Is there a function for every possible "path"?

[Jurgen M]

Has any path/line/shape/contour function? And how find function of complex "path"?

for example this?

 

[jedishrfu]

Well since your function backtracks on itself then no there is no equation f(x) that can represent it.

Do you know why?

However, if you parametrize both x and y with some variable called t then maybe you can piece together a vector function to imitate the path above.

 

[FactChecker]

A path is best parameterized. You can parameterize the above path in the form of $P: (x(t),y(t)), 0\le t \le1$.

 

[fresh_42]

This isn't a function in the sense of $f(x)=y$. But you can "walk" along the path as a function $[0,1]\longrightarrow \mathbb{R}^2.$ In this case, you have to glue all partial paths to a single long one.

 

[Jurgen M]

Well since your function backtracks on itself then no there is no equation f(x) that can represent it.

Do you know why?

However, if you parametrize both x and y with some variable called t then maybe you can piece together a vector function to imitate the path above.

No.Why?

What is procedure to find f(x) of some given complex path that is not backtracks itslef?

 

[fresh_42]

What is procedure to find f(x) of some given complex path that is not backtracks itslef?

You have to distinguish between the path, and the relation $\{(x,f(x))\}$. What do you want to talk about? One is a function $\mathbb{R} \longrightarrow \mathbb{R}^2$ and the other a subset of $\mathbb{R}^2$.

As long as this isn't answered, this thread will be a mess.

 

[FactChecker]

No.Why?

What is procedure to find f(x) of some given complex path that is not backtracks itslef?

It can not be in the form of f(x) because, for a single value of x, there are multiple values of y that are matched with it on the path. Which y-value should f(x) be?
On the other hand, suppose you parameterize the path in the form of $(x(t), y(t)), 0 \lt t \lt 1$. For any value of $t \in [0,1]$ you have exactly one (x,y) value.

 

[Jurgen M]

You have to distinguish between the path, and the relation $\{(x,f(x))\}$. What do you want to talk about? One is a function $\mathbb{R} \longrightarrow \mathbb{R}^2$ and the other a subset of $\mathbb{R}^2$.

As long as this isn't answered, this thread will be a mess.

Slowly.
First I don't understand this math terms..

What is difference between "path" and f(x)?

 

[jedishrfu]

For any value x there is one and only one y value. When the graph backtracks as yours does then for a given x there are more than one y values.

https://en.wikipedia.org/wiki/Function_(mathematics)

thats why you have to think of it more like the flight of a bumblebee whizzing around in 2D or 3D space and using the time t as the independent variable otherwise you'll get stung.

 

[fresh_42]

Slowly.
First I don't understand this math terms..

What is difference between "path" and f(x)?

I answered this in post #4. A path is a walkthrough of the line you drew (every point has a unique timestamp), from a starting point to an endpoint. If you consider $f(x)$ as a function graph, then it is a) not a function because there are points $x$ with two different values $f(x)$, and b) a subset of the plane. Such a subset can be considered as a relation. Relations allow ambiguities, functions do not.

 

[mcastillo356]

It's a relation, as far as I know (I am speaking without background; but I find interesting this thread) You are never express algebraically that path without a huge effort.

 

[fresh_42]

It's a relation, as far as I know (I am speaking without background; but I find interesting this thread) You are never express algebraically that path without a huge effort.

That's a guess. Therefore post #6. As long as everybody chooses for himself what the OP might have meant, as long will this thread be doomed to confusion.

 

[FactChecker]

It's a relation, as far as I know (I am speaking without background; but I find interesting this thread) You are never express algebraically that path without a huge effort.

You might be surprised at how well Fourier series can approximate such things. See this (especially the first few minutes).

 

[jbriggs444]

It's a relation, as far as I know (I am speaking without background; but I find interesting this thread) You are never express algebraically that path without a huge effort.

We distinguish between a "formula" and a "function".

A "function" is a mapping from elements in a "domain" to elements in a "range" such that each element in the domain is mapped to exactly one element in the range.

A "formula" like $\sin x$ or $\sqrt{t}\ \hat{i} + t^2\ \hat{j}$ can let you specify a function without having to write down a complete table for the mapping.

My guess (*nods to @fresh_42*) is that you are asking whether for every path you can draw with an ideal pencil on an ideal piece of paper there is a finite formula that describes that path with perfect fidelity.

The answer to that question is no. Unless I am misremembering, the cardinality of the set of continuous ideal paths is the cardinality of the continuum and the cardinality of the set of formulas is aleph null.

 

[fresh_42]

The answer to that question is no. Unless I am misremembering, the cardinality of the set of continuous ideal paths is the cardinality of the continuum and the cardinality of the set of formulas is aleph null.

If only a picture is given, then we can get arbitrary close with formulas and piecewise-defined paths.

This is the next problem with the question: How is that scribbling given? How accurately do we have to be?

 

[FactChecker]

Just to be clear, when a path is parameterized by a variable $t \in [0,1]$, like $P: f(t) = ( x(t), y(t) ), t \in [0,1]$, that is a function from $[0,1]$ to $\mathbb{R}^2$.

 

[fresh_42]

Just to be clear, when a path is parameterized by a variable $t \in [0,1]$, like $P: f(t) = ( x(t), y(t) ), t \in [0,1]$, that is a function from $[0,1]$ to $\mathbb{R}^2$.

Yes, and $G:=\{(x,y)\in \mathbb{R}^2\,|\,y=f(x)\}\subseteq \mathbb{R}^2$ is a relation, and in the case of $f$ as in post #1 not a function.

 

[WWGD]

I think this question is equivalent to whether every path can be parameterized. It seems that if it's rectifiable, it may be parameterized by arc length.

 

[jbriggs444]

I think this question is equivalent to whether every path can be parameterized. It seems that if it's rectifiable, it may be parameterized by arc length.

It is something akin to cheating, but even if the path cannot be parameterized as a continuous function over the reals, one can nonetheless unambiguously associate a real number t with each (x,y) coordinate in the graph (e.g. by interleaving the digits in the decimal expansions) and then declare f(t) = (x,y) and set the domain to the set of t values that are used. It is unlikely that this function will be continuous, but it will be a function from [a subset of] $\ \mathbb{R}\$ to the set of coordinates of points with pencil marks.

 

[WWGD]

It is something akin to cheating, but even if the path cannot be parameterized as a continuous function over the reals, one can nonetheless unambiguously associate a real number t with each (x,y) coordinate in the graph (e.g. by interleaving the digits in the decimal expansions) and then declare f(t) = (x,y) and set the domain to the set of t values that are used. It is unlikely that this function will be continuous, but it will be a function from [a subset of] $\ \mathbb{R}\$ to the set of coordinates of points with pencil marks.

How about just using the ortho projections to the respective axes as the coordinates. Wonder if this is specific enough for the op?