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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)
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)
Basically, our teacher is trying to convince us that which is the set of all infinitely differentiable functions with compact support on a given set , is dense in . 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 , there exists a sequence such that in . My question is how does this theorem prove that is dense in ? 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 ? A Real Kaiser ( talk) 04:52, 25 February 2008 (UTC)
Duh, I seem to have forgotten that little fact. Thanks! A Real Kaiser ( talk) 05:52, 25 February 2008 (UTC)
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 ) 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)
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)
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.
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).
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)
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)
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..
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)
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)
Mathematics desk | ||
---|---|---|
< February 24 | << Jan | February | Mar >> | February 26 > |
Welcome to the Wikipedia Mathematics Reference Desk Archives |
---|
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages. |
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)
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)
Basically, our teacher is trying to convince us that which is the set of all infinitely differentiable functions with compact support on a given set , is dense in . 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 , there exists a sequence such that in . My question is how does this theorem prove that is dense in ? 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 ? A Real Kaiser ( talk) 04:52, 25 February 2008 (UTC)
Duh, I seem to have forgotten that little fact. Thanks! A Real Kaiser ( talk) 05:52, 25 February 2008 (UTC)
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 ) 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)
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)
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.
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).
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)
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)
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..
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)
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)