# Spring 2016, problem 11

### Comments

Call the equation from the problem statement "the PE". The functions g(x)=x and h(x)=x2 are easily checked to solve the PE. |

$\begin{array}{l}{\rm{We calculate successive terms of Newton's forward differencing formula }}f\left( x \right) = f\left( 0 \right) + \sum\limits_{n = 1}^\infty {\left( {\begin{array}{*{20}{c}}x\\n\end{array}} \right)} {\Delta ^n}f\left( 0 \right)\\{\rm{Suppose that }}f:Q \to Q{\rm{ is a function satisfying the constraint in the problem}}{\rm{.}}\\{\rm{Let }}w = x = y = z = 0.{\rm{ Then }}9f\left( 0 \right) = 8f\left( 0 \right).{\rm{ Therefore }}f\left( 0 \right) = 0.\\{\rm{Let }}x \in Q{\rm{ be arbitrary and let }}w = 0,y = z = 1.{\rm{ Substituting into the constraint}}\\3f\left( {x + 2} \right) + f\left( x \right) + f\left( {x + 1} \right) + f\left( 2 \right) + f\left( 1 \right) + f\left( 1 \right) + f\left( {x + 1} \right) = 2f\left( {x + 1} \right) + 2f\left( {x + 2} \right) + 2f\left( 2 \right) + 2f\left( {x + 1} \right)\\{\rm{ and solving for }}f\left( {x + 2} \right)\\f\left( {x + 2} \right) = 2f\left( {x + 1} \right) - f\left( x \right) + f\left( 2 \right) - 2f\left( 1 \right)\\{\rm{Therefore}}\\{\Delta ^2}f\left( x \right) = f\left( {x + 2} \right) - 2f\left( {x + 1} \right) + f\left( x \right) = f\left( 2 \right) - 2f\left( 1 \right)\\{\rm{Since }}x \in Q{\rm{ is arbitrary }}{\Delta ^2}f\left( {x + 1} \right) = {\Delta ^2}f\left( x \right),{\rm{ Then}}\\{\Delta ^3}f\left( x \right) = {\Delta ^2}f\left( {x + 1} \right) - {\Delta ^2}f\left( x \right) = 0\\{\rm{By induction on }}n \ge 3\\{\Delta ^{n + 1}}f\left( x \right) = {\Delta ^n}f\left( {x + 1} \right) - {\Delta ^n}f\left( x \right) = 0.\\{\rm{In particular }}{\Delta ^n}f\left( 0 \right) = 0{\rm{ for all }}n.{\rm{ Substituting into Newton's forward difference formula}}\\f\left( x \right) = f\left( 1 \right)\left( {\begin{array}{*{20}{c}}x\\1\end{array}} \right) + \left[ {f\left( 2 \right) - 2f\left( 1 \right)} \right]\left( {\begin{array}{*{20}{c}}x\\2\end{array}} \right) = \alpha x + \beta {x^2}{\rm{ where }}\alpha {\rm{,}}\beta \in Q.\\{\rm{Conversely if }}f\left( x \right) = \alpha x + \beta {x^2}{\rm{ with }}\alpha ,\beta \in Q{\rm{ then it is routinely verified that }}f{\rm{ satisfies the constraint}}{\rm{.}}\end{array}$

f(x)=−x 2 +6x−8 We can see function is a polynomial and the domain of polynomial function is real number. ∴ x∈R f(x)=−x 2 +6x−8 =−(x 2 −6x+8) =−(x 2 −6x+9−1) =−(x−3) 2 +1 Maximum value of −(x−3) 2 would be 0 ∴ Maximum value of −(x−3) 2 +1 would be 1. ∴ f(x)∈(−∞,1] We can see from the given graph that function is symmetrical about x=3 and the given function is bijective. So, x would be either (−∞,3] or [3,∞) The correct option which satisfy A and B both is: A=(−∞,3] and B=(−∞,1]

In this question the origin is taken to be at a harbour and the unit vectors i and j to have lengths of 1 km in the directions E and N. A cargo vessel leaves the harbour and its position vector t hours later is given by r1 = 12ti + 16tj. A fishing boat is trawling nearby and its position at time t is given by r2 = (10 - 3t)i + (8 + 4t)j. Here is the

The possible functions are exactly those of the form $f(x) = \alpha x^2 + \beta x$ with $\alpha,\beta \in \mathbb{Q}$.

Proof:

Call the equation from the problem statement "the PE". The functions $g(x) = x$ and $h(x) = x^2$ are easily checked to solve the PE. Since the PE is linear, every linear combination $f(x) := \alpha h(x) + \beta g(x) = \alpha x^2 + \beta x$ will solve the PE. It remains to be proved, that all solutions are of this form.

So let $f$ be any solution of the PE. Letting $w=x=y=z=0$ in the PE gives us $f(0) = 0$. Letting $(w, x, y, z) = (a, a, a, (z-3)a)$ for some $z \in \mathbb{Z}$ and $a \in \mathbb{Q} $ gives:

$f(za) = \frac{1}{3}(2f(3a) - 3f(2a)) + 2 f((z-1)a) -f((z-2) a$. [Eq.1]

Now we keep $a \neq 0$ fixed. We define $d_z:= f(za) - f((z-1) a)$ and $C := \frac{2}{3}f(3a) - f(2a)$ and rewrite [Eq.1] as:

$d_z = d_{z-1} + C$. The general solution to this recursion is $d_z = C z + q$ with $q \in \mathbb{Q}$. But, since $f(0) = 0$, we have:

$f(nx) = \sum_{i=1}^n d_i = \frac{C}{2} n (n+1) + q n = \frac{C}{2} n^2 + (\frac{C}{2} + q) n$ for every $n \in \mathbb{N}$ [Eq.2].

The solution [Eq.2] of the recursion [Eq. 1] can be extended routinely back to negative $n$. Define $\alpha := \frac{C}{2}$ and $\beta := \frac{C}{2} + q$. We rewrite:

$f(za) = \alpha z^2 + \beta z$ for all $z \in \mathbb{Z}$.

Since $a$ was arbitrary, for a given $r \in \mathbb{N}$ there exist $\alpha_1$ and $\beta_1$, such that:

$f(z \frac{a}{r}) = \alpha_1 z^2 + \beta_1 z$.

Now pick an $s \in \mathbb{Z}$ and let $z := sr$. Then we have: $f(sa) = \alpha_1 r^2 s^2+ \beta_1 r s = \alpha s^2 + \beta s$.

Comparing coefficients: $\alpha_1 = \frac{\alpha}{r^2}$ and $\beta_1 = \frac{\beta}{r}$. So:

$f(\frac{z}{r} a) = \frac{\alpha}{a^2} [ \frac{az}{r} ]^2 + \frac{\beta}{a} \frac{az}{r}.$

But as we go through all $z \in \mathbb{Z}$ and $r \in \mathbb{N}$ the number $\frac{az}{r}$ goes through all rationals. Therefore:

$f(q) = \frac{\alpha}{a^2} q^2 + \frac{\beta}{a} q$ for all $q \in \mathbb{Q}$, as desired.