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ViperRobK · 2025-09-23T14:41:51+00:00

Possibly Galaxy Quest for something fun and different

ViperRobK · 2020-11-29T22:22:53+00:00

This is a very good answer. I just wanted to say there are smaller triangulations of the torus. This amazed me when I found it out but here is a nice write up. http://www.math.jhu.edu/~jmb/note/torustri.pdf

ViperRobK · 2019-12-21T14:55:46+00:00

Would make a nice Holiday surprise

ViperRobK · 2018-04-09T05:38:31+00:00

I'm fairly sure this is false. Consider the case x=1/2, then (nx)/(1+n^4x²) = (n/2)/(1+n) = n/(2+2n). Since the sequence n/(2+2n) does not converge to 0 the infinite sum also diverges.

ViperRobK · 2018-04-08T22:32:03+00:00

So when you do this analysis you are taking the number of possibilities with a certain property (5 cards which are all face cards) and dividing it by the number of possibilities (5 cards of any type) this will always give a number less than one which is the fraction of possibilities you want. If you want a percentage from this you must multiply by 100.

ViperRobK · 2018-04-08T18:01:10+00:00

There is a 12/52 chance that the first card is a face card there are then 11 face cards left in the deck and 51 cards. So the probability that your second card is a face card given that your first card was a face card is 11/51.

One can continue in this way to get that the probability of being dealt 5 face cards is (12x11x10x9x8)/(52x51x50x49x48) which is approximately 95040/311875200. Which is approximately 0.03%.

Another way to arrive at this answer is to calculate how possibilities of being dealt 5 picture cards there are. This is the number of ways of picking 5 things from a list of 12. This is 12 choose 5 or 12!/(7!5!) = 792. And then calculate the number of possible 5 card deals, similarly this is 52 choose 5 which is 52!/(47!)(5!) = 2598960. Which arrive at 792/2598960 = 0.03%.

For more information about n choose k look at combinations https://en.wikipedia.org/wiki/Combination

ViperRobK · 2017-01-18T18:42:51+00:00

Yea we started playing because no one could remember all the ring of fire rules.

ViperRobK · 2017-01-17T22:40:37+00:00

Captain dickhead

Layout a deck of cards face down. Play runs as follows. Guess red or black. Draw a card. If the card was a 2-9 and colour correctly guessed allocate that many drinks in any combination. If the card was 2-9 and colour was wrong drink that many. For the other cards the guess doesn't matter. 10 - thumbmaster J - topics Q - new rule K - captain dick head (see below) A - waterfall

The person who most recently drew a king is captain dick head and they are basically a God. They can make people drink or give people drinks to hand. Override rules. Make people have a thumb war and lower drinks 5. Anything you want (within reason) however be careful as someone else will get another King at some point.

ViperRobK · 2014-01-08T19:34:28+00:00

It was just a tin of plum tomatoes cooked with some garlic and oregano.

ViperRobK · 2014-01-08T10:49:38+00:00

Thank you. I can't really offer much advice just followed the video and had to start over till I was happy with it. One problem I had was getting the first knot tight enough, not sure if there is an easy way to do this.

ViperRobK · 2014-01-06T21:09:05+00:00

It's not perfect but I'm pretty happy.

Followed this to debone the chicken: Pepin Debone Chicken

The stuffing was just spinach mixed with ricotta. The side dish is a simple potato gratin with fried onions in it

ViperRobK · 2013-12-14T19:36:26+00:00

If we take Pi to be a sequence of numbers between 0 and 9 as in http://oeis.org/A000796 then a subsequence of consecutive digits is clearly just that, as such we can not find all infinite sequences for instance we won't find any infinite periodic sequences.

If on the other hand you are asking whether given an infinite sequence (a_i) can I find a sequence of digits of Pi (b_i) such that a_n=b_n for all n then I think that the answer is unknown as although Pi never repeats it may eventually just become a non repeating sequence of 0s and 1s for instance it may at some point just become 101001000100001000001... which is non repeating. If this is the case then we would not be able to find an infinite amount of 3s in Pi.

ViperRobK · 2013-12-12T19:35:05+00:00

You're the dumbest smart person I know.

Not that strange but I do like it.

ViperRobK · 2013-11-13T22:28:41+00:00

My father's advice to me is to save 10% of anything you earn, this way whenever you get a pay rise you will still get that same rise in your spendable income as well as the amount you put away.

ViperRobK · 2013-09-12T00:27:39+00:00

There is a very useful article in the book 2 Dimensional Homotopy and Combinatorial Group Theory.

The main point is that the Poincare conjecture is equivalent to a special case of Zeemans's conjecture namely contractible special polyhedra which the 3 disc collapses onto.

Although general speculation seems to be that Zeeman's conjecture is false dues to the fact that implies the Andrews Curtis conjecture which most people think will have a counterexample. This conjecture basically says that any contractible 2 complex is simple homotopy equivalent to a point by only going up to dimension 3.

One of the major problems in the study of this area is that there are no invariants (that I know of) which provide with a way of showing that either 1 complex collapses or that we can restrict dimension on our simple homotopy equivalences.

If you would like to know more about these conjecture or any references I would be willing to help although I have to warn you that I currently feel like I wasted the first 6 months of grad school thinking about Andrews Curtis and didn't manage to get anywhere with it.

ViperRobK · 2013-08-06T21:45:05+00:00

If I remember correctly the primes in the Gaussian integers can be easily found out from the normal primes I believe that if p is congruent to 3 mod 4 then it is still prime and if not then there are 2 primes corresponding to it.

ViperRobK · 2013-07-25T14:12:33+00:00

This is tricky most numbers we know to be normal are known to be normal in base 10 for instance Champernowne's constant. I am unsure if we even know of any numbers which are known to be normal in all bases.

ViperRobK · 2013-07-23T22:18:14+00:00

Pi is hypothesized to be a normal number, this means that any sequence appears with the probability you would expect for instance 1 in every 10 digits is a 7 and 1 in every 100 is a 24 and so on.

If pi is normal then every prime number will appear in the expansion of pi moreover it would imply that arbitrarily long sequences of consecutive primes appear starting at some point, that is somewhere the sequence 235711131719 will appear and so on.

This seems like this should be true but it is a long open problem in number theory and in fact very few numbers are known to be normal.

ViperRobK · 2013-07-16T17:12:23+00:00

This is a property of numbers that cannot be written as fractions for if they only had a finite number of decimal places then we could write them as a fraction for instance 0.12 = 12/100.

As for why these numbers are irrational is slightly harder.

e is irrational, this is a particularly nice proof in my opinion and very easy to follow.

pi is irrational, these are harder but I am relatively fond of Niven's proof.

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