# Wikipedia:Reference desk/Archives/Mathematics/2008 April 9

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# April 9

## Gallons and cubic inches

I took my first calculus course about 40 years ago, and some of the things to stick with me over the intervening years include a few of the "handy constants" that come up repeatedly in story problems. Water weighs 62.5 (or 125/2) pounds per cubic foot, for example. Another is, 231 cubic inches in a gallon.

Recently, I had cause to think about that one again — specifically, how it's derived. The only way I can think to derive that from other known units is convert 1 gallon to 3.785 liters, to 3,785 ml, and divide by the cube of 2.54. This works due to what you might call the "bridge" formula, that 1 ml = 1 cc.

There's got to be something equivalent in English units, yes? How can you get from pints, quarts, and gallons to cubic inches and cubic feet? What's the "bridge" conversion in English units?

Thanks, Danh 63.231.153.74 (talk) 01:40, 9 April 2008 (UTC)

According to our gallon article, the wine gallon, or Queen Anne's gallon, was originally the volume of a cylinder six inches high and seven inches wide. If you use the approximation 22/7 for π, that comes out to 231 cubic inches, which is now the exact legal definition of the U.S. gallon. --Trovatore (talk) 03:41, 9 April 2008 (UTC)
The Imperial (UK) gallon is legally defined as 4.54609 litres (≈ 277.42 cu in). If a gallon is taken as the volume of 10 lb of water, the density in lb per cubic foot is a bridge relating gallons and cubic feet, and hence cubic inches.—81.154.107.33 (talk) 21:16, 10 April 2008 (UTC)

## Integration problem

How do you evaluate

${\displaystyle \int _{0}^{b}{\sqrt {x^{2}+y^{2}}}\,dx}$

--Yanwen (talk) 03:20, 9 April 2008 (UTC)

Try this: List_of_integrals_of_irrational_functions. It might actually be the first one on the list... Someletters<Talk> 03:28, 9 April 2008 (UTC)
Here's one hint - unless you've been given some kind of relationship between x and y, you would have to assume that y is a constant. Confusing Manifestation(Say hi!) 04:50, 9 April 2008 (UTC)
You can do this by trigonometic substitution. Whenever the integrand contains a2+x2 (where a is a constant), you can substitute x=a tanθ to obtain a2sec2θ. Visit me at Ftbhrygvn (Talk|Contribs|Log|Userboxes) 12:08, 10 April 2008 (UTC)

## A Logical Question

Hi, one of my frienda recently asked me this question and challanged me to solve it. Here's the question- There is a 5-digit number. the first digit represents the number of 0's in the number. The second digit represents the number of 1's in the number. the third digit represents the number of 2's in the number, the fourth digit represents the number of 3's in the number and the fifth digit represents the number of 4's in the number. we have to find this number. Can anyone help me out on this as i'm really not able to figure out the answer... —Preceding unsigned comment added by GK ROCKS (talkcontribs) 03:40, 9 April 2008 (UTC)

Start by setting some limits for each digit. For example, if the 5th digit were 2, then there must be 2 4s elsewhere in the number, meaning that there would be at least 8 digits. So the 5th digit is either 0 or 1, and it shouldn't take you long to rule one of those out as well. Confusing Manifestation(Say hi!) 04:53, 9 April 2008 (UTC)
If you can work that one out (and I seem to have found a unique solution), try the 10-digit case for a challenge. Confusing Manifestation(Say hi!) 04:56, 9 April 2008 (UTC)

If you are confused by such questions, note that it is a finite problem. There are only 90000 5-digit numbers, and you could check them one by one. I don't suggest that you do it unless you know how to program a computer, but it gives you peace of mind to understand that this is basicly a question of identifying a small subset of a finite set. Then you may make observations like that of ConMan above to reduce the set that needs to be tested. Finally a solution may be identified together with a short and elegant proof that this is the only solution possible, and then you are satisfied. But when everything else fails, try brute force. Bo Jacoby (talk) 07:08, 9 April 2008 (UTC).

See self-descriptive number (but wait until you have solved your problem or given up, because it contains the answer). Gandalf61 (talk) 12:33, 9 April 2008 (UTC)
Even brute force isn't likely to be THAT difficult. The sum of the digits must equal the number of digits in the number (right? ;-), so you don't have THAT many combinations to test. Find all the sums of natural numbers that add to N and then test the unique permutations of those numbers. For example, a 4-digit such number must be a permutation of one of the following sums: 0+0+0+4, 0+0+1+3, 0+0+2+2, 0+1+1+2, 1+1+1+1. Looking at these, you can eliminate most of them before even permuting. For example, there isn't 4 of any digit in 0+0+0+4 (nor is there a place to count 4s), so it's out. --Prestidigitator (talk) 20:31, 10 April 2008 (UTC)
That's a good method (for small n, at least). Interesting, there are 2 4-digit numbers and only 1 5-digit number. Any ideas on how to determine in advance how many numbers there will be for a given n? At least, any way to tell if there will be a unique answer. --Tango (talk) 20:52, 10 April 2008 (UTC)

## what is the smallest finite (positive) number?

What is the smallest number that is still finite and not infitessimal? —Preceding unsigned comment added by 79.122.13.205 (talk) 17:25, 9 April 2008 (UTC)

I mean, something with a name, like one trillionth. —Preceding unsigned comment added by 79.122.13.205 (talk) 17:26, 9 April 2008 (UTC)
No such number exists, as there is a regular method of constructing such prefixes. — Lomn 17:42, 9 April 2008 (UTC)
Consider the following, because for any small finite number a and another number b such that ${\displaystyle a there exists a c such that ${\displaystyle a, then there cannot be a number b>0 such that there does not exist a c. This is do the fact that the difference of a-b is nonzero, (as is forced by the inequality I gave above). It seems your understanding of the concept of infinity may be inaccurate. Infinity is not a number, and it doesn't display the mathematical properties that your probably familiar with from algebra. For example ${\displaystyle \infty +1=\infty }$, what finite number can you think of that does that???! (See, there isn't one. Infinity is the concept of an unbounded value, where as any finite number is bounded. That is to say that if you represented a finite number as a long sum of 1s ${\displaystyle 1+1+1...+1}$, it would terminate, whereas infinity does not terminate, there is no last 1 on the sum ${\displaystyle 1+1+1...}$.) The relation here is that your asking about ${\displaystyle {\frac {1}{n}}}$ for larger and larger n. A math-wiki (talk) 20:40, 9 April 2008 (UTC)
One centillionth (10−303) is probably the smallest with a name you can find in a dictionary.[1]  --Lambiam 23:20, 9 April 2008 (UTC)
wikt:googolplexth --tcsetattr (talk / contribs) 04:57, 10 April 2008 (UTC)
One graham ... th :D Black Carrot (talk) 23:35, 11 April 2008 (UTC)
Although, the famed nanograhamth would be smaller still. Black Carrot (talk) 23:36, 11 April 2008 (UTC)

## A real discovery or a trivial fact?

I've discovered a theorem, but I don't know whether it's trivial. This theorem is of the type: "there is...such that for every...there is...", and it's about real polynomials and real intervals. This theorem states as follows: "there is a function F from the set of natural numbers to the set of real polynomials, such that for every real interval, there is a natural number n such that the polynomial F(n) has a root in the interval". Do you think it's trivial? Eliko (talk) 20:50, 9 April 2008 (UTC)

It's pretty trivial. Let ${\displaystyle (q_{n})_{n\in \mathbb {N} }}$ be a denumeration of the rational numbers. Let ${\displaystyle F(n)=x-q_{n}}$. Every interval contains a rational number ${\displaystyle q_{n}}$, and it is a root of ${\displaystyle F(n)}$. -- Meni Rosenfeld (talk) 20:57, 9 April 2008 (UTC)
Yep. It's a pretty trivial consequence of the countability of the rational numbers and their denseness in the real numbers. --Tango (talk) 21:01, 9 April 2008 (UTC)
Sorry, I forgot to add another condition: the polynomial coefficients determined by the function F must be defined by an arithmetical formula (depending on a given natural number n). i.e. this theorem states now as follows: "there is a function F (which can be presented as an arithmetical formula) from the set of natural numbers to the set of real polynomials, such that for every real interval, there is a natural number n such that the polynomial F(n) has a root in the interval".
No denumeration of the rational numbers - can be presented by an arithmetical formula (depending on a given natural number n), right?
Eliko (talk) 21:42, 9 April 2008 (UTC)
Well, that depends on what you mean by an arithmetical formula. If you allow things like floors and logs you can have something close enough to a denumeration for my proof above to work (e.g. ${\displaystyle r_{n}={\frac {n}{2^{\lfloor \log _{4}n\rfloor -1}}}-3{\sqrt {n}}}$). If you only allow the basic 5 (+, -, *, /, ^) then this is not possible, but then again, I doubt the theorem holds. If the theorem is correct (perhaps with more relaxed conditions you will have to define), it may be interesting but probably not ground-breaking. -- Meni Rosenfeld (talk) 21:54, 9 April 2008 (UTC)
"Log" is tolerable, but the floor function is not (though it's rather simple). let's say we allow the "basic" 7 (+, -, *, /, ^, !, Σ). If you want to add the "Log" I wouldn't resist, though it's not the preferable way. Eliko (talk) 10:18, 10 April 2008 (UTC)
${\displaystyle F(n)=x\prod _{k=1}^{n^{2}}\left(x^{2}-{\frac {k^{2}}{n^{2}}}\right)}$
so that F(n) has 2n2+1 roots at intervals of 1/n from -n to n. To ensure there is a root in the interval [a,b] take n greater than max(|a|, |b|, 1/|a-b|). Does that fit your conditions ? Gandalf61 (talk) 21:59, 9 April 2008 (UTC)
Is there a reason for saying "|a-b|" and not just "b-a"? --Tango (talk) 22:31, 9 April 2008 (UTC)
No, just laziness. Gandalf61 (talk) 22:43, 9 April 2008 (UTC)
Is there a reason for saying: ${\displaystyle x\prod _{k=1}^{n^{2}}\left(x^{2}-{\frac {k^{2}}{n^{2}}}\right)}$, and not simply: ${\displaystyle \prod _{k=-n^{2}}^{n^{2}}\left(x-{\frac {k}{n}}\right)}$, i.e. without the x before the summation sign and without some power signs? Eliko (talk) 10:18, 10 April 2008 (UTC)
No reason - it's exactly the same thing. My format just came from thinking about truncating the infinite product representation of sin(nx). Gandalf61 (talk) 11:53, 10 April 2008 (UTC)

The existence of such a function isn't very surprising, but I've been trying to think about whether or not the behavior described above is typical. Topologize the space of polynomials by thinking of them as finite sequences of real numbers (with the sup norm, say). You can now put a box topology or a product topology on the space of functions from the natural numbers to the polynomials (since the space of functions from the naturals to anything is really just a countable cartesian product). In this new topological space, is the set of functions like the one described above dense? Open and dense? What if we pick different topologies (I picked these pretty arbitrarily)? 134.173.93.127 (talk) 23:50, 9 April 2008 (UTC)

In the product topology -- yes, it's dense, because having specified (up to a tolerance) any finite number of the polynomials in the sequence, you still have room to specify the rest so as to get in the set. No, it's not open; in fact the complement is dense (having specified any finite number of the polynomials, you can specify the rest so as to stay out of the set). However it is the intersection of countably many open dense sets, and thus comeager -- the set of all sequences that has some polynomial with a root in a given basic open neighborhood is open dense. --Trovatore (talk) 08:32, 10 April 2008 (UTC)
Thank you for the topological analysis. Eliko (talk) 10:18, 10 April 2008 (UTC)
Oh, I should say: I was sort of implicitly assuming that polynomials of different degrees were treated separately. If every open set containing a polynomial contains polynomials of arbitrarily high degree, then my argument about the complement being dense doesn't work. That doesn't strike me as a very natural topology -- but it might be the easiest interpretation of the thing about sup norm. --Trovatore (talk) 07:27, 11 April 2008 (UTC)
I'm pretty curious about the original theorem, at this point. How's it go? Black Carrot (talk) 06:56, 10 April 2008 (UTC)
At this point, the original theorem would go this way:
"there is a function F (which can be presented as an arithmetical formula whose operation-signs are just the basic seven: +, -, *, /, ^, !, Σ) from the set of natural numbers to the set of real polynomials, such that for every real interval, there is a polynomial F(n) having a root in the interval".
Eliko (talk) 10:18, 10 April 2008 (UTC)
I'm thinking you might have some trouble in that you have not defined the interval very specifically. Couldn't we consider degenerate intervals such as ${\displaystyle [\pi ,\pi ]}$ that would require that the polynomial have an irrational root? That could pose a problem given the allowed operations. If it were defined as an open or half-open interval, on the other hand, it would have to include more than a single point and would thus allow for solutions with only rational roots. --Prestidigitator (talk) 19:50, 10 April 2008 (UTC)
If it hasn't been clear yet - then let me make it clear now: degenerate intervals (having one point only) are excluded. In other words, by "interval" I mean an "open interval". Eliko (talk) 20:55, 10 April 2008 (UTC)
${\displaystyle F\left(n\right)=\sum _{k=0}^{2^{n}}{\frac {\left(-1\right)^{k}\left(nx\right)^{2k}}{\left(2k\right)!}}}$ should work. -- Meni Rosenfeld (talk) 17:37, 12 April 2008 (UTC)
Thanx, but unfortunately, you haven't proven that for every real interval there is a natural number n such that F(n) has a root in the interval. Eliko (talk) 19:28, 12 April 2008 (UTC)
Right, I haven't proven it, but I think it's pretty obvious. The general idea is that for each n, ${\displaystyle F(n)}$ is a truncation of the Taylor series of ${\displaystyle \cos(nx)}$. For n large enough we can take some interval (say [-n, n]) and bound the error in it, and use this to show that for every root of ${\displaystyle \cos(nx)}$ in this interval, there is a root of ${\displaystyle F(n)}$ nearby. Then it is almost identical to Gandalf's suggestion. -- Meni Rosenfeld (talk) 08:23, 13 April 2008 (UTC)
Your formula is much better than Gandalf's, because it uses the basic seven operation signs only; However, Gandalf, who (like you) confessed he had been inspired by the trigonometric functions, succeeded to prove his formula by purely algebraic (non-trigonometric non-analytic) considerations. Do you think you can do the same with your (better) formula, i.e. prove it by purely algebraic (non-trigonometric non-analytic) considerations? Eliko (talk) 11:12, 13 April 2008 (UTC)
We should avoid terms such as "better" here. My formula was designed to conform to the list of operations you have allowed, which is quite arbitrary - if you ask me, log is much more basic than summation or factorial, and if you include it then the product (which Gandalf used) is equivalent to summation. Unfortunately, I doubt there is any non-analytic way to prove my formula satisfies the criterion.
I have another suggestion: If we agree to let ${\displaystyle x!}$ mean Γ(x+1), then for any non-integer x, we have ${\displaystyle {\frac {(nx+n^{2})!}{(nx-n^{2})!n^{2n^{2}+1}}}=\prod _{k=1-n^{2}}^{n^{2}}\left(x-{\frac {k}{n}}\right)}$. We can thus agree that the LHS is a notation for the RHS. It is thus a formula which only uses the allowed operations, and it is easy to prove its validity. -- Meni Rosenfeld (talk) 11:37, 13 April 2008 (UTC)
Your new suggestion could have been satisfactory, had the "basic" factorial sign been allowed to mean the Gamma function. Unfortunately, by the "basic" factorial sign I mean its "basic" meaning (i.e. a function which gives back an integer for any integer), rather than its analytic meaning (i.e. a function which gives back a non-algebraic number for a rational non-integer number).
To sum up, in your opinion: there is (probably) no function F (which can be presented as an arithmetical formula whose operation-signs are just the following seven: +, -, *, /, ^, !, Σ) from the set of natural numbers to the set of real polynomials, such that one can prove by purely algebraic (non-trigonometric non-analytic) means that for every real interval there is a polynomial F(n) having a root in the interval". Did I correctly describe your apparent opinion? Eliko (talk) 12:18, 13 April 2008 (UTC)
I wouldn't, at this point, go as far as saying there is no such function. Sums and factorials allow quite a bit of flexibility, and there might be some suitable algebra-friendly formula. It will just have to be more clever than my cosine truncation.
Even if in reality there is no such function, proving that something cannot be proven with a certain method is very difficult. I'm not sure why you would mind an analytic proof, though. -- Meni Rosenfeld (talk) 13:22, 13 April 2008 (UTC)
OK, Let's avoid talking about all of the possible functions, since it might really be rather difficult to prove that there is "no" such a function (provided there's really no such a function). Instead, Let's look at a specific function: the imaginary part of the polynomial ${\displaystyle \left(x+i\right)^{n}}$. Note that by the Binomial Theorem it is very easy to extract this imaginary part - using the basic seven operations only, and now my theorem can relate specifically to this constructive instance for the function F, and will go as follows: "For every real interval there is a polynonial: ${\displaystyle \left(x+i\right)^{n}}$ whose imaginary part has a root in the interval". Note that this new constructive version of my theorem is easily proven by trigonometric means, i.e. by simply selecting ${\displaystyle x=\cot \left(\pi q\right)}$ out of the interval - for some rational q, and using De-Moiver's Formula. However the formula ${\displaystyle \left(x+i\right)^{n}}$ looks very naive, purely algebraic (based on two basic operations only: +,^), it's imaginary part being able to be based on my "basic seven operations" (like your formula). Therefore, one might expect that it may be possible to prove this (new constructive version of my) theorem by purely algebraic (non-trigonometric non-analytic) means, just as Gandalf has done with his function (which is not based on the basic seven oprerations). Do you think it's possible? Eliko (talk) 22:40, 13 April 2008 (UTC)
I doubt it. Gandalf's function could be worked with algebraically because its roots could be expressed with an algebraic formula. To express the roots of your function you need the cotangent function, which is transcendental. -- Meni Rosenfeld (talk) 23:30, 13 April 2008 (UTC)
1. "transcendental"? The cotangent function gives back algebraic values - rather than transcendental ones - as long as we select rational multiples of π as the arguments. Note that I've chosen ${\displaystyle x=\cot \left(\pi q\right)}$ out of the interval - for some rational q.
2. Would you agree that the following new version of my original theorem is (apparently) not trivial at all: "It's algebraically-provable that for every real interval there is a polynonial ${\displaystyle \left(x+i\right)^{n}}$ whose imaginary part has a root in the interval"? or in other words: "It's algebraically-provable that every real interval contains a point x having a natural number n such that ${\displaystyle \left(x+i\right)^{n}}$ is a real number"?
Eliko (talk) 08:48, 14 April 2008 (UTC)
1. The values it gives (for rational multiples of π) are indeed algebraic, but the function itself is transcendental because there is no polynomial ${\displaystyle P(x,y)}$ such that ${\displaystyle P(x,\cot x)=0}$ (compare ${\displaystyle f(x)={\sqrt {x}}}$ which solves the polynomial ${\displaystyle y^{2}-x}$).
2. Have you succeeded in proving it? Otherwise it is probably incorrect. If it is correct, it's not trivial (for some sense of the word "trivial") but I don't find it terribly interesting, either.
-- Meni Rosenfeld (talk) 10:11, 14 April 2008 (UTC)
I haven't said it's not transcendental; it really is, just as the function ${\displaystyle \left(-1\right)^{x}}$ is transcendental. However, I can't find any clear link - between its being a transcendental function - and the possible existence of algebraic proof which deals with algebraic values supplied by such a function for suitable arguments. Just think about it: the theorem stating that "the transcendental function ${\displaystyle \left(-1\right)^{x}}$ supplies a real number for every integer" - is provable by trigonometric means as well as by purely algebraic (non-trigonometric non-analytic) means - although the function is transcendental, because this theorem deals just with algebraic values supplied by such a function for suitable arguments, so due to the same reason - why wouldn't one be able to algebraically prove the theorem: "every real interval contains a point x having a natural number n such that ${\displaystyle \left(x+i\right)^{n}}$ is a real number", although the function is transcendental, because this theorem deals just with algebraic values supplied by such a function for suitable arguments... Eliko (talk) 11:00, 14 April 2008 (UTC)
[outdent] I didn't say it's impossible to work algebraically with a transcendental function. I only said that in my opinion, Gandalf's function was so easy to work with because of the simple algebraic representation of its roots, and that I wouldn't bet on the existence of a similar argument for ${\displaystyle (x+i)^{n}}$. -- Meni Rosenfeld (talk) 15:42, 14 April 2008 (UTC)
You "wouldn't bet on the existence of a similar argument"? me too, so if our common guess is correct then this shows that any algebraic proof for the new version of my theorem - wouldn't be trivial. Eliko (talk) 15:57, 14 April 2008 (UTC)
I suppose you could say that... -- Meni Rosenfeld (talk) 16:17, 14 April 2008 (UTC)
And let me take it one step further, and add that the very absence of any (trivial) algebraic proof for the purely-algebraic function ${\displaystyle (x+i)^{n}}$ is itself non-trivial, since this algebraic function looks very naive, and involves two basic operations only (+,^). Eliko (talk) 16:34, 14 April 2008 (UTC)
I'd say it isn't quite trivial, in the sense that if you used this fact to prove something else, you'd want to give justification. On the other hand, by itself it doesn't seem particularly deep. Do you have a direction you want to take this (i.e. why were you thinking about this)? kfgauss (talk) 07:49, 12 April 2008 (UTC)
Well, the original version is trivial in the sense that anyone can find a solution after half a minute of thinking. -- Meni Rosenfeld (talk) 17:37, 12 April 2008 (UTC)
I meant the revised version. Appropriately truncating the power series of sin(nx) and using Taylor's inequality etc was what I could come up with when I wrote the above, and it does take a bit of work.
I'll also mention that, in some sense, allowing factorials as above isn't really so strange. Each coefficient (i.e. of x^n) is itself a polynomial in n and the characteristic function of a set of the form [k,\infty), which seems more arithmetic than using logs. In any case, why one wants such a formula to exist or what extensions of this formula to other settings there are is more important than such considerations. kfgauss (talk) 20:09, 13 April 2008 (UTC)
I agree with you. The importance of my theorem is its applications. Eliko (talk) 22:40, 13 April 2008 (UTC)