# Can you help me build a simple formula?

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• Jul 27th 2010, 10:53 PM
Flightline
Can you help me build a simple formula?
I'm not in school, and I'm not looking for any answer to any homework assignment. I have something I'm doing on my own, and I wanted to see if anyone comes up with the same formula, or close to it, as I did for the following.

Take a graph: draw a line so that $x=y$, which we know is a $45^o$angle and a $135^o$ angle from $(0,0)$ heading in a positive and negative direction on to infinity.

Now, let's say that the line goes to $(1,1)$ then $(2,2)$ then $(3,3)$ etc. What formula would also extend the line to $(-1,-1)$ $(-2,-2)$ $(-3,-3)$ ...ect., as the line extends positively?

In other words, you start from $(0,0)$ and the vectors extend positively, and negatively in proportion to each other. When the vector goes from $(0,0)$ to $(1,1)$ to $(2,2)$ it then extends to at the same time to $(-1,-1)$, $(-2,-2)$, etc.

I understand people have their own work and may not want to be sidetracked with this, but I do appreciate anything anyone has to suggest. My guess is this is a well-known formula, and I'm just don't know well enough to know it. Nevertheless, I want to get this simple formula down before proceeding on with my work.

Thank you very much in advance for anything.

Ed
• Jul 27th 2010, 10:57 PM
yehoram
Hello !

The same formula is suitable $\ \ y=x \ \$
• Jul 27th 2010, 10:57 PM
pickslides
$y=x$ goes through all the points mentioned above
• Jul 27th 2010, 11:14 PM
Flightline
Thank you, I thought about that, that simply X=Y would do, but that produces an infinite line in both directions. I was looking for a formula, or call it a model, if you will, that indicats that as the line goes to 1,1 it also goes to -1,-1. I have a formula I'd like to show here, but I'm not sure how, just yet. But I will try using TEX, and do a bunch of edits and see what happens. Here it goes:

$
\left(f_(_x_)= x | x=\sum\limits_{n=0}^{\infty}n+1\right)\Longrightar row \left(f_(_x_)= x | x=\sum\limits_{n=(-1)}^{\infty}n-1\right)$
• Jul 28th 2010, 12:13 AM
Flightline
OK, above is the formula I have come up with.
Jeez...talk about a crash course in LaTex(Headbang)
In your opinion, what is wrong with this formula? does it do what I am wanting it to do?

I don't know. I think I don't need the summation symbols. I'm not summing an infinite series for (x,y). I'm just trying to say that if (x,y) is such and such a coordinate, then the line moves from zero as well to the negative of that coordinate.

I should have waited to post this until I worked with it. Oh well, it was good LaTex practice.

Again, I realize this takes up your time, so any comments I appreciate.(Hi)
• Jul 28th 2010, 12:35 AM
yeKciM
$\displaystyle (\forall x,y \in \mathbb{R} ) (y=x)$

it's just a line, and for every x witch goes from $\displaystyle -\infty$ to $\displaystyle+\infty$ u'll have $\displaystyle y=x$ ....
• Jul 28th 2010, 03:47 AM
HallsofIvy
Quote:

Originally Posted by Flightline
Thank you, I thought about that, that simply X=Y would do, but that produces an infinite line in both directions. I was looking for a formula, or call it a model, if you will, that indicats that as the line goes to 1,1 it also goes to -1,-1. I

I'm not certain what you mean by "as the line goes to 1,1 it also goes to -1,-1". A line doesn't "go" to one point or another, it just is there. Perhaps what you mean is a set of functions: $f_X(x)= x$ for $-X\le x\le X$. Or a function of two variables: $f(x,t)= x$ for $-t\le x\le t$.
• Jul 28th 2010, 08:52 AM
Flightline
Thanks for the input, and now I wish I had waited before posting about this until I worked it out better. Nevertheless, I know what $x = y$ means, or at least what it looks like. It's a line that runs infinitely in both directions from $(0,0)$, i.e., $\{...-3,-2,-1,0,1,2,3...\}$

What I want to show is that there is a force starting at point $(0,0)$. If the force increases in magnitude in one direction, there is a corresponding increase in the other direction. Thus if the increase in magnatude is represented by a line going from $(0,0)$ to $(1,1)$ in the first quadrant of the graph, the line also increases in the third quadrant of the graph from $(0,0)$ to $(-1,-1)$.

$x=y$ is not the formula that illustrates this, or at least it doesn't seem that way to me. Because as has been pointed out, $x=y$ or other forms of saying the same thing such as $\displaystyle (\forall x,y \in \mathbb{R} ) (y=x)$ merely stand for a line that goes from infinity to infinity along that vector.
• Jul 28th 2010, 09:01 AM
SpringFan25
edit wrong answer
• Jul 28th 2010, 09:35 AM
Flightline
After some more thought, I came up with this. What do you think?

$
\left(f_(_x_)= x | x=n+0)\Longrightarrow \left\left(f_(_x_)= x | x=0-n)$

If $y=x$ such that $x = \{1,2,3,...\}$; then $y=x$ such that $x=\{-1,-2,-3...\}$
• Jul 28th 2010, 09:57 AM
SpringFan25
this is all very confusing. Why are you trying to find an equation to draw 2 vectors at once?

If your objection to y=x is that it is unbounded, then you can simply add a boundary
$y=x : -c < x < c$

Alternatively, graph the vectors seperately:
Vector 1: (c,c)
Vector 2: -(c,c)
• Jul 28th 2010, 11:23 AM
bondesan
Flightline,

Maybe you should see this blog post - the equation of a line segment. I think this is what you're looking for.

The Equation of a Line Segment

I hope it helps.
• Jul 28th 2010, 12:30 PM
Flightline
Quote:

Originally Posted by SpringFan25
this is all very confusing. Why are you trying to find an equation to draw 2 vectors at once?

If your objection to y=x is that it is unbounded, then you can simply add a boundary
$y=x : -c < x < c$

Alternatively, graph the vectors seperately:
Vector 1: (c,c)
Vector 2: -(c,c)

Thanks, but I don't think that will work for what I'm trying to model. I'm trying to model the idea that as a vector progresses one way, it causes an equal and opposite progression the other way. As to why I'm doing it, well, I'm not sure. I think it represents a concept in a larger concept I'm working on, but since it's probably all a delusion of grandeur anyway, I'd rather not embarrass myself by talking about it.(Blush)
• Jul 28th 2010, 12:32 PM
Flightline
Quote:

Originally Posted by bondesan
Flightline,

Maybe you should see this blog post - the equation of a line segment. I think this is what you're looking for.

The Equation of a Line Segment

I hope it helps.

I actually think that article will help me a lot in my understanding of what I'm trying to do. Thank you very much for leading me to it. I'm going to give it a good read in the bathtub--the place where all great ideas are born!
• Jul 28th 2010, 01:00 PM
bondesan
Hahaha, good one. You are welcome.