Wikipedia:Reference desk/Archives/Mathematics/2007 June 29

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June 29

What do you call this really large number?

What do you call a number that, strictly speaking, is finite, but is so large that it may as well be infinite? For example, if a book has 500 000 letters in it, there are 500 000 ^ 26 possible books that could exist, and the number is technically finite because I can't print (500 000 ^ 26) +1 without repeating myself, but in practice that number is larger than all the atoms in the universe, right?

Thanks --Duomillia 00:34, 29 June 2007 (UTC)

Actually, that's $26^{500000}$, which is a much larger number. And I think you're asking for a name for numbers that "can't be written in decimal notation in this Universe". Never heard any names for those. — Kieff | Talk 00:57, 29 June 2007 (UTC)

I've seen bigger :P In any case, while there may or may not be some kind of specific term, there's a lot of information on large numbers at the article on, well, large numbers. Confusing Manifestation 03:09, 29 June 2007 (UTC)

I want to point out that saying really large numbers "may as well be infinite" is asking for trouble. One of the biggest problems people have with infinite things being counterintuitive is that they expect infinity to behave like an arbitrarily large number. For instance, see Infinite Monkey Theorem. It's worth remembering that under the traditional construction of the counting numbers, any number you think of is negligibly small with respect to almost every number (that is, except for a finite number of numbers). 209.189.245.114 19:23, 29 June 2007 (UTC)

About the Infinite Monkey Theorem: I always heard it as "an infinite number of monkeys typing for an infinite amount of time" will eventually create one of Shakespeare's works. The Wikipedia article says a (i.e., one) monkey typing for an infinite amount of time will produce Shakespeare. Which is it? Thanks. (JosephASpadaro 19:48, 29 June 2007 (UTC))

This isn't my area, but here goes. The math in the article doesn't really deal with monkeys, it just uses them as a metaphor for an infinite random string of letters (or ASCII characters, or whatever). One idealized monkey (i.e. one that types randomly), or an infinite number of idealized monkeys do the trick. The notion of "almost surely" is also important here. 209.189.245.114 20:13, 29 June 2007 (UTC)

The second paragraph in Infinite Monkey Theorem says: "Variants of the theorem include multiple and even infinitely many typists, and the target text varies between an entire library and a single sentence". I don't know which variant is most common, but only one monkey is needed. PrimeHunter 20:15, 29 June 2007 (UTC)

Infinitely many monkeys do allow one to strengthen the result however. If we give inifinitely many monkeys a typewriter each (and they hit the keys uniformly and independently at random), then almost surely infinitely many monkeys type the complete works of Shakespeare straight off with no junk beforehand. Hell, it's almost sure that every finite string of symbols is the first thing written by infinitely many of the monkeys. Algebraist 20:36, 29 June 2007 (UTC)

A follow up on the infinite monkeys: something I've often wondered about that. How can we say that monkeys typing at random are capable of producing a certain text, because as soon as they hit to be or not to be etc they are now producing non-random text, right? --Duomillia 14:27, 30 June 2007 (UTC)

Sure, and a coin that gives fifty heads in a row isn't a fair coin. But if you flip a coin 1000000 times, it seems reasonably likely that somewhere in there will be a string of fifty heads. And a monkey typing "to be or not to be" seems unlikely, but if it's in the form "awiugnipq3u4ng9 8u apyes3ui4n9 hq3489 iuto be or not to beo[uq23-94hello8tj u9n9entropy35ny" (emphasis for clarity, not part of actual output) then it's more likely just a small part of the random noise that happens to have a structure. As long as every possible string of text has an equal possibility of occurring, the IMT says you'll eventually get the text you want - you just have to be searching through the random text looking for a pattern, something our brains happen to be very good at. Confusing Manifestation 14:44, 30 June 2007 (UTC)

Not really. I mean, we spotted that bit easily, especially with the ephasis you kindly provided, but we'd be most likely to catch things that really stood out like entire plays. Proofreading them then, to make sure that, for instance, Hamlet didn't have a single letter out of place anywhere, is really better done by machine. After all, the theory does suggest that you'd hit just about every misprint and partial rendering of the play before you got to the real thing. Black Carrot 15:38, 30 June 2007 (UTC)

By the way, the probability of getting fifty consecutive heads is one in 1,125,899,906,842,624. This is so unlikely it has in all likelyhood never happened with any aproximately fair coin. You'd only get about twenty consecutive heads with a million flips. — Daniel 16:00, 30 June 2007 (UTC)

So, back to this Infinite Monkey Theorem for a minute. Let's say that we use this variant:

(A) If you have an infinite number of monkeys and an infinite amount of time, one of these monkeys will eventually replicate the text of Hamlet.

Now, say that I change the wording of the above statement slightly to read:

(B) If you have an infinite number of monkeys and an infinite amount of time, an infinite number of these monkeys will eventually replicate the text of Hamlet.

Is that second statement (B) mathematically sound / accurate / "safe" to assert? In other words, eventually, each and every one of those monkeys (not just one individual monkey) will replicate Hamlet ... or no ...? Thanks. (JosephASpadaro 17:36, 30 June 2007 (UTC))

Well, it depends. First, you have to understand that there's no absolute guarantee that any monkey will write Hamlet; all you can say is that it will happen with probability 1, or almost surely. That's not the same as absolute certainty.

Second, it depends on how many monkeys. There are different sizes of infinity. If you have a countably infinite collection of monkeys, and you give them infinite time, then with probability 1, every single one of them will have produced Hamlet. If you have more monkeys than that, it gets more complicated. --Trovatore 17:42, 30 June 2007 (UTC)

[Edit conflict]You have actually stated 4 variants here, which I will rephrase here ("almost surely" means "with probability 1"): If you have an infinite number of monkeys and an infinite amount of time, then...

1. almost surely one of these monkeys will eventually replicate the text of Hamlet.
2. almost surely an infinite number of these monkeys will eventually replicate the text of Hamlet.
3. each and every monkey will, almost surely, eventually replicate the text of Hamlet.
4. almost surely each and every monkey will eventually replicate the text of Hamlet.

To illustrate the difference between 2 and 4, note that infinitely many natural numbers are even, and yet not each and every natural number is even. The first three variants are certainly correct. As for variant 4, I haven't thought it through but it may depend on the cardinality of the set of monkeys. -- Meni Rosenfeld (talk) 17:57, 30 June 2007 (UTC)

It does depend on the cardinality, in the sense that for a sufficiently large collection of monkeys, the question is no longer amenable to any techniques in probability theory that I've ever heard of -- say, if you have more monkeys than the cardinality of the continuum.

If Martin's axiom holds, then the answer would be the same for any number κ of monkeys where κ is less than the cardinality of the continuum. That's because MA implies that Lebesgue measure is κ-additive for any such κ.

If Hugh Woodin is correct with his ideas about Ω-logic, then I think it follows that (1) MA holds and (2) the cardinality of the continuum is $\aleph _{2}$. In that case, with $\aleph _{1}$ monkeys, you still get that almost surely, every single one of them types Hamlet. --Trovatore 19:55, 30 June 2007 (UTC)

To go back to my original follow up, after reading the article on Infinite Monkey Theorem, I realise that I must have been quibbling over the exact meaning of randomly produced. If I have a monkey or a fair 26-sided dice, the sequence algemgboen is as equally probable as to be or not to be. If our monkeys produce the sequence, (to quote the Simpsons) It was the best of times, it was the... it's still a random sequence because there is no way to predict whether the next few letters will be worst of times or blursrt of times. Right? --Duomillia 03:11, 1 July 2007 (UTC)

Right. --mglg_(talk) 03:42, 1 July 2007 (UTC)

Pi

Here is a question that relates to the number pi. Every once in a while, you hear about an individual who has pi memorized to x places (say, 700 places). (1) Why on earth do people do that? (2) Is there any type of "trick" that enables them to perform these seemingly impossible feats? I thought that pi did not have repeating patterns in its numbers, so I can't imagine a useful trick or pattern that assists their memorization of so many numbers. Any insight is appreciated. Thanks. (JosephASpadaro 07:49, 29 June 2007 (UTC))

It's memorized for no other reason than a sense of achievement and for showing off. There's a lot of memory techniques, one is to say walk around your house and memorize a short segment for each piece of furniture. If you can remember a few numbers each time you think of a piece of furniture, you can remember the whole length, because you'r intimately familiar with what furniture is next to each other. EverGreg 08:55, 29 June 2007 (UTC)

It is best to avoid terms like "no other reason than...", as every person may have his or her own reasons for memorizing digits of the number. Another reason I can think of is simply appreciation for the importance of the number, combined with the belief that its essence is embodied in its sequence of decimal digits - and thus that knowing more digits represents more intimate knowledge about the number. Of course, the reasoning behind this motivation is questionable, as the essence of the number is in its exact value and not in any finite sequence of digits in an arbitrary base, but it is a possible motivation nonetheless.

As for your other question - I'll take a wild guess that the total of all the phone numbers you know by heart easily exceeds a hunderd digits. Adding other information you know by heart, not necessarily numeric, can amount to the equivalent of tens of thousand of digits. And if we include all the information that is stored in your brain - pictorial and the like - my guess is that it can be equivalent to billions of digits. So there's no question that the human mind is more than capable to store a large number of digits. Additionally, some people are better at remembering imprecise information like images, while others excel at recalling exact data like digits - and of course, some people just have a signficantly better memory than others. You can guess which kind the people who memorize myriads of digits of pi are. -- Meni Rosenfeld (talk) 11:17, 29 June 2007 (UTC)

Why does anyone want to win an eating contest, or an Olympic medal? It's not unusual to be attracted to a challenge, especially when we find out we're good at it. As for how, different people choose different methods, but breaking the sequence into manageable pieces seems to be universal. Keep in mind that an actor playing Hamlet memorizes many words, and a concert pianist memorizes even more notes. The story of one master pi memorizer, Hideaki Tomoyori, is told here, which may help. --KSmrq^T 12:10, 29 June 2007 (UTC)

A common technique to memorize long numbers is to compose a song or story such that the number of letters in Nth word equals the digit in the Nth place. Here's an example for π to 32 digits (3.141592653589793238462643383279). --TotoBaggins 20:00, 29 June 2007 (UTC)

Now I will a rhyme construct,
3   1 4    1 5     9
By chosen words, the young instruct.
2  6      5      3   5     8
Cunningly devised endeavour,
9         7       9
Con it and remember ever.
3   2  3   8        4
Widths in circle here you see,
6      2  6      4    3   3
Sketched out in strange obscurity.
8        3   2  7       9

Pi is known to be an irrational number, so the digits have no repeating pattern that holds forever. Pi#Open questions says: "The most pressing open question about π is whether it is a normal number". There is no known pattern which can help memorize many digits. And computing many digits when you need them is far too hard to do in the head, so you have to memorize what might as well have been a completely random sequence of digits. There are mnemonic methods to help with that. PrimeHunter 20:07, 29 June 2007 (UTC)

Daniel Tammet recited 22,514 digits a few years ago. He claims that (due to the screwed-up wiring in his head) he sees numbers as colors and shapes, and things like pi become entire landscapes. So, he memorized it the same way anyone else would memorize unfamiliar terrain. Black Carrot 15:31, 30 June 2007 (UTC)

"Pi" by Kate Bush is a fucking good tune and an excellent way to remember the numbers of the decimal expansion of π.

Yeah, as a few folks here have alluded to, if you imagine that each digit refers to a musical note, for instance, there are lots of people who have memorized huge musical works; the difference being that if the digits are viewed as a more or less random series it's much harder to memorize them, whereas if there is some sort of pattern to them, it's much easier. Gzuckier 16:19, 2 July 2007 (UTC)

Funny, the very opposite strikes me as true. In the first case, you are merely memorizing the numbers of pi. In the second case, you are memorizing the words/pictures/musical notes/associated symbols ... then making the transition to its parallel pi value ... then reciting the pi value. Seems like a 3 or 4 or 5 step process, as opposed to a 1 step process. Look at the example that TotoBoggins gave (a few posts above). First, you have to memorize the poem/song. Then, you have to count the digits of each word. And only then do you recite the digits of pi. And this is repeated 32 times. In other words, mentally, this is the thought process: "OK, the next word in the poem is "construct". OK, the word "construct" has, let's see, c-o-n-s-t-r-u-c-t, so that's, let's see, 1-2-3-4-5-6-7-8-9 letters in total. OK, so the sixth digit of pi is a "9". Sheesh. 32 times. Seems like a lot of steps, where it could just be one-step (memorize the 32 digits of pi). But, I do see the points being made here. Thanks. (JosephASpadaro 18:00, 2 July 2007 (UTC))

I agree wholeheartedly, that kind of system is roundabout and inconvenient. The problem is, it works. Even though it takes a lot of work to translate, and by the time you're done you've risked losing your place dozens of times, it's still easier to remember than random digits. And it scales well, which the simpler approach tends not to. Black Carrot 04:19, 4 July 2007 (UTC)

Well, on a related note ... how is it that "all of us" (generally speaking) have literally dozens of phone numbers stored in our brains? Those are random numbers. (JosephASpadaro 23:47, 4 July 2007 (UTC))

A number of reasons. First, I don't have dozens of phone numbers memorized. I don't think I even know one dozen by heart, and those that I do know I had to learn the hard way, by repeating them over and over again until they stuck. I didn't know my current phone number for months after I got it, for that reason. Second, I doubt there are very many people who do know dozens, even if they know dozens of people. If there were, why would they bother to equip phones with address books? Third, even 5 dozen = 60 times the length of a modern phone number, 10 digits, = 600. I think it's generally agreed that memorizing a single set of 1000 ordered random digits (say, the first 1000 digits of pi) for long-term use isn't that hard. Not that many people bother with it, but if they wanted to, they'd find it fairly manageable. It's when you get into the tens or hundreds of thousands of digits that very few people can handle it, and those that can tend to need the assistance of a memory system. BTW, I love the way you phrased that. Quotes around "all of us" to distance yourself from it, "generally speaking" to indicate that it might not apply to broad swathes of the population, plus "literally" to show you still mean business. Classic. Black Carrot 20:42, 5 July 2007 (UTC)

You caught all that (not so) subtle phrasing, huh? (JosephASpadaro 01:07, 9 July 2007 (UTC))

additional mathematics

Ang, Bakar and Chandran are friends and they have just graduated from a local university. Ang works in a company with a starting pay of RM2000 per month. Bakar is a sales executive whose income depends solely on the commission he receives. He earns a commission of RM1000 for the first month and this commission increases by RM100 for each subsequent month. On the other hand, Chandran decides to go into business. He opens a cafe and makes a profit of RM100 in the first month. For the first year, his profit in each subsequent month is 50% more than of the previous month.

In the second year, Ang receives a 10% increment in his monthly pay. On the other hand, the profit made by Chandran is reduced by 10% for each subsequent month

1(a)how much does each of them receive at the end of the 1st year?(2 or more methods are required for this question) (b)what is the percentage change in their total income for the 2nd year compared to the 1st year?comment on the answers. (c)Ang, Bakar and Chandran, each decided to open a fixed deposit account of RM10000 for 3 years without any withdrawal.

-Ang keeps the amount at an interest rate of 2.5% per annum for duration of 1 month renewable at the end of each month.

-Bakar keeps the amount at an interest rate 3% per annum for duration of 3 months renewable at the end of every 3 months.

-Chandran keeps the amount at an interest rate of 3.5% per annum for a duration of 6 months renewable at the end of every 6 months.

(i) Find the total amount each of them will receive after three years.

(ii) Compare and comment on the different in the interests received. If you were to invest RM10000 for the same period of time, which fixed deposit account would you prefer? Give your reasons.

FURTHER EXPLORATION

2(a)When Chandran's first child, Johan is born, Chandran invested RM300 for him at 8% compound interest per annum. He continues to invest RM300 on each of Johan's birthday, up to and including his 18th birthday?

(b)If Chandran starts his investment with RM500 instead of RM300 at the same interest rate, calculate on which birthday will the total investment be more than RM25000 for the first time —The preceding unsigned comment was added by 60.48.49.251 (talk • contribs).

For the third time, we will not do your homework assignment. Please stop asking. -- Meni Rosenfeld (talk) 13:11, 29 June 2007 (UTC)

Not again! Make it stop! Make it go away! 211.28.129.209 14:29, 29 June 2007 (UTC)

That is beyond my abilites. You still have 3 wishes left if you desire. --GTPoompt(talk) 16:10, 29 June 2007 (UTC)

To be fair, the last response he got was, "Come back when you at least have started some of the problems." I think we can safely assume that some startage has occurred, though the details are admittedly fuzzy at the moment. Of course, it's against policy to help with this many homework-like questions at once, and it's equally against policy to help with homework questions when the questioner hasn't demonstrated they'd done their utmost to work it out themselves. That said, you might try your luck at the linear or exponential growth pages. Black Carrot 15:23, 30 June 2007 (UTC)

No, the last response he got was my own, "Wikipedia's reference desk is not a service to do people's homework, so you should not waste what little time you have asking us to do so. However, if there is anything you do not understand and wish us to clarify, we will be more than happy to help.". For some reason, he is still under the assumption that the refdesk is, in fact, a service to do people's homework. -- Meni Rosenfeld (talk) 15:28, 30 June 2007 (UTC)

Uh, not unless I missed your response; Black Carrot is correct. —The preceding unsigned comment was added by 203.49.208.227 (talk • contribs).

By the way, please sign your posts with four tildes. If you don't understand what that means, the instructions at the top of the page will explain. Black Carrot 06:34, 1 July 2007 (UTC)

Well, either I've gone crazy and remember saying things that I actually didn't, or you didn't look good enough. Here's a nice trick: Your web browser should have a text search feature. As I have copied ny text verbatim, you can copy a small piece of it, paste it to the search window, and set it to search up (since that text is above where we are now) - you will find my response. With probability 1, the person to whom I have responded there is the one who has posted these particular exercise here and above. -- Meni Rosenfeld (talk) 07:49, 1 July 2007 (UTC)

I see what you mean. My bad. Still, the last response he was aware of suggested he repost later. I'm pretty sure they didn't intend him to repost it verbatim, though. Ah, well. Black Carrot 07:39, 2 July 2007 (UTC)

Assuming probability 1, that is. —The preceding unsigned comment was added by 203.49.208.227 (talk • contribs).

Awareness is a subtle issue. We currently have no indication that the OP knows he should check his questions on this page for responses, so it is possible that he is unaware of any of the responses. Assuming he does do this check, however, he should be aware of both.

By the way, the failure of some anons to sign their posts might be a source of confusion; I've added some unsigned tags. -- Meni Rosenfeld (talk) 10:22, 2 July 2007 (UTC)

A couple of calculus questions

Hi, all, hope you can help me with a couple of things.

Firstly, how do I go about evaluating this?

$\int _{-\infty }^{\infty }e^{-x^{2}}\,dx$

I know what the answer is from looking at a page of integrals, but I'm more interested in the process.

Also, how do you differentiate something in the form of, say, $3^{x}$ ? I tried using the chain rule to give ${\frac {d}{dx}}3^{x}={\frac {d(3^{x})}{d3}}\times {\frac {d3}{dx}}=x(3^{x-1})\times 0$, but clearly the answer is not 0 for all x, as this implies. Neither of these are homework, but I don't mind working through them myself if you give me some hints. Thanks, 80.169.64.22 19:45, 29 June 2007 (UTC)

The first example has no antiderivative, except one artifically constructed just for the purpose, the error function. But the second example is easy once we rewrite it as exp(x log(3)), then apply the chain rule. Very creative, your derivative with respect to 3; but it has the fatal flaw that 3 is not a variable! --KSmrq^T 20:18, 29 June 2007 (UTC)

Check out the limits on the first question; we want a specific definite integral. The standard technique for this one is to square the integral and then transform from cartesian co-ordinates to plane polars. Algebraist 20:32, 29 June 2007 (UTC)

I suspect the OP may not be familiar with the tools involved; I will therefore sketch a calculation based on Algebraist's suggestion, and invite requests for anything that needs to be explained further:

$\left(\int _{-\infty }^{\infty }e^{-x^{2}}\ dx\right)^{2}=\int _{-\infty }^{\infty }e^{-x^{2}}\ dx\int _{-\infty }^{\infty }e^{-y^{2}}\ dy=\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }e^{-x^{2}-y^{2}}\ dx\ dy=$

Changing to polar coordinates, $x=r\cos {\theta }$ and $y=r\sin {\theta }$, which has Jacobian $r$:

$=\int _{0}^{2\pi }\int _{0}^{\infty }e^{-r^{2}}r\ dr\ d\theta =\int _{0}^{2\pi }{\frac {1}{2}}\ d\theta =\pi$

$\int _{-\infty }^{\infty }e^{-x^{2}}\ dx={\sqrt {\pi }}$

-- Meni Rosenfeld (talk) 21:03, 29 June 2007 (UTC)

KSmrq: you can differentiate 3, or any other constant (you should know this), it can be considered a function f(x) = 3. The problem is that we are talking about exponentials, rather than the differentiability of a constant function.

• Differentiating 3 is something else than differentiating with respect to 3.--P.wormer 07:32, 30 June 2007 (UTC)

On the second (differentiation) problem, you might consider implicit differentiation. Consider $\ln y=x\ln 3$. Thus

${\frac {y'}{y}}=\ln 3$

Substitute y and solve for y-prime. All the best (and great questions!), --TeaDrinker 21:13, 29 June 2007 (UTC)

${\frac {d}{dx}}\ a^{x}=a^{x}ln(a)$

for constant and positive a.

For the first question, check Meni Rosenfeld's rather elegant method for evaluating it.

It's not mine. I was in the middle of trying to asymptotically evaluate an integral in the complex plane when Algebraist gave a gentle reminder of the "right" way to do it. -- Meni Rosenfeld (talk) 10:38, 30 June 2007 (UTC)

Thanks, guys. I was able to do the second one easily after converting it to exp(x ln(3)), but you're right in that I'm not familiar with the techniques for the second one. I follow Meni Rosenfeld's solution up until $=\int _{0}^{2\pi }\int _{0}^{\infty }e^{-r^{2}}r\ dr\ d\theta$. If someone could explain how to get that from the preceding step for me, I'd be grateful. Thanks to all who have replied, 80.169.64.22 18:06, 30 June 2007 (UTC)

I figured that would be the hard part :). This requires some relatively extensive prior knowledge to understand fully, but I'll do my best to make it at least sound reasonable. You are perhaps familiar with integration by substitution, which can be written as:

$\int _{\phi (a)}^{\phi (b)}f(x)\ dx=\int _{a}^{b}f(\phi (t))\phi '(t)\ dt$

Somewhat informally, and disregarding some necessary conditions, this can also be writeen as: If $x$ is our original variable, $t$ is a variable to which we want to change, $A$ is the set of values for $x$ over which we wish to integrate, and $B$ is the set of corresponding values of $t$, then we have:

$\int _{A}fdx=\int _{B}f{\frac {dx}{dt}}\ dt$

A similiar trick can be used when we are integrating a function of two (or more) variables, over some set which is a subset of the plane. The formula is similar, but the absolute value of the determinant of the Jacobian matrix takes the place of the derivative. In our case, we are integrating the function $f=e^{-x^{2}-y^{2}}$ over the entire plane, $A=\mathbb {R} ^{2}$. We do this by changing the orginal variables $x$ and $y$ to the new variables $r$ and $\theta$, given by the formulae $x=r\cos {\theta }$ and $y=r\sin {\theta }$. This is a valid transformation, since every point (other than the origin) given in Cartesian coordinates $(x,y)$ can also be given uniquely in polar coordinates $(r,\theta )$ where $x=r\cos {\theta }$, $y=r\sin {\theta }$, $r>0$ and $0\leq \theta <2\pi$. So, for $x$ and $y$ to cover the entire plane, $r$ must go from 0 to $\infty$ and $\theta$ must go from 0 to $2\pi$. We therefore have $B=[0,\infty ]\times [0,2\pi ]$. Now, the Jacobian determinant is:

${\begin{vmatrix}{\frac {\partial x}{\partial r}}&{\frac {\partial x}{\partial \theta }}\\{\frac {\partial y}{\partial r}}&{\frac {\partial y}{\partial \theta }}\end{vmatrix}}={\begin{vmatrix}\cos {\theta }&-r\sin {\theta }\\\sin {\theta }&r\cos {\theta }\end{vmatrix}}=r\cos ^{2}\theta +r\sin ^{2}\theta =r$

And finally, since we have $r^{2}=x^{2}+y^{2}$, we can write:

$\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }e^{-x^{2}-y^{2}}\ dx\ dy=\int _{A}e^{-x^{2}-y^{2}}\ dx\ dy=\int _{B}e^{-r^{2}}r\ dr\ d\theta =\int _{0}^{2\pi }\int _{0}^{\infty }e^{-r^{2}}r\ dr\ d\theta$

I hope this clarifies things a bit. If this still doesn't make any sense, perhaps it's better to let this particular problem rest until you have sharpened your multivariate integration skills. -- Meni Rosenfeld (talk) 18:43, 30 June 2007 (UTC)

Using polar coordinates, matrices, set theory and limits to evaluate an integral, and coming up with an answer as beautiful as √π. Is there a good article on "multivariate integration"?

Thanks, Meni. I pretty much get it now. This problem does seem way beyond what I've learned so far, though. 80.169.64.22 15:52, 1 July 2007 (UTC)

Multivariable calculus offers a general introduction to the field, and Multiple integral contains information more specific to what we have done here. -- Meni Rosenfeld (talk) 16:21, 1 July 2007 (UTC)

Whoops, looks like I missed a bit. I guess from the last bit of the original solution that $\int _{0}^{\infty }e^{-r^{2}}r\ dr\$ evaluates to 1/2, but I'm not sure how that comes about, as it seems to be of the same form as the original problem, unless there's something about the multiple integrals I'm not getting. 80.169.64.22 20:58, 1 July 2007 (UTC)

No, that step has nothing to do with multivariate integrals. It just happens that the extra factor of r, which is present here but not in your original problem, makes things much easier. Just use substitution, like Meni explained above, with $t=r^{2}\,$. Then ${\frac {dt}{dr}}=2r$, so that $r\,dr=dt/2$, which gives us $\int _{0}^{\infty }e^{-r^{2}}r\,dr\ =\ \int _{0}^{\infty }e^{-t}\ dt/2\ =\ 1/2$. --mglg_(talk) 21:40, 1 July 2007 (UTC)

Sometimes, knowing that something is solvable is all that is required in order to solve it. As long as the OP believed that evaluating this integral has anything to do with multivariate calculus, he would be consumed by doubt and refrain from tackling the problem head-on. Once reminded that this can be solved with the tools he is familiar with, I believe he would have figured it out eventually (assuming, of course, that he is familiar with simple substitutions).

This is also a perfect example that evaluating a more complicated expression is often easier than a simple one. With integrals in particular, pay attention if the integrand includes both some expression and its derivative (or a constant multiple of it - as in here, $r^{2}$ and $r$). Such cases invite a change of variables. -- Meni Rosenfeld (talk) 08:56, 2 July 2007 (UTC)

Ok, I can do it with the t=r^2 substitution. I guess that's the final piece in the solution. Thanks to everyone who's helped, especially to Meni for making concepts new to me so clear. 80.169.64.22

Glad to be of assistance :) -- Meni Rosenfeld (talk) 22:11, 2 July 2007 (UTC)

For the derivative function, the power rule can only be applied when the power on top is a constant, which is not so in your case(it is x instead). Furthermore, when you change variable, the expression should be in terms of the variable you're differentiating with respect o only. But the constant function f(x)=3 is not injective(information-preserving), so no inverse function exist and you're pretty much doomed there. Now as for the integral, there is a way to understand the added r intuitively. dx dy is the Infinitesimal area of a small rectangle of width dx and height dy. Now, in polar coordinate, we consider the area of a small sector of circular shell, which is approximately a rectangle(when the angle is small) and its area is about dr times arc length. Since the angle is in radian, the arc length is rdθ, so we end up with r dr dθ. --Lemontea 09:14, 5 July 2007 (UTC)