Erratic quadratic

Steve says: When I made this problem I wondered how people might approach its solution. In my notes, I recorded that "This quadratic does the trick $y= 2(x-1)(x+1) -1$" and was delighted to see Herschel's and Ben's wonderful analyses by the least squares method improving upon this. I include both of these in full.

 

Both Ben, from Kenny, and Herschel, from the European School of Varese, realised that minimising the squared distance from the points reduced to an (involved!) exercise in calculus (Ben provides the details in his write-up). Both of the solvers discovered the 'ideal' solution

 $$

f(x)= \frac{100}{49}x^2-\frac{20}{7}

$$

The least squares analysis needed to find this points towards the sort of thing that goes on at university, so very well done for thinking about this!

Read Ben's full, and very detailed solution, which I include as full pdf.

 

Herschel gave a lovely account of the thought process:  

 

At first, when considering this problem, the only simple approach seems to be trial and error - experimenting with different values for $a$, $b$ and $c$ in $y=ax^2+bx+c$.

Since the points are symmetric about the vertical $y$-axis, we can reduce this to $y=ax^2+c$, so that our parabola is also symmetrical about the $y$-axis. One could try various combinations of $a$ and $c$ and search for a good combination, but there is no guarantee of a simple solution. Also, we need a way to measure the distance from a parabola to a point, which isn't easy if you don't know what the parabola is in the first place!

 

Mathematicians trying to fit a parabola to a set of data often use the "Least squares" method. Instead of trying to find the perpendicular (and usually diagonal) distance between the point and the curve, we will measure only the vertical distance. The distance between a point and the parabola could be positive or negative depending on whether the point is above or below the parabola. We want to find the sum of all these distances (errors) so we first make them all positive by squaring them (hence the name "Least squares" - we try to minimise the sum of the square of the distances). We can actually form a set of equations that use the Least Squares principle. Using some differentiation and calculus (which I won't delve into now), they can be reduced to a set of linear equations:

$$ 3c+14a=20\quad\quad 7c+49a=80 \quad\quad 28b=0$$

The third of these equations tells us straight away that $b=0$, although we already knew that! Solving the other two equations give us: $a=\frac{100}{49}, c=-\frac{20}{7}$ Thus, our "ideal" quadratic according to this method would be

$$

y = \frac{100}{49}x^2 -\frac{20}{7}

$$

Of course, we could round this to $a=2$ and $c=-\left(2+\frac{2}{3}\right)$, or even further to $c=-3$, so that we get a nice and simple final result of $y=2x^2-3$. Some might say that we could've just found this by trial and error, and indeed, they would be right - we would have skipped all the calculus and algebra and got to the answer much sooner! But where's the fun in that, eh? :)  

 

 For the second part of the question Ben looked at finding the smallest distances between the points and curve, and realised that the perpendicular distances are what are needed. 

 

The smallest distance will lie along the normal of the line at a point where the normal also crosses the point given. In other words, a point on the line where the normal to that part of the line crosses the given point is the closest part of the line. So that means that first we must work out the derivative of $f(x)$; We can then include this in the formula for a straight line to work out the normal at each point and denote this as $g(x)$:

 

$$

g(x) = \frac{100}{49}x^2_0 -\frac{49}{200 x_0}x -\frac{3657}{1400}

$$

Ben then found the normals which pass through the points and solved numerically to show that:

 ... it may not be close for each predicted y-coordinate, but it does get very close for the iterated points and never exceeds being $0.2$ units away from any of the points given.

 

Ben rather cunningly also found another interesting solution by considering functions which pass exactly through the points and noting that splicing together two quadratics yields a solution.   

As well as noticing that $f(x) = f(-x)$ I noticed another interesting property of this function (assuming the function goes through all the points). Imagine that it was a standard $x^2$ graph when $x> 0$, then the x values needed to pass through these points would be $0, 2, 4$. If you match these up with the ones we are given and call the standard ones $u$ and the given ones $x$ then you'll notice a basic pattern emerge if we map $x\rightarrow u$: $1\rightarrow 0, 2\rightarrow 2, 3\rightarrow 4, \dots, x\rightarrow 2x-2$. Therefore

$$

u=2(x-1) \mbox{ if } x> 0

$$

If you do this with $x< 0$ then the pattern becomes

$$

u=2(x+1) \mbox{ if } x< 0

$$

This means that u can be written as a function of $x$ (ie $u=g(x)$):

$$

g(x) = 2(x-1) \quad(\mbox{ if } x> 0)\quad;\quad 2(x+1) \quad(\mbox{ if } x< 0)

$$

This function returns the value of u, which we can then square to get the actual value of the function. Or mathematically

$$

f(x)=g(x)^2

$$

We can also simplify the function $g$ using the sgn function which returns the sign of the variable. In other words $sgn(+5)=+1$, $sgn(-123)=-1$, $sgn(0)=0$. Also note that $sgn(x)^2 = 1$, unless $x=0$. This simplifies $g(x)$ to

$$

f(x)=2(x-sgn(x))

$$

This can then be substituted into f(x) and simplified to give

$$

f(x)=4(x-sgn(x))^2 = 4x^2-8|x|+4

$$

This can be expressed as

$$

f(x)=4x^2-8\sqrt{x^2}+4

$$

This formula works, but breaks down when asked what $f(0)$ is, and its not really a proper quadratic, but other than that it can get all the right results and is 0 units away from every point.