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[Vector]

Because I'm sort of using this as a personal mathematics happiness thread, I was reminiscing about the past and ended up blurting out: "God, I want to write some functions!"

I highly recommend that anyone who feels the same way should try out computability or category theory.  I'm going to go dig into some pleasant work tomorrow morning

[MaximumZero]

Just got to trig functions yesterday. Soh Cah Toa indeed!

[Virex]

Somehow I think the dutch mnemonic (Sos Cas Toa) sounds better.

[ed boy]

Imagine that you have two dice, one with A sides, and one with B sides.

If you roll both dice, what is the chance that the number gotton from the dice with B sides is greater than the number gotten from the dice with A sides?

[Rysith]

Imagine that you have two dice, one with A sides, and one with B sides.

If you roll both dice, what is the chance that the number gotton from the dice with B sides is greater than the number gotten from the dice with A sides?

I, being a computer science person, would do that with case analysis:

Consider each possible result from the die with B sides
  For each result, compute the probability that the roll on A is lesser. Thus, if you are looking at 3 and A=6, then you have a .5 chance.
  But, since each result is only a possible result, divide by the probability of rolling that result.
  Add the probability of each result being greater than the roll on A, multiplied by the probability of getting that result, to a running total.

now, your running total is your answer.

Spoiler: example (click to show/hide)

Suppose A=4 and B=6, with A->X meaning that die A rolls the number X
For A->1, our probability is 0 (no rolls of B will be less than 1)
For A->2, our probability is 1/6 (one of six possible rolls of B). Add 1/6 * 1/4 (probability of rolling a 2) = 1/24 to the total
For A->3, our probability is 2/6 (two of six possible rolls of B). Add 2/6 * 1/4 = 2/24 to the total
For A->4, our probability is 3/6. Add 3/24 to the total

The total is now 6/24 = 1/4, the total chance that a four-sided die will roll higher than a 6-sided die.

We can also now see that what we really need is

Code: [Select]

sigma(1..B-1)/(B*A)
when A < B. It gets slightly more complicated when A > B, since the maximum contribution of any one result is 100%, so you need

Code: [Select]

sigma(1..B-1)/(B*A) + ((A-B)*B)/(A*B)

[Virex]

Assuming that each die is numbered 1, 2, 3.... n-1, n with n = A for die A and n = B for die B, we can say:


Each value on both A and b has a chance of 1/A and 1/B respectively.
For each value rolled on A, the amount of values on B that is higher then the given value is B-R_A with R_A the value rolled on A.
That means that for a given value of R_A, the chance that R_B will be higher is (B-R_A)/B = 1 - R_A/B for R_A < B and 0 for R_A > B.
The chance of rolling each specific value of R_A is 1/A, so for each possible value of R_A the chance that it occurs and that R_A < R_B is (B-R_A)/(A*B) for B > R_A and 0 for B < R_A.
Now to calculate the total chance we just add the chances for R_A = [1, 2.... A-1, A] together. 

[ed boy]

The die would be numbered from 1 to n, though. A six-sided dice, for example, has numbers {1,2,3,4,5,6}.

I was working on it a bit, and I got the following:

either A=B, A<B, or A=B.
let the values that the dice give be called a and b.

If A=B, then the chances are as follows:
a=b has a A^-1 chance of happening.
a>b and a<b each have an equal chance of happening, so ¹/₂(1-A^-1).

If A<B, then we can write B=A+K.
b will have a ^A/_B chance of being less than or equal to A, and a ^K/_B chance of being  greater than A. In this case, there will be the following chances:
a=b has a (^A/_B)(A^-1)=B^-1 chance of happening.
a>b has a (^A/_B)(¹/₂(1-A^-1)=^A-1/_2B=¹/₂(^A/_B-B^-1) chance of happening.
a<b has a ^A-1/_2B+^K/_B=^A-1/_2B+^2(B-A)/_2B=1+^-A-1/_2B=¹/₂(2-B^-1-^A/_B) chance of happening.

Note that this works if  A=B.

So, if you rearrange A and B into P and Q, where P is greater than or equal to Q, you get:
p=q has a P^-1 chance of happening.
p<q has a ¹/₂(^Q/_P-P^-1) chance of happening.
p>q has a ¹/₂(2-P^-1-^Q/_P) chance of happening.

[lordnincompoop]

I've hit a bit of a snag. How would you do this:
     
(90r⁶+28r⁵+2r⁴+45r³+5r²)/(9r+1)

[Vector]

Look up synthetic division, also known as Horner's Algorithm =)

[eerr]

I've hit a bit of a snag. How would you do this:
     
(90r⁶+28r⁵+2r⁴+45r³+5r²)/(9r+1)

Oh thats a toughie.

But it can be written, with uh, synthetic division?

[lordnincompoop]

Oh gawd that confused my poor brain even more. This is probably more basic maths than you guys are used to.

Apparently I'm supposed to do something like this:

      _____10r⁵____________________________
9r+1|90r⁶+28r⁵+2r⁴+45r³+5r²
       90r⁶+10r⁵
       _______
              18+2r⁴

But that didn't work out, as you can see.

[Vector]

Yeah, and then you keep on going until the end.  Keep going!  Keep going!

[Jim Groovester]

Long division.

But I'll make it simple for you 'cause I can already see how it can be factored because I'm awesome like that and don't you forget it. Just by regrouping things a little bit:

((90r⁶ + 10r⁵) + (18r⁵ + 2r⁴) + (45r³ + 5r²))/(9r + 1)

And this equals:

10r⁵ + 2r⁴ + 5r²

Oh gawd that confused my poor brain even more. This is probably more basic maths than you guys are used to.

Apparently I'm supposed to do something like this:

      _____10r⁵____________________________
9r+1|90r⁶+28r⁵+2r⁴+45r³+5r²
       90r⁶+10r⁵
       _______
              18+45r³

But that didn't work out, as you can see.

What are you doing not dropping the r⁴ term.

[Virex]

I've hit a bit of a snag. How would you do this:
     
(90r⁶+28r⁵+2r⁴+45r³+5r²)/(9r+1)

Code: [Select]

10r5 / 90r6 + 28r5 + 2r4 + 45r3 + 5r2 \ 9r+1
       -90r6 - 10r5 =


10r5+2r4 / 18r5 + 2r4 + 45r3 + 5r2 \ 9r+1
          -18r5 - 2r4 =


10r5+2r4 + 5r2 / 45r3 + 5r2 \ 9r+1
                -45r3 - 5r2 =
                0


So (90r6+28r5+2r4+45r3+5r2)/(9r+1) = 10r5+2r4 + 5r2

[Heron TSG]

Okay, so I have this problem that I can't seem to figure out.

Bobby is playing on a baseball diamond where each side is 90 feet long, and the field is a square. Bobby always runs at 16 feet per second. When Bobby is 30 feet away from first base, running to second, how quickly is his distance to third base changing?

I know that it has to do with taking derivatives, but these new story problems just don't make any sense to me. Can anyone help explain how to do these?