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February 25

1-1

let a,b positive integers.

Is

f(a,b)= 3^a + 10^b

1-1? —Preceding unsigned comment added by 71.97.4.194 (talk) 00:46, 25 February 2008 (UTC)[reply]

If you mean one-to-one, then yes, I think it is. Black Carrot (talk) 01:11, 25 February 2008 (UTC)[reply]

g(a,b) = 2^a + 3^b fails to be 1-1. g(3,1)=11=g(1,2). It seems like we could do something similar for f, but the numbers get too big. —Preceding unsigned comment added by 71.97.4.194 (talk) 01:25, 25 February 2008 (UTC)[reply]

If 3^a+10^b=3^c+10^d, then 3^a-3^c=10^d-10^b. Investigate the properties of differences between powers of 3 and differences between powers of 10. Does that help? —Bkell (talk) 04:48, 25 February 2008 (UTC)[reply]

Yes, I tried that, to no avail (yet). It did make me think of using the pigeonhole principle. *How*, I'm not quite sure. 192.91.253.52 (talk) 13:20, 25 February 2008 (UTC)[reply]

I was unable to solve the problem, though I approached it in the way Bkell suggested. If a>c then d>b, so I factored it as 3^c*(3^(a-c)-1) = 10^b*(10^(d-b)-1). You get congruences like a=c mod OrderMod( 3, 10^b ), and b=d mod OrderMod( 10, 3^c ) so if b or c is nonzero, this reduces the possible solutions, and in fact you can apply the same argument again with much larger moduli. However, induction does not quite work, because I could not rule out b and c being zero (especially not inductively). When the problem is changed to the g(a,b), in fact very early there is a solution with b or c being 0, g(2,0)=5=g(1,1). While one might decide the arguments are not allowed to be 0 for some reason, during the induction you have to allow that case. At any rate, there are no small solutions for f (0<= a,b,d <= 1000). JackSchmidt (talk) 15:43, 25 February 2008 (UTC)[reply]

The original problem states that the exponents are positive integers. —Bkell (talk) 16:33, 25 February 2008 (UTC)[reply]

My idea, while not fully fleshed out, is that a difference between powers of 3 is always going to be a string of 0's and 2's when written in base 3, and a difference between powers of 10 is always going to be a string of 0's and 9's when written in base 10. I think there's a contradiction lurking somewhere: it feels like it's going to involve a claim that in order to write a difference between powers of 3 as a sum of things that look like 9000…0 in base 10, you're going to have to use a particular 9000…0 twice. Sorry for the roughness of this idea; it's really just a hunch and needs a lot of work still. —Bkell (talk) 16:40, 25 February 2008 (UTC)[reply]

Consider this, if 3^a+10^b=3^c+10^d, then you can rewrite it in most cases as 3^f=((10^h-1)*10^g)/(3^j-1), which must be an integer. Is it possible for this to be an integer, though? (hint: look at 10^h-1 and 3^j-1 under the proper modulus). GromXXVII (talk) 19:33, 25 February 2008 (UTC)[reply]

Sure it can be an integer. For example, g=3, j=2 makes ((10^h-1)*10^g)/(3^j-1) an integer for every nonnegative h. JackSchmidt (talk) 19:42, 25 February 2008 (UTC)[reply]

I haven't come up with a good way to prove it yet, but I have another direction you might go in. If you write down a given power of 3, the question is whether you can subtract 1 from one digit, and add 1 to another, and get back a different power of 3. If you consider it base 3, the question is whether given a power of 101, you can subtract 1 from one digit and add 1 to another and get back another power of 101. Without carry, powers of 101 are pascal's triangle, which gives a clear enough pattern that it may be possible to prove it false. Black Carrot (talk) 20:50, 25 February 2008 (UTC)[reply]

I think I've got it. Take it mod 5, 4, and 20, in that order. The first shows that a=c(mod4), the second shows that that's only possible if either b or d is 1, and the third shows that that can't happen. It's one-to-one. Black Carrot (talk) 07:22, 26 February 2008 (UTC)[reply]

Unfortunately, simply looking mod finitely many numbers cannot prove the result. For instance, it is possible for f(a,b) = f(c,d) mod 4, mod 5, and mod 20 (simultaneously), without a=c, b=d: more explicitly, f(1,2) = f(9,10) mod 4, mod 5, and mod 20. JackSchmidt (talk) 16:45, 26 February 2008 (UTC)[reply]

Interesting problem. – b_jonas 10:59, 29 February 2008 (UTC)[reply]

Approximation by Smooth Functions

Basically, our teacher is trying to convince us that $C_{0}^{\infty }(\Omega )$ which is the set of all infinitely differentiable functions with compact support on a given set $\Omega$ , is dense in $L^{2}(\Omega )$ . Now, I know that one of the definitions of a set A being dense in another set B is that the closure of A must be B. Which is the same as saying that B contains A and all of its limit points. So he gave us a theorem in class saying that, for any $f\in L^{2}(\Omega )$ , there exists a sequence $\phi _{n}\in C_{0}^{\infty }(\Omega )$ such that $\phi _{n}\rightarrow f$ in $L^{2}(\Omega )$ . My question is how does this theorem prove that $C_{0}^{\infty }(\Omega )$ is dense in $L^{2}(\Omega )$ ? All we have done is shown that every square integrable function f is a limit of a sequence of smooth functions with compact support. We have shown that each f is a limit point but how do we know that those are all the limit points? What if there is a limit point of smooth functions with a compact support outside $L^{2}(\Omega )$ ?A Real Kaiser (talk) 04:52, 25 February 2008 (UTC)[reply]

L^{2}(\Omega )

is a closed set - it contains its own limits. Bo Jacoby (talk) 05:36, 25 February 2008 (UTC).[reply]

Duh, I seem to have forgotten that little fact. Thanks!A Real Kaiser (talk) 05:52, 25 February 2008 (UTC)[reply]

The above comment obscures the point at hand. It is not enough to talk about a set being open or closed; you have to specify the topology you are talking about. Every set is closed in its own subspace topology. Here, the question is whether or not the closure of

C_{0}^{\infty }(\Omega )

in the metric topology of

L^{2}(\Omega )

is

L^{2}(\Omega )

. In this topology, it is impossible for the closure of

C_{0}^{\infty }(\Omega )

to be bigger than

L^{2}(\Omega )

. Perhaps the above post was referring to the fact that

L^{2}(\Omega )

is complete (is this true actually? ahh...), which would mean that however you embed

L^{2}(\Omega )

, the closure of

C_{0}^{\infty }(\Omega )

will be exactly

L^{2}(\Omega )

. However, that would not be a requirement for saying that the one is dense in the other. If

A\subset B\subset C

, to say that A is dense in B is to say that the closure of A, in B's subspace topology, is all of B. It would not matter if B were not closed in C, because in the B subspace topology, the closure of A cannot exceed B. (Can someone with more experience please check my post for correctness? Thanks...) J Elliot (talk) 06:08, 25 February 2008 (UTC)[reply]

L^{2}(\Omega )

is a normed space, the norm being the integral of the absolute square. This norm defines the topology. In this topology the set

L^{2}(\Omega )

is closed. A function outside

L^{2}(\Omega )

has no well defined distance to a a function within

L^{2}(\Omega )

, so it is not a limiting function of a sequence of functions in

L^{2}(\Omega ).

Bo Jacoby (talk) 14:15, 25 February 2008 (UTC).[reply]

The sentence "In this topology [on a space X] the set X is closed" is one of the axioms of a topology, and thus is never a meaningful observation. As was noted above, the only meaningful thing that we can say in this direction (and indeed it is very meaningful) is that L²(Ω) is a normed space (and thus more generally a uniform space), and with respect to this structure L²(Ω) is complete. In particular, this implies that whenever any metric space V is isometrically embedded into L²(Ω) (we might take V=C^∞₀(Ω) with the induced square-integral norm), then any Cauchy sequence in V has a limit in L. Thus it is the completeness of L²(Ω), rather than the meaningless "closedness", which implies that the completion of C^∞₀(Ω) [with the L²-norm] is L²(Ω). This is not necessary to show that C^∞₀(Ω) is dense in L²(Ω), however. Tesseran (talk) 05:26, 27 February 2008 (UTC)[reply]

Kaiser, you should take some algebra classes next semester, because my analysis books had been getting nice and dusty before you started posting this semester. :)

BTW, you may seriously want to prove that if f_n are C^oo with compact support, and they form a cauchy sequence under the L^2 norm (so they "converge to something"), then there is some number C such that the L^2 norm of all f_n is bounded by C. In other words, if something is an L^2 limit of compact functions, then it is L^2. This should be easy to prove, other than the fact that it is posed randomly on the net. Once you do this, some of the previous posters talking about completeness should make more sense.

This idea is actually pretty important, even though some people might think my question is silly. The point of using the L^2 norm is to control the L^2 norms of limits. As long as everything in the sequence was L^2, and the sequence converges (is cauchy) according to the L^2 norm, then the limit is L^2. This is why you need to use the Sobolev norms in PDE. You want to take a sequence of approximate solutions to a PDE (or solutions to some approximately equal PDE), and know that the limit is still a solution. If you use L^2 for that, you will often fail, but if you use H^1 (L^2 of the function and its derivative), you will have much better luck, since the limit function at least has an L^2 derivative. JackSchmidt (talk) 15:57, 25 February 2008 (UTC)[reply]

I think, Kaiser, that your original post misses the point a bit. B containing A and its limits is nothing like sufficient for A being dense in B (consider

A=[0,1],B=\mathbb {R}

). Instead, what you want is that

\forall x\in B\;\exists \{y_{n}\}\subseteq A\;\lim _{n\rightarrow \infty }\|y_{n}-x\|=0

(with some metric appropriate to B, which is where topology comes in). The theorem you were given establishes precisely this fact. --Tardis (talk) 16:08, 25 February 2008 (UTC)[reply]

Jack, you know what, you are absolutely correct. I am going for a degree in Applied Math and I incorrectly assumed (in my early childhood as a lower division Math student) that Applied Math and Pure Math were two disjoint sets. So I decided to stay away from Pure courses which is exactly why now I have huge holes in my understanding of real, complex, functional analysis, and algebra. I was already thinking about an algebra course next semester but it seems vital that I must take it. Well, lol, don't put those books away. This doesn't mean that I am going to stop asking questions. There are plenty more where these come from. I know that all of you don't have to help me but I really appreciate it. Thanks everyone! :)
Ok, so let me recap. L^2 is the space of all square integrable functions. It has a norm defined on it (the usual L^2 norm). It has an inner product defined on it and it is complete with respect to this norm. Therefore L^2 is a Hilbert space. Being complete means that any Cauchy sequence of L^2 functions converges to an L^2 function. So there are no limit points of L^2 that are outside L^2. Furthermore, if I want to show that the set of all smooth functions (members of $c^{\infty }$ ) is dense in L^2, then it will suffice to show that if I take any square integrable function, I can always find a sequence of smooth functions that converges to that square integrable function. This is the same as showing that any square integrable function can be approximated arbitrarily well by smooth functions. It is also not possible for a sequence of smooth functions to converge outside the L^2 space because smooth functions are a subset of square integrable functions and the set of all square integrable functions is closed. Is this correct, guys? I want to make sure that my line of reasoning is correct.A Real Kaiser (talk) 20:04, 25 February 2008 (UTC)[reply]

GMAT data sufficiency problem

A homeowner must pick between paint A, which costs $6 per liter, and paint B at $4.50 per liter. Paint B takes one-third longer to apply than paint A. If the homeowner must pay the cost of labor at $36 per hour, which of the two paints will be cheaper to apply?

(1) The ratio of the area covered by one liter of paint A to that covered by one liter of paint B is 4:3.

(2) Paint A will require 40 liters of paint and 100 hours of labor.

I first encountered the problem in Arco Math Workbook (2000). Kaplan's 2005 version is similar, but it's got the same problem and explanation, which I cannot agree with:

"Statements (1) and (2) taken together are not sufficient to answer the question. To make an intelligent decision, we need to know which requires more paint and how much more, how long each will take, and we need some info on their labor costs...Using both statements together, we still cannot find the labor costs."

It seems rather clear to me how to find the labor costs.

First, the two statement taken together mean that it'd take 40*4/3 or 160/3 L of paint B.

Second, statement (2) says that paint A is applied at a rate of 40L per 100 hours, or 2/5 L/hr. - but the original problem says it would take a third longer for paint B, which means that paint B is applied at a rate of 2/5 * 3/4 L/hr. or 3/10 L/hr.

This last conversion was the most difficult for me, but you can see it's just multiplying A's rate by 4/3 hr/L.

Thus, it will take (160/3) / (3/10) or 1600/9 hours to use the 160/3 L of paint B to cover the house.

Therefore we know how long it takes. Given the unit labor cost in the original problem, we can figure out how much labor costs for either paint. We also know how much paint we need and how much it costs. Why then would the problem be unsolvable so long as we have statements (1) and (2)?

Imagine Reason (talk) 07:26, 25 February 2008 (UTC)[reply]

(1) and (2) on their own aren't enough, you need to information in your first paragraph as well - is that what it means? --Tango (talk) 11:42, 25 February 2008 (UTC)[reply]

The first paragraph is a given, while the two statements are optional. I'm saying that if we have the two statements (the first paragraph is always there), we can solve the problem, but the review books say we can't even then. —Preceding unsigned comment added by Imagine Reason (talk • contribs) 17:14, 25 February 2008 (UTC)[reply]

Well, it's hard to comment on what the review book says without seeing it, but the problem you've posted is certainly solvable in the way you came up with. Black Carrot (talk) 06:32, 26 February 2008 (UTC)[reply]

Clarity

Working in dollars throughout...

Focusing on the given paragraph ONLY... We assume that equal amounts (L) of paint in litres are required no matter whether paint A or paint B is used and that if H hours are needed for paint A, then 4H/3 hours are needed for paint B.

Total cost of using paint A is 6L + 36H and the total cost of using paint B is 4.5L + 48H

Setting these equal to each other gives 6L + 36H = 4.5L + 48H and a solution of L = 8H

Therefore numerically, if L > 8H then paint B is cheaper and if L < 8H then paint A is cheaper.

Conclusion is that additional information is needed.

Incorporating ONLY statement 1 with the given paragraph... We need to make an assumption that the reduced volume of paint A needed (3/4 of the volume required by using paint B) results only in a cost saving on the paint, and NOT that less paint equals less hours of labour: after all, the same area needs covering and we know it takes a third longer with paint B. If this assumption is false and that less paint ALSO equals less labour hours then we have a slightly different problem.

Relative to the L litres required for paint A, paint B requires 4L/3 litres which now costs 6L (dollars).

Case 1.1 Less paint only required.

The expression for the cost of paint A remains 6L + 36H which in all cases is less than the cost for paint B of 6L + 48H.

Paint A is cheaper.

Case 1.2 Less paint and 3/4 of the previous amount of labour hours required (4/3 times the amount are now needed for B compared to A), i.e. relative to the H hours required for paint A, paint B requires 4/3 × 4/3 = 16/9 times the hours (cost = 36H × 16/9 = 64H).

The expression for the cost of paint A remains 6L + 36H which in all cases is less than the cost for paint B of 6L + 64H.

Paint A is even cheaper.

Statement 2 is irrelevant in determining which paint is cheaper to apply; it just gives the information to calculate the ACTUAL costs. All the problem requires though is a general 'which is cheaper', not by how much. So, as far as I can read it from the information provided, choosing paint A is cheaper in all cases if statement 1 is included with the given paragraph; statement 2 being irrelevant, as is the request for any additional information.

However, all this is blindingly obvious from the simple point that statement 1 means that the effective price of paint B is $6.00 per litre, identical to that of paint A. B takes longer to apply at an additional cost of $12 per hour for case 1.1 and $28 per hour for case 1.2 above based on the number of hours needed to apply paint A.

The only ambiguity is in calculating ACTUAL costs for paint B depending on case 1.1 or 1.2 above. Paint A costs $3840: Paint B either costs $5040 or $6640.

Please don't hesitate to contact me if you feel I have made any error in reading this situation! AirdishStraus (talk) 11:23, 26 February 2008 (UTC)[reply]

I agree that the wording is ambiguous, but I'm pretty sure it is saying that a gallon of paint B takes a third longer to apply than a gallon of paint A. Regardless, you're right--we only need the first statement to answer the question. I think that's actually what I got the first time I saw the problem, actually, but somehow I let it go. Now I've seen the same nonsensical explanation in a second book, I'm not happy. Imagine Reason (talk) 23:36, 26 February 2008 (UTC)[reply]

aquestion about closed surface

i am wondering about something,can we find or create aclosed surface where there all of the points inside that surface has adifferent distances to any of the all points of the surface?i mean here we cannot finde apoint inside the space of that surface has the same distance to at least 2points on the surface?this surface should not be for example like asphere because the center of the sphere is apoint that has asame distances to all of the points on the spherer`s surface.i hope my words is clear.thank you.Husseinshimaljasimdini (talk) 12:04, 25 February 2008 (UTC)[reply]

If I understand the question correctly, then I would think not, since you can just choose any two points on the surface and pick the midpoint between them. Well, if the surface isn't convex, you couldn't choose *any* two points, but there there will still be some points - just choose any line that passes through the interior of the region bounded by the surface and the intersection points of that line with the surface should do. I think as long as the surface does bound a region with a non-empty interior, there will be points in the interior equidistant to points on the surface. --Tango (talk) 13:41, 25 February 2008 (UTC)[reply]

Interesting question - I think the answer is no. I'll explain in two dimensions, you can easily expand the answer to three dimensions for surfaces..

Suppose I start my asymmetric ring at point A with distance r from the 'centre'.. then increasing r as I turn around the surface eg using polar coodinates (radius,angle) - the first thing I notice is that if I increase r with angle I cannot every decrease it since that would return r to a previously used value.. but to get back to r at angle=0 I must decrease.. Both can't be true so it's impossible..

Such a shape would be a spiral - spirals cannot be closed surfaces.. Did that make sense?83.100.158.211 (talk) 15:00, 25 February 2008 (UTC)[reply]

Even a spiral loses: take the midpoint of a small enough chord that it doesn't intersect the next layer (so as to avoid questions of containment). --Tardis (talk) 15:54, 25 February 2008 (UTC)[reply]

Didn't understand that - the spiral r=e^angle is single valued in r for all values of angle ?83.100.158.211 (talk) 15:59, 25 February 2008 (UTC)[reply]

The OP's request was for an open region

\Omega \in \mathbb {R} ^{n}\ni \forall x\in \Omega \,\not \!\exists (y,z)\subseteq \partial \Omega \;\;y\neq z\land \|y-x\|=\|z-x\|

. It's not enough for the "unique distances" clause to hold for one contained point. --Tardis (talk) 16:50, 25 February 2008 (UTC)[reply]

((For the spiral described above every point is at a unique distance, - am I wrong?))

You've confused me now - I understood the OP asked for a closed surface (hence not a spiral) - but said that excluding that 'closed surface' clause a spiral would be such a shape (if only in 2d) I see that a spiral fails for a single unique point in 3D ie a spiral surface has more than one point with a given radius. Was that what your original point was saying?83.100.158.211 (talk) 17:12, 25 February 2008 (UTC)[reply]

cf open region with Closed manifold - they asked for closed manifold eg 'closed surface'83.100.158.211 (talk) 17:42, 25 February 2008 (UTC)[reply]

I interpreted the question as taking place in

\mathbb {R} ^{3}

where the closed manifold (ie. closed surface) bounds an open region. --Tango (talk) 17:59, 25 February 2008 (UTC)[reply]

still a bit confused about the 'chord' explanation - the answer is a definate 'no' anyway -

another way to look at it is to note that the closed surface must be (at least) double valued (of r in angle) - to be closed (in 2D) {and gets worse in 3d ie infinite values with same r for (angle1,angle2) could go on about being able to draw a loop/ring on the surface of a 3D closed surface at any point }

- which of course means it cannot be single valued as requested. are we answering the same question??? at least we are getting same answer83.100.158.211 (talk) 18:19, 25 February 2008 (UTC)[reply]

What function are you saying is multi-valued,

r(\theta )

or

\theta (r)

? I assume the latter, since that's what we need, but your terminology is a little confusing (to me, at least). The OP (in the 2D case - I'm pretty sure he's actually asking about the 3D case, since you can't have a closed surface in

\mathbb {R} ^{2}

) requires that

\theta (r)

be multi-valued for some choice of origin within the interior of the region bounded by the surface. That's just a restatement of the question, though, it doesn't prove anything. --Tango (talk) 18:39, 25 February 2008 (UTC)[reply]

Yes I wrote "(of r in angle)" is should of course have read "doubled valued in angle for a given r" - my mistake that was very unclear.

\theta (r)

is multivalued for not true.. or single valued if the original construct was possible - I think I was showing that when

\theta (r)

=singlevalued the surface cannot be enclosed .

Clearly for a given radius r there must be only one value of theta (2D) or (combiantion) theta,mu (3D) that gives that radius ie if r=fn(theta) then the function theta=fn^-1(r) must have a single value of theta for that value of r eg for a ellipse there are 4 values of theta that give a specific r, so I would have said that (in the case of the ellipse) theta is multivalued.

and multivalued = not possible83.100.158.211 (talk) 19:51, 25 February 2008 (UTC)[reply]

My argument for it's impossibility was of this kind:

given that the rate of change of radius with angle is x

x cannot be zero (since this gives a 'flat' region - ie two points have same r)

so x is either positive or negative

x cannot change sign since this would require x=0 at some point.

So x is either always positive or negative.

BUT r=fn(x) = fn(x+2pi) (for a continuous surface) (a full rotation) (equation A)

equation A cannot be true since d fn(x)/dx is not zero and never changes sign.

Therefor assuming that the example function is single valued ie x=fn^-1(r) is single valued for r (PROBABLY DID STATE THIS THE OTHER WAY ROUND at some point sorry)

Then fn(x) <> fn(x+2pi) - it's not a closed ring.

This expands easily to 3D since for a closed 3D surface any plane section through it (through the 'centre') must be a closed ring.

So single valued in x=fn^-1(r) and 'closed ring' are mutually exclusive.

I've expanded a bit on the original explanation (including the differentials) - but it's the same un-proof. I guess I wrote single valued for fn(x) somewhere before when it should have been fn^-1(r) - that would account for any confusion you are getting.

(whether r=fn(angle) is multivalued (or not) (ie non-convex surfaces) doesn't affect this proof)83.100.158.211 (talk) 20:06, 25 February 2008 (UTC)[reply]

I think x=0 is allowed, but only instantaneously, and you still couldn't have a change of sign (it would have to be a point of inflexion), so your basic conclusion is correct. Dealing with the 3D case by taking the intersection with a plane is a good idea (although, pedantically, the intersection must be a union of closed rings, but that doesn't affect anything). --Tango (talk) 20:13, 25 February 2008 (UTC)[reply]

Yes, forgot about an inflexion -good point.83.100.158.211 (talk) 20:18, 25 February 2008 (UTC)[reply]

Geometry Help: Mysterious Equation

So, I'm here studying for a test and one of the formula's in my notes makes no sense to me. I have completly forgotten how to use it and I don't have any examples. The formula is V=BiT or V=BLT. Either or.. does anyone know what the formula stands for.. —Preceding unsigned comment added by 80.148.24.98 (talk) 21:26, 25 February 2008 (UTC)[reply]

I expect we're going to need some context. What's the section it's in about? Is there any other mention of V, B, i/L and T near it? --Tango (talk) 21:39, 25 February 2008 (UTC)[reply]

No, nothing at all.. We're on a unit about volume and surface area of rectangular prisms and pyramids..... —Preceding unsigned comment added by 80.148.24.98 (talk) 21:42, 25 February 2008 (UTC)[reply]

I think you'll have to ask your teacher. I don't see any way we can identify one formula in isolation. The volume of a prism is Volume=Base area * Length, that gives you V=BL, but I don't know what the T could be. --Tango (talk) 22:38, 25 February 2008 (UTC)[reply]

The volume of a pyramid is base*height/3, so maybe that extra factor has something to do with the third letter. On the other hand, the volume of a prism is also length*width*height. So yeah, your teacher's probably the only person who knows what she meant. Or one of your classmates. Black Carrot (talk) 22:42, 25 February 2008 (UTC)[reply]

T = tallness? —Bkell (talk) 23:48, 25 February 2008 (UTC)[reply]

What I meant here, I guess, was "Volume = Breadth × Length × Tallness", though "Tallness" is a strange word to use for that. Maybe BLT? ;-) —Bkell (talk) 23:53, 25 February 2008 (UTC)[reply]

Perhaps it's just bad handwriting? V=bh (volume = base area x height). Imagine Reason (talk) 23:49, 25 February 2008 (UTC)[reply]