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[Christes]

Well, I'm just happy I remembered some linear algebra since, you know,  I suck at it generally.

[Miggy]

...
I guess if a real matrix has exclusively complex eigenvectors, the proof falls apart. But then comes the question: Is that actually possible? I'm positively sure I remember something like that being impossible, but looking through my literature again I can't find out where it's mentioned.
...

I think that this question is easy to answer by demonstration
0 1
-1 0
it's eigenvalues are i and -i and eigenvectors are (-i,1) and (i,1).

Thanks a bunch! That helped.

This thread made me happy today.

Well, I'm just happy I remembered some linear algebra since, you know,  I suck at it generally.

I'm happy that I got my question answered and got a bit smarter. Oh, and I understood Putzer's algorithm today which means another evil vanquished. More or less got all of the nasty stuff out of the way so now I just need to memorise all of the little stuff so I don't botch up on some silly misspeak. Exams should be all-right!

[Vector]

Well, I'm just happy I remembered some linear algebra since, you know,  I suck at it generally.

Haha, yeah.  Me too.  Worst grade I have ever gotten in anything: Honors Linear Algebra class.


On the original subject, though, I think you should check around in your book.  IIRC there's a weaker version of Schur's, unnamed, which linear algebra texts classically prove first.  I'm almost certain of it.

[Christes]

Anyone here know anything about GAP?  I need a simple command to test if an integer is a perfect square.  Currently I'm checking it with:

Code: [Select]

IsInt(Tau(n)/2)=false
Which is way inefficient (though cool in a theoretical way ).  There's probably a simple command I'm missing.

[Another]

((Int(Sqrt(n)))^2-n)!=0 should be reasonably efficient. I am not familiar with GAP syntax though.

If you really want to optimize - you could probably perform a few hand-made tests for divisibilities into smallest primes and their powers to quickly weed out ~99% of cases before performing Sqrt() or any other non-linear function.

[Grek]

Probably a reasonably simple question, but one that I don't have the calculus for none the less:

What is the volume of the space between the plane defined by y=0, the plane defined by y=1 and the curve y=k_x(1-x²)=k_z(1-z²), where k_x and k_z are both constants, and x, y and z are variables?

It's for a probability problem that I'd rather not get into if it's not relevant to the caluculation of the volume.

[Another]

What is the volume of the space between the plane defined by y=0, the plane defined by y=1 and the curve y=k_x(1-x²)=k_z(1-z²), where k_x and k_z are both constants, and x, y and z are variables?

I presume that you mean the volume of a figure made by rotating said curve around the Y axis, right? Or was it a typo, meaning the surface y=k_x(1-x²)+k_z(1-z²)?

In the first case x=(1-y/k_x)^0.5,z=(1-y/k_z)^0.5, V=Int[0;1](pi*(x^2+z^2)) dy = pi*(2-1/2(1/k_x+1/k_z)).

In the second case V=Int[S₁](1-k_x*(1-x^2)-k_z(1-z)^2) dx dz + Int[S₂](1-0) dx dz where S₁ is where 0<k_x(1-x²)+k_z(1-z²)<1 and S₂ is all (x,z) points inside S₁ where k_x(1-x²)+k_z(1-z²)<0. For 0<k_x+k_z<1 V=pi/(k_x*k_z)^0.5*((k_x+k_z)^2/2-(k_x+k_z)+0.5). It is also a part of a paraboloid, stretched along Z by (k_x/k_z)^0.5.

[Christes]

((Int(Sqrt(n)))^2-n)!=0 should be reasonably efficient. I am not familiar with GAP syntax though.

If you really want to optimize - you could probably perform a few hand-made tests for divisibilities into smallest primes and their powers to quickly weed out ~99% of cases before performing Sqrt() or any other non-linear function.

I actually didn't check back here.  My code was efficient enough to do what it needed to do, but that should work.  Of course, GAP doesn't have a sqrt command, but it does have RootInt, which returns the floor function of the square root, so it would have to look like:

Code: [Select]

RootInt(n)^2=n
Where RootInt returns the floor function of the square root.

[Bauglir]

This D&D problem is eating my brain. Okay, so, there are abilities that let you reroll dice. My first question was, "Given that I can roll 2 twenty sided dice (or 2d20, in the common notation), what are the chances of any particular result?" Well, after way too much time, I solved that one (it turns out a general formula for 2 dice is P=(2n-1)/x^2 where P is the probability of rolling n on an x-sided die). But then I started wondering about the even more general case of arbitrarily many dice. I worked out P=(3n^2-3n+1)/x^3 for that, and while I have a functional algorithm for adding additional dice, it gets ever more annoying to apply with each die. So, can anyone help me work out the general case?

The algorithm treats the rolls as a series of sequential rolls, since this should be indistinguishable from rolling them all simultaneously (since it is assumed you cannot decide to stop rolling until the sequence completes). The above formulae are simplified versions of the results from below, which may be more informative. When I say "term" below, I'm referring to a number (obtained by multiplying other numbers) added to another term; I am pretty sure I probably deviate from the real definition of the word.

Spoiler: Algorithm mentioned above (click to show/hide)

1. Begin with the chance of rolling any given number (1/x).
2. To add 1 die;
   a. Add a new term to the end of the series of terms equal to the value of the current last term, multiplied by the probability that the number (or higher) has not already been rolled ((n-1)/x).
   b. Multiply all terms EXCEPT the term added in step a by the probability that the die being added will not roll higher than n (n/x).
3. Repeat 2 until enough dice have been added.
4. Divide the lot by x to the power of the number of dice rolled.

I think a and b were the reasonings behind what I did, but I'm tired and I worked it out a while ago, so I'm not 100% sure. I know the expressions mentioned in those steps are accurate, I'm just not currently lucid enough to think through whether or not the justifications I provide for them here actually match up with what they represent.

EDIT: Added a crucial step 4 to the algorithm that I forgot about because it's easy.

[Christes]

I'm a little unclear on some of the words you're using.  What does each "term" represent?  The rolls of dice?  Or probabilities?

You're multiplying them, but also listing them in sequence and that is confusing me.

An example would be helpful - say the probability of rolling a 9 on a 3d6.

[ed boy]

First, I'd like to say that your current algorithms are wrong (or I have misunderstood them). Your two-dice version should be following a Triangular distribution.

I've had a stab at it:

Spoiler: maths (click to show/hide)

Let's try thinking about it a different way.

You have N dice. Each dice has p₁^-1,p₂^-1,...,p_N^-1 sides (and hence a probability p_i of rolling the value i).

It has a maximum value of ΣP_i^-1, and a minimum value of N.

So for the values outside that range, it has a probability of 0.

We will assume that the dice all have the same number of sides (p^-1), and so anyone set of results for the dice follows a multinomial distribution. However, this occurs when the dice a distinguishable. So you want to find all the different distinguishable arrangments that are desireable, and sum their probabilities.

If you want a total T, then the probability for any one suitable result (the first dice gives r1, the second gives r2, etc where Σr_i = T) is:

T!Π(^pr_i/_ri!)
T!Π(p^r_i)Π(¹/_ri!)
T!p^Σr_iΠ(¹/_ri!)
T!p^TΠ(¹/_ri!)

Now you sum that over all the possible arrangements of r_i. I would do it now, but I'm having some trouble hunting down the formula for that.

[Bauglir]

Well, to clarify, it looks something like this (using my earlier variables):

1/x
(1/x)(n/x)+(1/x)((n-1)/x)
(1/x)(n/x)(n/x)+(1/x)(n/x)((n-1)/x)+(1/x)((n-1)/x)((n-1)/x)

Simplifying these yields the short formulae above (I've checked the second by writing a dice roller program to roll 2d20 a million times and choose the best of each pair, but I haven't had time to write the third yet). I was referring to each of the number summed as a "term", which like I said, is probably wrong.

So, that example asked for can't happen; on 3d6, the highest individual die result is a 6. But if you wanted the probability of that, it's (1/6)(6/6)(6/6)+(1/6)(6/6)(5/6)+(1/6)(5/6)(5/6), or about 42% (if I did it right, which if ed boy is correct, I may well not have).

Huh, looking above, I obviously forgot to state in the quotation "... and choose the best". Derp.

@ed boy
I followed your explanation until the multinomial distribution link. Unfortunately, wikipedia's math articles are denser than I can parse. Any chance at an explanation of that?

[Another]

I was quite puzzled at your formulas before the "... and choose the best" words. 2d20 is commonly well defined as the sum.

Probability of getting N in k rolls of 1dN is P_N=1-(1-p)^k, where p=1/N. Probability that N-1 will be the maximum is P_N-1=1-(1-2*p)^k-P_N. Iterative formula is P_l=1-(1-(N-l+1)*p)^k-P_l+1-P_l+2-...-P_N.
 
From the other side P₁=p^k and you can construct an "ascending" iterative formula P_l=(l*p)^k-P_l-1-P_l-2-...-P₁.

I would probably check this for a few arbitrary N, k and l versus a brute-force "roll a virtual die a few million times" data first if I were to apply it anywhere. I remember finding all probabilities of (4d6 - minimum) some years ago and plotting a nice graph of that.

[ILikePie]

1/mx = 4m/(x + 4m - 2)

How do I solve for x?

[Darvi]

Scrap that. Can't think straight.

It's a simple second degree equation with a parametre though. Easy enough.