On January 11 2019 02:01 Simberto wrote:Furthermore, what you are counting are not the prime numbers up to 32, but, as the article clearly states, Pi(x) - Pi(Sqrt(x)) + 1.

Preceding the formula is the statement "A more elaborate way of finding pi(x) is ..". That seems pretty clear that it should be a formula to find pi(x), which is as stated in the beginning of the article a function to count the prime numbers less than or equal to x. I'm not sure how I'm misunderstanding that.

It is a way of finding pi(x). The way is to use basic equation solving to turn

N = pi(x) - pi(sqrt(x)) + 1

into pi(x) = N + pi(sqrt(x))-1

which, given that you can calculate N and clearly know pi(sqrt(x)) since you used all of the primes smaller than sqrt(x) to calculate N, is a way of finding pi(x)

On January 11 2019 02:41 Simberto wrote: It is a way of finding pi(x). The way is to use basic equation solving to turn

N = pi(x) - pi(sqrt(x)) + 1

into pi(x) = N + pi(sqrt(x))-1

which, given that you can calculate N and clearly know pi(sqrt(x)) since you used all of the primes smaller than sqrt(x) to calculate N, is a way of finding pi(x)

Okay, I think I get it. But I don't think it's useful for what I need to do. Thanks!

On November 21 2018 03:36 DarkPlasmaBall wrote: When I was doing my master's in math education, I took an awesome course called Problem Solving. We met once per week, for a three-hour block of time, and our professor would give us exactly one new problem to work on. We'd collaborate and explore the problem, tinker with a bunch of different strategies, try to solve the problem in as many ways as possible, consider additional extensions to the problem, and then reflect on and analyze the entire process.

My favorite problem in that class- which has come to be my favorite math problem of all time- is The Doubling Problem. Today and tomorrow I'm giving it to my high school math students (as it's right before Thanksgiving break, so I'd rather have the students play around with an interesting problem like this instead of trying to force down a final lesson and risk them losing focus).

The Doubling Problem is remarkably simple to explore, as it's just based on addition, multiplication, and moving around digits of a number. Thanks to its low mathematics entry point, I have plenty of students (and not just honors/ AP/ high-level students) who are making progress in solving the problem (or have already solved it).

Without further ado, I challenge you to solve The Doubling Problem (please spoiler your answers):

Pick a positive whole number. We’re going to apply a special rule to this number: take the last digit of that number (the ones’ digit) and move it to the front of the number. For example: 1234 becomes 4123 because the 4 is moved to the front, 567 becomes 756, 9002 becomes 2900, etc. Can you find a positive whole number such that, when you apply this rule to it, the resulting number is double the original number?

if there is a solution depends on the base I think. for base2 there should be none (if you don't allow leading 0s at least - else you have 01 -> 10). Does anyone have a clue if it is solvable for all bases bigger than a certain value or if iit is solvable for some and not solvable for others and if there is a pattern which influences that?

Imho, those numbers should exist for all bases >2. Lets again take a quick look at the algorithm I also mentioned earlier, but adjust it to baseN:

We assume the last digit to be

p_0 = d,

and then continue:

r_0 = 0 d_i = (p_i*2)%N

r_(i+1) = (p_i*2)/N (/ used in the integer sense) p_(i+1) = d_i+r_(i+1)

We can abort the algorithm when we get p_x == p_0 = d and r_x == r_0 = 0.

Now this isnt necessarily terminating for any d. But since p_i is in {0,...,N} and r_x in {0,1} there is only a finite number of combinations possible, so there must be cycles at some point, where p_k == p_l && r_k == r_l && k < l obviously, we can now state that for the lowest such k and l it holds true that for all m>k p_m == p_(m+l-k) && r_m == r_(m+l-k)

This means we have a number of cycles. If there is a single m>k, where r_m == 0, we can take p_m as last digit of our number and generate a valid number in regards to the original question.

So all we are required to do is to proof, that in this cycle there is guaranteed to be a position where r_m != 1. And this is rather trivial (I will make some leaps here, but it should be obvious, if not I can detail this) as the only cycle where r_m can stay permanently at 1 is the cycle of p_k = N-1 and r_k = 1. But as in this case it is necessary that p_(k-1) = N-1 and r_(k-1) = 1. Now this violates the rule of k being the lowest k to build such a cycle. And as we originally started with a different pair (d,0), it is clear that we can never enter this cycle -> there must be another cycle. -> There is a solution.

PS: There is the additional condition, that p_(m-1) != 0. But it also can be shown, that this is only a problem for N=2, as in all other cases simply taking m+1 produces a valid solution.

Yeah, that was like watch that one movie and get lost in them for three hours. They're all so pretty. I saw a couple videos ages ago, and had no idea he continued to produce them at the same quality.

Some of them are on topics that I understand and problems I mechanically solve, but never thought about 2d or 3d interpretations of the rules.

I have been given the question "show that sigmoid(-x) + sigmoid(x) = 1

I am not sure.... what to show ? Do I need to know how the limits of x going to infinity and negative infinity, or going to zero, will make this equation go to 1?

(I can do that, it's just that another question asked the upper bounds and lower bounds of the question so i already showed the limits... so it just confuses me a bit)

Well, did you plug in the expressions with the e power? What it does in the limit doesn't matter because it should be true for all x, not x going to plus or minus infinity.

On February 11 2019 05:19 travis wrote: aand now I have another math question

sigmoid function is sigmoid(x) = 1/(1+e^(-x))

I have been given the question "show that sigmoid(-x) + sigmoid(x) = 1

I am not sure.... what to show ? Do I need to know how the limits of x going to infinity and negative infinity, or going to zero, will make this equation go to 1?

(I can do that, it's just that another question asked the upper bounds and lower bounds of the question so i already showed the limits... so it just confuses me a bit)

Differentiate that function and observe its derivative’s properties.

On February 11 2019 06:04 mahrgell wrote: You show that for ANY x the equation holds true. So just transform the equation until you have the same on both sides of the equal sign.

1984's Rescue On Fractalus is being remade. The planet is a giant fractal. Pretty cool how unique an experience they were able to create in 1984 on such limited hardware. I think it was playable on theAtari 400. It definitely worked on the Commodore 64. I think the Atari 400 only had 16K of memory while the "high powered" C64 had an amazing 64 Kilobytes of RAM! If this game only worked on the Atari 800 then it is not as big of a technical feat.

in 1984 1st person space games were not very good. Using a fractal as the planet surface is such a cool ballsy move. Man, back in the day LucasArts was pretty damn cool. Leave it to Disney to shut them down.

On January 15 2019 00:57 travis wrote: You guys may enjoy this

On November 18 2018 21:48 Nebuchad wrote: Hi guys, I still suck at math

I'm using a program that calculates equity for different ranges in poker (hold'em). Here's the data I have:

Number of tries (exhaustive): 403835094432

Range 1 wins: 142599616520 Range 1 ties: 10945429968 Range 2 wins: 140980819160 Range 2 ties: 11040244316 Range 3 wins: 108972219224 Range 3 ties: 2243569640

Range 1 wins %: 35,3113% Range 1 ties %: 2,7104% Range 2 wins %: 34,9105% Range 2 ties %: 2,7338% Range 3 wins %: 26,9843% Range 3 ties %: 0,5556%

Range 1 equity: 36,5978% Range 2 equity: 36,2087% Range 3 equity: 27,1934%

I'm trying to determine how the program got the equity from the data.

Obviously you get the win equity by going (number of wins*100)/number of tries, and the tie equity by going (number of ties*100)/number of tries.

My understanding is that the equity percent is (wins%) + (part of ties% that are ties with only one player/2) + (part of ties% that are ties with both other players/3) - (because when you win you get all the money, when you tie with one player you get half and when you tie with both you get a third)

I run into an issue because the program doesn't let me know how often you tie with one player and how often you tie with both players.

I've been playing around with the numbers and I got the right results.

I added wins for all players and (tie/2) for all players. I get slightly over 100%, because some of the ties should have been /3 instead of /2.

I took the amount that was above 100% and did that *2, then /3.

Then I substracted that result from the number of ties for one player

Then I did number of wins for that player + (that result/2), and I got the right answer. I don't understand why. Shouldn't I also have had to divide an amount by 3?

Hi again

Similar to the last one, but significantly more annoying, I got this:

Number of tries (exhaustive): 1267892619840

Range 1 wins: 791976687662 Range 1 ties: 12482408790 Range 2 wins: 159949046721 Range 2 ties: 10525724277 Range 3 wins: 159949046721 Range 3 ties: 10525724277 Range 4 wins: 135556038238 Range 4 ties: 15394856484

Range 1 wins %: 62,4640% Range 1 ties %: 0,9845% Range 2 wins %: 12,6153% Range 2 ties %: 0,8301% Range 3 wins %: 12,6153% Range 3 ties %: 0,8301% Range 4 wins %: 10,6914% Range 4 ties %: 1,2142%

Possibilities are:

Range 1 wins = a Range 2 wins = b Range 3 wins = c Range 4 wins = d Range 1&2 tie = e Range 1&3 tie = f Range 1&4 tie = g Range 2&3 tie = h Range 2&4 tie = i Range 3&4 tie = j Range 1&2&3 tie = k Range 1&2&4 tie = l Range 1&3&4 tie = m Range 2&3&4 tie = n Range 1&2&3&4 tie = o

a, b, c and d are known.

Because range 2 and range 3 have the same stats, I also know that e=f, i=j and l=m

Because of the specifics of those ranges, I also know that e=f=0, k=0 (every time range 1 ties with range 2 or range 3, it must also tie with range 4).

I'm getting:

Range 1 ties => g/2+l/3+l/3+o/4=0,9845

Range 2 ties => h/2+i/2+l/3+n/3+o/4=0,8301

Range 4 ties => g/2+i/2+i/2+l/3+l/3+n/3+o/4=1,2142

From this I get that h/2 = (100-a-b-c-d)-1,2142 = 0,3996

I can also add all of the ties and get:

g+2i+2l+n+o = 3,0598

And from there I have trouble getting the other five variables, and I'm not sure if it's because I suck at this or because I just don't have enough equations to solve...

Unless you have additional information, i don't think you can conclude:

Because range 2 and range 3 have the same stats, I also know that e=f, i=j and l=m

Because of the specifics of those ranges, I also know that e=f=0, k=0 (every time range 1 ties with range 2 or range 3, it must also tie with range 4).

Just because two ranges have the same win percentages does not mean that they are the same hand. Neither does this necessarily mean that they have the same combination tie percentages e and f (etc....), nor that always only one of them ties with 1.

You could have a distribution where 2 often ties with 1, but never with 4, and 2 often tie with 4, but never with 1 which delivers the same distribution.

Other than that, if you have problems solving something like that, put it into a matrix and use gaussian elimination. That is an algorithm that always solves a solvable set of equations. It's gonna be a bit bit annoying with that many coefficients, but will tell you the solution if there is a clear one, and the set of solutions if there isn't.

Because range 2 and range 3 have the same stats, I also know that e=f, i=j and l=m

Because of the specifics of those ranges, I also know that e=f=0, k=0 (every time range 1 ties with range 2 or range 3, it must also tie with range 4).

Just because two ranges have the same win percentages does not mean that they are the same hand. Neither does this necessarily mean that they have the same combination tie percentages e and f (etc....), nor that always only one of them ties with 1.

You could have a distribution where 2 often ties with 1, but never with 4, and 2 often tie with 4, but never with 1 which delivers the same distribution.

Other than that, if you have problems solving something like that, put it into a matrix and use gaussian elimination. That is an algorithm that always solves a solvable set of equations. It's gonna be a bit bit annoying with that many coefficients, but will tell you the solution if there is a clear one, and the set of solutions if there isn't.

Yeah I worded that poorly, I do have additional information. Range 2 and 3 contain the same hands so it's true that e=f etc., and the way the ranges are constructed it's impossible for range 1 to tie with range 2/3 without also tying with range 4, so those informations were correct.

I have a assigned question to determine the last two digits of 99^14 by using modulus 100 so I know 99 is congruent -1 (mod 100) which means 99^14 is congruent (-1)^14 (mod 100) which means the last two digits must be 01 because -1^14 = 1