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That's why I clearly said for small mod.
No, you said "Computing has replaced thinking."
My question: How can I re-create the above polynomial using the finite field coordinate locations?
You can't.
First off, four points determine a cubic polynomial in the real numbers. So consider the polynomial which passes through
(0, 4 + 10*251), (1, 163 + 251), (2, 1888 + 5*251), and (3, 6685 * 251)
And remember that 4 points uniquely determine a cubic. This cubic must be different than the one you wrote, and so these points are not enough to determine your polynomial.
I'm counting 6, and I believe my enumeration covers all cases (up to equivalence).
Computing has replaced thinking.
Let's see you go compute all groups of order 8633 up to isomorphism.
Principle of Expediency: A good answer today is better than a great answer tomorrow.
Why are both not an option?
In any case, I'll wager the time it took for me to find those results was much shorter than the time it would take you to write a program.
Normally for a small mod like that I would just plow through the 64 possibilities of
My post reduced it to 15 possibilities.
First: all x,y that are divisible by 9 are solutions.
This is what ZHero meant by (0,0). Anything congruent to a solution will of course be a solution.
ZHero, the way you typically do these problems is you eliminate groups of cases till there is a small enough number to check by hand. For example, (x, 0) and (0, y) will not be solutions for x and y nonzero. Further, there are no solutions for x=1 or y=1.
No solutions where x+y = 9 except for x=3 and y=6 (or vice versa), and this is a solution.
It sounds like you don't have many tools, for example the correspondence between covering spaces and subgroups of the fundamental group. If that's the case, then I don't know how you'd ever prove you have all. Think about the behavior at the center point of the figure 8. In a neighborhood around the preimage of this point, your cover must at the same way. All other points are uninteresting because lines will lift to lines (remember, you're looking locally).
Break it down. First solve the following problem:
If z and w are any non-negative integers, how many solutions are there to:
z + w = n
Where n is any integer.
Once you know this, we can use this solution because we just need to the number of solutions to:
z + w = 20 - x - y (= n)
So we can split it up into cases for each pair (x,y) with x > y.
Have you heard of a Universal cover?
What definition? The epsilon-delta definition?
Number 2 is wrong. Should that be x approaching infinity?
If so, jeez... whats the point o.o
To give 1 point to people who at least understand the question.
Hint:
Before you start considering sequences, it is best to say what we mean by the "size of infinity". Of course when we say this, we mean the size of an infinite set.
So we can begin with something you know: the size of a finite set. For example: {a, b, c} and {1, 2, 3}. These sets have the same size. You could say that their size is three, but the genius Cantor discovered that you don't need numbers to talk about the relative sizes of these sets. We say that two sets are the same size if you can pair elements together so that:
(a) You don't use a single element twice and (b) every element is used.
For example, a pairing from the sets above is: {a, 1}, {b, 2}, {c, 3}
It should be obvious that if two finite sets have the same size, then you can pair their elements in this way. And if course, if two finite sets can be paired, then they have the same size.
Now here is the kicker: When it comes to infinite sets, we can't describe their sizes by a number. But we can still try to pair elements. For mathsyperson's example, we have the natural numbers and the even natural numbers. Here is a pairing:
{1, 2}, {2, 4}, {3, 6}, {4, 8}, {5, 10}, ..., {n, 2n}, ...
I use every element from each set only once, and I used every single element from each set. Therefore, the natural numbers have the same size as the even natural numbers. Now this may seem odd, because the even numbers are properly contained inside the natural numbers, but infinity is odd, so that's ok.
Cantor proved that you can't pair the integers with the real numbers. No such pairing can exist. This is known as Cantor's Diagonal Argument. This is why we say the real numbers are bigger than the integers, there is absolutely no way to pair them.
It is a piece-wise definition. I was being a bit informal before, so here is the actual definition:
The function
Is differentiable at the origin, but it's derivative is not continuous there.
Δ here is symmetric difference of sets?
Use Bézout's identity to prove the proposition.
All functions? They need not be continuous?
There are two points you should start with, and they aren't hard to find. What are they?
What I find a bit interesting is that the module structure of an abelian group doesn't add any information, but makes the above proposition more straightforward.
y must be in the set {-3, -2, -1, 0, 1}
It should be easy from here. To come up with this restriction, use my previous post #6.
When divided by 3, y can not leave a remainder of 2.
Substitute x = 1-y-z into the 2nd equation, and you get
This gives the infinitely many solutions for z=1, and y can be chosen arbitrarily with x = 1-y-z. It also proves that if z is not 1, then 3 must divide z.
Do you need all solutions? There is an obvious one you can see.
What is f(1)? What about f(3)?
This is not quite a useful invariant, but it may illustrate the purpose that invariants serve.
Let A and B be two finite sets of integers. Then each set has an invariant: odd or even, if the sum of their numbers is either odd or even. As with all invariants, this can tell you when two sets are not the same: a set which adds up to an even number can never be the same as a set which adds up to an odd number. Of course it can not tell you when two sets are the same as 2 + 8 = 6 + 4.