Time being the Fourth Dimension VS others

[jinydu]

Well, I did learn some things about vectors in high school, but it was the final unit of the final year (thus, the people who wrote the curriculum probably considered it to be even more advanced than integration by parts).

We spent the bulk of the time covering vectors in 3D. For instance, we learned how to derive the equation of a plane, given 3 non-collinear points. We covered things like intersection of planes, lines and the distance between parallel planes and lines. Unfortunately, we weren't really shown the connections with other areas in mathematics.

What do you mean by C^n being a vector in n-dimensional space? I always tought of C as being 2-dimensional. So shouldn't it be 2n-dimensional space?

[houserichichi]

Well fret not, you'll get your fair share of vectors in both introductory linear algebra and vector/multivariable calculus (both usually covered first year university).

When referring to C^n I was thinking ahead of myself - I meant n-tuples with entries in the complex field as opposed to the reals. And yeah, it would take 2n dimensions to plot such objects which is why we don't bother trying after we pass n=1. :wink:

What level of maths are you sitting at now anyways? Seems like you've got enough passion for the both of us Buy yourself a pocket protector and then you and I can start hanging out.

[jinydu]

Right now I'm a freshman undergraduate at UCLA, just finishing my first quarter (my last final exam is tomorrow!). This quarter, I took Differential Calculus. I knew most of the material already, although there were 2 topics I had never learned before: Complex Exponentials and Second-Order Linear Differential Equations.

In my spare time, I like to read math articles in books and online. Although they are interesting, I find that nothing beats classroom learning, in terms of my level of familiarity and understanding of concepts.

And of course, I also enjoy doing my own math "work". For example, during high school, I found a way to get formulas for:

1^n + 2^n + 3^n + 4^n + 5^n ... + m^n

For each (positive integer) value of n, I derived two formulas: One in terms of combinations and the other in terms of a polynomial of degree n+1. Furthermore, I managed to prove the formulas all the way up to n = 7. But later, I found out that someone had beaten me to the idea (although possibly not using the same approach) by over 370 years: http://mathworld.wolfram.com/FaulhabersFormula.html

I've also done other things. For an example of something I've done the last few months, you can look at the "Tetration and Large Numbers Thread" on this forum.

[PWrong]

Hey, What a coincidence, I found a formula for exactly the same thing just last week!

My approach eventually used matrices. I started with a polynomial of degree n+1, and then came up with a matrix equation to find the coefficients. It's really the same as Faulhaber's formula, (I hadn't seen that before), but I think it looks more elegant because the matrices are all based on pascal's triangle.

I just graduated from high school a few weeks ago, but I've hardly done anything apart from maths since recovering from leavers week (I'm not sure what you call it, but it involves mostly drinking) But I can't wait until uni now

Mostly I've just been trying to figure out the "top-function", which is an idea I came up with for the tetration thread.

[quickfur]

[...]And of course, I also enjoy doing my own math "work". For example, during high school, I found a way to get formulas for:

1^n + 2^n + 3^n + 4^n + 5^n ... + m^n

For each (positive integer) value of n, I derived two formulas: One in terms of combinations and the other in terms of a polynomial of degree n+1. Furthermore, I managed to prove the formulas all the way up to n = 7. But later, I found out that someone had beaten me to the idea (although possibly not using the same approach) by over 370 years: http://mathworld.wolfram.com/FaulhabersFormula.html

I've also done other things. For an example of something I've done the last few months, you can look at the "Tetration and Large Numbers Thread" on this forum.

All this talk about math reminds me of some of the stuff I used to do. One time, I wrote a program to calculate a "subfactorial": SubFac(n) is defined to be the smallest number divisible by all integers 1 to n. It coincides with the factorial for the first few numbers, then loses out pretty quickly. I found out that SubFac increases in leaps: it'd jump by a large factor when you pass a prime, and sometimes it stays still when you go through several consecutive numbers with factors that have already occurred before, before it jumps again at the next new factor.

I also found a derivation of the formula for the n'th Fibonacci number using matrices and eigenvalues. My main idea was to think of a 2x2 matrix as a "machine" operating on 2D vectors which are "registers". Each cycle of the "machine" is a left-multiplication by the same matrix. The idea is to find the value of the vector after N iterations of this "machine". This is done by devising the matrix so that it's diagonalizable, so then you can "telescope" to the N'th iteration of the "machine" just by taking the power of the diagonal elements. (Of course, later on I found out that this is essentially just finding the orbit of the linear operator corresponding with the matrix.) I noticed that placing a 1 in an off-diagonal position makes the matrix "copy" one register to another, and "copying" multiple registers into one obviously adds their values together. So by this, I devised a matrix that, when repeatedly multiplied to the column vector [1 1], yielded [F_n+1 F_n+2], where F_n is the n'th number in the Fibonacci series. This matrix turns out to be diagonalizable, so after suitable derivation of the eigenvalues and diagonalizing the matrix, out popped the golden ratio numbers, and when you extract one of the linear equations from the result, you get the formula for the n'th Fibonacci number.

[Rkyeun]

There is some credibility to time being a spatial dimension of sorts.

Your velocity through the four dimensions must be C. If you go faster in realmspace, you slow down in time.

This suggest some kind of strange gravitational link that you are evading by approaching C and putting yourself in orbit around whatever force is pulling us to the future.

I'm not sure what this means exactly, though.

[PWrong]

All this talk about math reminds me of some of the stuff I used to do. One time, I wrote a program to calculate a "subfactorial": SubFac(n) is defined to be the smallest number divisible by all integers 1 to n. It coincides with the factorial for the first few numbers, then loses out pretty quickly. I found out that SubFac increases in leaps: it'd jump by a large factor when you pass a prime, and sometimes it stays still when you go through several consecutive numbers with factors that have already occurred before, before it jumps again at the next new factor.

That sounds interesting. So SubFac(n) = Lcm(1,2,...,n)? Is there a simple formula?

I often look at numbers with lots of factors using a kind of grid to show factors graphically. On the first row, a dot exists at every point, because every number is divisible by 1. On the second row, every second column is a dot. On the nth row, a point exists at every nth column. I think there's an example somewhere on mathworld, but I can't find it.

Code: Select all


6     .     .     .     .     .     .     .     .     .     .     .     . 
5    .    .    .    .    .    .    .    .    .    .    .    .    .    .    . 
4   .   .   .   .   .   .   .   .   .   .   .   .   .   .   .   .   .   .   .
3  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .
2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.................................................................................
                                                            ^


With this kind of graph, you can identify special numbers simply by looking at them. A prime occurs when a column is empty, apart from 1. Numbers like factorials and subfactorials appear as an unbroken upward line. Every straight line like this appears in the middle of a "flower".
This means there are two straight diagonal lines either side of the column. (The flower at 60 is shown by the ^ above)

For instance, 6!=720 is divisible by 2,3,4,5 and 6.
But look at the numbers next to it.
720-1 = 719, is prime. 720-2 =718, divisible by 2. 720-3 = 717, div. by 3,
and so on. It works up to six, on both sides of the number (except 721 isn't prime, but it has no factors less than 6)

The pattern has an interesting structure of flowers at the bottom, things that look like parabolas higher up, and finally, straight diagonal lines.

Your second idea is interesting too. Is this what you mean?

Code: Select all

[0 1]^n [1]  =  Phi *F[sub]n+2[/sub]
[1 1]     [1]             F[sub]n+2[/sub]


How do you go from there to get the formula?

[houserichichi]

There is some credibility to time being a spatial dimension of sorts.

Your velocity through the four dimensions must be C. If you go faster in realmspace, you slow down in time.

If you have mass you can't travel at the speed of light. When you increase your speed your mass increases. GR says the more mass the more space curves. You'd simply be making a shorter trip for yourself is all - time still flows on straight.

[quickfur]

All this talk about math reminds me of some of the stuff I used to do. One time, I wrote a program to calculate a "subfactorial": SubFac(n) is defined to be the smallest number divisible by all integers 1 to n. It coincides with the factorial for the first few numbers, then loses out pretty quickly. I found out that SubFac increases in leaps: it'd jump by a large factor when you pass a prime, and sometimes it stays still when you go through several consecutive numbers with factors that have already occurred before, before it jumps again at the next new factor.

That sounds interesting. So SubFac(n) = Lcm(1,2,...,n)? Is there a simple formula?

Yes, although it's not one that lends itself well to direct computation. But it is closely related to the graph you describe below.

I often look at numbers with lots of factors using a kind of grid to show factors graphically. On the first row, a dot exists at every point, because every number is divisible by 1. On the second row, every second column is a dot. On the nth row, a point exists at every nth column. I think there's an example somewhere on mathworld, but I can't find it.

Code: Select all


6     .     .     .     .     .     .     .     .     .     .     .     . 
5    .    .    .    .    .    .    .    .    .    .    .    .    .    .    . 
4   .   .   .   .   .   .   .   .   .   .   .   .   .   .   .   .   .   .   .
3  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .
2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.................................................................................
                                                            ^


With this kind of graph, you can identify special numbers simply by looking at them. A prime occurs when a column is empty, apart from 1. Numbers like factorials and subfactorials appear as an unbroken upward line. Every straight line like this appears in the middle of a "flower".
This means there are two straight diagonal lines either side of the column. (The flower at 60 is shown by the ^ above)

Exactly. So with this graph, the way you compute SubFac(n) is by creating a counter for each prime you encounter, say you call the counter C_p for the prime p, setting the starting value to 1. For each composite number you encounter, you set the counter for each prime factor to the maximum of C_p and the exponent of p in the factorization of the composite number.

I know this is not a very efficient way of computing SubFac(n), but I haven't found a better way to do it yet.

Also, w.r.t. to a formula for SubFac(n), you can derive it from the description above. It's basically factorizing all the numbers from 1 to n-1, and for each prime p, taking the maximum exponent that occurs in all of these factorizations. I was going to actually type out the formula, but phpBB doesn't like my math formatting so I guess I'll just describe it instead.

Your second idea is interesting too. Is this what you mean?

Code: Select all

[0 1]^n [1]  =  Phi *F[sub]n+2[/sub]
[1 1]     [1]             F[sub]n+2[/sub]


How do you go from there to get the formula?

Yes this is the idea. However, the matrix you have there isn't diagonalizable directly; what you need to do is to swap the two rows. This will still give you Fibonacci numbers on the right hand side of the equation, just in reverse order. The resultant matrix can be diagonalized by finding the eigenvalues, and then factoring it into PDP^-1, where D is a diagonal matrix and P is the matrix that diagonalizes the original matrix (I forget what the term is). Since (PDP^-1)ⁿ = PDⁿP^-1, and powers of diagonal matrices are obtained simply by taking the powers of each diagonal entry, you can easily rewrite the matrix equation as two (independent) linear equations involving F_n and F_n+1, which will give you the formula for the Fibonacci series.

[wendy]

Were time the fourth dimension, we should not be able be able to experience it. For once time is set in space, we are just a stack of cards. This is our yesterday, this our today, this our tomorrow.

We have a peception of time other than the three dimensions we live in, and if we want to, even in our mind-journies to the fourth dimension, talk and experience, then making time the fourth dimension would deny us that.

Think of the einstein-space-time as a kind of graph paper for measuring relativity, and not some real description of the world.