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March 24

balanced product of two modules

Latest comment: 19 years ago8 comments2 people in discussion

let R be a ring, M a right R module, N a left R module

a balanced product (P,f) consists of an abelian group, a map f from ${\displaystyle M\times N}$  to P, such that f(u+v,w)=f(u,w)+f(v,w) f(u,v+w)=f(u,v)+f(u,w) f(va, w)=f(v,a w)

Is this a general definition? It is used in my course before introducing tensor products? Why can't I find balanced products anywhere else, wikipedia and the rest of the internet included? Evilbu 15:16, 24 March 2006 (UTC)Reply

This is what Rotman calls an R-biadditive function (not bilinear because you can't pull the a out front). The purpose of the definition is that these are the morphisms that fill in the universal property for the tensor product, which sounds like this: for all abelian groups P and biadditive functions f: M×N → P, there exists a unique group homomorphism M⊗_RN→P such that the obvious diagram commutes. R-biadditivity is what replaces bilinearity which is used in the maps in the case of tensor products of vector spaces (or indeed, modules over commutative rings). But I'm a little confused by the way your definition is phrased. I think this should be a property for maps to have, the phrase "balance product" makes it sound like you're doing some operation to M and N, but for starters, your balanced product is not unique (even up to isomorphism), so it can't be viewed that way. Anyway, if our tensor product article weren't so crappy, it would explain all this. -lethe ^{talk +} 17:01, 24 March 2006 (UTC)Reply

OK, actually we have something at Tensor product of modules, these maps are called bilinear maps there. -lethe ^{talk +} 17:25, 24 March 2006 (UTC)Reply

Thanks, after we define balanced product, we define morphisms between two of them (P1,f1) and (P2,f2) it has to be a group morphism g of P1 to P2 and for all m, n in M and N g(f1(m,n))=f2(m,n)

a tensor product is then a balanced product such that there is a unique morphism to each balanced product

Now it is supposed to be obvious (without even proving existence of one) that the elements f(m,n) if (P,f) is a tensor product, generate the abelian group P now why is that? Evilbu 17:56, 24 March 2006 (UTC)Reply

Well, I'm not sure, but lemme take a stab. Let G be the range of f in P. The idea is that G also satisfies the universal property, and so is isomorphic, so the range is all of P, so those elements generate P. Explicitly, then if i: G → P is the inclusion map, I have the a map q: M×N→G such that iq=f. Then let r be the unique map P→G which the universal property promises us. So now I have a map qr: P→P, which satisfies fqr=f. But the identity map also satisfies that equation, and so by uniqueness, that map is the identity. I guess we do a similar argument for rq, and then we're done. This seems pretty standard for proofs involving universal properties, but I've never described universal properties as “obvious”, so maybe that‘s not the proof we‘re looking for? I dunno. -lethe ^{talk +} 18:22, 24 March 2006 (UTC)Reply

thanks but as i understand you used universal property what actually did you use then as i understand, you do not think this is such a trivial question( my syllabus works that way :balanced product, then morphisms between them, then tensor product, then this exercise)

I'm sorry, I don't understand what you're asking here. Are you asking whether I used the universal property? The answer is: yes, of course I did. The universal property characterizes the tensor product, if you don't use it, you've written a wrong proof. Shall I explain the step where I used the universal property again? It's like this, the universal property guarantees me a morphism from the tensor product to any group. The image of f is such a group, so the universal property guarantees a map to it. But the equations that hold for all these maps, along with the uniqueness, make the image isomorphic to the entire tensor product. -lethe ^{talk +} 19:45, 24 March 2006 (UTC)Reply

You know, given that you're not supposed to use use the explicit construction of the tensor product (I think that's what you meant by "without even proving existence of one"), the only way to proceed is with the universal property. I think it's probably just a case of your professor using the word "obvious" too liberally. -lethe ^{talk +} 20:12, 24 March 2006 (UTC)Reply

forgive me for insisting but I am quite interested allow me to continue for a moment with my not so general terminology if (Z,f) is a tensor product , let P be the subgroup of Z generated by the image of f ( in general, you can't say the image is a group right?) now (P,f) will be a balanced product as well so according to my definition of tensor product there is a unique morphisms p from Z to P such that p(f(x,y))=f(x,y)

this is all correct right? so how exactly should I proceed now?

Yes, and I guess you're right about the image not being a subgroup, that's my mistake, oops. So since P is a subgroup of Z, there is an inclusion map i:P→Z and there is a morphism q: M×N→P such that iq=f. You've chosen to use the same symbol for f and for q, but I prefer to use different symbols; they are different morphisms, since they have different codomains. This morphism q is just f restricted to its range. OK, so ip is a morphism from Z to itself, and we see that it satisfies ipf=iq by the universal property, which is equal to f by the property of subgroups. Thus ipf=f. Now since Z is itself a balanced product, then there exists a unique morphism id from Z to itself such that idf=f. So we have ipf=f and idf=f. But the universal property says that the morphism which makes the diagram commute must be unique, so we must have ip=id. We have to do something similar to show that also pi=id, and then we know that P and Z are isomorphic. -lethe ^{talk +} 06:03, 25 March 2006 (UTC)Reply

Far too hard.

Latest comment: 19 years ago6 comments4 people in discussion

If 45^x = 1 (mod 56), what does x equal? Is it

1. 56186
2. 29629586
3. 29629608
4. 29629633

?

Please, how do I go about this question? Thanks Reffies. --Dangherous 17:58, 24 March 2006 (UTC)Reply

Simply trying each value with the aid of a computer shows that the answer is 29629608. Are you looking to solve it by hand? Fredrik Johansson 18:19, 24 March 2006 (UTC)Reply

There's probably a number theory theorem which makes this easier, but a possible way is this: Calculate small powers of 45 modulu 56: For x=0, 1, 2... You'll get 45^x = 1, 45, 9, 13, 25, 5, 1, ... . So 45^6 = 1, therefore 45^(6n)=1. So x should be divisible by 6. The answer is 29629608. -- Meni Rosenfeld (talk) 18:22, 24 March 2006 (UTC)Reply

This isn't an answer to the problem (Meni essentially nailed this for our purposes), but this strikes me as very reminiscent of Fermat's Little Theorem or Euler's Totient Theorem, maybe something there would help.

Yes, it seems Euler's theorem tells us that 45^24 = 1 (mod 56). At worst, this narrows down our search - In this particular case, it can immediately give a solution since 29629608 is divisible by 24. -- Meni Rosenfeld (talk) 19:15, 24 March 2006 (UTC)Reply

i think the point is that you have to chinese remainder theorem 45^x=1 mod 56 is equivalent with the set of equations

{45^x=1 mod 8, 45^x=1 mod 7} or {5^x=1 mod 8,3^x=1 mod 7} now you see immediately this can only happen if x is true without calculations that require computers

My calculations didn't require computers. But I agree that the method you present may be easier. -- Meni Rosenfeld (talk) 19:33, 24 March 2006 (UTC)Reply

Using the Euclidean algorithm we can quickly determine that the greatest common divisor of 45 and 56 is 1, so these two numbers are coprime.

• GCD(45,56) → GCD(11,45) → GCD(1,11) → 1

We can also easily find the prime factorization of 56 as 2³7. Thus we can easily compute the Carmichael function,

• λ(56) = lcm(2^{3−2,71−1(7−1)) = lcm(2,6) = 6.}

Now we know a very small power of 45 is congruent to 1 modulo 56, either 2 or 3 or 6; and we can quickly eliminate 2. This tells us the correct exponent must be a multiple of 3 (or 6), so we can use the ancient trick of casting out nines:

• 5+6+1+8+6 → 2+6 → 8
• 2+9+6+2+9+5+8+6 → 4+7 → 1+1 → 2
• 2+9+6+2+9+6+0+8 → 4+2 → 6
• 2+9+6+2+9+6+3+3 → 4+0 → 4

The only number divisible by 3 is 29629608, so that must be the correct exponent.

Warning: A multiple choice question like this surely only appears in an educational context, where the person taking the test is assumed to have absorbed some relevant instruction to support answering this quickly. We have no way of knowing what you are expected to know. For example, perhaps you should know Euler's theorem. ;-) --KSmrq^T 19:58, 24 March 2006 (UTC)Reply

conference graph (spectrum of strongly regular graphs)

Latest comment: 19 years ago1 comment1 person in discussion

let G be a strongly regular graph. ${\displaystyle srg(v,k,\lambda ,\mu )}$  assume nontriviality : do not allow it to be disconnected or to have a disconnected complemenent it has k as eigenvalue once then, and two other eigenvalues

when ${\displaystyle 2k\neq (v-1)(\mu -\lambda )}$  all eigenvalues are integers

apparently the conference graphs , those are graphs of this form

${\displaystyle srg(2k+1,k,{\frac {k}{2}}-1,{\frac {k}{2}})}$  , are the only ones that don't have integer eigenvalues

but how to see this , essentially there are four parameters here that are restricted by two equations

${\displaystyle 2k\neq (v-1)(\mu -\lambda )}$  and ${\displaystyle (k-1-\lambda )k=(v-1-k)\mu }$  (this holds for all strongly regular graphs)

So where does the third restriction come from, as I see only one parameter in a conference graph?

I already requested a conference graph page on wikipedia The problem is that on the internet, google etc. think you talk about conferences ON graphs :)

Evilbu 23:37, 24 March 2006 (UTC)Reply

March 25

Circle

Latest comment: 19 years ago5 comments3 people in discussion

Lets say a circle, with diameter D was placed on the corner of a wall. Obviously, there would be a gap between the wall and the circle. The question is what is the diameter of the largest possible circle that can fit in this gap. Using Pythagorean Theorem, I found that the length of the gap would be ${\displaystyle (D^{2}-D)/2}$ . However, I have to contend with the gap between this hypothetical circle and its gap between the wall. The length of this gap I called x. I wrote the diameter of the second circle as ${\displaystyle [(D^{2}-D)/2]-x}$ . However, I cannot solve for x. I realised that the new gap would follow the same rule (${\displaystyle (D^{2}-D)/2}$ ). So the new equation would be:

${\displaystyle {\frac {({\frac {D^{2}-D}{2}}-x)^{2}-({\frac {D^{2}-D}{2}}-x)}{2}}={\frac {D^{2}-D}{2}}}$ 

With that, I was hoping to bring all the x's to one side. However, this proved troublesome. Can you help. Thanks.

I get a different result for the initial gap: (sqrt(2)-1)D/2 = 0.2071D, approximately. Check your work and then see if you can proceed using this value. StuRat 17:23, 25 March 2006 (UTC)Reply

Yes, the gap should be linear in D, i.e. not involve D². Also, I can't figure out what you mean by "the gap between this hypothetical circle and its gap between the wall". How many circles are there? —Keenan Pepper 17:37, 25 March 2006 (UTC)Reply

Maybe this crude diagram will help:

**********************************************************
*    *           *                                 +
*  *               *                       +
* *                 *                 +
*                                 +
**                   *         +
*                           +
**                   *   +
*                      +
* *                 *+
*  *               *
*    *           *
*          *   +
*            +
*           +
*         
*         +
*       
*       +
*    
*    
*     +

StuRat 18:16, 25 March 2006 (UTC)Reply

Yes, I forgot to square root. Sloppy mistake. Thanks.

I'm not sure I follow the problem statement or any of the reasoning here. Assume a unit circle for simplicity. A diagonal line from the corner to the circle center has length √2, so the corner of the square is √2 − 1 away from from the nearest point on the circle, where the diagonal crosses. However, the diagram makes clear that the sought center cannot be halfway between.

Instead, the sought center C will be at a point where a horizontal (or vertical) line to C is the same length as the distance from C to the circle. Let C be at (r,r), which places the circle contact at r+r¹⁄_√2 in x and y. But we already know the coordinates of the circle contact, namely (√2 − 1)¹⁄_√2 in x and y. Thus the equation to solve is

${\displaystyle r(1+1/{\sqrt {2}})={\frac {{\sqrt {2}}-1}{\sqrt {2}}},}$ 

which gives

${\displaystyle r={\frac {{\sqrt {2}}-1}{{\sqrt {2}}+1}},}$ 

or 3−2√2 (approximately 0.17). In general this would be scaled by the radius of the given circle. --KSmrq^T 22:10, 25 March 2006 (UTC)Reply

That's the answer I got, too, although by a different method:

As stated previously, the gap distance from the large circle to the corner is approximately 0.2071D. The total of the large circle diameter and the large gap is therefore approximately 1.2071D. This makes the ratio of the large gap distance to this total approximately 0.2071D/1.2071D = 0.1716. If we assume the same ratio for the small gap to the small circle, then the size of the small gap is approximately 0.1716(0.2071D) = 0.0355D. So, if we subtract the small gap from the large gap, we get approximately 0.2071D - 0.0355D = 0.1716D. This is the diameter of the small circle in terms of the diameter of the large circle. StuRat 23:53, 25 March 2006 (UTC)Reply

HOW MUCH SHOULD I PUT INTO A SAVINGS ACCOUNT TO HAVE A BANK MAKE AUTO MATIC WITHDRAWALS

Latest comment: 19 years ago3 comments3 people in discussion

IF I AM PAYING A MORTGAGE OF $225.00 PER MONTH FOR 15 YRS ,,HOW MUCH MONEY SHOULD I PUT IN THE BANK TO HAVE THE BANK PAY OUT THE PAYMENTS?ASSUMING I'M GETTING 3% INTEREST WHILE MY MONEY IS IN THE BANK,,,AT THE END OF 15 YEARS MY ACCOUNT SHOULD BE ZERO..

CAN YOU CALCULATE THIS???

THANK YOU,,,JAMES

If I understand the question correctly, it's something like 32232$. But if this is a real-world problem, you should discuss it with a financial consultant instead. -- Meni Rosenfeld (talk) 18:10, 25 March 2006 (UTC)Reply

What? Give me the name of your bank if you have a mortgage of only $225 a month and you get 3% interest on savings... You must have a mortgage on a doghouse! KWH 04:43, 27 March 2006 (UTC)Reply

F = $32,580 dollars

I = 0.03 / 12 = 0.0025 (or 0.25% per month)

T = 12 * 15 = 180 months

Y = $225 dollars payment per month

8. The family court orders you to give your former wife a payment of $500

  made at the end of each year for 3 years. The interest is 6.7% per year.
  You would like to buy out the obligation by paying her a lump sum
  instead. How much should you pay in the lump sum?

  Interest    I = 0.067 per year
  Term        T = 3 years
  Payment     Y = $500

  Solution:

      Y *  ( 1 - 1/(1+I)^T )
  F = ----------------------
                I

      $500 ( 1 - 1/1.067^3 )
    = ----------------------
              0.067

      $500 ( 1 - 1/1.21477 )
    = ----------------------
              0.067

      $500 * 0.17680
    = ---------------
          0.067

    = $1319.38

 Proof:

 The amount you need to pay is equivalent to an amount which you put
 into an interest account which your wife can withdraw the payment
 at the end of each year. The account should be empty when she made
 the last payment withdraw.

 Since she makes 3 withdraws, assume the amount needed at the start of
 the period to pay for each withdraw are F1,F2 and F3.
 Thus we have

  let x = (1 + I)

  F1 * x   = Y
  F1 = Y / x

  F2 * x^2 = Y
  F2 = Y / x^2

  F3 * x^3 = Y
  F3 = Y / x^3

  Therefore the amount you need to put into the interest account at the
  start of the period is

  F = F1 + F2 + F3
    = Y ( 1/x  +  1/x^2  + 1/x^3 )
    let r = (1/x)
    = Y ( r + r^2 + r^3)

in general
 F = Y * ( r^1 + r^2 + ... + r^T )

     Y * ( r - r^(T+1) )
   = -------------------
           1 - r

     Y * ( 1/x - 1/x^(T+1) )
   = -----------------------
           1 - 1/x

     Y * ( 1 - 1/x^T )
   = -----------------
           x - 1

     Y * ( 1 - 1/(1+I)^T )
   = ---------------------
           (1+I) - 1

     Y *  ( 1 - 1/(1+I)^T )
   = ----------------------
               I

Ohanian 09:53, 27 March 2006 (UTC)Reply

March 26

Subject: naming points on graphs (house numbers on twisty roads)

Latest comment: 19 years ago1 comment1 person in discussion

Let's say that:

>> Lane 256 Alley 34 House 7 means approximately:
>> we travel 256 units and turn right on to Lane 265; then
>> we travel  34 units and turn right on to Alley 34; then
>> we travel   7 units and there is the house, on the left.
>> 
>> Telephone pole 22 L 33 R 6 means
>> we travel to pole 22, follow the left fork wire 33 poles, then fork
>> right for 6 poles and arrive at our goal pole.
>> 
>> ( jidanni.org/geo/house_numbering/ )

P> What is your question?  Do you want to assign house numbers?  Do you
P> want to build a cross references from those numbers to spatial
P> ___location?  Do you want to compute the distance from one to another?
P> Do you want to uniformly store house numbers like that in a software
P> data structure?

I guess I just want to find which part of graph theory deals with naming points and segments on graphs the closest.

e.g., http://en.wikipedia.org/wiki/Glossary_of_graph_theory doesn't ever mention (systematic) ways to name each segment and point... That's what I want: what's the branch of math that deals with names for points in twisty graphs... maybe they have a better way of naming the two houses on the image on jidanni.org/geo/house_numbering/mountain_en.html

I want to know if there are even smarter ways of naming such points, or how folks go about naming points on graphs.

Yes, we would use nice grid coordinates if we were in a flat city, but we have twisty hilly roads.

--Dan Jacobson, jidanni.org

Looks like you copied the above discussion from another bulletin board, email exchange, or something like that. It would still be possible to number each ___location by the XY cartesian coordinates. That type of label would not be useful in describing the path you take to find the ___location, however. The general term for describing the distance along a curve is it's parametric coordinate. I'm not sure what you would call it when you chain multiple parametric coordinates together, though, perhaps a parametric graph ? Parametric coords in 2D use parallel curves on a surface, however, which doesn't sound like what you have here. (I think the article here might be called curvilinear coordinates instead of parametric coordinates, but it 's written in such an inaccessible way (no pics !), that I find it hard to understand.) StuRat 00:19, 26 March 2006 (UTC)Reply

topology

Latest comment: 19 years ago1 comment1 person in discussion

prove or disprove:A and B are path connected subsets of a space X and intersection of A and B closure is non-empty.A union B is path connected.

check out the topologist's sine curve. -lethe ^{talk +} 11:16, 26 March 2006 (UTC)Reply

March 27

l² tail

Latest comment: 19 years ago4 comments3 people in discussion

It's damn obvious, but I can't remember how to prove that the tail of an l² sequence goes to zero.

That is, given ${\displaystyle x={\left(x_{i}\right)}_{i=1}^{\infty }x_{i}\in \mathbb {C} }$  where ${\displaystyle \sum _{i=1}^{\infty }{|x_{i}^{2}|}<\infty }$ , to show that ${\displaystyle \lim _{n\rightarrow \infty }{\sum _{i=n}^{\infty }{|x_{i}^{2}|}}=0}$ . Confusing Manifestation 11:04, 27 March 2006 (UTC)Reply

It follow almost by definition: If the first limit exists, per definition ${\displaystyle \forall \epsilon >0:\exists n\in N:\forall m>n:|\sum _{i=1}^{m}{|x_{i}^{2}|}-\sum _{i=1}^{\infty }{|x_{i}^{2}|}|<\epsilon }$ . But ${\displaystyle |\sum _{i=1}^{m}{|x_{i}^{2}|}-\sum _{i=1}^{\infty }{|x_{i}^{2}|}|=|\sum _{i=m+1}^{\infty }{|x_{i}^{2}|}|}$  so we have (n'=n+1) ${\displaystyle \forall \epsilon >0:\exists n'\in N:\forall m>n'|\sum _{i=m}^{\infty }{|x_{i}^{2}|}-0|<\epsilon }$ , which gives you your other limit. Rasmus (talk) 11:49, 27 March 2006 (UTC)Reply

What Rasmus wrote is true, of course, and answers your question. But it might be helpful to point out that there's nothing special about ${\displaystyle l^{2}}$ ; this is just a statement about convergence of infinite sums. The following more general theorem is true: If ${\displaystyle a_{i}>0}$ , then

${\displaystyle \sum _{i=1}^{\infty }a_{i}=L<\infty }$  implies ${\displaystyle \lim _{n\to \infty }\sum _{i=n}^{\infty }a_{i}=0}$ . As you see, this gives you the result you need without using anything about ${\displaystyle l^{2}}$ .

The proof is pretty straightforward. If we define ${\displaystyle S_{N}=\sum _{i=1}^{N}a_{i}}$ , then the assumption of the theorem is that ${\displaystyle \lim _{N\to \infty }S_{N}=L}$ . But then it is easy to see that ${\displaystyle \lim _{N\to \infty }|L-S_{N}|=0}$ , and ${\displaystyle L-S_{N}=\sum _{i={N+1}}^{\infty }a_{i}}$ , so we are done. --Deville (Talk) 18:48, 27 March 2006 (UTC)Reply

Thanks for both answers. I knew it was so obvious as to practically follow from the definition, but for some reason the actual steps eluded me. Confusing Manifestation 00:12, 28 March 2006 (UTC)Reply

Height of Parallelogram

Latest comment: 19 years ago3 comments2 people in discussion

I am working on producing a spreadsheet and have hit a road block in some geometry. I am trying to upload my diagram but something isn't working right so I'll need to explain it verbally first.

I have a parallelogram, with theta in the lower left hand corner, by starting at theta and traveling clockwise I have B and then the horizontal leg A. Theta and A are my inputs. I can also even determine F, which is perpendicular to B and terminates at angle BA, but I don't really see a way that it helps me out. On lower leg A, if we go to it's far right end and draw a verticle line we get a right triangle with leg d, B, and H. I had a hunch that the quantity (A-d) relates to theta. So in AutoCAD I set A = 1 and varied theta from 0-90 in increments of 10 degrees, converted into radians and plotted the values against (A-d). Found using linear regression with a third polynomial equation, [-0.0698(theta)^3 + 0.3564(theta)^2 - 0.031(theta) + 0.44 = (A-d) ] resulted in a R^2 = 1, so someone please set me straight if I am way off target here but I assumed I since the R^2 = 1 everything was fine and I would be able to multiply the equation by left hand said of the equation by A and I would be able to get (A-d) do some geometry and have the Height pop out. (This part may get confusing and I can set up a junk hotmail account, post it, and if anyone is interested in seeing how I did this in excel email me and I'll send it to you). But I took two arbitrary values of A varied theta, got more equations by from trend lines, divided each term by A and compared the terms to the first equation and they came out reasonably the same......I thought I was good to go but when I do it in AutoCAD....it doesn't really work......don't know if I'm just spinning my wheels or if there really is a way to determine the height of a parallelogram from by given input. But doesn't there just seem a way that theta can be a direct function of (A-d)???

You fail to state a problem. Name your inputs. And what is each: lengths, angles, line segments? Likewise, what is your desired output? Stream-of-consciousness babble doesn't help. Also, please sign your post using four tildes, which the wiki software automagically converts to your id and a time stamp. --KSmrq^T 18:44, 27 March 2006 (UTC)Reply

See that with the same imput data : an angle and a segment lenghth only, you have different parallelogram shapes. Make sure you said everything; type in a simple drawing for us like those :

       #####        A
      #   #     ##########
    A#   #     ##########
    #   #     ##########
  θ#####    θ##########

Is what you're looking for H=Bsin(theta)? TimBentley (talk) 00:48, 28 March 2006 (UTC)Reply

Nevermind, I think there's no way to determine the area from your inputs. TimBentley (talk) 00:54, 28 March 2006 (UTC)Reply

Multivariable Max/Min Problem

Latest comment: 19 years ago5 comments3 people in discussion

I do Independent Study math, so it's not graded - I make up my own assignments. Anyway, here's a problem I can't quite figure out:

Find the local max/min values and saddle points of the function:

${\displaystyle f(x,y)=(x^{3})y+12x^{2}-8y\,\!}$ 

I take the first partial derivatives and equate them to 0 (I can do that part ;):

${\displaystyle F_{x}=3(x^{2})y+24x=0\,\!}$ 
${\displaystyle F_{y}=x^{3}-8=0\,\!}$ 

Then I can find one set of points, (2, −4): x=2 (from ${\displaystyle F_{y}}$ ) → y=−4 (plugged into the other partial derivative)... but then I'm stuck. It's a tiny roadblock but I can't seem to figure it out -_-

Thanks. – ugen64 22:44, 27 March 2006 (UTC)Reply

Well, you seem to have solved most of the problem correctly to me. What's left to do? Deciding whether this critical point is a maximum, minimum or critical point, I guess? For that, use the Hessian. If the matrix is positive definite, it's a minimum, negative definate, it's a maximum, or indefinite, saddle point. One criterion for 2x2 matrices is that they're definite if their determinant is positive and indefinite if their determinant is negative. So it goes like this:

${\displaystyle H={\begin{pmatrix}f_{xx}&f_{xy}\\f_{yx}&f_{yy}\\\end{pmatrix}}={\begin{pmatrix}6xy+24&3x^{2}\\3x^{2}&0\\\end{pmatrix}}}$ 

from which I observe that the determinant is –9x², which is –144 at x=2. This is negative, so I conclude that f(2,–4) = 112 is a saddle point (and there are no extrema since this function is linear in y). Hope that helps. -lethe ^{talk +} 23:12, 27 March 2006 (UTC)Reply

See, the back of the book (answer) says that there is a 2nd critical point... but I can't figure out how to find it. Guess I could have made that clearer :) Thanks for that part of the answer, though. – ugen64 01:15, 28 March 2006 (UTC)Reply

Well, uh... maybe it's supposed to be a complex function? There are three roots over the complexes... that doesn't seem right though. And you're expecting only 1 more point, not 2. Maybe there's a mistake in the text. I'll bet the function isn't supposed to be linear in y. -lethe ^{talk +} 02:16, 28 March 2006 (UTC)Reply

One guess is that the second term should use y, not x; namely,

${\displaystyle f(x,y)=x^{3}y+12y^{2}-8y\,\!}$ 

This would have real extrema at (0,¹⁄₃) and (2,0), and the two Hessian signatures differ. --KSmrq^T 03:07, 28 March 2006 (UTC)Reply

March 28

China's dollar reserves

Latest comment: 19 years ago5 comments3 people in discussion

(Not sure where to ask Macroeconomics question, so I will try here.) I read in The Economist that China has >$800 billion in reserves. Why doesn't China spend this money on sorely needed infrastructure projects, rather than saving it? I.e., what marginal benefit exists for such savings? Lokiloki 02:23, 28 March 2006 (UTC)Reply

The benefit isn't in having the savings, but in buying and selling it. Buying and selling dollar reserves allows them to manipulate the exchange rate of the currency. By buying dollars (i.e. selling yuan), they can decrease the value of the yuan. This helps the economy by making it cheaper for other nations to buy goods from China. If the yuan gets too cheap (making it overly expensive for China to buy US [and other foreign] goods), they can sell the reserves, driving up the value again. For such transactions to have any noticeable effect, the quantities must be massive. Superm401 - Talk 03:36, 28 March 2006 (UTC)Reply

Thanks very much for your answer... but since China artificially sets the exchange rate at a fixed rate, how does the buying and selling play a role? In that there is a "ghost" exchange rate that is unfixed? Lokiloki 05:22, 28 March 2006 (UTC)Reply

The reserve system is how they peg the exchange rate. If they just set rates and ignored the market, people would resort to the black market. I found a helpful Forbes article. Superm401 - Talk 21:27, 28 March 2006 (UTC)Reply

India has the same problem. Lots of foreign exchange reserves, and a sore need to develop infrastructure. It doesn't spend the money for infrastructure, in part because suddenly introducing so much money in the economy will lead to hyperinflation, and the poor will find it difficult to afford things. deeptrivia (talk) 20:16, 31 March 2006 (UTC)Reply

On Construction of Ramanujan Graphs (LPS Constructions)

Latest comment: 19 years ago1 comment1 person in discussion

Optimal expander graphs are known as Ramanujan Graphs. Explicit constructions for them is given in the Ramanujan Graphs by LPS. I have a doubt in the construction of them. While using PSL or PGL, how they are different from SL and GL. It has been said that PSL is quotient group with out zero transform. Can any one expalin me more about how to construct them? Any full example available with any one? --Guru 03:45, 28 March 2006 (UTC)Reply

Cube

Latest comment: 19 years ago6 comments5 people in discussion

A giant lamington cake in the shape of a cube.How many small lamingtons have 0,1,2 or 3 sides iced if the cake was cut into 64 and 125 cubes?

They all have four sides. Isn't this obvious? It is like asking: "I have 64 cubes. How many of the cubes have 0, 1, 2, or 3 sides?" --   Mac Davis] ⌇☢ ญƛ. 10:17, 28 March 2006 (UTC)Reply

ahem "iced sides" (judging by our lamington article, this is a cake with icing on all sides including the bottom and hence a good candidate for a maths question to do with vertices, sides and so on; given the geographical specialisation, I also deduce that our questioner is from Oz or New Zealand). On the assumption that this is homework (but fun homework that I enjoyed overgeneralising), I shall give the solution for a cut into n³ smaller cubes (n>=2; for n=1 we trivially have one six-sides-iced piece) and then our questioner will at least have to do a bit of algebra to get the answers for his problem:

3 sides iced: always 8

2 sides iced: 12(n-2)

1 side iced: 6(n-2)²

0 sides iced: (n-2)³

The proof that these add up to n³ is left to the reader. --Bth 10:49, 28 March 2006 (UTC)Reply

A quick note: your math seems to assume that the bottom of the cake is iced, which is rare (to say the least) in my experience. On the other hand, I have no idea what a "lamington" cake is vis a vis other cakes. — Lomn Talk 15:33, 28 March 2006 (UTC)Reply

Appears to be one of Australia's national treasures. Yep, all six sides are iced. --jpgordon∇∆∇∆ 22:45, 28 March 2006 (UTC)Reply

Indeed so, but it's an assumption I explicitly stated with a link to the article about lamington cakes here on Wikipedia to justify it. --Bth 07:33, 29 March 2006 (UTC)Reply

You wretched creatures who know not the Glorious Lamington are all invited to my place next weekend for the first annual International Wikipedians Lamington Festival. Seriously though, I promise you they taste much, much better than Vegemite. (That's not a slur on Vegemite - I love it - but I know Americans just don't get it.) JackofOz 14:53, 29 March 2006 (UTC)Reply

Cost of Iraq War per person

Latest comment: 19 years ago7 comments5 people in discussion

A recent study indicated the total cost of the war is 1 to 2 trillion dollars depending on how long it lasts. Going with the 1 trillion figure and figuring that the US population is close to 300 milllion--how much would the cost be per person?

The answer is 2.50, but not necessarily in current US dollars. Sheesh. [1] --KSmrq^T 07:48, 28 March 2006 (UTC)Reply

Answering the maths question first of all ... a trillion (in this context) is a million million, so you want to divide a million million dollars by 300 million. Answer is $3,333 and some cents - say $3,000 in round numbers.

Now ask yourself a few questions. Is the "total cost" in this estimate all paid by the US, or is it an international cost (other countrries are paying financial and human costs for the Iraq occupation too) ? Is it an estimate of the cost so far, or of the "final" cost when everything is finished (and if so, how does the estimater define "finished") ? And is it useful to divide by the total US population, or would it be more meaningful to divide by the number of US tax payers, which is more like 100 million ?

Anyway, whichever way you cut the numbers, it is clear that the Iraq war and occupation have cost far more than originally envisaged, both financially and in human lives. But financial considerations are unlikely to sway the opinion of those who take a moral stance on the issue. Those who believe that the war was a justifiable way of achieving its objectives will still believe that this is true at almost any cost. Those who believe the war was unjustified will not agree it would have been the right thing to do if only it had cost less. Gandalf61 09:12, 28 March 2006 (UTC)Reply

You guys are monsters. The cost of human life is priceless. Have you no dignity? :P --   Mac Davis] ⌇☢ ญƛ. 10:12, 28 March 2006 (UTC)Reply

It also might be important to note that all of this is not taken from current tax income, but a huge chunk is in loans (considering that there is a sizable deficit in the US right now). You could argue that that would mean that the taxpayers would pay even more than that, just over a significantly longer time. Oskar 20:58, 28 March 2006 (UTC)Reply

Gandalf61, those seem like extreme positions but they're not even entirely separate. The war turns out to be the right thing to have done in both positions, and the human and monetary costs are reduced to mere quibbles. Sorry, but you don't speak for me. JackofOz 13:54, 29 March 2006 (UTC)Reply

JackofOz - what a strange thing to say ! Of course I wasn't trying to speak for you - whatever gave you that impression ? The point I was trying to make (speaking only for myself) was that someone who takes an absolute moral position on the Iraq war one way or the other is unlikely to be influenced by estimates of its cost. Gandalf61 14:26, 29 March 2006 (UTC)Reply

Then I have inaccurately divined your meaning. for which my apologies. I should explain, I've re-read your post, and I think your logic baffles me. A will believe X, and B will not believe Y. Which is confusing enough. By "B will not believe Y", I inferred you were implying B should believe Y, hence my indignation. Then Y turns out to be modified form of X anyway. The message I keep on getting is that there is only one serious attitude to the Iraq war worth having, that of support for it. I accept that was not what you were intending to say, but that's what comes thru. To me. (Maybe it's just me. I'm having a funny week.)  :-)) JackofOz 14:47, 29 March 2006 (UTC)Reply

Factorization

Latest comment: 19 years ago2 comments2 people in discussion

Hi,I am a new person to this website and i am getting a lot of information from you guys.

I am an IGCSE student that studies in Britian and am studying at grade 10.

I have 4 question to ask you about from the IGCSE.

First question is that you give me brief information about the sine and cosine rule. Secondly, What is factoization of expression, and also it's definition. Thirdly, The simplification of long expressions with exponents. Last question is that can you give me information about the sets.

Thank you very much for your help and i hope that it will come back to you one day.

We have some nice little articles on the sine and cosine rules, which are basic identities in trigonometry.

I'm not quite sure what you mean by "factorization of expression", but am guessing it refers to, eg, writing down expressions like ${\displaystyle n^{2}+5n+4=(n+1)(n+4)}$ , which is useful for solving polynomial equations where such factorisations exist (in general it can't always be done). We have an article on that too. "The simplification of long expressions with exponents" I think just refers to using powers notation to avoid hand cramp, eg ${\displaystyle n\times n\times n\times n\times n\times n\times n}$  can just be written ${\displaystyle n^{7}}$ . Sets are a big topic that in many ways underpin large chunks of mathematics itself, so you'll need to be a bit more specific there. --Bth 11:52, 28 March 2006 (UTC)Reply

The significance of the law of sines and the law of cosines is that you can use them to solve a triangle if you know two sides and an angle (law of cosines) or two angles and a side (law of sines). – b_jonas 12:13, 28 March 2006 (UTC)Reply

Help

Latest comment: 19 years ago5 comments3 people in discussion

How would I figure this out 1⁻¹. I can't remember how to figure the answer if if the power is negative. Thank you for the help. ILovEPlankton 16:36, 28 March 2006 (UTC)Reply

${\displaystyle a^{-b}={\frac {1}{a^{b}}}}$ , so e.g. 2⁻¹=1/2. Kusma (討論) 16:47, 28 March 2006 (UTC)Reply

So 2⁻² would be 1/4? ILovEPlankton 17:00, 28 March 2006 (UTC)Reply

Precisely. You got it. Kusma (討論) 17:09, 28 March 2006 (UTC)Reply

Remember that addition in the exponents equals multiplication in the base, so x^ax^b = x^a+b. In the case where b = −a, the exponent simplifies to 0 and the answer to 1; therefore additive inverse (negative) in the exponent implies multiplicative inverse (reciprocal) in the base. As always, reciprocation is only meaningful if the base (here, x) is nonzero. --KSmrq^T 20:04, 28 March 2006 (UTC)Reply

I have been wondering if everyone has the concept of order in coombinations backwards.

Latest comment: 19 years ago6 comments5 people in discussion

I did a little experiment. I tested the Combination and Permutation functions.

P(n, r) = n! / (n - r)!

C(n, r) = n! / ((n - r)!r!)

Let's say that for each one I'm using 4 (n) items in set of 3 (r).

P(4, 3) = 4! / 1! = 4! = 24

C(4, 3) = 4! / (1! * 3!) = 4

Well everywhere I go, it says that the Combination returns the number of combinations of items in which the order does not matter. Well tell this to the following sets.

(1 2 3), (1 2 4), (1 3 4), (2 3 4)

Notice that it only has 4 items (n), sets of 3 items (r), 4 sets (C), and it IS ordered.

Thats not all though. They always say that Permutation returns the number of ordered pairs, but I found that I can get the same number of unordered sets with all the same variables.

(1 2 3), (1 2 4), (1 3 2), (1 3 4), (1 4 2), (1 4 3), (2 1 3), (2 1 4), (2 3 1), (2 3 4), (2 4 1), (2 4 3), (3 1 2), (3 1 4), (3 2 1), (3 2 4), (3 4 1), (3 4 2), (4 1 2), (4 1 3), (4 2 1), (4 2 3), (4 3 1), (4 3 2)

Again notice that it only has 4 items (n), sets of 3 items (r), 24 sets (C), and it IS NOT ordered.

You think it stops there? Wrong. It is the same for the functions with repetition. I'm not going to go into detail about them though. I have already talked long enough.

I was just wondering about this, and though somewhat might look into it.

Matt DeKok

I think you've reversed "ordered" and "unordered". I don't see a problem. — Arthur Rubin | (talk) 18:10, 28 March 2006 (UTC)Reply

Yeah, it's just confusion with how you're reading the terms. For combinations, (1 2 3) is not order dependent. All of the order-dependent permutations of that combination [(1 2 3), (2 1 3), (3 2 1), etc] reduce to order-independent (1 2 3).

You could also write the combinations as (1 3 2), (4 1 2), (4 1 3), (4 3 2) and it would be the same set -- standard ordering just seems to make the most sense. — Lomn Talk 19:05, 28 March 2006 (UTC)Reply

Anytime you find yourself wondering if "everyone has ⟨whatever⟩ backward", stop and use a little common sense. Almost certainly you are the one who has it backward. --KSmrq^T 20:09, 28 March 2006 (UTC)Reply

KSmrq, this is good advice for anyone in any context, but especially in math. Let me say that I could have heeded this advice many times in my younger days and obtained more optimal outcomes :-)--Deville (Talk) 13:34, 29 March 2006 (UTC)Reply

Let me see if I can rephrase this. "Ordered" means that you count different orders as different things. "Unordered" means that you count different orders as merely rearrangments of the same thing. I think what you're seeing is that "Unordered" only includes one particular order listed, but it actually includes all orders since they're the same thing. Similarly, "Ordered" includes all possible orders because each order is different. Think unordered=committee (all spots are equally powerful) and ordered=hierarchy (different offices have different powers), and ask yourself how many committees vs. how many arrangements into offices. --Geoffrey 05:46, 30 March 2006 (UTC)Reply

Here's a way to remember: When we shuffle a deck of cards, we want to randomly choose a permutation, preferably with each of the 52! = 52×51×50×⋯×3×2×1 orderings equally likely. But when we look at the cards dealt to us, we only care about the combination, not the order in which the cards come to us.

Now for a little fun. Assuming a simple mathematical model of a riffle shuffle, how many shuffles are required to have confidence that the deck is well-shuffled? One way to analyze this is using a Fourier transform on the group of permutations. It can be shocking to learn how poorly shuffled a deck is after as many as four shuffles. Of course, all bets are off (literally) if a magician is shuffling the deck—especially if that magician happens to be Persi Diaconis, author of Group representations in probability and statistics. --KSmrq^T 10:15, 30 March 2006 (UTC)Reply

Inequality

Latest comment: 19 years ago7 comments4 people in discussion

Does the inequality: (x^(t))(y^(1-t))<=(t)(x)+(1-t)(x) hold? with 0<=t<=1 and x,y > 1 Also any proof? Oh and by "<=" I mean less than or equal to. Thanks Qeee1 17:24, 28 March 2006 (UTC)Reply

I'll describe a hint from which you should be able to get the answer. But the real reason it works is that the function ${\displaystyle f(t)=x^{t}y^{1-t}}$  is convex.

Fix any ${\displaystyle x>1,y>1}$ . Define the functions ${\displaystyle f(t)=x^{t}y^{1-t}}$  and ${\displaystyle g(t)=tx+(1-t)y}$  (I'm assuming that's what you meant above). Compute ${\displaystyle f(0),g(0)}$  and compare, compute ${\displaystyle f(1),g(1)}$  and compare. Finally, compute ${\displaystyle f''(t)}$ , and what does this tell you?--Deville (Talk) 18:11, 28 March 2006 (UTC)Reply

Take the logarithm on both sides and use convexity of logarithm. Kusma (討論) 18:17, 28 March 2006 (UTC)Reply

Concavity, of course. Kusma (討論) 18:42, 28 March 2006 (UTC)Reply

You mean

${\displaystyle x^{t}y^{1-t}\leq tx+(1-t)\color {Red}y}$ ?

Yes. See, for example inequality of arithmetic and geometric means. (written before I saw Deville's and Kusma's answers. Grumble.) — Arthur Rubin | (talk) 18:24, 28 March 2006 (UTC)Reply

Thanks, I used Kusma's method. (I have problems differentiating, and I had already checked AM-GM generalisations) But also thanks to both of you too. I'm impressed at the speed of the response.Qeee1 20:09, 28 March 2006 (UTC)Reply

Oh and yes I meant Y, just to be clear.Qeee1 20:10, 28 March 2006 (UTC)Reply

Graphing

Latest comment: 19 years ago9 comments6 people in discussion

Hi there,

I am looking for a good freeware software to plot graphs. I looked through the Graph-Plotting Software Page but was very unsure...does anyone of you know a good programme. I do need it for school calculus, so it should be able to do soem analysis.

Thank you --165.165.228.18 19:16, 28 March 2006 (UTC)Reply

It may be more than you need, but the freeware Maxima computer algebra software has a variety of graphing options, and much more. --KSmrq^T 20:18, 28 March 2006 (UTC)Reply

Octave, a clone of matlab, has a nice graphing system. It's essentially a frontend over gnuplot which is a graphing software usable by itself too. – b_jonas 20:52, 28 March 2006 (UTC)Reply

I have no experience to use any packages on computer graphic. Only Maple and Matlab are available to me paid by my University but I never try them. My simple question is how do I produce a JPG-file, say sample.jpg of the function joining (-1,0) to (0,1), from (0,1) to (1,0) and set all other values to be zero. Can I use formula such as \sin nx/(\pi x) for n=1,5,10 on the same output? I can start Maple on MacIntosh with a double click, then what? How do I draw a circle inscribed in a triangle? Do I have to study a lot from their users' manual (RTFM) before I can use Maple or Matlab? If successful, I can use TEX-graphicx to paste this file as part of a PDF-file. Thank you in advance. Twma 02:03, 29 March 2006 (UTC)Reply

For geometric constructions of the inscribed circle variety, a specialized package like C.a.R. [2] (which is written in cross-platform Java) is likely to be simpler than learning Maple or MATLAB. For more general illustrations, including plots, Maple or gnuplot are powerful with only a modest learning curve. For those comfortable around TeX, MetaPost may be your cup of tea. For 3D illustrations, the task itself tends to be challenging, regardless of the software used. --KSmrq^T 07:25, 29 March 2006 (UTC)Reply

Microsoft PowerToys Graphing Calculator is pretty good at basic graphing, easy to use, and technically freeware, but it's Windows XP only. And no analysis. If you can figure out Maxima or Octave, those are useful too. --Geoffrey 05:42, 30 March 2006 (UTC)Reply

Not exactly a graphing calculator, but a quality cross-platform tool for graphical mathematics, is GrafEq. Its specialty is implicit functions, so the results are raster images. --KSmrq^T 07:07, 30 March 2006 (UTC)Reply

Actually I think trying to use the POVRay renderer might be a good idea, of course you shall have to invest yourself quite a lot but you'll then be able to produce amazingly beautiful 2D and 3D graphs. Sure worth the time. Xedi 13:26, 30 March 2006 (UTC)Reply

Yes, raytracing can produce lovely mathematical illustrations. [3] Yet I repeat my caution that 3D illustrations are a significant challenge, much harder than 2D. --KSmrq^T 01:10, 31 March 2006 (UTC)Reply

trigernometry

Latest comment: 19 years ago3 comments3 people in discussion

a radius of a circle is 15cm. find the length of the arc of the circl intercepted by a central angle of 3pie/4radians. (leave in terms of pie)

mmm... pie.... -lethe ^{talk +} 22:10, 28 March 2006 (UTC)Reply

It says at the top of the page to DO YOUR OWN HOMEWORK! But in spite of that I decide to give you the formula:

${\displaystyle {x^{\circ } \over 360^{\circ }}={y \over 2\pi r}}$ 
${\displaystyle x^{\circ }}$ =degree measure of the arc

${\displaystyle y\,\!}$ =arc leangth

there is an algebraic way, but I find this way easier. (updated by myself schyler 00:01, 29 March 2006 (UTC))Reply

I'm guessing the "algebraic way" is just:

${\displaystyle {\frac {x}{2\pi }}={\frac {y}{2\pi r}}=}$ 

${\displaystyle x={\frac {y}{r}}=}$ 

${\displaystyle xr=y\,\!}$ 

${\displaystyle x\,\!}$ -radian measure of arc

${\displaystyle y\,\!}$ -arc length

The two equations are equivalent. Superm401 - Talk 01:16, 30 March 2006 (UTC)Reply

Pi and the Universe

Latest comment: 19 years ago2 comments2 people in discussion

On a disk the same radius as the observable universe what is the minimum number of digits to describe an arc the length of a ... pick something interesting. The width of a finger, a human hair, a hydrogen atom. I was just reading the Pi article and thought this might be good to add.Trieste 14:57, 29 March 2006 (UTC)Reply

Well, you don't use π to calculate just any arc length, rather you use it to calculate the entire perimeter of a circle or semicircle. But you want some upper bound on how many digits of π one might need to talk about even cosmological distances, right? So let's say you wanted to calculate the surface area of our Hubble volume. I think our Hubble volume is 80 billion lightyears across. That's 13.4 billion years old, times the speed of light, adjusted for the cost of expansion. About 10^27 meters? So in order to get the surface area accurate to, say, a meter squared (absurd, given the error bars on the Hubble volume, which might be 50%), then you need maybe 52 digits of π (double 27, since we're squaring). This is a rather silly calculation, doesn't have much physical relevance; nevertheless, I think it's a nice example to show how completely physically irrelevant knowing a million digits of π is. Doing so is mostly for testing computer algorithms or hardware, or for testing mathematical notions about the distribution of digits, and not for knowing the perimeter of a circle more accurately. -lethe ^{talk +} 15:01, 29 March 2006 (UTC)Reply

Yes about the relevance. That is what I was hoping to illustrate e.g. one only needs x digits of π to accurately calculate the area of this period . in relation to a sphere of radius the mean distance of the earth from the sun. Or how about what you can calculate with 5,10,20,30 etc digits of precision of π

Functions with latex

Latest comment: 19 years ago3 comments3 people in discussion

Hi guys,

is there any packet for LaTeX to draw and display functions in a LaTeX Document, such that u enter a formula and parameters and it will draw it into the document.

Thank you --Da legend 15:54, 29 March 2006 (UTC)Reply

If I understand you correctly, you want to draw graphs and display them in a LaTeX document. Probably the easiest way to do that is to draw them with a package like gnuplot, export them as embedded postscript, and embed those in the LaTeX document. --04:12, 30 March 2006 (UTC)

I seem to recal there is some picture drawing capabilities in TeX/LaTeX, (might be a add on) and the syntax was horible. Irrelavant for wikipedia as its not supported. --Salix alba (talk) 08:06, 30 March 2006 (UTC)Reply

Haha I remember people trying to do analytical geometry with LaTeX, they must have spent week to do a triangle !

What if "Functions with latex" was a crossword definition ... --DLL 21:06, 30 March 2006 (UTC)Reply

Dow Jones Average

Latest comment: 19 years ago2 comments2 people in discussion

I have been trying to learn "how the $11,000+" amount is calculated. I have looked up the 30 stocks and added the stock price but doesn't come close to $11,000. How is this number derived on a daily basis?

Thanks, Bill

It is a "price-weighted" ratio. See the article Dow Jones Industrial Average. --LarryMac 22:39, 29 March 2006 (UTC)Reply

You add up the stock prices and then divide the total by a term called the divisor, which is currently about 0.125. Whenever there is a change in the constituent stocks that make up the index, the value of the divisor is adjusted so that the index value immediately after the change is the same as its value immediately before the change. This avoids artificial jumps in the index value and ensures that the index has continuity. For more details see [4]. Gandalf61 11:42, 30 March 2006 (UTC)Reply

March 30

Function for making 1,2,3,4,5,6 then any number

Latest comment: 19 years ago5 comments4 people in discussion

When I was studying the problem of induction at University the lecturer wrote '1,2,3,4,5,6' on the board, and asked what number came next. The general consensus was '7'. He then wrote out a simple function that, when we worked out the next number, it turned out to be something in the hundreds. Depending on a certain variable in the function, it would give a series of successive integers followed by any number you specify. Does anyone know what this function is? Purely out of interest, no hurry. Phileas 04:40, 30 March 2006 (UTC)Reply

Here's one easy way: The function ${\displaystyle f(x)=x+(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)(x-k)}$  will always return 1-6 for inputs 1-6, but for any other value, the output will depend on k. (The function basically works by adding y=x to an expression with roots at 1 through 6.) In particular, to get any given output at 7, use k=(847-f(7))/120. Is this kinda what you saw? --Geoffrey 05:40, 30 March 2006 (UTC)Reply

That looks pretty familiar, yeah! I'm not certain that that's the same one for sure, but it definitely answers my question. Thanks! Phileas 05:54, 30 March 2006 (UTC)Reply

I'd be more impressed if someone could write a recursive sequence whose first 6 terms were one through six and whose seventh was anything. OEIS has nothing of the like. -lethe ^{talk +} 06:03, 30 March 2006 (UTC)Reply

Try

${\displaystyle f(1)=1;f(n)=1+f(n-1)+\left\lfloor {\frac {f(n-1)}{6}}\right\rfloor (A-7)}$ 

where A is anything. Replace the floor expression with the polynomial with roots 1-5 and adjust constants accordingly for a slightly hairier but "continuous" expression. Fredrik Johansson 10:34, 30 March 2006 (UTC)Reply

Uncle Scrooge's wealth

Latest comment: 19 years ago7 comments5 people in discussion

In the 1952 story Back to the Klondike, Uncle Scrooge claims that he has 3 cubic acres of cash in his Money Bin. "Cubic acres" is an unusual measurement for volumes so I need to ask: How much is 3 cubic acres in cubic metres?

Also, the depth gauge in the bin shows a depth of 99 feet. Is that reasonable? Thuresson 16:27, 30 March 2006 (UTC)Reply

Well, it doesn't quite make sense on first glance. Since "acre" is a measurement of area, it's not clear what a "cubic acre" should be in the first place. But, for example, if we make the following definition: a cubic acre is a cube each of whose face has an acre of area, it then makes sense to talk about the volume of a cubic acre. According to Acre, an acre is 4046.8564224 m². A square which has an acre in area would have to have side length a square root of that, roughly 63.615 m. A cube whose side lengths are each 63.615 m would have a volume of 257 440 m³ and three of these would have a volume of 772 321 ³. That's a lot of cubic yards.

As for depth, if the height is 99 feet, that's roughly 33 m which is about half of 63.615 m. Thus a cubic acre with depth 99 ft would have to have a base area of about 2 acres. Big bin.--Deville (Talk) 17:15, 30 March 2006 (UTC)Reply

Cubic acres make perfect sense—as a measurement of six-dimensional hypervolumes. I would take the comic-book usage as a joke, not intended to have any precise meaning. --Trovatore 17:30, 30 March 2006 (UTC)Reply

Thanks for the answer. The weight of all this cash (mostly coins?) must be massive. Thuresson 17:36, 30 March 2006 (UTC)Reply

Here's a trivial but cute-sounding exercise that occurred to me: How many square gallons in a cubic acre? --Trovatore 23:27, 30 March 2006 (UTC)Reply

Interestingly, even Google can do it. Superm401 - Talk 00:37, 31 March 2006 (UTC)Reply

Hehe, that made my day. SanderJK 13:55, 31 March 2006 (UTC)Reply

How manys suns would fill the sky.

Latest comment: 19 years ago4 comments3 people in discussion

How many suns would it take to fill the sky,I know the size of the sun is about 1/2degree, and the sky is basically half a sphere.But what is the formula to work it out step by step. Looking at the sky I tryed to visalize how suns it would take,the answer must be quiet large,but it is the maths that I have find hard.

Well, the question could use a more-precise formulation. Do you have to be able to see a solar surface in every direction, with no gaps in between? If so, then some of the suns are going to have to overlap.

If we ignore this point, though, we should be able to get a quick-and-dirty approximation easily enough. Half a degree is about 0.01 radian, so the area of the solar disc should be about 0.0001 R², where R is the distance to the Sun. (Aside—can that be right? Seems too much.) That's taking the Sun to be a square; you'd multiply by π/4 to get the area of a circle, so we'll remember that the answer might be off by 30% or so.

Now the area of the entire hemisphere is 2πR², so you'd need about 60,000 Suns if they were squares. Closer to 80,000 if you could distribute circular area to fill the gaps, which of course you can't, so maybe guesstimate 100,000 allowing overlaps. --Trovatore 19:32, 30 March 2006 (UTC)Reply

I just want to add that How manys suns would fill the sky. would be an excellent first line for a poem. Could you be channeling E. E. Cummings? --Trovatore 19:38, 30 March 2006 (UTC)Reply

And yes, the average diameter of the sun is (roughly) 0.0093 radians, which is 0.53 degrees. -- Meni Rosenfeld (talk) 08:20, 31 March 2006 (UTC)Reply

I see Trovatore used 2πR². That is for the whole hemisphere, not the visible amount of sky. An interesting way to find the circumference of the visible sky would be looking at Image:Anticrepuscularpano.jpg and comparing the size of the sun with the length of the photograph. Assuming the photograph is full-circle. --   Mac Davis] ⌇☢ ญƛ. 08:52, 31 March 2006 (UTC)Reply

March 31

Approximation by polynomials

It is well known that for every m-times continuously differentiable real function f on R^n, every compact subset K of R^n, and every e>0, there is a polynomial p on R^n such that for every multi-index ${\displaystyle \alpha }$  with ${\displaystyle |\alpha |<m+1}$  and every x in K, we have ${\displaystyle |\partial ^{\alpha }f(x)-\partial ^{\alpha }p(x)|<e}$ .

On page 155 of Topological vector spaces, distributions and kernels by F. Treves, Academic 1967, there is a proof using extension of f to a complex entire function on C^n.

Most of the standard undergraduate textbooks in numerical analysis AVAILABLE to me do not deal with this specialized topic.

Question: Is there any proof which is internal to real analysis? Any references?

Thank you in advance. Twma 07:18, 31 March 2006 (UTC)Reply

For the case m=0 this can be done with Bernstein polynomials; I think no results from outside real analysis are used. A concise discussion online is Shadrin's lecture notes on Approximation Theory; you need Lectures 2 and 3. I think this is also treated in M.J.D. Powell, Approximation Theory and Methods, Cambridge University Press, 1981. My guess is that extension to general m is quite possible. -- Jitse Niesen (talk) 08:56, 31 March 2006 (UTC)Reply

The m=0 case goes by the name of the Stone-Weierstrass theorem. -lethe ^{talk +} 14:13, 31 March 2006 (UTC)Reply

Probability Question

Latest comment: 19 years ago3 comments3 people in discussion

Suppose an event e has two equally likely outcomes, x or y. Suppose e has occurred k=5 times in succession, each with outcome x. What is that probability that the 6th outcome of event e will be x? --Simian1k 18:37, 31 March 2006 (UTC)Reply

It's still 50:50. Truly random events have no memory. The common belief that a Y will be more likely after a run of Xs comes from a misinterpretation of the law of large numbers. —Keenan Pepper 18:53, 31 March 2006 (UTC)Reply

Be careful not to overstate the facts. It is possible for "truly random" events to be correlated. We explicitly assume that, say, coin flips are independent of each other. A Markov process, however, is also random. --KSmrq^T 22:45, 31 March 2006 (UTC)Reply

Infinite Sum

Latest comment: 19 years ago3 comments3 people in discussion

What is the value of the series: ${\displaystyle s=\sum _{i=0}^{\infty }(-1)^{i}*a=a-a+a-a+\ldots }$ ? Thanks in advance, Mickey 195.93.60.7 19:30, 31 March 2006 (UTC)Reply

The traditional definition of an infinite sum says that this does not have a sum. It is, however, summable by the methods of Cesaro and Borel, and they give answers of a/2. -lethe ^{talk +} 19:41, 31 March 2006 (UTC)Reply

Someone asked a closely related question over at AskMetafilter some time ago, and I'll just point you to the many interesting comments there. Chuck 23:17, 31 March 2006 (UTC)Reply

Brachystochrone and Snell's Law

Latest comment: 19 years ago2 comments2 people in discussion

In the article on the brachystochrone Snell's law is given as sin theta divided by velocity equals Cste. What is Cste ? --204.69.190.14 19:38, 31 March 2006 (UTC)Reply

Snell's law says that n sin θ is a constant. So sin θ/v is a constant. I expect that "constant" is what is meant by "Cste". Looking at the history of the article, I see that that section was copied from the french, which might explain its funny terminology. I don't know how they would say "constant" in french, or how they would abbreviate it, so that's a bit of speculation on my part. Based on this assumption, I'm going to fix the article. -lethe ^{talk +} 19:55, 31 March 2006 (UTC)Reply

A cruel set of differential equations

Latest comment: 19 years ago1 comment1 person in discussion

I have to solve the following set of differential equations numerically:

${\displaystyle \psi ''=f_{1}(\psi ,\phi ,\theta ,\psi ',\phi ',\theta ',s)}$ 
${\displaystyle \phi ''=f_{2}(\psi ,\phi ,\theta ,\psi ',\phi ',\theta ',s)}$ 
${\displaystyle \theta ''=f_{3}(\psi ,\phi ,\theta ,\psi ',\phi ',\theta ',s)}$ 

Boundary conditions:

${\displaystyle \theta (0)=\pi /2,\phi (0)=0,\psi (0)=0}$ 
${\displaystyle \theta '(L)\sin \phi (L)-\psi '(L)\sin \theta (L)\cos \phi (L)=M_{1}/EI_{x}}$ 
${\displaystyle \theta '(L)\cos \phi (L)-\psi '(L)\sin \theta (L)\sin \phi (L)=M_{2}/EI_{y}}$ 
${\displaystyle \phi '(L)+\psi '(L)\cos \theta (L)=M_{3}/GJ}$ 

I have three 2nd order equations, and 6 boundary conditions, so I should be able to solve this. The derivatives are with respect to s. Functions f1, f2, f3 are a bit complicated. How can I find solutions to this problem numerically to find ψ(s), θ(s) and φ(s)? I have Matlab at my disposal. Any help will be greatly appreciated. deeptrivia (talk) 19:55, 31 March 2006 (UTC)Reply