Finding a solution to $f(x+1) - f(x) = (x+1)^2$

Question

I was trying to find an analytical solution for

$$ f(x+1) - f(x) = (x+1)^2 $$

So after hours of trying and algebraic manipulation, calculus, taylor series, I found that the best way was to seek a general solution in the form

$$ f(x) = ax^3+bx^2+cx+d $$

Now with some algebraic manipulation manipulation

$$ f(x) = a((x+1)-1)^3+b((x+1)-1)^2+c((x+1)-1)+d \\ f(x) = a((x+1)-1)((x+1)-1)^2+b((x+1)^2-2(x+1)+1)+c(x+1)-c+d \\ f(x) = a((x+1)-1)((x+1)^2-2(x+1)+1)+b(x+1)^2-2b(x+1)+b+c(x+1)-c+d \\ f(x) = a(x+1)^3-3a(x+1)^2+3a(x+1)-a+b(x+1)^2-2b(x+1)+b+c(x+1)-c+d \\ a(x+1)^3+b(x+1)^2+c(x+1)+d-3a(x+1)^2+(3a-2b+c)(x+1)-a+b-c \\ f(x) = f(x+1)-3a(x+1)^2+(3a-2b+c)(x+1)-a+b-c \\ f(x+1)-f(x)=3a(x+1)^2-(3a-2b+c)(x+1)+a-b+c $$

This seems to be in the right direction for the solution, and also yields an interesting result that is $d$ can take any value, kinda intuitive tho.

Now I want the $(x+1)$ term to vanish, I can achieve that with choosing $a$,$b$,$c$ to satisfy $3a-2b+c=0$, also $3a$ needs to be $1$ and at last, $a-b+c$ should be also zero.

$$ 3a = 1 \\ 3a-2b+c = 0 \\ a-b+c=0 \\ $$

Solving this on my trusty casio yields:

$$ a = \frac{1}{3} \\ b = \frac{2}{3} \\ c = \frac{1}{3} \\ $$

This suggests that the function

$$ f(x) = \frac{1}{3}x^3+\frac{2}{3}x^2+\frac{1}{3}x+d $$

Satisfies my equation $f(x+1) - f(x) = (x+1)^2$, but unfortunately it does not.

$$ \text{assuming} ~~ d=0 \\ f(1) = \frac{4}{3} \\ f(2) = 6 \\ f(2)-f(1) \neq 2^2 $$

I needed graphical representation to debug this problem, so here is a desmos graph

And indeed current solution is just barely off for a certain reason, that I can't figure out for the last couple of hours.

So what is the reason? Did I miss something? Thanks.

Answer 1

$$ f(x+1) - f(x) = (x+1)^2 $$ without knowing more information about the function f(x), we can't solve the original equation exactly. However, we can make some educated guesses about what f(x) might look like. For example, we might guess that f(x) is a polynomial function of degree 3 or higher. We could then try plugging in various polynomial functions of degree 3 or higher and see if they satisfy the equation. but for the sake of simplicity, we can assume that f ( x ) is of 3rd degree as you did, as follows

$$ f(x) = ax^3+bx^2+cx+d $$

By evaluating f (x+1)

$$ f(x+1) = a(x+1)^3+b(x+1)^2+c(x+1)+d \\ $$ $$ f(x+1) = ax^3+3ax^2+3ax+a+bx^2+2bx+b+cx+c+d \\ $$ Expanding f (x+1) - f (x) and simplifying ,we get : $$ f (x+1) - f (x) = 3ax^2+(3a+2b)x+(a+b+c) $$ By equating the previous equation to the original one: $$ 3ax^2+(3a+2b)x+(a+b+c) = x^2+2x+1 $$ we can now compare the coefficients to get $$ 3a =1 $$ $$ 3a+2b=2 $$ $$ a+b+c=1 $$ Solving this system of equations, we get the following: $$ a = \frac{1}{3}$$ $$ b = \frac{1}{2} $$ $$ c = \frac{1}{6} $$ Therefore, $$ f(x) = \frac{1}{3} x^3 + \frac{1}{2} x^2 + \frac{1}{6}x $$ Note that: the value of d doesn't matter as it will vanish due to the equations being subtracted from each other, so any constants will vanish.

Answer 2

HINT…You could make it a lot simpler for yourself if, having started with your general cubic expression for $f(x)$, you then simplified $f(x+1)-f(x)$ and obtained $$3a(x+1)^2-(x+1)(2b-3a)-5a+3b+c$$

From this you can deduce the correct values for $a, b$ and $c$ as given by Goncalo in the comments.

Answer 3

In fact we can describe explicitly the general solution to the functional equation.

First, any nonconstant polynomial function $p(x) = a_n x^n + \cdots$, computing the difference gives $$p(x + 1) - p(x) = n a_n x^{n - 1} + \cdots,$$ and in particular $\deg[ p(x+1) - p(x)] = \deg p(x) - 1$, so any polynomial function $p$ satisfying the condition $$p(x + 1) - p(x) = (x + 1)^2$$ must be cubic: $$p(x) = a x^3 + b x^2 + c x + d.$$ Substituting this ansatz in the condition gives $$3 a x^2 + (3 a + 2 b) x + (a + b + c) = x^2 + 2 x + 1,$$ and comparing like coefficients gives a linear system in $a, b, c$ with a unique solution, $$a = \frac{1}{3}, \qquad b = \frac{1}{2}, \qquad c = \frac{1}{6} .$$ So, the polynomial solutions are $$p(x) = \frac{1}{3} x^3 + \frac{1}{2} x^2 + \frac{1}{6} x + d, \quad d \in \Bbb R .$$ Notice that for $d = 0$, $p(x) = \frac{1}{6} x (x + 1) (2 x + 1)$, which for $n \in \Bbb N$ is the sum of the first $n$ positive squares---which isn't surprising given the original recurrence!

Now, suppose there were another solution $f(x)$ of the functional equation, which we write as $p(x) + g(x)$. Substituting in the condition gives $$(p(x + 1) + g(x + 1)) - (p(x) + g(x)) = (x + 1)^2 ,$$ but $p(x + 1) - p(x) = (x + 1)^2$, so equivalently $g$ is a solution of the "homogenized" equation $$g(x + 1) - g(x) = 0, $$ and this condition is exactly that $g$ is periodic with period $1$. The constant function $x \mapsto d$ has this property, so the general solution of the functional equation is $$\color{#bf0000}{\boxed{f(x) = \frac{1}{3} x^3 + \frac{1}{2} x^2 + \frac{1}{6} x + g(x)}} ,$$ where $g$ is periodic with period $1$, hence the set of solutions forms an infinite-dimensional affine space. Just as well, for any function $h : [0, 1) \to \Bbb R$, we can set $g(x) = h(x - \lfloor x \rfloor)$.

We can use this description to build pathological solutions to the original functional equation. For example, taking $$g(x) = \chi_{\mathbb Q}(x) := \left\{\begin{array}{ll}1, & x \in \Bbb Q \\ 0, & x \not\in \Bbb Q \end{array}\right. ,$$ produces a solution $f$ that is continuous nowhere.