Math Trick With Functions

Okay, this is a relatively simple mathematical concept, but there is a trick a friend of mine stumbled upon, and I was wondering if anyone could tell me whether this can be widely applied or even if you guys have ever heard of this before..?

Supposing you have a table with the domain & range of a function, and you are left to find the function on your own..

Domain: 0,1,2,3

Range: 1,4, 7, 10

Obviously, the domain is X and the range is Y.

The trick I am wondering about is that if you look at the domain, you'll see it's going up by one. So, while the domain is going up by one, the difference between the ranges is what you are multiplying x by.

So: Y = 3x + 1

If you look at the domain/range I have posted above, you'll see that all the ranges are in intervals of 3. So, as long as you are using whole numbers, does finding the intervals of the ranges help you determine the equation used to turn out those results?

#473457

Wouldn't it be easier to just write the equation? I usually don't bother with finding the difference. I just write the equation for the line or whatever it may be.

However, in theory, it should work. But it's only dependant on the one, and only works if it is completely linear.

#473458

Well, it would be easier to just write the equation, but what my math teacher does is he gives us a graph, and asks us to make a chart with Domain/Range, Write a verbal rule, and make the equation for the function.

What I was asking is would this be a correct method by which to find the equation, I wasn't sure if it only worked for SOME problems.

#473467

i think it's easier to not think about stuff like that.

#473475

eh? you lost me after the second line :confused:

#473477

Where is Kyle? He is the Math Wizard in A-S

#473485

That's pretty much just a complicated (i.e. verbose) verbal description of the definition of slope... ( y2 - y1) / (x2 - x1), where the difference between x2 and x1 in your example is 1.

-NC

#473487

Indeed. Newtons interpolating polynomial comes to mind.

#473500

I like that equation, but I would let Macintosh do it...

#473769

Now you are in my field of expertise. I will try not to lose too many people :-) That "theory" is flawed, and here is why. Take your theory, and apply it as a graphed equation and you will see that if the vertex has both parent sets, then the range is transversal. Please excuse the font, as there is no other way to illustrate this on the fly. Put another way, consider this:

Take any factorization relative to your discovered "theory", such as n-1=m1m2... m2k into integer factors mj all greater than one; define sets M1,M2,... ,M2k by Mi={ tm1m2... mi-1: t=0,1,... ,mi-1} and write A=M1+M3+...M2k-1, B=M2+M4+...M2k . The partitionable graph, with vertices v0 , v1,..., vn-1, has alpha=m1m3...m2k-1 and omega=m2m4...m2k ; with subscript arithmetic modulo n, its stable sets of size alpha are the n sets {vj+a:a in A} with j=1,2, ... ,n and its cliques of size omega are the n sets {vj-b:b in B} with j=1,2, ... ,n. Your theory will therefore not specify the graph completely: for instance, if n=10 and m1=m2=3, then A={0,1,2}, B={0,3,6}, and so each vi may or may not be adjacent to vi+5. Therefore, you are suggesting that it has both of the properties, m1=m3=...=m2k-1=2 and m2=m4=...=m2k=2; How can that be? It can't. Now, let's revise your theory, and say that the graph has precisely one of these properties; this would prove that, by your theory, any such graphed equation with alpha(G)>2 and omega(G)>2 has a small transversal. By using the intervals of the ranges as you mentioned, this theoretical graph would contain an induced C5 (without indifferent pairs of vertices). However, it could be solved if expressed as a P4-structure. Using the "x" and "y" from your theory, let's say that the one natural invariant that interpolates between these two extremes is the P4-structure of x; this invariant is defined as the hypergraph whose vertex-set y is vertex-set of x and whose edges are the subsets of y that induce P4's in x. For example, here are two graphs with the same P4-structure, {a,b,c,d}, {b,c,d,e}, {c,d,e,f}, {d,e,f,a}, {e,f,a,b}, {f,a,b,c}: a b o--------o a b c / o---------o---------o / | | | / | | | f o o c | | | / | | | / | | | / o---------o---------o o--------o d e f e dSince P4 is a self-complementary graph, the P4-structure of x has the requisite properties to solve your equation. Of course, it is always just easier to do it the right way instead of trying to use mathematically incorrect "tricks" a friend showed you.

#473809

Well with a linear equation in 2 variables it is possible, and that's because your graph will be linear.

So technically that is actually what you're doing. Like NC said, linear equations in 2 variables simply define a line.

y = mx + c

Where m is the slope of the line given by (y2-y1)/(x2-x1) and c is the intercept of the line on the y axis (the value of y when x = 0)

So basically with a linear equation in 2 variables, the function is one-one meaning that each element in the range has a unique pre-image in the domain. Also the elements in both sets will be in arithmetic progression.

Therefore there is nothing to do but write the equation

y = mx + c

Find m by using the formula I mentioned (using consecutive elements of the set)

Find c by checking the value of y when x = 0

And then multiplying both sides of the equation by the denominator (if any) in the m fraction.

This applies strictly to a linear equation in 2 variables.

#473811