Wikipedia:Reference desk/Archives/Mathematics/2006 November 21

From Wikipedia, the free encyclopedia

< Wikipedia:Reference desk | Archives | Mathematics

Mathematics desk
< November 20	<< Oct \| November \| Dec >>	November 22 >

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

1 November 21

[edit] November 21

[edit] Differential equation- Temperature

I am doing differential equations in calc. and there was a problem where the rate of change of the temperature of coffee was modeled by:

y=-0.1(y-7)

in minutes and it started at 190 and the room was 70, how do i find the temperature after ten minutes? --205.154.39.10 16:25, 21 November 2006 (UTC)

This looks like Newton's Law of Cooling. I think the equation that you should have is

$\frac{dy}{dt}=-0.1(y-70)$ . From here, you can use the formula on that page to find your answer. --HappyCamper 17:40, 21 November 2006 (UTC)

Using the equation above, separate the variables so you have $\int\frac{dy}{y-70} = \int{-0.1}\,{dt} + c$ . Then solve, and get rid of the constant using the fact that

y (0) = 190

. You'll then be able to find

y (10)

to get your solution. Readro 21:04, 21 November 2006 (UTC)

[edit] Monty Hall problem (undeleted)

Monty Hall Problem

Any chance of getting the page Monty Hall problem wiped, deleted or corrected? I am assuming the answer is no, not in this place. Still I must ask.83.100.253.24 21:18, 20 November 2006 (UTC)

Imagine an encyclopedia in which whenever a reader didn't like an article, they could erase it. Would that be an encyclopedia worth reading? Melchoir 21:27, 20 November 2006 (UTC)

Attempting to do this would be like trying to take out a machine gun nest with a toothpick, any volunteers?83.100.253.24 21:46, 20 November 2006 (UTC)

User:Kieff (→Monty Hall Problem - removed non-question (possibly trolling) - this is a wiki, if you feel something must be corrected, do it.)

It's not a non-question - the monty hall page is mathematical nonsense - not a troll. Whether it has value as it stands (as perhaps User:Melchoir seems to be suggesting is another question).87.102.16.174 17:49, 21 November 2006 (UTC)

Hypothetical metaphors aside, I'm not sure why you would want to the delete the article, or what specifically you're looking to "fix". It's basic conclusion is correct (that you have a better chance of winning the prize if you switch to the curtain Monty Hall didn't open). But if you found a minor error somewhere in the article, your best bet is to point it out on that article's talk page. Dugwiki 22:18, 20 November 2006 (UTC)

I think that you are wrong, and that the majority of people still reading the Monty Hall talk page might be difficult to reason with. That's why I've brought it up here.

From the problem itself:The chances of the prize being behind either of the two remaining doors (after one non-winning door has been eliminated) is/are equal. The Monty Hall page goes to great lengths to prove the wrong answer true. Does this matter?87.102.16.174 17:49, 21 November 2006 (UTC)

The chances are not equal, and the article is correct. If you don't believe me, try it with someone. Take bets. Record the outcome. You will find that it is statistically speaking it is even, Alex is right and Brett is wrong. - Rainwarrior 18:18, 21 November 2006 (UTC)

I don't get it. there are two doors, the prize is behind one - statistically speaking I expect to break even.87.102.16.174 18:40, 21 November 2006 (UTC)

Well, your expectations don't take into account the information gained between the first and second steps. If you've read the article and you don't trust its analysis, then play the game 30 times or so with someone, and record the results. The correct probability will become apparent. - Rainwarrior 18:46, 21 November 2006 (UTC)

What information is gained? As far as I can see the probability of winning increases from 1/3 to 1/2. Nothing more.

You can read the article or its talk page which has plenty of explanation of this, but as I've said, you really need to try out the game yourself before you continue arguing against the facts. There are several simulators available online linked from the website [1] [2] [3], and if you don't trust them you can do it yourself with some dice a pencil and paper. - Rainwarrior 18:59, 21 November 2006 (UTC)

This is nonsense. The reasoning on the page is wrong. Can't you reason?87.102.16.174 19:04, 21 November 2006 (UTC)

I think the best analogy is to look at this similar example. Let's say the there were a million curtains, only one of which has a car. You pick curtain number 1. Monty says "Ok, you can keep that curtain, or ..." then he opens ALL the other curtains except for curtain number 237,486 and shows that none of the other curtains had a car. He continues "or you can have what is behind curtain 237,486". Which curtain do you think is more likely to have the car: the one you picked at random initially, or the curtain that was left over out of all the other 999,999 choices when Monty opened everything else up? Obviously the answer is that your curtain still has only a one in a million chance of having the car, so you should switch.

The situation is very similar with just three curtains. You pick a curtain at random initially and have a 1/3 chance of getting the car. Monty opens up all but one of the remaining curtains, so the chance that the curtain he didn't open has the car is 2/3. Dugwiki 19:08, 21 November 2006 (UTC)

For those with a programming bent, it's quite easy to code up a simulation of the Monty Hall problem and run it 3 million times. It should become clear that switching is actually beneficial. I see it this way: 2/3 of the time, your first guess was wrong - that much is uncontroversial. Every time your first guess is wrong, you'll get a car by switching, because he's narrowed the places the car could be down to one other. Only the 1/3 of the time that your first guess was correct is it a good idea to stick with it. -GTBacchus^(talk) 19:17, 21 November 2006 (UTC)

Let's try a different way - this time there are only two curtains; one has a car, the other a goat behind it. You can pick one. Which curtain should should pick out of the two? This is the same as the monty hall problem after a non winning curtain has been eliminated.87.102.16.174 19:23, 21 November 2006 (UTC)

No its not, as there is a 1 in 3 chance it is behind one of the original curtains, so the, and a 1 in 2 chance it behind one of the remaining curtains, so you switch from a 33:66 chance to a 50:50 in your favour. Philc _T^E_C^I 19:27, 21 November 2006 (UTC)

On removing one curtain the odds of your first answer being right change from 1 in 3 to 1 in 2. Is this a source of confusion?87.102.16.174 19:32, 21 November 2006 (UTC)

I'm sorry, but no. Try it, seriously. 2/3 of the time, your initial guess is wrong, and that makes it worth it to switch to what's been revealed as the only other possible non-goat curtain. You can't change the probability of something that's already happened. Really - try it. -GTBacchus^(talk) 19:34, 21 November 2006 (UTC)

I'm sorry you don't understand or choose to be stupid. The probability of you being right does change. Are you drunk or something?87.102.16.174 19:42, 21 November 2006 (UTC)

It turns out that I'm not drunk; thanks for asking. I was hasty to say that you can't change the probability of an event that's already happened. Dugwiki is quite correct below that it hinges on the fact that Monty knowingly reveals everything that you didn't pick and that isn't a car. He's not acting at random, but based on your original choice. Try this: there are three curtains: A, B and C, with a car behind C and goats behind A and B. You choose A, B or C with equal probabilities. If you choose A, Monty opens curtain B, and you win if you switch to C. If you choose B, Monty reveals goat A, and you win if you switch to C. Only the 1/3 of the time that you chose C in the first place is it in your interest to stick with your first choice.

If that doesn't work for you, I'm not going to try any more explaining until you try it for yourself. Run an experiment, and convince yourself of the truth of the matter. I have done it, and it worked. If you don't do it, then you're in the same logical position as the churchmen who refused to look through Galileo's telescope to see the moons of Jupiter, because they were already convinced they didn't exist. Ok? -GTBacchus^(talk) 21:06, 21 November 2006 (UTC)

By the way, a particularly good way to convince yourself is to play Monty, and try it on others. Keep a record of what percentage of switchers win over time, and what percentage of non-switchers win over time. See how that goes, and then argue with reality if you're still not convinced. -GTBacchus^(talk) 21:09, 21 November 2006 (UTC)

For an example of probabilty changing try this: You are asked to pick a card at random from a deck of 52 cards (all the cards,one of each). You win if you pick an ace. The probability of this is 4/52. Someone else then reveals the other 51 cards. Say in this example there are 4 aces shown in the revealed cards. The probability of winning is now 0 since you did not pick an ace. This shows that probabilty can change.87.102.16.174 19:50, 21 November 2006 (UTC)

The point you're missing is the Monty doesn't randomly pick curtains to open. Rather, he opens ALL the curtains other than yours that don't have the car. If Monty opened curtains at random, then it would be a completely different situation. So what's happening in this problem is that Monty is explicitly telling you "if you didn't pick the car, then the car is over here behind this curtain you didn't pick out of all the curtains you could have picked." Thus the partition is your curtain versus the set of all the other curtains, and obviously the set of curtains besides yours is more likely to have the car. In your card example, it would be Monty is picking and choosing which cards to reveal. He's not picking the cards randomly, as in the example you provided, but is intentionally keeping an "ace up his sleeve" if he has one. Dugwiki 20:00, 21 November 2006 (UTC)

Is this really the place to continue arguing? Either he's a troll pretending not to understand the article, or he really doesn't understand it. If the latter, perhaps there is more to do to make the article even more easily understandable, but I don't see how repeating the same arguments here will persuade someone who refuses to be persuaded. And, 87.102.16.174, please be civil. —David Eppstein 19:44, 21 November 2006 (UTC)

To help explain it think of it with much larger odds, with a deck of cards, you have to pick the ace of spades, so you pick a card, but your not shown it, there is a 1 in 52 chance that this is the right card, and a 51 in 52 chance it is in the deck still, if he takes away all of the rest of the deck apart from one card, and tells you that either this card, or the card you picked isthe ace of spades, there is still a 51 in 52 chance it is the other card, and still a 1 in 52 that you picked it right the first time. Just because he takes away cards doesnt mean that he changed the probability of an event that has already happened, therefore the chance that you picked right the first time is and will remain 1 in 52, even after the rest of the deck has been fiddled. I cannot change to 1 in 2 chance that you picked it right firsttime, as for this to be true, half the deck would have to be ace of spades. Philc _T^E_C^I 21:26, 21 November 2006 (UTC)

I know that y'all are trying to be helpful here but it seems that User:83.100.253.24/User:87.102.16.174 is gettin' mighty plump with all of the feeding going on. ;-) --hydnjo talk 22:15, 21 November 2006 (UTC)

That may be, but I'm happy to have just learned what I now think is the best way to explain the Monty Hall problem. If someone truly (or trolly) won't see, the best thing is to insist that they go play Monty, and try it on their friends. By experiencing Monty's side of the deal rather than the contestant's I think it should become quite clear, and if nothing else, the law of large numbers always wins. -GTBacchus^(talk) 22:24, 21 November 2006 (UTC)

In that case you may enjoy this little side-dish (well at least it isn't green bean casserole. ;-) --hydnjo talk 22:54, 21 November 2006 (UTC)

Yeah, it's just surprising how many people aren't convinced by that argument either. There's something about actually experiencing the game first-hand, as Monty, that no argument can convey. That is a good summary that you've got there, of the "many doors argument". (I like it when good arguments have names. :)) -GTBacchus^(talk) 23:01, 21 November 2006 (UTC)

[edit] inverse secant

I was asked to prove

$\arcsec\left(x\right) = \int_x^1 \frac 1 {|z| \sqrt{z^2 - 1}}\,\mathrm{d}z, \quad x > 1$

how do I do this? --68.126.2.136 04:16, 22 November 2006 (UTC)

The first thing that occurs to me is to use the Fundamental Theorem of Calculus. -GTBacchus^(talk) 23:37, 21 November 2006 (UTC)

If you need a little more of a push, are you familiar with the derivative of the arcsecant function? It's $\frac{\mathrm{d}}{\mathrm{d}x} \arcsec (x)\,=\,\frac{1}{|x|\,\sqrt{x^2-1}}$ .

I'd be surprised if you could prove this; the limits of integration appear to be reversed. And given that those limits are positive, the absolute value is not necessary. So you want to prove

$\int_1^x \frac 1 {z \sqrt{z^2 - 1}}\,\mathrm{d}z = \arcsec\left(x\right); \qquad x > 1.$

It is important to note that this is an easier problem than "Find a closed-form result for the integral." That is why others have suggested using the fundamental theorem of calculus. However, this sounds like a homework assignment, and we cannot know what facts the class has already covered that you might be expected to use in the proof. Also, the rules of this page allow us to guide you, but prohibit giving a full answer. --KSmrq^T 06:29, 22 November 2006 (UTC)

The formula cited is exactly the one found in our article Inverse trigonometric function and apparently copied from there. ~~Note that the arcsec function is monotonically decreasing for each of the parts where it is defined and real-valued (x ≤ −1 and x ≥ 1), so its derivative should be negative (possibly −∞).~~ --Lambiam ^Talk 09:00, 22 November 2006 (UTC)

Come, let us calculate together, from first principals, backwards. Define

$\begin{align} x &{}= \sec y \\ &{}= \frac{1}{\cos y} \\ &{}= (\cos y)^{-1} \end{align}$

Since d cos z is −(sin z)dz, and du^a is au^a−1du for nonzero integer a, we immediately find

$\begin{align} \frac{dx}{dy} &{}= -(\cos y)^{-2} (-\sin y) \\ &{}= \frac{\sin y}{\cos^2 y} \end{align}$

This leads to the derivative of arcsec x by inversion, used in two senses. First, using the functional inverse we have

$y = \arcsec x . \,\!$

Second, using the reciprocal we have

$\frac{dy}{dx} = \frac{\cos^2 y}{\sin y} .$

The right-hand side is expressed in terms of y, but we can rewrite it in terms of x by noting two facts:

$\cos y = \frac{1}{x} ,$

and

$\sin y = \sqrt{1-\cos^2 y} .$

This gives us our desired formula for the derivative of arcsecant.

$\begin{align} \frac{dy}{dx} &{}= \frac{\cos^2 y}{\sin y} \\ &{}= \frac{1/x^2}{\sqrt{1-\frac{1}{x^2}}} \\ &{}= \frac{1}{x^2 \sqrt{\frac{x^2-1}{x^2}}} \\ &{}= \frac{1}{\frac{x^2}{\sqrt{x^2}} \sqrt{x^2-1}} \\ &{}= \frac{1}{|x| \sqrt{x^2 - 1}} \end{align}$

(Alert readers will have noticed that we quietly resolved sign ambiguity using principal values.) From this the solution to the given problem is a small step. --KSmrq^T 01:21, 23 November 2006 (UTC)

Categories: Reference Desk Archives

Wikipedia:Reference desk/Archives/Mathematics/2006 November 21

From Wikipedia, the free encyclopedia

Contents

[edit] November 21

[edit] Differential equation- Temperature

[edit] Monty Hall problem (undeleted)

[edit] inverse secant

Views

Navigation

Interaction

Search