# Two graphs, one relation - Pt. III

## March 27, 2021 | 7 minutes, 17 seconds

# A Little Exploration

## Recap

By this point, we've established the following, glorious, but otherwise limiting equation (which is okay!):

\[ \left|y-A(x)\right| = \hat{A}(x)\]

We did this in two steps: we derived the main equation and assumed a condition; we then proved the condition for our equation. (As incidental as it seemed to be!) This provided us with the ability to graph two functions via one equation, with the ability to retain intersections between them both, provided that one function was above another.

A forewarning to readers - this post is largely oriented around the examples and implications of our relation, as well as taking it a bit further. It is heavily recommended you play with this yourself via Desmos or GeoGebra, both amazing wonderful services I've linked. Also see WolframAlpha if you're super interested.

# Examples

So really, where are we leading with this? What's the point of this, how can we use this? It's sort of funny, really. The restrictions imposed by our equation actually create some really funky properties.

Up to this point, we used linear functions. We didn't provide any examples, because it's simple to see a running average between two lines - but now let's switch tact. Instead of assuming we have functions that this *always* holds for, that we can easily see the average line that exists, let's pick functions that it doesn't, ones that have intersections.

## The most trivial example: \(g(x) = f(x)\)

This will yield us \(\left|y-f(x)\right| = 0\). This is actually quite boring; our condition for our functions' existences is \(f(x) = f(x)\). In essence, our average is the function we can choose, and we *always* plot it. Not very fun, honestly.

A note for curious readers: Desmos almost always fails to graph this. It took me a while to figure out why. Can you? You can make use of GeoGebra for this, here.

## The (almost) most trivial example: \(g(x) = -f(x)\)

This will yield us the very 'clean' equation: \(\left|y\right| = f(x)\). Our equation tells us that this will only exist for which \(f(x) \geq 0\).

We can pick some very simple functions for \(f(x)\). Such as \(x^2\). This isn't very interesting - our function always exist and graphs both \(x^2\) and \(-x^2\). What about \(x\)? Well it's more interesting here - it's simply a 'sideways' absolute value function (when we're talking in terms of \(y=\left|x\right|\). If we return to \(x^2\) - what happens if we shift it down by say, 1, to give us \(x^2 - 1\)? Well here's more interesting - the equation creates an ellipse-like closed shape (which is actually two restricted parabolas), \(\left|y\right|=x^2-1\) (what happens if you do \(\left|y\right|=1-x^2\)? What does this mean?). Desmos is actually fairly handy here, by the way. Not promising all functions will work, though!

By the way - you can use this to visualise \(x = y^2\). Given \(y = \sqrt{x}\) or \(y = -\sqrt{x}\) and they both satisfy our restrictions, and fit into this example here particularly well, you can graph \(|y| = \sqrt{x}\). (Also by the definition of the square root of something squared. Note when you do this - you're actually going \(y^2 = x\) -> \(y^2 = \sqrt{x}^2\) -> \(\left|y\right| = \left|\sqrt{x}\right|\) -> \(\left|y\right| = \sqrt{x}\) because the square root function is inherently always positive).

This absolutely simple relation between our functions will actually be a super handy tool in a future post, particularly.

See if you can have a go at creating your equations with this relation! If you try out trigonometric functions, you're bound to get some *really* fun results!

## Any other function

You can plug in any other function you want and explore with this. Most will intersect. All the more fun! (Be careful with domains and all that though. \(ln(x)\) doesn't like it when you try and plug in \(-1\) expecting it to yield something not crazy)

# Following on

## Final extension

Back in our original post, we assumed \(f(x) \geq g(x)\). Here, let's do the opposite, and keep it in terms of \(A(x)\) and \(\hat{A}(x)\) for the sake of simplicity.
Let \(g(x) \geq f(x)\). Our average line remains the same, but the conjugate of it is the *negative* of our original conjugate.

This means we have the equations:

\[ \left|y-A(x)\right| = \hat{A}(x)\]

and

\[ \left|y-A(x)\right| = -\hat{A}(x) \]

What does this mean? Let's call our second equation a patch. It 'patches' our first equation because it covers everywhere in the plane where \(g(x)\) is greater than \(f(x)\). If there's nothing to patch, it simply won't exist. If there is, then it'll sort of cover it up.

But it won't do that on its own. What we need to do is combine the equations, in the same manner we did with \(y\) but this time in terms of \(\hat{A}(x)\).

Notice we have \(\hat{A}(x) = \left|y-A(x)\right|\) and \(\hat{A}(x) = -\left|y-A(x)\right|\). Applying our original relation, we obtain \(\left|\hat{A}(x)\right| = \left|y-A(x)\right| \implies \left|y-A(x)\right| = \left|\hat{A}(x)\right|\).

Interesting. Let's square it - \(\hat{A}(x)^2 = (y-A(x))^2\). This doesn't create any extra solutions. Solving:

\[ (y-A(x)-\hat{A}(x))(y-A(x)+\hat{A}(x))=0\]

Using our null factor law here (and here it ties in with the start of our first post!), we get that either \(y = A(x) + \hat{A}(x) = f(x)\) or that \(y = A(x) - \hat{A}(x) = g(x)\). Firstly - no \(A(x)\) remains, which is important. Secondly, this is *always true*. This means that, in some way, our *patched* equation implies the null factor law, and vice versa. Very cool!

## A nice summary

So we have several nice take-away points here:

- We can graph two restricted functions via one equation (and, in fact, 4, as by our final patch we actually graph 4 individual parts!).
- If we graph two unrestricted functions via our patched equation, it must imply the null factor law and vice versa.

And one very succinct and beautiful summary provided by a friend:

If you want to graph f and g simultaneously via y, for example, the distance between y and the average of f and g is always going to be half the distance between f and g.

And so concludes a 3-part post series. I will, however, being making use of this in the future (how much I don't know, but at least one). It's been an amazing adventure, through coincidence and dubious graphing programs like Desmos, trial and error and eventually something cool to make use of.

Please feel free to email me about any questions you may have about this :). I'll try and help as much as I can. This is, after all, my hobby and honestly an obsession at this point. ~Ally.