Wikipedia:Reference desk/Archives/Mathematics/2009 January 21

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January 21

solving for radius?

Okay, first off this is not a homework question, this is my own misguided attempt to prove to a friend Steampunk-style armored blimps are impractical at best.

I'm trying to prove armoring a blimp would make the size requirements massive. I looked up the average armor on a WWII vintage tank, using that as baseline. Given a 7850 kg/m^3 density of steel, and helium's lifting capacity of about 1kg/.45M^3 of gas (stat derived from 16Cf/Lb value from an old game book I have, may not be accurate and DOES NOT account for altitude, but it's a good starting point I think), I get this equation to find the radius required radius to reach neutral buoyancy in air= ((4.188 * R^3) / .453 ) – ((12.56 * R^2) *863.5) = 0

My problem is this, I don't know how to begin to solve that one other than trial and error. Online graphers usually barf at the form even when I try to plot two graphs (one for weight based on surface area and one for lift based on volume, attempting to find the intersection, they still balk at it, any way to solve this by hand?

EDIT: if it helps I think I can also reduce the problem to (((4.188 * R^3) / .453) / 863.5) /(R^2)= 12.56

Edit to the edit: AHA! the altered version of it could be solved, but some checking would be nice.

the solution appears to be R = 1173.12m or roughly a 1.2km sphere. Discounting any payload of course, I'm not sick enough to take that one on with a 10-foot pole.

69.210.43.106 (talk) 05:33, 22 January 2009 (UTC)

Diophantine

where can i find the proof to this problem?::

x² + y² + z² = 3xyz has infinitely many integer solutions.

(the left hand sides are sums of squares not cubes)

could anyone please give me a link to a proof or tell me how i would go about solving this problem?? and is it possible to find a general form of these solutions? Please help me!! (and no, this is not homework) Johnnyboi7 (talk) 00:59, 21 January 2009 (UTC)

Searching google for the formula led me to this [1], where a comment claims they're called Markov numbers. It turns out we have an article on them.Black Carrot (talk) 03:20, 21 January 2009 (UTC)

Some additional facts. The binary tree strucure comes from this observation: suppose (a,b,c) is a solution. If you fix (a,b), then z=c is a root of the polynomial in z of second degree, z²-3abz+(a²+b²). Since 3ab is the sum of its roots, the other root is 3ab-c, so another solution of the diophantine equation is (a,b,3ab-c); and (a,3ac-b,c) and (3bc-a,b,c) as well, that we obtained if we changed one of the other two coordinates. From the easy solution (1,1,1) then it springs a whole binary tree of solutions: and it is not difficult to prove that all (positive) solutions are found this way without repetitions, that is the graph is really a tree: there are no cycles. The reason is that with the only exception of the "root"=(1,1,1), any solution (a,b,c) has a unique larger coordinate; if you change that one, you lower the norm of the solution and you step down towards the root; if you change one of the other two coordinates you always increase the norm and you climb up in the tree (so you can't happen to reach the same solution from below in two ways, because this means two descending paths from it. One the other hand, given a solution different form the root, you can always go down, so you have to reach the root). This first picture is very elementary to prove (try it); however shortly after one also encounters some open problems, as told in the quoted article. --PMajer (talk) 19:23, 21 January 2009 (UTC)

Integrals ∫_a∏f(x_n)dx_n

How do you do ∫_a∏f(x_n)dx_n where a is something like "parrelalogram" or "cube" of something like that?----The Successor of Physics 11:07, 21 January 2009 (UTC) —Preceding unsigned comment added by Superwj5 (talk • contribs)

WJ, if you mean e.g. an integral

$\int _{I\times J}f(x)g(y)dxdy$

this is simply the product of the two integrals:

$\textstyle \int _{I}f(x)dx\cdot \int _{J}g(y)dy$.

Say that we assume that both f and g are of class $L^{1}$. Otherwise, as alternative hypoteses, we may assume both measurable and positive (so the integral is not necessarily finite but it could be $+\infty$). Again the equality holds, with the usual convention $0\cdot \infty =0$). If you want these are consequences of Fubini-Tonelli theorem, but you can also prove it directly. This generalizes of course to the case of several functions. PS. See below for a very clear explanation. A famous application you should be interested in, is the computation of the Gaussian integral. --PMajer (talk) 14:09, 21 January 2009 (UTC)

Incidentally, if you change your signature to use the standard wikilink syntax [[User:Superwj5|...]] instead of the external link syntax [http://en.wikipedia.org/wiki/User:Superwj5 ...], it could prevent SineBot from thinking you do not sign your posts. — Emil J. 16:11, 21 January 2009 (UTC)

Start by looking at the inside integral:

$\int _{I}\int _{J}f(x)g(y)\,dx\,dy=\int _{I}\left(\int _{J}f(x)g(y)\,dx\right)\,dy$

The inside integral is

$\int _{J}f(x)g(y)\,dx.$

Observe that

• g(y) does not depend on x, i.e. g(y) remains CONSTANT as x runs through the set J.
• g(y) is a FACTOR, i.e. it is multiplied by the function being integrated.

Constant factors can be pulled out:

$g(y)\int _{J}f(x)\,dx.$

Next look at the outside integral:

$\int _{I}\left(g(y)\int _{J}f(x)\,dx\right)\,dy=\int _{I}g(y)\cdot \blacksquare \,dy$

We've replaced the whole inside integral with the black square, and the important things to notice about that is that

• It does not depend on y, i.e. it remains CONSTANT as y runs through the set I.
• It is a factor.

Constant factors can be pulled out:

$\int _{I}g(y)\cdot \blacksquare \,dy=\blacksquare \cdot \int _{I}g(y)\,dy.$

Now we have

$\int _{J}f(x)\,dx\cdot \int _{I}g(y)\,dy.$

I.e. it's a product of the two separate integrals.

The same thing works with more than two. Michael Hardy (talk) 00:16, 22 January 2009 (UTC)

Thanks!The Successor of Physics 11:06, 22 January 2009 (UTC)

See also Fubini's theorem. --PST 13:26, 26 January 2009 (UTC)

What do you call a + or - sign (to indicate positive or negative numbers)?

I'm not sure if this question is best for the Language desk or the Math desk, but I'm hoping that my odds are better on the Math desk. What do you call a positive or negative sign? Is there an umbrella term for these two symbols? EDIT: To clarify, I'm referring to the insignificant + and - sign in the contect of indicating a positive or negative number, not addition or subtration.

A Quest For Knowledge (talk) 15:17, 21 January 2009 (UTC)

Yes: sign. — Emil J. 15:26, 21 January 2009 (UTC)

"Sign" seems too general purpose. A insign dollar sign is a sign but in this case I'm only dealing with + and -.

The word "sign" has several different meanings. In one meaning it may denote any funny character, but in the mathematical meaning it denotes the positivity or negativity of a number, or for that matter, the + and − characters used to represent positive and negative numbers. — Emil J. 16:21, 21 January 2009 (UTC)

They are also unary operators, but they aren't the only ones. StuRat (talk) 15:38, 21 January 2009 (UTC)

Stu, just to clarify the context of my question, I'm referring to + and - to indicate a positive or negative number, not addition or subtration. To me, "Operator" implies action (addition/subtraction), not value. A Quest For Knowledge (talk) 15:43, 21 January 2009 (UTC)

A number is normally considered "positive". To indicate a negative number, put a minus sign before it. For example: "10" - this is considered positive because it's a number bigger then zero. "-10" - this is considered negative because it's a number lower then zero. M1N (talk) 15:48, 21 January 2009 (UTC)

Yes, but is there an umbrella term that encompasses both of these symbols (but only these two symbols)? A Quest For Knowledge (talk) 15:50, 21 January 2009 (UTC)

Yes, and it's been given to you - unary operator. Rather than dismissing it out of hand because of what you think the word "operator" implies, you could read the first sentence of the article, which states "a unary operation is an operation with only one operand, i.e. an operation with a single input". --LarryMac | Talk 15:56, 21 January 2009 (UTC)

The article also states factorials, absolute value, power operations symbols are unary operators. In my case, I'm only referring to just the + and - sign. (To my layman's mind, an operator performs an action whereas positive and negative signs indicate value. These seem to my layman's mind to be two distinct concepts. I'm a software developer and the distinction is analogous to a method and a property.) A Quest For Knowledge (talk) 16:10, 21 January 2009 (UTC)

Whether or not an operator needs to be seen (as I agree the word seems to imply) as active is a little philosophical, but −5 can certainly be seen as the result of the act of negating the number 5. -- Jao (talk) 16:34, 21 January 2009 (UTC)

I don't think you are going to get anything more specific. As LarryMac says, StuRat is correct when he says + and - in the context of positive and negative numbers are examples of unary operators; when the same symbols are used in the context of addition and subtraction they are binary operators. Nearest specific umbrella term is "sign" used in its mathematical sense, as in signed number representations and sign function. Gandalf61 (talk) 16:00, 21 January 2009 (UTC)

Yep, "sign" is definitely the word that is used. It's usually clear from context what is meant. --Tango (talk) 16:04, 21 January 2009 (UTC)

Unfortunately, the context in which I'm using it also includes a dollar sign. So I can't use the term 'sign'. Based on the responses so far, it appears as if there is no term that will work for me. A Quest For Knowledge (talk) 16:16, 21 January 2009 (UTC)

What is the context? Black Carrot (talk) 16:47, 21 January 2009 (UTC)

Can't you specify "dollar sign" or "currency symbol" when you mean the dollar sign? --Tango (talk) 16:55, 21 January 2009 (UTC)

I guess the problem is, you can't very well call it a "sign sign." How about "plus or minus sign"? Joeldl (talk) 17:06, 21 January 2009 (UTC)

It's really not too big of a deal. I'm a software developer and I was reviewing some old code and there's a variable name that I never liked. Basically, it validates a string of characters to ensure it's a valid number. Digits 0-9 are always allowed. There are properties that allow (or dissallow) decimal points, dollar signs and positive and negative signs. One of my variable names is foundPosNegSign. I was just wondering if I could come up with better variable name. I don't want to use the term 'sign' (by itself) because I think that might be confused for the dollar sign which the code also checks. I thought of maybe "value sign" but I don't want to be inventing new terminalogy.

       Dim asciiValue As Integer          'ASCII value of each character in Text
       Dim foundDecimal As Boolean        'indicates if a valid decimal separator is found
       Dim foundPosNegSign As Boolean     'indicates if a valid positive or negative sign has been found
       Dim foundCurrencySymbol As Boolean 'indicates if a valid currency symbol has been found
       Dim isInvalid As Boolean           'indicates if Text contains any invalid data

Honestly, it's not really too big of a deal. A Quest For Knowledge (talk) 17:16, 21 January 2009 (UTC)

As a programmer myself, I long ago decided that variable names can't always fully describe the values they contain. The best you can do is give the variable a name that reminds you of what it's for, with comments included to describe all the gory details. In your case, calling the variable "foundSign", and keeping your current comment, seems more than adequate. StuRat (talk) 05:11, 22 January 2009 (UTC)

The computer language APL has the matter better sorted out I think. There is a minus sign which is part of the number as in ^–123, it is not a unary operator. There is also the unary operator of negation as in -x, and the binary operation subtraction which also uses the minus sign as in x-y. In normal use the minus sign at the beginning of a number simply gives the sign of the number and is not an operator. Dmcq (talk) 18:18, 21 January 2009 (UTC)

---

the actual answer to your problem* is to use the word signum or sgn. So you should say

Dim foundSgnMark As Boolean

 * where your problem is "I don't want to use the term 'sign' (by itself) because I think that might be confused for the dollar sign which the code also checks."

82.120.227.136 (talk) 21:27, 21 January 2009 (UTC)

Math, via Civil Engineering

Civil engineers calculate stormwater (sometimes) using a so called "rational method". The rational method is attributed within the document "TRANSACTIONS,AMERICAN SOCIETY CIVIL ENGINEERS (INSTITUTED 1852) VOL. LXV DECEMBER,.........

.......... NEW YORK,PUBLISHED BY THE SOCIETY,1909"

to an Emil Kuiching b. 1889.

My question is about Emil Kuiching. Who was he? Here is the link to above referenced document: http://www.archive.org/stream/transactionsofam65amer/transactionsofam65amer_djvu.txt —Preceding unsigned comment added by HughCMasonJR (talk • contribs) 17:02, 21 January 2009 (UTC)

I can't find much on him, but he does have a scholarship named after him as item number 10 here: [2]. StuRat (talk) 04:57, 22 January 2009 (UTC)

Transcentdental numbers

I remember reading somewhere that only one of e^(e^2) and pi^e (or some pair of such numbers) is transcendental. Does anyone know of any work with a result like that? My memory fails me. --Fusionshrimp (talk) 21:26, 21 January 2009 (UTC)

I think it probably is a famous example of a non-constructive proof of "there exist irrational numbers a and b such that a^b is rational", see the example in that article.Dmcq (talk) 22:54, 21 January 2009 (UTC)

I remember reading somewhere something like: At least one of $e^{\pi }$ and $\pi ^{e}$ is transcendental (or possibly irrational). Chances are good they both are, but I don't think anyone has proven it yet. Could that be what you're thinking of? --Tango (talk) 23:38, 21 January 2009 (UTC)

Are you thinking of the (easy) result that at least one of e+π and eπ is transcendental? Algebraist 23:41, 21 January 2009 (UTC)

It's entirely possible... --Tango (talk) 23:44, 21 January 2009 (UTC)

if it's so easy, could you tell us here? —Preceding unsigned comment added by 82.120.227.136 (talk) 01:33, 22 January 2009 (UTC)

Suppose both e+π and eπ were algebraic. Then (e-π)²=(e+π)²-4eπ is algebraic, so e-π is algebraic. But if both e+π and e-π are algebraic, then so are both e and π. --Trovatore (talk) 01:50, 22 January 2009 (UTC)

Another way of say the same thing: e and π are the two roots of x²-(e+π)x+eπ = 0. So if e+π and eπ were algebraic, then so are e and π. Algebraist 13:27, 22 January 2009 (UTC)

It might be pointed out that this proof uses the result that the field of algebraic numbers is algebraically closed. -- Jao (talk) 16:17, 22 January 2009 (UTC)

e^π is known to be transcendental. Algebraist 23:43, 21 January 2009

(UTC)

Hmm.. no it wasn't those, I'm familiar with those results. I was going through possible research projects and I remembered some either-or (not both) proof from some book or journal. Thanks for all the replies though. --Fusionshrimp (talk) 04:45, 22 January 2009 (UTC)

It's a long shot, but perhaps you're thinking of the following argument that irrational to the power irrational can be rational: consider ${\sqrt {2}}^{\sqrt {2}}$. Either this is rational, in which case you are done, or it is irrational, in which case you can raise it to the power ${\sqrt {2}}$ and get an irrational raised to an irrational power being rational. 163.1.148.158 (talk) 17:43, 22 January 2009 (UTC)

Memorizing mathematics is not the way to go: even with more complex results. Intutively thinking might help instead of relying on memory, or, with easy theorems, it is better to understand. --PST 13:29, 26 January 2009 (UTC)