Wikipedia:Reference desk/Archives/Mathematics/2009 October 7

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October 7

Integration by Substitution

I have a problem with integrating a certain function. I realise it can also be done as it is in Anton, Bivens and Davis Calculus, but I am always interested in seeing if a thing can be done in many different ways. The diagram on the link is not very clear. Its base equals √( 4 - x²), its height is equal to x, and the hypotenuse is 2.

[IMG]http://img194.imageshack.us/img194/7444/integration.png[/IMG]

If you have any advice, please do not hesitate. Thanks, The Russian.C.B.Lilly 02:11, 7 October 2009 (UTC) —Preceding unsigned comment added by Christopher1968 (talk • contribs)

You should not have cos(y) under the square root sign in the last integral of your solution. Things come out the way they should if you fix that, I think. 70.90.174.101 (talk) 06:55, 7 October 2009 (UTC)

The annoying constant 4 is eliminated by setting x = 2y. Then √(4−x²)dx = √(4−(2y)²)d(2y) = √(4−4y²)2dy = √(4)√(1−y²)2dy = 4√(1−y²)dy. The integration limits x=0 and x=2 becomes y=0 and y=1. So you have

$\int _{x=0}^{x=2}{\sqrt {4-x^{2}}}dx=4\int _{y=0}^{y=1}{\sqrt {1-y^{2}}}dy$

So (by renaming the dummy variable y into x) your problem is reduced to computing

$\int _{0}^{1}{\sqrt {1-x^{2}}}dx$

which I am sure you can do by setting x = sin(y). Bo Jacoby (talk) 10:23, 7 October 2009 (UTC).

The 4 isn't annoying. You just substitute $x=2\sin y$ in the first place instead of doing 2 different substitutions. I think the integral becomes $4\cos ^{2}y\,dy$ then. StatisticsMan (talk) 13:57, 7 October 2009 (UTC)

Yes, but beginners might prefer to take one little step at a time. If you think geometrically you will identify the integral as the area of one fourth of a circle with radius = 2, and then the result (1/4)π2² = π appears immediately. Bo Jacoby (talk) 17:53, 7 October 2009 (UTC).

Perhaps, but it's just a standard rule they teach in Calculus 2. If you have a^2 - x^2, you substitute x = a sin t. StatisticsMan (talk) 20:29, 7 October 2009 (UTC)

Or you just look at the inside back cover of your calculus textbook and it gives you the answer. While it is good to know how to do the integrals, a lot of people just look up (or learn) the final answer for all those inverse trig integrals. --Tango (talk) 01:56, 8 October 2009 (UTC)

Thanks for that. This will be very useful. The Russian. 202.36.179.66 (talk) 04:39, 8 October 2009 (UTC)

solving mathematical problems

A function is represented by f(x)=(x-2)²+4. Find the inverse of the function. —Preceding unsigned comment added by Proffartduny (talk • contribs) 13:14, 7 October 2009 (UTC)

Email address removed, as per comments at the top of this page. Also, this looks like a homework question. Have you attempted solving it yourself? —  Tivedshambo  (t/c) 13:20, 7 October 2009 (UTC)

As a hint, try expanding it into the form ax²+bx+c=0, then solving as a quadratic. —  Tivedshambo  (t/c) 13:23, 7 October 2009 (UTC)

Actually, don't do that, it only makes the problem more complicated. Just invert the operations one by one, starting with moving the 4 to the left-hand side. — Emil J. 13:28, 7 October 2009 (UTC)

Well, you could do that if you like doing things the easy way... ;-) —  Tivedshambo  (t/c) 15:01, 7 October 2009 (UTC)

I'm sure he do.--Gilisa (talk) 15:31, 7 October 2009 (UTC)

You can't know that there will even be an inverse until you specify a domain for the function. For example, if you take the domain to be the finite field with 4 elements, your function is its own inverse. More to the point, if you take the domain to be the interval (-10, -9) in the real line, then the inverse will be given by a different formula than if you take the domain to be the interval (9,10), if you follow the method suggested by EmilJ. — Carl (CBM · talk) 15:34, 7 October 2009 (UTC)

Indeed. Ordinarily I would assume the domain is the real numbers (if it were something unusual, the OP would have specified and the usual options are real and complex numbers. If it were complex, I would have called the variable z, so I'll assume reals) however there is no (unique) inverse over the real numbers... --Tango (talk) 15:47, 7 October 2009 (UTC)

• Keep in mind that (x − 2)² + 4 ≥ 4 for all real numbers x, i.e. ƒ : R → [4,∞). So if you want to find a real number x such that ƒ(x) < 4 then you're going to come unstuck. Even if you try to solve y = ƒ(x) for x as a function of y then you'll meet problems. You'll get two solutions coming from the ±√ in the quadratic formula. The two solutions have different images. One of them is a function g : [4,∞) → (−∞,2] and the other is a function h : [4,∞) → [2,∞). I guess that you might be able to cobble these two partial inverses together since R = (−∞,2] ∪ [2,∞); although I can't, as yet, see how. The problem comes from the fact that your ƒ isn't injective. If it were then you could find, (well, know that there exists) an inverse ƒ⁻¹ : [4,∞) → R. ~~ Dr Dec (Talk) ~~ 22:46, 7 October 2009 (UTC)

I'm not sure why you would try to combine them (and I can't see how you usefully would - you could change the domain of the inverse to $[4,\infty )\times \mathbb {Z} _{2}$ and then have half of it map by one inverse and half by the other, but I can't think of an application for such a function - it wouldn't strictly speaking be an inverse), it would be easier to just arbitrarily choose one. Obviously the union of the images is the whole real numbers since that is the domain of the original function - that isn't a useful piece of information. --Tango (talk) 02:05, 8 October 2009 (UTC)

I'm amazed by the amount of discussion that can erupt from such a basic homework problem. --COVIZAPIBETEFOKY (talk) 03:29, 8 October 2009 (UTC)

Probability question

I have 8 letters that I can put into 12 slots with repeats. what is the probability that one sequence of 12 selected at random has all 8 letters in it?

Also do the same for 10 slots. —Preceding unsigned comment added by 98.77.188.120 (talk) 22:50, 7 October 2009 (UTC)

• Take a look at the combinations article, and in particular this subsection. This should give you the total number of ways of distributing your eight letters amongst the 12 letter boxes. Then you need to work out how many of these distributions have the property that all of the eight letters are in one letter box, or another. Once you have these two number then getting hold of the probability just requires a little more arithmetic. ~~ Dr Dec (Talk) ~~ 23:01, 7 October 2009 (UTC)

• I read it differently. We're not choosing 8 elements from 12. We're distributing (with repeats) 8 elements among 12 available slots. The number of ways to do that is 8¹². Zain Ebrahim (talk) 11:15, 8 October 2009 (UTC)

• I read it assuming that the order of posting didn't matter. You are choosing eight things from 12, with repeats. Each time you put a letter in a letter box you're choosing a letter box. You make eight choices by posting your eight letters. That gives 75,582 ways of posting your eight letters into 12 letter boxes. If the order matters then there are 12⁸ = 429,981,696 ways to do it: there are twelve letters boxes to choose from for your first letter, twelve for your second, …, twelve for your eighth and final letter. ~~ Dr Dec (Talk) ~~ 11:38, 8 October 2009 (UTC)

You are choosing eight things from 12, with repeats: this would not make any sense, since then all 8 letters would be present in every sequence, i.e., the probability the OP is asking for would be trivially 1. Moreover, the OP explicitly describes the random choice being made as "one sequence of 12". We are choosing 12 things from 8, with repeats. — Emil J. 12:17, 8 October 2009 (UTC)

• Why doesn't "choosing eight things from 12, with repeats" make any sense? Let me rephrase what I thought was the OP's question: You go shopping and you need to buy eight tins of dog food. There are twelve types available. What is the probability that you pick all the same type of dog food? You are making eight selections from 12 varieties, and you can repeat, i.e. buy the same type more than once. Maybe I'm misunderstanding what the OP was asking, but I'm quite sure that my answer makes sense. It's just a re-wording of the doughnut example in this subsection. Posting the letters into letterboxes corresponds to making choices of dog food/doughnuts, e.g. putting a letter in letterbox 1 corresponds yo choosing a sugarcoated doughnut, etc. Anyway, seeing as the OP says that the order is important then it's all academic. ~~ Dr Dec (Talk) ~~ 13:52, 8 October 2009 (UTC)

Well, I'm not quite sure what the OP wants either. However, as I read it, the question was whether the chosen sequence contains all 8 letters; in the dog-food example, this would translate to something like What is the probability that your purchase includes all eight tins. This obviously does not make much sense, so that's how I meant it. Now I understand that you read also the rest of the question in a completely different way than I thought. PS: is there any particular reason why you keep making your posts one-element unnumbered lists? — Emil J. 14:48, 8 October 2009 (UTC)

• Replied at your talk page ~~ Dr Dec (Talk) ~~ 17:37, 8 October 2009 (UTC)

It seems fairly clear to me that the OP has 8 varieties of dog food available and wants to choose 12 tins at random (uniformly). The question is then what is the chance that he has chosen at least one of each variety. I don't see how the description can be interpreted in any other way. --Tango (talk) 21:40, 8 October 2009 (UTC)

• But why?! The OP has eight letters and they go into any of the 12 letterboxes. Putting one letter into a given letterbox is making a choice between the 12 letterboxes. The OP has to put eight letters into these 12 letterboxes, so he has to make a choice between the 12 letterboxes eight times, i.e. make eight choices from 12 with possible repeats. ~~ Dr Dec (Talk) ~~ 21:48, 8 October 2009 (UTC)

In other words, the OP wants the probability that a uniformly random function from 12 to 8 is a surjection. Now, counting surjections is a bit tricky. It can be computed using the inclusion-exclusion principle, where n = 8, and A_i is the set of sequences which avoid the ith letter. Unless I'm mistaken, the answer is $8^{-12}\sum _{k=0}^{8}{\binom {8}{k}}(-1)^{k}k^{12}={\frac {\ 50\,093\,505}{536\,870\,912}}\approx 0{\cdot }0933$. — Emil J. 12:45, 8 October 2009 (UTC)

order matters —Preceding unsigned comment added by 98.77.188.120 (talk) 23:44, 7 October 2009 (UTC)

Permutations, then. Intelligentsium^review 01:32, 8 October 2009 (UTC)

With this one it seems you have to be careful about the way the question was phrased. "One sequence of 12" means: on the first trial you choose one of the eight letters; on the second trial you choose one of the eight letters (which could be the same as the one chosen on the first trial); on the third trial you choose one of the eight (which could be the same as the one chosen on the first or second trial), etc., through 12 trials. So it does appear to ask for the probability that a random function from {1, 2, 3, ..., 12} to {1, 2, 3, ..., 8} is surjective. Michael Hardy (talk) 13:28, 8 October 2009 (UTC)

• The more I read the OP: "I have 8 letters that I can put into 12 slots with repeats. what is the probability that one sequence of 12 selected at random has all eight letters in it?" the less I understand it. This is almost certainly due to my weak and fragile mind. Take the first part of the OP: "I have 8 letters that I can put into 12 slots with repeats." Well, if you have 8 letters that you can put into 12 letterboxes then you need to make eight actions, i.e. put each of the eight letters into one of 12 letterboxes, maybe putting more than one letter into the same letterbox. What does it mean to say "What is the probability that one sequence of 12 selected at random has all 8 letters in it?" This second part seems to mean that you choose 12 letterboxes at random and you look to see if any of them have all eight letters in it. So here's what I think we're being asked: You post your eight letters into your 12 letterboxes, maybe putting more than one letter in each letterbox, and then you randomly point to letterboxes 12 times, possibly pointing to the same letterbox more than once. What is the probability that one of the letterboxes that you have pointed at had all of the letters in it?! I give up; my head's spinning. ~~ Dr Dec (Talk) ~~ 20:13, 8 October 2009 (UTC)

So we better phrase a formula for the connection between your edits here and the rate of your head spinnings.--Gilisa (talk) 21:16, 8 October 2009 (UTC)

• If the number of my edits is n ∈ N then my head is spinning with angular frequency ω where

$\omega (n)=(n!)^{n!}\ .$ ~~ Dr Dec (Talk) ~~ 21:33, 8 October 2009 (UTC)

($\scriptstyle \omega (0)>0$) . Bo Jacoby (talk) 08:17, 9 October 2009 (UTC).

Yeah, my head was spinning before I made my first post... ~~ Dr Dec (Talk) ~~ 22:05, 10 October 2009 (UTC)

What's then the inner radius of your cranium, given that your brain isn't affected by centrifugal force. --84.221.209.79 (talk) 00:50, 11 October 2009 (UTC)

I understand the question like this: "I have 8 letters, say A B C D E F G and H, that I can put into 12 slots with repeats , producing the 8¹² = 68719476736 possible twelve letter words AAAAAAAAAAAA AAAAAAAAAAAB AAAAAAAAAAAC ... and HHHHHHHHHHHH. What is the probability that one of these words, selected at random, has all 8 letters in it?" If the number of words having 8 different letters is N, then the probability is 8^-12 N. Emil J did it. Michael Hardy explained it too. Bo Jacoby (talk) 08:17, 9 October 2009 (UTC).

Ah, I get you. Letters as in a unit of the alphabet and not a written letter, and slots as in places and not letterboxes. ~~ Dr Dec (Talk) ~~ 21:35, 10 October 2009 (UTC)

Yes, that interpretation makes sense. The English and French languages use the same word (letter and lettre) for those two concepts, unlike say German (Buchstabe / Brief), and Danish (bogstav / brev), and Latin ( littera / epistula). Bo Jacoby (talk) 18:07, 11 October 2009 (UTC).

A Stirling number of the second kind S(n,k) gives the number of ways to partition an n element set into k nonempty subsets. Every surjection from an n element set to a k element set gives such a partition, and two functions give the same partition if and only if one of them can be obtained as the other composed with a permutation of the k element set. Thus, if you multiply S(12,8) by 8!, you will get the number of surjections from an 12 element set to an 8 element set. — Carl (CBM · talk) 11:28, 2009-10-9 (UTC)

When I did the calculation, I obtained:

S(12,8) = 159,027

8! = 40,320

Number of surjections: 6,411,968,640

Number of functions (= 8¹²): 68,719,476,736

Probability that a uniformly randomly chosen function is a surjection: 0.0933

Unsurprisingly this agrees with what EmilJ obtained. — Carl (CBM · talk) 11:55, 9 October 2009 (UTC)

The second occurrence of 6,411,968,640 should read 68,719,476,736. — Emil J. 12:22, 9 October 2009 (UTC)

Thanks, I changed it. — Carl (CBM · talk) 12:34, 9 October 2009 (UTC)

In case the OP's a novice like me, you can do this by applying the inclusion-exclusion principle (or, at least, the thinking behind it) directly to this problem without worrying about counting the surjections and whatnot (experts, please check).

1. There are 8¹² ways to arrange the letters into the slots and some of these arrangements don't use all 8 letters. We can calculate the number of arrangements that use 7 or fewer letters and subtract that from 8¹².
2. The number of ways to choose i letters is ${\binom {8}{i}}$ and for each of these choices, there are i¹² possible arrangements. For integers i = 1, 2, ... 7, define $N_{i}={\binom {8}{i}}i^{12}$
3. N₇ takes care of the seven letter combinations (i.e. arrangements that contain exactly seven letters) but it overcounts the combinations that contain fewer than seven letters.
4. Let's consider the six letter combinations. In N₇, there are 8 choices of seven letters and each of those have 7 choices of six letters. That's 56 choices of six letters. But there are only 28 choices of six from the original eight so we've overcounted in N₇ by 28 choices of six letters. Since N₆ counts 28 six letter combinations, N₇ - N₆ takes care of the six letter combinations.
5. Let's consider the five letter combinations. In the original eight letters, there are 56 five letter combinations. In N₇, there are 8 choices of seven and each of those have 21 choices of five - that's 168 choices of five. In N₆, there are 28 choices of six letters and each of those have 6 choices of five letters - that's 168 too. So N₇ - N₆ excludes all the five letter combinations. We need to add them back with N₅.
6. So the number of arrangements that use 7 or fewer letters is N₇ - N₆ + N₅ - N₄ ...
7. This is what Emil's expression above calculates.
8. I believe this is essentially a more user-friendly version of Emil's answer above.
9. User-friendliness is a good thing.

I hope this helps. Zain Ebrahim (talk) 14:06, 9 October 2009 (UTC)

If by "user friendliness" you mean "detailed solutions", I don't think that is a goal of the reference desk. The primary goal is to steer people towards the resources they need to solve the problem, just like the "reference desk" at a library. In this case, EmilJ gave a link to inclusion-exclusion principle, and went beyond the call of duty by actually computing the numeric answer. I pointed out that there is a different way to solve the problem, by using Stirling numbers of the second kind, and that like any other correct method it would give the correct answer. In the end it is the responsibility of the person asking the question to follow up on the links, or ask for clarifications if they are unable to see what's going on after a reasonable amount of effort. — Carl (CBM · talk) 14:21, 9 October 2009 (UTC)