Wikipedia:Reference desk/Archives/Mathematics/2007 January 6

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January 6

An odds question

This seems like it would be simple, and it probably is, but I don't know the answer. If I were to roll a six-sided die six times, what are the odds it will come up "3" one of those times? I know if I roll it once, the odds are 1:6, but how do I expand that? My gut reaction is that the odds for six rolls would be 6:6, but that can't be right, because it's not a certainty I'll get a 3. On the other hand, it's definitely more likely I'll get a 3 if I roll the die six times than if I roll it once (at least ahead of time; I'm aware that if I roll it five times and don't get a 3, I'm back to 1:6). So how would I calculate this? —Chowbok ☠ 03:14, 6 January 2007 (UTC)

Reason backwards. For one roll, the odds are 5:6 against rolling a "3". To not get a "3" in six rolls, it must not happen each time. So the odds against are (⁵⁄₆)⁶. Thus the odds for at least one "3" turn out to be slightly less than 2:3. --KSmrq^T 03:30, 6 January 2007 (UTC)

Aha! That's very clever. Thanks! —Chowbok ☠ 03:38, 6 January 2007 (UTC)

Note that this gives the odds of rolling a "3" one or more times. The odds of rolling a "3" exactly once would be a bit lower. StuRat 20:52, 6 January 2007 (UTC)

Golden Ratio

I was wondering, is the altitude of a right triangle somehow related to the golden ratio (phi)? I've been trying to find some kind of relation by extending the altitudes of the right triangle but have not arrived at the golden ratio. Thanks for any input that you might have. --Proficient 18:13, 6 January 2007 (UTC)

For any positive value there is a right triangle whose altitude has that length. For positive values less than 1, the same applies for the ratio between the altitude and the base. The golden ratio appears when you have specific angles in your triangle, such as 36° and 72°.  --Lambiam^Talk 19:28, 6 January 2007 (UTC)

All right. Thanks, but that is not quite the answer I was looking for. --Proficient 20:11, 6 January 2007 (UTC)

The answer is no. Side lengths or side ratios of right triangles can vary freely, subject only to the Pythagoras identity (a²+b² = c²). Your quest is doomed to futility, or irrelevance. --KSmrq^T 05:18, 7 January 2007 (UTC)

0.333...=1/3?

Does 0.333...=1/3 it's my belief that 0.333... does not exactly equal 1/3 because there is no point at which 0.333... becomes 1/3. It gets closer and closer as it extends but it never exactly reaches 1/3. What do you think?

Firstly, 0.333... is a number, not a process. It doesn't "get closer and closer" to 1/3; its value doesn't change over time. Secondly, 1/3 is a real number and as such has a decimal representation. 0.333... is the only candidate. --Huon 18:36, 6 January 2007 (UTC)

You say it's the only candidate but it's still only an aproximation. It's not exact, is it and if so prove it.--71.234.101.173 19:02, 6 January 2007 (UTC)

See 0.999.... -- mglg_(talk) 19:14, 6 January 2007 (UTC)

I think you are using flawed logic. --Proficient 20:02, 6 January 2007 (UTC)

The proof presented as algebraic proof can be modified to prove that 0.333... = 1/3. --TeaDrinker 20:05, 6 January 2007 (UTC)

We have an article on recurring decimals, with this as the first example. Using standard real numbers and the standard interpretation of decimal expansions, the equality is exact. Although decimal expansions have many practical advantages in computations when compared to ratios of integers, they do have this one dramatic complication. Only fractions like ³⁄₅ or ¹³⁄₁₆ or ⁷⁄₁₀, where the denominator contains only factors of 2 and 5, have terminating expansions. All the rest, of which ¹⁄₃ is the simplest example, have recurring expansions. Perhaps this is one reason decimal expansions were slow to catch on. However, it was known in Ancient Greece that some important numbers cannot be given by a ratio of integers; the diagonal of a unit square is a famous example. For a time continued fractions seemed like an appealing alternative; and, indeed, they do have special virtues. (The length of that diagonal is a repeating continued fraction.) But eventually mathematics discovered that almost all numbers, what we now call real numbers, cannot be written simply no matter what system we adopt. So for theoretical work we prefer exact abstract definitions, while for most practical work we use approximate decimal expansions in which we truncate the recurring expansion. If we truncate 0.333… after, say, 42 repetitions of "3", the truncated number is strictly less than ¹⁄₃. Likewise, if we round 0.666… after 42 repetitions of "6", the rounded number is strictly greater than ²⁄₃. But do not confuse these approximations with the full decimal expansions, for which equality is exact. --KSmrq^T 06:02, 7 January 2007 (UTC)

Radians

is a π radian 180^o? so 2π would be 360^o? 81.129.12.4 19:46, 6 January 2007 (UTC)

That is correct; see radian.  --Lambiam^Talk 19:54, 6 January 2007 (UTC)

For that correct answer you deserve a nice slice of pi. StuRat 20:43, 6 January 2007 (UTC)

integro differential equation

what is integro differential equation

An integro-differential equation is an equation in which an unknown function appears under an integral sign operation, and also in an expression subjected to a differential operation. For a very simple example, take

$f'(x)=1+\int _{0}^{x}uf(u)\,du.$

This example happens to be easily transformable into a straightforward differential equation, but in general that is not possible. See also Integral equation.  --Lambiam^Talk 05:27, 7 January 2007 (UTC)

And so now you're thinking, "When will I ever need a beast like that?" And the answer is, you're probably using something like this all the time in the form of a PID controller. For example, the hard disk drive of a computer has a moving head, which quite likely is positioned by a controller that depends on a integro-differential equation.

That said, most of the famous equations of physics will be written either in differential form, or in integral form, but not in integro-differential form. So it may be some time before you meet one of these beasts "in the wild". --KSmrq^T 06:19, 7 January 2007 (UTC)

I belive I've heared a presentation about a model that used such a beast. It was something about tires of vehicles. – b_jonas 17:24, 7 January 2007 (UTC)

Calculus Question - Optimization

Hello,

I have an interesting calculus question that I found in an old textbook. I would like to attempt this question on my own; however, I do not know where to start. Here is the question:

You are allergic to pollen, however you are forced to live on a straight line between two pollen sources. One source is four times as strong as the other. The amount of pollen (A) that reaches you is calculuated by A = (kS)/R where k is a constant, S is the strength of the source and R is the distance from the source. If the distance between the two sources is 1 km, where must you live on this line to suffer the least allergic reaction.

This question is amazing. It truly shows the useful nature of calculus. If I can remember from my calculus courses in university you need to take a derivative and make it equal to 0 to find the greatest or least points. However, the above equation has so many variables that you would probably need another equation. Any tips on solving this would be great!

Thanks Rick

You have it exactly right, however you don't need another equation. You're looking for dA/dR, which you can compute easily assuming both k and S are constant. The minimum will occur at a point where dA/dR = 0. Remember there are two sources, so the function A will be a sum of the function you have. -anonymous6494 06:44, 6 January 2007 (UTC)

You should learn the very simple argument that shows a minimum or maximum of a continuously differentiable function can only occur under two circumstances: (1) you cannot move past a boundary, or (2) the function is not changing. The second condition becomes the formal necessity that the derivative be zero. But be careful; if you seek a minimum and find a zero derivative, you may have found a maximum or even an inflection. If possible, check the derivative of the derivative (the second derivative of the function) to be sure it is positive.

This pretty example places us on a line, so the only input variable is a number between 0 and 1. Let's agree that d = 0 places us at pollen source P₀, and d = 1 places us at P₁. (Thus d is the distance from P₀ in kilometres.) Also, let's make source P₁ the stronger one. The amount of pollen, A(d), is the sum of what reaches us from both sources. Notice that, between the two sources, as we move closer to one we move farther from the other; so we expect there will be a position that gives an optimal balance. However, if we are not so constrained, we can move far away from both sources, in either direction along the line.

Notice that we are given neither k nor S, which hints that the optimal location may not depend on these values. (Or, our answer will be a symbolic function of these quantities.) The amount of pollen at the optimum location between the sources will depend on these, but we are not asked for that. --KSmrq^T 08:48, 6 January 2007 (UTC)

Please do not shuffle the sections out of the chronological order in which they were created.  --Lambiam^Talk 04:43, 7 January 2007 (UTC)

Quadratic simultaneous equations

x-2y=1

x²+y²=29

i subbed in x=2y+1 into x²+y²=29 to get 4y²+4y+2+y²=29 which simplified i got as 5y²+4y+2=29

Trying to solve this i ended up 5y²+4y-27=0, but now i cant factorise it into to brackets to find the y value as it is from a non-calculator paper. Pleasae help-i thought i had it correct until i tried to factorise it.

Thanks in advance86.129.218.230 21:24, 6 January 2007 (UTC)

Your error seems to be in the expansion of (2y + 1)² 80.169.64.22 21:34, 6 January 2007 (UTC)

how is it wrong? —The preceding unsigned comment was added by 86.129.218.230 (talk) 21:42, 6 January 2007 (UTC).

(2y + 1)² = 4y² + 4y + 1. You should use the quadratic formula (one of the solutions is non-integer). Clarityfiend 21:52, 6 January 2007 (UTC)

AHAH. i now get 5y²+4y-28. this then Factorises to (5y+14)(y-2)which then gives y=2 or y=-2.8???? I think this is right now all i need to do is find the x values. have i got it right so far??? 217.42.215.97 01:39, 7 January 2007 (UTC)

Yes.  --Lambiam^Talk 05:39, 7 January 2007 (UTC)