Wikipedia:Reference desk/Archives/Mathematics/2007 March 9

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March 9

vectors

i am having trouble with vectors and cable tension problems. Can anyone explain to me how the relationship between force, and tension works, (example; two cables attatched to a weight with the same vertical component Please help, I have no idea what to do!!!! -Meliss

If you have two cables and the same weight then the strain on each will be half what it would have been with only one cable. If you increase the weight you increase the cable tension. Think outside the box 13:05, 9 March 2007 (UTC)

Also, if the block is not accellerating, the net force will be zero. —The preceding unsigned comment was added by 199.197.125.65 (talk) 17:01, 9 March 2007 (UTC).

Dice odds question

if i roll two 8-sided dice and only look at the higher of the two, what is the average result i'll get? What if I roll 3 8-sided dice and only look at the highest die, what will the result be on average?--Sonjaaa 11:45, 9 March 2007 (UTC)

You might look at order statistic; although I don't know if that information will help you compute it. It might be easier if you just went through all the outcomes on a computer and computer the average, or if you did a lot of random trials and estimated it probabilistically. --Spoon! 11:59, 9 March 2007 (UTC)

I computed it in QBasic and got 4.5 for 1d8, 5.8 for 2d8 and 6.5 for 3d8.--Sonjaaa 12:41, 9 March 2007 (UTC)

For n 8-sided dice the average of the highest die is

$8-{\frac {(1^{n}+2^{n}+...+7^{n})}{8^{n}}}$

For two dice the average of the highest is 8-(140/64), which is 5 13/16. For three dice the average of the highest is 8-(784/512), which is 6 15/32. Gandalf61 14:18, 9 March 2007 (UTC)

Gandalf's answer looks incorrect. For instance, looking at the case when n=1, his formula says the average result of one eight sided die is 7 7/8. But in fact the average of a single 8-sided die is 4.5 . Dugwiki 16:21, 9 March 2007 (UTC)

How did you get 7 7/8? Looking for the case when n=1 Gandalf's formula says the average is $8-{\frac {(1^{1}+2^{1}+...+7^{1})}{8^{1}}}$$=8-{\frac {(1+2+...+7)}{8}}$$=8-{\frac {28}{8}}$$=8-3.5=4.5$ --CiaPan 16:34, 9 March 2007 (UTC)

...and the formula I gave also agrees with Sonjaaa's QBasic results for n=2 and n=3. Note that the sum of powers inside the brackets is always from 1ⁿ to 7ⁿ. It's fairly straightforward to derive the formula - not rocket science.Gandalf61 17:03, 9 March 2007 (UTC)

Oops, misread it, my bad. Dugwiki 17:17, 9 March 2007 (UTC)

Although it's not "rocket science", the derivation looks like it might not be obvious to someone asking the original question. So for reference, here's a way to derive the average -

Rolling n 8-sided dice labelled $d_{1}$ through $d_{n}$, first consider that the number of possible rolls where the highest result is at most a given number x is $x^{n}$ (each of the n dice can independently be any of 1 through x). Therefore for x>1 the number of possible rolls where the highest result is exactly x would be the number where the highest is at most x minus the number where the highest is at most (x-1); ie $x^{n}-x^{n-1}$. Thus the average of the highest die rolls is the sum of all possible results divided by $8^{n}$, which is -

${\frac {1^{n}+2(2^{n}-1^{n})+3(3^{n}-2^{n})+\dots +8(8^{n}-7^{n})}{8^{n}}}$

$={\frac {(1^{n}-2(1^{n}))+(2(2^{n})-3(2^{n}))+\dots +(7(7^{n})-8(7^{n}))+8(8^{n})}{8^{n}}}$

$={\frac {-1^{n}-2^{n}-\dots -7^{n}+8(8^{n})}{8^{n}}}$

$=8-{\frac {1^{n}+2^{n}+\dots +7^{n}}{8^{n}}}$

Hope that helps. Sorry about the confusion before. Dugwiki 18:00, 9 March 2007 (UTC)

To help you see this visually

Consider rolling two 8 sided dice.

The number of rolls where the highest result is 1 or less

[1,1] which is consistent with $x^{n}=1^{2}=1$

The number of rolls where the highest result is 2 or less

[1,1],[1,2]

[2,1],[2,2] which is consistent with $x^{n}=2^{2}=4$

The number of rolls where the highest result is 3 or less

[1,1],[1,2],[1,3]

[2,1],[2,2],[2,3]

[3,1],[3,2],[3,3] which is consistent with $x^{n}=3^{2}=9$

The number of rolls where the highest result is 3 = $x^{n}-x^{n-1}=3^{2}-2^{2}=9-4=5$

[1,3],[2,3],[3,1],[3,2],[3,3]

211.28.238.164 21:42, 9 March 2007 (UTC)

Does the formula $8-{\frac {(1^{n}+2^{n}+...+7^{n})}{8^{n}}}$ work for dice of any size? In other words, for d20s it would be $20-{\frac {(1^{n}+2^{n}+...+19^{n})}{20^{n}}}$?

Intersection of plane and sphere

A sphere is intersected by a plane. How can one calculate the surface area of the parts thus formed, given the radius of the sphere and the distance from the center to the plane surface, and ignoring the "cut" surfaces of each part? MS —The preceding unsigned comment was added by 74.12.28.148 (talk) 12:12, 9 March 2007 (UTC).

See Spherical cap. --Salix alba (talk) 12:34, 9 March 2007 (UTC)

You need to use the "Tennis Ball" theorem which is quite a surprising result. Imagine a tennis ball, that just fits inside a cylindrical tin. Then cut across the tennis ball and cut across the tin at the same time with two cuts that are parallel to the bottom of tin. You will get a piece of tennis ball and piece of tin and amazingly they have exactly the same area. You will now find it quite easy to work out the area of the piece of tin. Geoffcobra 23:49, 10 March 2007 (UTC)

Amusing name; I don't recall hearing this called the "tennis ball theorem", but it fits. (No pun intended!) I was going to suggest looking at the tomb of Archimedes. He proved the relationship between sphere and cylinder without the benefit of tin cans or tennis, and it was his proudest result (which is why he wanted it depicted on his tombstone). The benefit for modern readers is that we can contemplate this without formal calculus.

We also have the hairy ball theorem, telling us we cannot smoothly comb the hairs on a tennis ball. So what is it with mathematics and tennis? The only mathematician I've seen play well with a tennis ball was Ron Graham, and he was using it to juggle. (In fact, he juggled five of them, with ease, starting from a one-handed behind-the-back toss.) --KSmrq^T 08:18, 11 March 2007 (UTC)

Is there any children's version of Mathematica ?

Is there a version for 12-14 year old kids? 211.28.238.164 23:55, 9 March 2007 (UTC)

There is a reasonably priced Mathematica for Students, as well CalcCenter (essentially a trimmed-down version of Mathematica). You might also want to consider a free alternative like Maxima. "12-14 years old" is not a meaningful description in this context, what features do you need? Phils 00:45, 10 March 2007 (UTC)