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[PsyberianHusky]

Anyone know anything about SPSS?
I am trying to make a histogram of a bunch of numbers in it, but I don't think I am providing both things needed to construct the histogram of the mean, mode and median.

I have a string of numbers and I believe what I need to complete the string is to somehow provide the position in the string the value exist at, but I am unsure if this is right.

[Vector]

What does SPSS stand for?

[PsyberianHusky]

Statistical-blah-blah-blah, Read:(Statistical Package for the Social Sciences(Aka people already bad at math))
It is a little bit like Microsoft Excel.

[Vector]

Oh, I see.  No, sorry, I can't help you there--I was wondering if it was some particular method I didn't know to do the usual thing.

[PsyberianHusky]

Right now I am getting "V5 is a string so histogram-cracker can not be produced" and I am mighty hungry for histogram crackers. I am betting right now I am only I recon the string I am feeding it the independent variable in the string of numbers and need to somehow give it a dependive variable to compare it too, I think? I have a shaky grasp on all of this.

Okay, I Figured it out, I had to cut and copy the list, The values where default being displayed to the 0th notation

[Christes]

Statistical-blah-blah-blah, Read:(Statistical Package for the Social Sciences(Aka people already bad at math))
It is a little bit like Microsoft Excel.

We had to use something like that in one of my undergrad linguistics classes.  It was amusing to use it while I was taking a stats class.  Unfortunately, that was a long time ago so I probably can't help you.

Unrelated: Topology Qual tomorrow!  I'm so screwed.

[ed boy]

FFFFFFFFFFFFUUUUUUUUUUUUUUUUUUUUUUU-

Find a general solution of the following differential equation:
x(dy/dx) = y + √(x² + y²)

What I've got so far:

v = x² + y² <--substitution
y = √(-x² + v)
dy/dx = 1/2(√(-x² + v))(-2x+ (dv/dx)) + 1/2(√(-x² + v))(-2x+1)(dx/dv)
dv/dx = -2x² + 2v +2√(-x² + v)√(v) + 2x + (2x+1)(dx/dv)

a) Which question should I be asking: Where did I go wrong? or What is dx/dv?
b) What's the answer to the question I should be asking?

Fucking differential equations and their eldritch ways.

Divide both sides of the equation by x
Take the x into the square root
Let w=y/x and substitute
Use the chain rule on dy/dx and hopefully you should be able to continue from there

[Bauglir]

This is a question similar to the one I asked a while ago about dice, and I've got this weird suspicion that it might be the same problem masquerading as a different one. Still, I'm trying to figure out if there's a formula for determining the probability of rolling X or higher on the Y highest of Z A-sided dice. I found an online tool that was helping me along, but it can't generate sufficiently high numbers of dice for my purposes (which require, at minimum, the odds for the 3 highest of 10 10-sided dice, and may go higher).

EDIT: No, I was wrong, for proper exploration of all my options, I need to be able to change how many dice you keep. However, A can be a constant of 10 if that simplifies things. This stuff is breaking my brain.

SECOND EDIT: If it's necessary, Y can be a constant of 3, though. That'd be good enough to get shit done.

[Grek]

The standard notation is XdYkN. Your question is P(XdYkN >= A). For X=10, Y=10, N=3;

P(10d10k3 >= 3) = 1.00
P(10d10k3 >= 4) = 0.9999999999
P(10d10k3 >= 5) = 0.9999999989000001
P(10d10k3 >= 6) = 0.9999999934000001
P(10d10k3 >= 7) = 0.9999998866000001
P(10d10k3 >= 8) = 0.9999993701000001
P(10d10k3 >= 9) = 0.9999977016000001
P(10d10k3 >= 10) = 0.9999907426000001
P(10d10k3 >= 11) = 0.9999699031000001
P(10d10k3 >= 12) = 0.9999195391000002
P(10d10k3 >= 13) = 0.9997899054000002
P(10d10k3 >= 14) = 0.9994970814000002
P(10d10k3 >= 15) = 0.9989093544000002
P(10d10k3 >= 16) = 0.9977125775000002
P(10d10k3 >= 17) = 0.9954366845000001
P(10d10k3 >= 18) = 0.9914221500000001
P(10d10k3 >= 19) = 0.9843497344000001
P(10d10k3 >= 20) = 0.9725121389
P(10d10k3 >= 21) = 0.9537060954
P(10d10k3 >= 22) = 0.9244678216
P(10d10k3 >= 23) = 0.8805924731
P(10d10k3 >= 24) = 0.8189000866
P(10d10k3 >= 25) = 0.7335500416000001
P(10d10k3 >= 26) = 0.6233553126000001
P(10d10k3 >= 27) = 0.4892514086000001
P(10d10k3 >= 28) = 0.34016728340000014
P(10d10k3 >= 29) = 0.18840359890000016
P(10d10k3 >= 30) = 0.07019082640000017

[Bauglir]

Would I understand how you got there if you explained it? What you've provided is helpful, because it means that I've raised A too quickly based on X (if that doesn't make any sense, I can explain in more detail), and thus need to recalibrate so that P never goes down from the combination of raising A and X, but doesn't go up too absurdly fast either. Thanks.

[Grek]

I got those numbers by running a simulation. As far as I know, there's not a reasonable way to calculate XdYkN problems formulatically. Would make a good disseration/thesis for some stats major some day.

[Bauglir]

... Curses. That explains Google's unhelpfulness, then. Thanks for your help, though. At least I know what I need to do now (build a simulation for whatever I need).

[Another]

The standard notation is XdYkN. Your question is P(XdYkN >= A). For X=10, Y=10, N=3;

To make sure I understand this the same way as you - XdYkN is the sum of N highest numbers out of X casts of an Y-sided die, right?

For N=1 I listed iterative analytical formulas in my reply to the previous question. Iterative formulas will give precise (up to rounding errors) probabilities and are best left for a simple computer program to crunch.

For a general case this matlab/octave program will compute P(XdYkN = A)*Y^X which are integer numbers and can be recoded in a language that supports arbitrary-precision arithmetic to get really big numbers. Standard floating point "double precision" will be exact up to about 15d10.

Spoiler (click to show/hide)

Code: [Select]

function R=P_XdYkN(X, Y, N, A)
%
R=0;
[high, done]=start_high(X, Y, N, A);
while(!done)
  R=R+combo(high, X, N);
  [high, done]=next_high(high, X, N, A);
end;

function R=combo(high, X, N)
%
n=1;
mult(n)=1;
for i1=2:N
  if high(i1)==high(i1-1)
    mult(n)=mult(n)+1;
  else
    n=n+1;
    mult(n)=1;
  end;
end;
R=1;
for i1=1:(n-1)
  R=R*nchoosek(X-sum(mult(1:(i1-1))), mult(i1));
end;
if high(N)>1
  R=R*comb_tail(mult(n), high(N), X-sum(mult(1:(n-1))) );
end;

function t=comb_tail(k, n, x)
%
t=0;
for i1=0:(x-k)
  t=t+nchoosek(x, i1)*(n-1)^i1;
end;
 
function [high, done]=start_high(X, Y, N, A)
%
if (A<N)||(A>N*Y)
  done=1;
  return;
end;
done=0;
A1=A;
for i1=1:N
  if A1>Y+N-i1-1
    high(i1)=Y;
    A1=A1-Y;
  else
    high(i1)=A1-N+i1;
    A1=A1-high(i1);
  end;
end;

function [high, done]=next_high(high, X, N, A)
%
done=1;
for i1=N-1:-1:1
  [high, done]=next_high1(high, X, N, A, i1);
  if done==0
    break;
  end;
end;
 
function [high, done]=next_high1(high, X, N, A, k)
%
done=1;
if high(k)-high(N)>1
  high(k)=high(k)-1;
  A1=A-sum(high(1:k));
  for i1=(k+1):N
    if A1>high(k)+N-i1-1
      high(i1)=high(k);
      A1=A1-high(k);
    else
      high(i1)=A1-N+i1;
      A1=A1-high(i1);
    end;
  end;
  done=0;
  break;
end;

function P_XdYkN_sample_print()
%
format long
S=1;
for i1=3:30
  disp(S=S-P_XdYkN(10, 10, 3, i1)/10^10);
end;


A check for the case P(10d10k3 >= A) confirms that all 10 digits for all A are the same as in one of the previous posts.

It is open to debate if a computer program that produces precise (in rational numbers) probabilities could be called a formula. For sufficiently large numbers there definitely should exist approximations properties of which should be easier to analyse.

[RedWarrior0]

Is it at all possible to take a non-integer derivative? You can take non-integer powers for example, as square roots.

Secondly, is it possible to find the "rate of change", so to speak, of a derivative? By this I mean the "derivative" of the transition from f(x) to f'(x) to f''(x) and so on.

[Christes]

1) Yes.  See fractional derivative.  I know basically nothing about it, though.  I know it also goes deeper if you really want to get into functional analysis.

This seems like a nice intro to it, but I haven't read the whole thing: http://www.xuru.org/fc/Intro.asp

2) Note that in order to even make sense of a "derivative" of a derivative, you're going to need to be operating in a space of functions rather than numbers.  You're also going to want to use the previous generalization of the derivative to get some sort of map from R to that space of functions.  You can probably get something from that, but I am most certainly not an analyst and this is merely blind musing on my part.