# Formalization - does this make sense?

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Hello.

I would appreciate some help in determining how much sense the following really makes. I don't know very much about "writing mathematics" so any advice is very welcome.

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Let $f:\mathbb{Z}T\rightarrow 0$ be a periodic function with period $T$ and let $g:\mathbb{Z}T\rightarrow \mathbb{R}$ be invertible. Then the zeros of the composition $f\circ g^{-1}:\mathbb{R}\rightarrow 0$ is the set of all $t$ for which $t=g\left(\frac{1}{2}Tz\right),z\in \mathbb{Z}$, is satisfied.

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That is it. I can explain more if necessary. I welcome all constructive feedback, positive and negative.

Thank you.

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Notation is hard to follow. A few confusions on my part:

* What is $T$?

* What can it mean for $f : \mathbb Z T \rightarrow 0$ to be periodic? It can't be anything other than the zero function. It's trivially periodic with every number in the domain being a period.

* Since $f$ is the zero function, $f \circ g^{-1}$ must be zero anywhere it's defined.

Can you clarify your intent here?

Edited by wtf

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T is sometimes used in matrix notation as a transformation matrix (like say you wanted to skew or shear a plane spanned by some linearly independent vectors). I would be very explicit.

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Thank you for your responses. I would be happy to clarify.

The function$f$ is meant to be a periodic function with period $T$ that has zeros for every element of its domain, being the integers times the constant $T$.

To give a solid example, let $f=sin\left ( \frac{2\pi }{T}x \right )$ and let $g=x^{2}$. Then $f(0)=0, f(T)=0, f(2T)=0,$ etc., so that all integer multiples of the period map to some zero of the sine function.

And then the composition $f\circ g^{-1}=f(g^{-1}(x))=sin\left(\frac{2\pi}{T}g^{-1}(x)\right)$ is an aperiodic wave with a variable period whereby the original periodic wave has been bent out of shape, the idea being to take a regular wave and to make it aperiodic.

Then we can describe the zeros of the composition as the points where $f$ is said to vanish. What are these values? They are all $x$ for which

$\frac{2\pi}{T}g^{-1}(x))=\pi z, z\in\mathbb{Z}$

which, when solved for $x$ yields

$x=g\left(\frac{1}{2}Tz \right )$.

The original goal was to communicate in a more formal manner. What could be done differently? I am thankful for your input.

(Sorry for my errors or typos, if any; I hope the idea is still clear despite.)

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The function$f$ is meant to be a periodic function with period $T$ that has zeros for every element of its domain, being the integers times the constant $T$.

Ok, $T\$is a constant. What kind of constant? Integer, real, complex? Quaternion maybe? I'm guessing it's a real but you should not make your readers guess. Better to say:

Let $T \in \mathbb R$.

When you say that the domain is the integer multiples of $T$, then that's all we know about f. As far as we know it's not defined anywhere else. But I don't think that's what you mean. Rather, I think you mean that $f$ is defined on ALL the reals and happens to be zero on integer multiples of $T$. Is that right? If so:

Let $f : \mathbb R \rightarrow \mathbb R$ with $f(nT) = 0\ \text{if}\ n \in \mathbb Z$.

Now is $f$ continuous? Differentiable? What else are we assuming about $f$? If you have any other unspoken assumptions, they should be stated clearly.

To give a solid example, let $f=sin\left ( \frac{2\pi }{T}x \right )$

You should write $f(x)$ on the left. Otherwise we don't know what the independent variable is. Is it $x$? Is it $T$? Or maybe it's $\pi$! Since $x$ is the independent variable, we write $f(x)$.

I didn't look at the rest but hopefully you see what needs to be clarified. The main thing is that the domain is ALL the reals, and the zeros happen to be integer multiples of $T$, assuming that's what you mean. And as a general principle, try to eliminate any ambiguity in the mind of the reader.

ps -- You have a subtle error in your logic. You have $g(x) = x^2$ and later you ask us to consider $g^{-1}$. But $g^{-1}$ does not happen to be uniquely defined. So you have to either tell us what $g^{-1}$ is, or show that your argument is independent of the choice of a particular one-sided inverse of $g$.

Maybe I better say a few words about a common point of confusion. The word inverse has two different meanings when it's applied to functions.

If a function $f$ is invertible, that means (by definition) that there exists some specific function that uniquely inverts it. (I'm omitting the technical details here). But $x^2$ is not invertible. Rather for each point in the range, there are either one or two points in the domain that get mapped to that point in the range; and we can create "a", not "the" inverse by making a choice from each of the inverse images of points in the range. But this will only be a one-sided or "partial" inverse.

So you can't just use the notation $g^{-1}$ because it's not uniquely defined.

Here's more than you ever want to know about it, but you should give this page a look:

Pay particular attention to the section on partial inverses.

Edited by wtf

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Great help there, wtf. +1