Random Thoughts about Math

Euler · 2004-05-13 23:33:19

I'm pretty sure if you replace "any prime number n" with "any number larger than 2" it is equivalent.

You mean "any integer larger than 2", but it is still not equivalent. The problem is that (3n+1)/2 does not equal 3(n/2)+1.

BTW, n = 3n+1 can not be applied a finite number of times to get a power of two for all integers. All powers of 2 greater than 2 are congruent to 0 mod 4. If n is congruent to 2 mod 4, then 3n+1 is congruent to 3 mod 4. If n is congruent to 3 mod 4, then 3n+1 is congruent to 2 mod 4. Therefore, any number that is congruent to 2 mod 4 or 3 mod 4 will never become a power of 2 no matter how many times 3n+1 is applied.

Mark Friedenbach · 2004-05-14 02:20:13

You mean "any integer larger than 2", but it is still not equivalent. The problem is that (3n+1)/2 does not equal 3(n/2)+1.
BTW, n = 3n+1 can not be applied a finite number of times to get a power of two for all integers. All powers of 2 greater than 2 are congruent to 0 mod 4. If n is congruent to 2 mod 4, then 3n+1 is congruent to 3 mod 4. If n is congruent to 3 mod 4, then 3n+1 is congruent to 2 mod 4. Therefore, any number that is congruent to 2 mod 4 or 3 mod 4 will never become a power of 2 no matter how many times 3n+1 is applied.

You are correct.

bolbuyk · 2004-05-15 05:55:19

http://en.wikipedia.org/wiki/Collatz_conjecture]Collatz conjecture(wikipedia)
Paul Erdos said "Mathematics is not yet ready for such problems." What does he know? I haven't given up.

Never realized this problem appears to be so serieus. at first glance it seems to be solvable on a tissue....

6-times numbers: english is not my native language, so I don't really know how to call it, but ment ais a number 6n, where n is a natural number.

Let's tell what I've also done to solve this (maybe it's a little bit inspiring):

- Calculate the computer this sequence and counting the odd and even numbers that were past. The result was graphiced and it was about triangle-like, repeated. I mean sloping up, falling down, and so on.

- Trying to work bottom-up, so start with 1, multiply by 2, and so on, Every 3x+1 was then 1 substr. and divided by 3 to get a new number. It was remarkble that the branches taht appeared differed after a certain amount of steps just 1!!

- Working with expected values. For big numbers, the expected value after one sequence is .5*3+.5*.5, assuming the probability of even and odd are equal. You can work this through for some sequences in order, but does this really proof anything?

When math is not ready for this stuff, am I ready for this?

John Creighton · 2004-05-16 10:47:27

I am surprised you can’t use induction. For instance we know it works for 1 and 2. Now assume it works for all numbers up to n and show this implies it works for all numbers up to n+1. Clearly if n+1 is even then it works for (n+1)/2 by the induction hypothesis. Now if n+1 is odd the next number is 3*(n+1)+1 which is even since:
n+1 odd -> n is even
and
(3*(n+1)+1)/2=(3/2)*n+2 but since n is even n is devisable by two therefore (3*(n+1)+1) is divisible by two and hence even. The next number in the sequence is therefore (3/2)*n+2.
If this number is even then the next number in the sequence is:
(3/4)*n+1
which is less then n thus by the induction hypothesis it approaches 1.

If n/2 is odd then we set n/2=2*m+1 Therefore:
(3/2)*n+2=3*(2*m+1)+2=3*m+5
Since 3*m+5 is odd we know m must be even.
The next number in the sequence is:
3*(3*m+5)+1=9*m+16. Since m is even the next number is
9*m/2+4=9*(n/2-1)/4+4=9/8*n+4-9/4=9/8*n+1+3/4
Again if this number is even we are done and if it is odd we are back to where we started. Notice that we seem to be getting closer and closer to n.
(3/4)*n+1>9/8*n+1+3/4
You would think that you could do something like bound every element in the sequence by some function of n and show that the bound is decreasing. Hence one of the elements of the sequence must be bounded under n and hence by the induction hypothesis we would know the sequence would converge.

John Creighton · 2004-05-16 12:55:17

I study electrical engineering but I think it is the math behind it that motivates . I enjoy seeing the application of such things as linear algebra, statistics and system theory. However, when studying engineering you often only learn what you need to know. It is driven by what is most obviously useful. But by just studying the math that has the most purpose, often the math with the most beauty and elegance is missed. I have recently started reading a book on group theory mostly for my own interest. Reading this book I remember how once it was routine to show if a function was one to one or onto but, because I haven’t been studying this stuff I have forgotten some of the basics. Group theory is very fascinating because it puts algebra into a much more general framework. It shows how seemingly unrelated things such as permutations, multiplication, addition and function composition are all related. It also makes good use of that prime number stuff in algebraic context.

John Creighton · 2004-05-21 22:13:27

I enjoyed the start of this thread when people talked about their dislike of math. For someone who has learned a fair amount of math I realize that what seems elementary to me many not be a concept others wish to spend the mental energy to grasp at a given moment.

For instance my sister has a major in physiology so she should have a fairly good technical background and I asked how many permutations there were in a rubix cube. She asked me what a permutation was so basically I said it is the amount of ways you can reorder a set of numbers.

She then proceeded to try to figure out the amount of permutations by exponents. She got a number that was much too low. Although I am not sure if this technique would have worked or not. I pointed out that a group of numbers can have many more permutations then you get by using simple power laws like this. I asked her how many ways can you rearrange four numbers. She said 16. I told her the answer was 24 (four factorial written 4!). Before, I proceeded to explain how n! grows much faster then 4^n (The number of possibilities of a dna chain of length n) she decided that the concept of a factorial too much additional information to grasp late at night. I’ll let her read Paul Davis “about time”. Books like that help interest people in the physics which may eventually influence some people to lean about the math behind it. She is still interested in learning more about math. She said she might review her old calculus book. I suggested she read linear algebra, or discrete math. I feel, that reviewing stuff you already learned is not as interesting as learning new stuff. A lot of textbooks review what you need to know, and If you still need more review you can always go back and just look at the relevant sections. Then again who knows. Maybe sometime I will reread my textbook on complex numbers.

Euler · 2004-05-21 22:31:47

I asked how many permutations there were in a rubix cube.

A rubix cube is 3x3x3 right? At first it seems like the answer should be 54!/(9!)^6, but I think that you can't get all of them due to limitation on how the cube can be manipulated. Also, you have to think about how you define permutations. If you rotate the whole cube is it a different permutation?

John Creighton · 2004-05-21 22:44:01

If you rotate the whole cube is it a different permutation?

I'm not sure.

Mundaka · 2004-05-22 01:38:17

John Creighton · 2004-05-22 10:50:09

I asked how many permutations there were in a rubix cube.

Why over 9!^6?. My hypothesis, any corner can be manipulated to the position of any other corner. There are 8 corners. That’s, 8! (40320) Permutations due to the corners. By similar logic, there are 6! (720) Permutations due to each center piece, and 8! Permutations due to the block that is between the two corners. So we have8!*8!*6! (1,170,505,728,000) Permutations in total. If we want to disregard permutations that are rotational equivalent we note that a cube can be rotated 4! Ways. Therefore the total amount of permutations is 8!*8!*6!/4! (48, 771, 072, 000)
Hmm, That number is too small, I wonder where I went wrong? I guess the corner pieces and the pieces between the corners must have more then one orientation for each position.

~Eternal~ · 2004-05-22 11:01:20

On the 0b101/0b10110/0b11111010100
I hence forth declare that sqrt(?4) and its relation to the acceleration of gravity on Earth to be purely ironic .

Palomar · 2004-05-22 11:08:09

I enjoyed the start of this thread when people talked about their dislike of math. For someone who has learned a fair amount of math I realize that what seems elementary to me many not be a concept others wish to spend the mental energy to grasp at a given moment.

*Well, it was nice of Shaun Barrett to mention that perhaps some of the issue for folks like me who aren't math whizzes refers back to school teachers. Like the ones who snapped, "Well what did I just get done telling you?" when you'd approach them with an uncertainty.

I have watched scientists (on documentaries, etc.) writing out huge mathematcial formulae on long blackboards with such ease and flow of the chalk...makes me envious.

I really don't know why higher mathematics is an obstacle for me. Perhaps because it's too abstract (for instance, I can prove 2 + 2 = 4; who can prove -- show me -- the square root of blah-blah is blah-blah?) It's just not tangible enough, I guess. :-\

::shrugs:: Again, the introduction of alphabetical letters into the mix, beginning in the 7th or 8th grade, really threw me. Explanations weren't adequately provided. I remember the math teacher writing a problem on the blackboard, giving another cursory explanation after the lesson, and telling us to figure out what "s" was. :-\ I'm like, you're kidding, right?

Whatever. The brainiacs in the class got the answers quickly enough -- a smile for a reward, the rest of us are limping along in the dust and then it's "Well, what did I just get done telling you?"

I think the seeming repetitive nature of it also "kills it" for me. Doing math always reminded me of my aunt trying to teach my sister and I to crochet (repetitive, over-and-over, boring movements) or like filing (in medical records offices, according to patient account numbers -- "fill-in" work I sometimes had to do years ago at a couple of other jobs, when the transcription work ran out).

And I never could find a practical use for "A train departing from St. Louis at 10:40 a.m., traveling 60 mph, will pass a train departing Little Rock at 11:50 a.m. traveling 45 mph at what time and where on the map?" type crapola.

--Cindy :laugh:

Josh Cryer · 2004-05-22 13:14:24

I think algebra should be introduced at a younger age, even before fractions (maybe around the same time as multiplication; I learned fractions after multiplication though, not sure how others did). Nothing deep, just basic variable usage. Children pick up quickly on variables, just throw a kid some LOGO and they'll pick up on it nicely. Math should be fun, that's the one thing.

On another note, I've been really interested in trinary algebra (base 3 algebra). Apparently base 3 is the most efficient way to represent data. I started to write a base 3 register machine in Scheme, but it's just one of those projects that I can never quite get done.

Euler · 2004-05-23 17:28:55

Why over 9!^6?. My hypothesis, any corner can be manipulated to the position of any other corner. There are 8 corners. That’s, 8! (40320) Permutations due to the corners. By similar logic, there are 6! (720) Permutations due to each center piece, and 8! Permutations due to the block that is between the two corners. So we have8!*8!*6! (1,170,505,728,000) Permutations in total. If we want to disregard permutations that are rotational equivalent we note that a cube can be rotated 4! Ways. Therefore the total amount of permutations is 8!*8!*6!/4! (48, 771, 072, 000)
Hmm, That number is too small, I wonder where I went wrong? I guess the corner pieces and the pieces between the corners must have more then one orientation for each position.

I was just looking at the faces, not the actual blocks. That is why I said that the actual number of permutations would be less. There are 3 ways each corner block can be rotated and 2 ways each side block can be rotated. So 8!*8!*6!*3^8*2^8/4!

who can prove -- show me -- the square root of blah-blah is blah-blah?

blah - blah = 0
sqrt(blah - blah) = sqrt(0)
sqrt(blah - blah) = 0
sqrt(blah - blah) = blah - blah Q.E.D.

I think the seeming repetitive nature of it also "kills it" for me. Doing math always reminded me of my aunt trying to teach my sister and I to crochet (repetitive, over-and-over, boring movements) or like filing (in medical records offices, according to patient account numbers -- "fill-in" work I sometimes had to do years ago at a couple of other jobs, when the transcription work ran out).

It sounds like you spent a lot of time learning arithmetic and not much learning math.

Apparently base 3 is the most efficient way to represent data.

I think I know how they decided that, though it really depends on what hardware you are using to sore the data. And, if you do not limit yourself to integer bases, then base e becomes the most efficient.

bolbuyk · 2004-05-24 06:41:14

I asked how many permutations there were in a rubix cube.
Why over 9!^6?. My hypothesis, any corner can be manipulated to the position of any other corner. There are 8 corners. That’s, 8! (40320) Permutations due to the corners. By similar logic, there are 6! (720) Permutations due to each center piece, and 8! Permutations due to the block that is between the two corners. So we have8!*8!*6! (1,170,505,728,000) Permutations in total. If we want to disregard permutations that are rotational equivalent we note that a cube can be rotated 4! Ways. Therefore the total amount of permutations is 8!*8!*6!/4! (48, 771, 072, 000)
Hmm, That number is too small, I wonder where I went wrong? I guess the corner pieces and the pieces between the corners must have more then one orientation for each position.

The center pieces cannot be displaced, they determine the color of a certeain side of the cube..

Possibilities due to the corners: 8!x8^3

Possibilites due to the other elements: 12!x12^2

If you like, multiply by 6x4 for different orientations of the whole cube (six possible side for to rest on, 4 different sides in each case to draaw on the front).

But the number you get is an overestimation, because of the fact not all these possibilities can be obtained, due to the rotation. I think the outcome will be:

8!x8^3x12!x12^2(x6x4)/2 (Last digit is an intuitive estimation to take into account the limited possibilities due to the rotations.)

bolbuyk · 2004-05-26 06:15:31

Calculating on:

8 corners on 8 possible locations: 8!
Every corner counts 3 different positions: 3^8 (and not 8^3!!)
12 between-blocks: 12!
Every 2 positions: 2^12

Not corrected for the fact that there's no manouvre to just turn one block. There will always be a second block that turns also. This is both the case for edges and the between-blocks.
Assuming the outcome has due to this last fact to be divided by 2*2 the outcome is:

40320*6561*479001600*4096/4=

129,756,009,823,469,568,000

The right answer appears to be:

43,252,003,274,489,856,000 which is exactly 1/3 of my calculation. Apparantly, there is a 3-factor due to the rotational limitations.

bolbuyk · 2004-05-26 06:22:27

OK, let's prove 1=2:

Assume x=y

Then xy=y^2

And also x^2-xy=x^2-y^2

Working this out x(x-y)=(x+y)(x-y)

Divide both sides x=x+y

So x=x+x -> x=2x -> 1=2 :band:

Euler · 2004-06-10 15:32:47

Mathematics professor Louis de Branges claims to have a proof of the Riemann hypotheisis, which has been one of the most imprtant unsolved problems in mathematics ever since it was formulated in 1859. If his proof is found to be correct, he will recieve a $1 million prize.

http://news.uns.purdue.edu/UNS/html4eve … tml]Purdue University press release

Hazer · 2004-06-11 13:56:28

Random thought about math

It's better than drugs;
It's better than sex.
What else could it be?
Dy/Dx

John Creighton · 2004-06-11 14:17:36

Mathematics professor Louis de Branges claims to have a proof of the Riemann hypotheisis, which has been one of the most imprtant unsolved problems in mathematics ever since it was formulated in 1859. If his proof is found to be correct, he will recieve a $1 million prize.
Purdue University press release

In the article the primes it relates this conjecture to the structure of prime numbers. Why is that the case? It is wierd how such a simple function can have so much struture.

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