$ is monic.

Claim 1: $f(n)=kn+d$

Proof: Define $s(k,n)=\sum_{i=0}^{n}k^i$. First we see that $s(k,n)$ is a polynomial of degree $n+1$, because we can then always look at $c\cdot(m+1)^l-c \cdot m^l$ and always choose $c$ like that for each the rate of change matches in the of $s(k,n)$. this implies that $\deg(f)=1$.

Claim 2: Let $k>0$ denote the degree of $g$, then $f(x)=\frac{x}{k+1}+\frac{k}{k+1}$

Since $g$ is monic we get for the leading coeefficient $s(m,n)$ is $\frac{1}{{{k+1}\choose{1}}}=\frac{1}{k+1}$. Furthermore for $n=1$ follows $ f(1)=1$. Hence the claim is proven.

Claim 3: $g(n)=g(1)\prod_{i=1}^{n-1}\left(\frac{k+1+i}{i}\right)$ Induction Claim 1: I've shown that $f(n) = kn + d$.

Claim 2: I've found that $f(x) = \frac{x}{k+1} + \frac{k}{k+1}$.

$f(n)g(n) - f(n-1)g(n-1) = g(n)$

$\left(kn + d\right)g(n) - \left(k(n-1) + d\right)g(n-1) = g(n)$

$kng(n) + dg(n) - k(n-1)g(n-1) - dg(n-1) = g(n)$

$kng(n) + dg(n) - kng(n-1) + kg(n-1) - dg(n-1) = g(n)$

$g(n) \left(kn + d - k(n-1) + k - d\right) = g(n)$

$k(n - n + 1 - 1) + k - d = 1$

$k(0) + k - d = 1$

$k - d = 1$

$k - d = 1$

$k - d = 1$

$d = k - 1$

$f(n) = kn + (k - 1)$

$d = k - 1$

$g(n) = g(1) \prod_{i=1}^{n-1}\left(\frac{k+1+i}{i}\right)$

$g(n) = g(1) \prod_{i=1}^{n-1}\left(\frac{k+1+i}{i}\right)$

$g(n) = g(1) \prod_{i=1}^{n-1}\left(\frac{kn + k}{i}\right)$

$g(n) = g(1) \prod_{i=1}^{n-1}\left(\frac{k(n+1)}{i}\right)$

$g(n) = g(1) \prod_{i=1}^{n-1}\left(\frac{k(n+1)}{i}\right)$

$g(n) = g(1) \left(\frac{k(n+1)}{1}\right) \left(\frac{k(n+1)}{2}\right) \left(\frac{k(n+1)}{3}\right) \ldots \left(\frac{k(n+1)}{n-1}\right)$

$g(n) = g(1) \frac{k(n+1)}{1} \cdot \frac{k(n+1)}{2} \cdot \frac{k(n+1)}{3} \cdot \ldots \cdot \frac{k(n+1)}{n-1}$

$g(n) = g(1) \cdot \frac{(k(n+1))^n}{(n-1)!}$

The main idea is if you sum a polynomial $G(n)=\sum_{i=1}^{n} g(i)=\sum_{i=1}^{n} i^k+a_{k-1}i^{k-1} \cdots a_0=\sum_{i=1}^{n} \left(i^n\right)+\sum_{i=1}^{n} \left(a_{k-1}i^{k-1}\right)\cdots \sum_{i=1}^{n} \left(a_0\right)$. but now only the sum $\sum_{i=1}^{n} \left(i^k\right)$ generates a poynomial of degree $k+1$ so also it is the only one which determines the leading coefficient. if you look at the rate of change of $\sum_{i=1}^{n} \left(i^k\right)$ it is $n^k$, which can only be produced by the highest deree term, so we look at $c\cdot (n+1)^{k+1}-c\cdot n^{k+1}$ the rate of change of the highest degree term. This is by binomial expansion $c\cdot {{k+1}\choose{1}}n^{k}+\cdots$. But $c\cdot {{k+1}\choose{1}}=1$ since else our polynomial does not match the growth of the sum. So $c=\frac{1}{{{k+1}\choose{1}}}=\frac{1}{k+1}$

Answer 1

Hint: $f(n) g(n) - f(n-1) g(n-1) = g(n)$, i.e. $$ (f(n)-1) g(n) = f(n-1) g(n-1) $$

Answer 2

I want to look at the sum $S_k(n)=\sum_{i=1}^ni^k=1+2^k+\cdots+n^k$ for $k\ge 0$ and $n\ge 1$. We only need the limiting behavior, so we avoid Faulhaber's formula and instead use the integral approximation to obtain $$\frac{1}{k+1}n^{k+1}=\int_0^nx^k\mathrm dx\le S_k(n)\le\int_1^{n+1}x^k\mathrm dx=\frac{1}{k+1}\left[(n+1)^{k+1}-1\right].$$ Let $\varepsilon_k(n)$ be the function given by $S_k(n)=(1+\varepsilon_k(n))\frac{1}{k+1}n^{k+1}$, then the above shows that $\lim_{n\rightarrow\infty}\varepsilon_k(n)=0$.

Now, let $f(x)=\sum_{k=0}^Da_kx^k$ and $g(x)=\sum_{k=0}^db_kx^k$ be the two polynomials, i.e. we have $$f(n)=\frac{1}{g(n)}\sum_{i=0}^ng(n),$$ and we assume wlog that $b_d=1$. Before we start, notice that we necessarily have $g(n)\neq 0$ for all $n\ge 1$ since otherwise the right hand side is not well-defined (or infinite in absolute, depending on your preference) and hence the equality cannot hold.

First, assume that $d=0$, i.e. $g(x)=b_0=1$ is constant. Then we have $$f(n)=\frac{1}{g(n)}\sum_{i=1}^ng(i)=\frac{n}{1}=n,$$ so we necessarily have $f(x)=x$ (since there is only one polynomial which agrees with the identity on an infinite number of points, namely the identity).

Now, let $d>0$. Notice that \begin{align*} \sum_{i=1}^ng(i)&=\sum_{k=0}^db_kS_k(n) =\sum_{k=0}^db_k(1+\varepsilon_k(n))\frac{n^{k+1}}{k+1}\\ &=\left(1+\varepsilon_d(n)+\sum_{k=0}^{d-1}\frac{(1+\varepsilon_k(n))(d+1)b_k}{(k+1)n^{d-k}}\right)\frac{n^{d+1}}{d+1}. \end{align*} Let $\varepsilon(n)$ be the function given by $\sum_{i=1}^ng(i)=(1+\varepsilon(n))\frac{n^{d+1}}{d+1}$, then the above shows that $\lim_{n\rightarrow\infty}\varepsilon(n)=0$. On the other hand, we have $$g(n)=\left(1+\sum_{k=0}^{d-1}\frac{b_k}{n^{d-k}}\right)n^d,$$ so the function $\delta(n)$ given by $g(n)=(1+\delta(n))n^d$ satisfies $\lim_{n\rightarrow\infty}\delta(n)=0$. But this means that the relative error in the numerator and denominator vanishes and we have $$f(n)=\frac{(1+\varepsilon(n))\frac{n^{d+1}}{d+1}}{(1+\delta(n))n^d},$$ so the function $\gamma(n)$ given by $f(n)=(1+\gamma(n))\frac{1}{d+1}n$ satisfies $\lim_{n\rightarrow\infty}\gamma(n)=0$. Further, we have $$1+\gamma(n)=\frac{\sum_{k=0}^Da_kn^k}{\frac{1}{d+1}n}=\sum_{k=0}^D(d+1)a_kn^{k-1},$$ which shows that $D=1$ since the left hand side tends to $1$, so the right hand side also has to tend to $1$, but it would be diverging for $D>1$. This also shows that $a_1=\frac{1}{d+1}$, again since the left hand side tends to $1$, and the right hand side tends to $(d+1)a_1$. This shows that $f(x)=\frac{1}{d+1}x+a_0$.

Next, we notice that $f(1)=\frac{g(1)}{g(1)}=1$ on the one hand, and on the other hand $f(1)=\frac{1}{d+1}+a_0$, so we get $a_0=1-\frac{1}{d+1}=\frac{d}{d+1}$. Hence, with $F_d(x)=\frac{1}{d+1}x+\frac{d}{d+1}$ we have $f=F_d$.

As we already noticed, we have $g(n)f(n)-g(n)=\sum_{i=0}^{n-1}g(i)=g(n-1)f(n-1)$. This gives $$g(n)=\frac{f(n-1)}{f(n)-1}g(n-1)=\frac{\frac{1}{d+1}(n-1)+\frac{d}{d+1}}{\frac{1}{d+1}n+\frac{d}{d+1}-1}g(n-1)=\frac{n-1+d}{n-1}g(n-1)$$ for $n\ge 2$ (using that $f(1)=1$ and that $f$ is increasing, so the denominator is positive). Induction yields \begin{align*} g(n)&=\prod_{k=2}^n\frac{k-1+d}{k-1}g(1) =\frac{(n-1+d)!}{d!(n-1)!}g(1)\\ &=\binom{n-1+d}{d}g(1)=\frac{g(1)}{d!}\prod_{k=0}^{d-1}(n-1+d-k). \end{align*} For any real $c\neq 0$ and any integer $d>0$ let $G_{c,d}$ be the degree $d$ polynomial given by $G_{c,d}(x)=\frac{c}{d!}\prod_{k=0}^{d-1}(x+d-1-k)$. Since $g$ and $G_{g(1),d}$ agree on infinitely many points, we necessarily have $g=G_{g(1),d}$. Since we assumed that $g$ is monic, we must have $c=g(1)=d!$. Notice that the roots of $g$ are $-(d-1),\dots,0$, so we also have $g(n)\neq 0$ for $n\ge 1$. From this proof it is also clear that $$f(n)=\frac{1}{g(n)}\sum_{i=0}^{n}g(i)$$ indeed holds for all $n\ge 1$. Notice that we also have $f=F_0$ and $g=G_{1,0}$ for $d=0$ by the above, so in general the desired pairs are $$\{(F_d,G_{c,d}):d\ge 0,c\neq 0\}$$ over integers $d$ and real numbers $c$.