Does there exist a number that makes this work?

Question

I am wondering if there is some kind of polynomial somewhere out there that makes the following functional equation true:

$y(x) = y(x-1) - \tfrac{1}{2}x.$ I don't normally work on this type of problem so I don't know what the standard procedure is. If I try testing say $a x^2$, then I can arrange that as $0 = a(x-1)^2-ax^2-\tfrac{1}{2}x.$ For some reason, instead of giving me scalar quantities in $a,$ wolfram gives me $a = x/(2 x^2 - 6 x + 2) $ and $x^2 + 1 \neq3 x.$ I just want a scalar, like $a = \sqrt{2}/2 - 1$ or something. Why isn't this working?

Answer 1

We have $$ y(x) = y(x-1) - \tfrac12 x. \tag1$$ You can substitute $x-1$ for $x$ and it must still be true: $$ y(x-1) = y(x-2) - \tfrac12(x-1). \tag2$$ Use Equation $(2)$ to substitute into equation $(1)$: $$ y(x) = y(x-2) - \tfrac12(x-1) - \tfrac12 x.$$

If we choose $x$ a positive integer, we can repeat this process as many times as needed to "unroll" the recursion:

\begin{align} y(x) &= y(x-1) - \tfrac12 x \\ &= y(x-2) - \tfrac12(x-1) - \tfrac12 x \\ &= y(x-3) - \tfrac12(x-2) - \tfrac12(x-1) - \tfrac12 x \\ &\vdots\\ &= y(0) - \tfrac12(1) - \tfrac12(2) - \cdots - \tfrac12(x-2) - \tfrac12(x-1) - \tfrac12 x \\ &= y(0) - \tfrac12 \sum_{t = 1}^x t. \end{align}

Now use the fact that $\sum_{t = 1}^x t = \frac{(x+1)x}{2},$ and we have that $$y(x) = y(0) - \tfrac14 x^2 - \tfrac14 x.$$

This is a polynomial, in fact a quadratic polynomial, but not a quadratic polynomial of the form $ax^2.$ That's why you don't get a solution when you assume the solution has to be $ax^2.$ There is no value of $a$ that will make up for the missing $\frac14 x$ term.

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Note that the derivation above assumes $x$ is a positive integer, so if we want the equation to be true for all real $x$ we still have to show it. But now that we have a formula, we can just try it for any real $x$:

\begin{align} y(x) - y(x - 1) &= \left(y(0) - \tfrac14 x^2 - \tfrac14 x \right) - \left(y(0) - \tfrac14 (x-1)^2 - \tfrac14 (x-1) \right) \\ &= - \tfrac14 x^2 - \tfrac14 x + \tfrac14 (x-1)^2 + \tfrac14 (x-1) \\ &= - \tfrac14 x^2 - \tfrac14 x + \tfrac14 (x^2 - 2x +1) + \tfrac14 (x-1) \\ &= -\tfrac12 x, \end{align} which (fortunately) is exactly what we need in order to establish the desired equation. (Until this verification step, we do not really know that the formula for $y(x)$ is correct; it is only an inspired guess up to that point.)

Notice that this works regardless of the value of $y(0),$ so the recursion does not define a unique function; it defines a set of quadratic functions in which you can set the constant term to anything you like.

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Another method, instead of working downward from a large positive integer, is to work upward from zero:

\begin{align} y(1) &= y(0) - \tfrac12(1). \tag3 \\ y(2) &= y(1) - \tfrac12(2). \tag4\\ \end{align}

Use $(4)$ to substitute into $(3)$:

$$ y(2) = y(0) - \tfrac12(1) - \tfrac12(2) . $$

Continue like that, using the equation for each value of $x$ in $y(x)$ to substitute into the equation for the next larger $x$:

\begin{align} y(3) &= y(2) - \tfrac12(3) = y(0) - \tfrac12(1) - \tfrac12(2) - \tfrac12(3). \\ y(4) &= y(3) - \tfrac12(4) = y(0) - \tfrac12(1) - \tfrac12(2) - \tfrac12(3)- \tfrac12(4). \\ \end{align}

By now you may see a pattern and be able to make a guess about a formula, which you can check using the same checking method as before.

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A third alternative: if you suppose initially that the formula is quadratic, then you only need to relate three values to each other, and this gives you three points on the curve relative to one of the points. If you are really inspired you might choose $x = -1, 0, 1$:

\begin{align} y(0) &= y(-1) - \tfrac12(0) = y(-1). \\ y(1) &= y(0) - \tfrac12(1) = y(0) - \tfrac12. \\ \end{align}

The equation $y(0) = y(-1)$ tells us that the graph of the function is a parabola with axis $x = -\tfrac12$ (halfway between $x=-1$ and $x=0$), so the equation has the form $$ y(x) = a \left(x + \tfrac12\right)^2 + h. $$

(This also explains why the answer cannot be of the form $y(x) = ax^2,$ because that would be a parabola with an axis at $x=0$.)

You can then use the fact that $y(1) = y(0) - \tfrac12$ to find the possible value of $a.$ The value of $h$ is arbitrary. You should still check that the formula works for all $x$ and not just these three values.

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I am not aware of any general method for solving all recursions of the form $y(x) = y(x-1) + f(x)$ for any known function $f.$ For $f$ a polynomial you will get a polynomial one degree higher (a fact that I did not use in the first two solution methods), but that's only one class of functions.

The first two methods in this answer can be summarized as "try a few consecutive values and see what happens", which is a technique you can try when you don't know anything better for a particular problem. It is not guaranteed to work but it does often help find a solution when a problem is not too hard.

Answer 2

Hint A polynomial with your property must be quadratic. indeed, if $z$ is the second derivative of $y$ then $z$ satisfies $z(x)=z(x-1)$. Only constant polynomials $z$ can satisfy this. So $z$ is constant and $y$ is quadratic. Now write $y(x)= ax^2+bx+c$ and try to find $a,b,c$.You have done that when $b=c=0$.

Answer 3

To solve $$ y(x)-y(x-1)=-\tfrac12x\tag1 $$ note that taking a difference (forward or backward) of a polynomial yields a polynomial of one degree less. For example, $$ x^2\overset{y(x)-y(x-1)}\longmapsto2x-1\tag2 $$ and $$ x\overset{y(x)-y(x-1)}\longmapsto1\tag3 $$ Using $(2)$, $(3)$, and the linearity of difference operators, we get that one solution to $(1)$ is $$ y(x)=\frac{-x^2-x}4\tag4 $$ The difference between any two solutions to $(1)$ must satisfy $$ y(x)-y(x-1)=0\tag5 $$ $(5)$ means the general solution to $(1)$ is $(4)$ plus a function with a period of $1$.

If you want a polynomial, the only periodic polynomial is a constant function. So there is the family of polynomial solutions $$ y(x)=\frac{-x^2-x+c}4\tag6 $$ where $c$ can be any constant.