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Suppose that n is odd composite. Then, Fermat assures us that it ay be written for some m and d integers. Suppose . Then we may show that . Equivalently, . For what moduli does a theorem of this form hold, and how do we lift from a statement for a small modulus to a statement about a larger modulus without a quadratic increase in the number of cases to be retained? (I.e. if we lift to the modulus 16, then depending on the residue of n we find that one of m and d is constrained to one value (mod 4) and the other is constrained to two values (mod 8) (that are not congruent (mod 4). If we then lift (mod 3) then we get two or four cases (depending on whether either of m or d can be congruent to zero (mod 3)), and using the Chinese remainder theorem to glue these cases to the cases derived from n (mod 16), we end up with four or eight cases -- some from residue classes (mod 12) and some from residue classes (mod 24).)
So, how do we encode the retained cases without creating an exponential explosion in the encoding of the retained cases and retaining the ability to perform additinoal lifting and additional applications of the CRT. -- Fuzzyeric 03:33, 12 September 2006 (UTC)
Require a method to calculate confidence intervals for weighted sums. The sum is of the form SUM wi xi and SUM wi = 1.
Thank You
Gert Engelbrecht Pretoria South Africa
need to know how many square centimeters are in a piece of tisse that measures 4cm X16cm --need the answer - not the formula. I am not a student needing assistance with homework ---
Can you draw a graph to show octahedral rotational symmetry
Many thanks!-- 82.28.195.12 20:17, 12 September 2006 (UTC)Jason
Hello. This is one of those problems which hits you hard when realise you don't know how to do it.
-So- much mathematics is based on the result that d/dx(e^x) is e^x, or written in a different way.. the integral of 1/x is ln(x). The question is, how do we prove this?
We can go back to first principles easily enough, and say that the derivative of a^x, as h tends to zero, is:
(a^(x+h) - a^x)/h
Factorise out a^x, and get:
a^x(a^h - 1)/h
Now, we know from basic calculus that differentiating this should go ln(a).a^x, so we're looking to show that the below limit is true:
(a^h - 1) / h = ln(a), as h tends to zero.
This doesn't look tricky, does it? But remember that we're trying to prove a result fundamental to calculus, so what we can use is limited (no pun), we can't use l'hopital's rule (which would give the right answer), as it relies on differentiating an exponential - our thing to be proven.
So, basically, I would really, really appreciate it if somebody could attempt to prove my limit is true, or comment that they could not (so I know the ability levels it's going to take).
Thank you, and remember: No circular reasoning! No proving that e^x differentiates to itself by assuming it in the first place!
Michael.blackburn 20:56, 12 September 2006 (UTC)
I found an easy solution:
Using definition
so iff
This would seem to require L'Hôpital's rule, but what do you all think?
M.manary
21:11, 12 September 2006 (UTC)
I also believe that TeaDrinker's solution will rely on a formula already using d/dx e^x, as Taylor series can ONLY be derived from that notion (try it yourself and see), so that solution is no-go. M.manary 21:15, 12 September 2006 (UTC)
M.manary, I think using L'Hôpital's rule is okay here, as long as we don't use it with exponentials. It can be proven from fairly basic principles. Michael.blackburn 21:17, 12 September 2006 (UTC)
I just found a proof that you can read at:
http://www.ltcconline.net/greenl/courses/106/ApproxOther/lhop.htm
so: Q.E.D.
M.manary
21:19, 12 September 2006 (UTC)
Okay, thank you (Very) much for the awnsers to the previous question. This one is probably a lot more obvious, yet through searching Wikipedia and the internet I've still failed to find a solution. My mathematics teachers were also uncertain.
Basically, the question is "Why radians?"
From thinking about it, it's obvious that the Calculus can only work with a single angle measure, and obviously it's the radian. But the best explaination I've recieved is that solving Taylor series of Trig. functions works in radians only... but those Taylor series surely required Radians in the first place.
Can -anybody- offer a sensible 'proof' or reasoning that you have to use 2pi 'units' per circle for the Calculus to work, without starting in radians in the first place?
Thank you again!
Oh, so.. if you used degrees for example, and arclength is.. kθ, so sin(θ) ~ kθ in degrees (where k isn't one), and so the results go icky. Is that it? Michael.blackburn 21:44, 12 September 2006 (UTC)
So the definition of a radian is that in a circle of radius r, a central angle of one radian will define an arclength of the circle that is exactly r long. If the radius of a circle is r, then there are (by circumference) 2π radii in the arclength of the circle. SO when we say an angle = 2 radians, it inscribes an arclength of 2r out of its circle.
The numbers actually don't work out that well, as you may notice, because we always have to say an angle is π/3 or 2π/7 radians, which really aren't that great numbers... M.manary 23:12, 12 September 2006 (UTC)
We are learning determinants and inverses in matrices in math. My teacher said that if you have a determinant of "0' then there is no inverse. I went on to say that in some remote field of mathematics, there is probably a way to fond the inverse of matrix with d=0. She said, "Maybe, but I doubt it. Why don't you look that up and tell us tomorrow." I took look at abstract algebra, and it was kind of confusing. I did some relevant Google searches, but to no avail. My question is: am I right? Is there, in some field of mathematics, a way to find the inverse of a matrix that has a determinant of 0? schyler 23:37, 12 September 2006 (UTC)
Please tell me if I've got this right, or is there a better formula I could use?
I have ten spot heights scattered irregularly over a retangular map. I intend to estimate the height of every point on this map so that I can create a contour map.
I am going to estimate the height of each point by the weighted average of all the spot heights. The weight I am going to use is the inverse of the distance squared.
In fact I am going to use a further refinement - instead of just using the square, I am going to estimate the exact power to use by disregarding each of the ten spot heights in turn, and finding what the power is that best predicts the disregarded spot height from the nine known spot heights, and then calculate the arithmetic average of these ten powers.
So my weights will be 1/d^n where d is distance and n is the power. My questions are-
a) is there any better (ie more accurate estimator) formula to use for the weights than 1/d^n?
b) should I use an arithmetic average of the ten powers, or some other kind of average?
c) is there any better approach I could use, even though the scheme described above is easy to program? Thanks 62.253.52.8 23:55, 12 September 2006 (UTC)
Thanks. Although Delaunay triangulation is attractive, I think it would be far too difficult to program - it would require several days I expect.
I did wonder if there would be any advantage in using weights based on formulas such as:
where x is a constant. I think the magnitude of x would determine to what extent the estimated local height was based on the average of all heights, rather than those of the nearer spot heights. Are there any better formulas I could use?
I don't actually know how I would estimate x and n by regression or otherwise - could any help me please? Thanks.
< September 11 | Mathematics desk archive | September 13 > |
---|
| ||||||||
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions at one of the pages linked to above. | ||||||||
|
Suppose that n is odd composite. Then, Fermat assures us that it ay be written for some m and d integers. Suppose . Then we may show that . Equivalently, . For what moduli does a theorem of this form hold, and how do we lift from a statement for a small modulus to a statement about a larger modulus without a quadratic increase in the number of cases to be retained? (I.e. if we lift to the modulus 16, then depending on the residue of n we find that one of m and d is constrained to one value (mod 4) and the other is constrained to two values (mod 8) (that are not congruent (mod 4). If we then lift (mod 3) then we get two or four cases (depending on whether either of m or d can be congruent to zero (mod 3)), and using the Chinese remainder theorem to glue these cases to the cases derived from n (mod 16), we end up with four or eight cases -- some from residue classes (mod 12) and some from residue classes (mod 24).)
So, how do we encode the retained cases without creating an exponential explosion in the encoding of the retained cases and retaining the ability to perform additinoal lifting and additional applications of the CRT. -- Fuzzyeric 03:33, 12 September 2006 (UTC)
Require a method to calculate confidence intervals for weighted sums. The sum is of the form SUM wi xi and SUM wi = 1.
Thank You
Gert Engelbrecht Pretoria South Africa
need to know how many square centimeters are in a piece of tisse that measures 4cm X16cm --need the answer - not the formula. I am not a student needing assistance with homework ---
Can you draw a graph to show octahedral rotational symmetry
Many thanks!-- 82.28.195.12 20:17, 12 September 2006 (UTC)Jason
Hello. This is one of those problems which hits you hard when realise you don't know how to do it.
-So- much mathematics is based on the result that d/dx(e^x) is e^x, or written in a different way.. the integral of 1/x is ln(x). The question is, how do we prove this?
We can go back to first principles easily enough, and say that the derivative of a^x, as h tends to zero, is:
(a^(x+h) - a^x)/h
Factorise out a^x, and get:
a^x(a^h - 1)/h
Now, we know from basic calculus that differentiating this should go ln(a).a^x, so we're looking to show that the below limit is true:
(a^h - 1) / h = ln(a), as h tends to zero.
This doesn't look tricky, does it? But remember that we're trying to prove a result fundamental to calculus, so what we can use is limited (no pun), we can't use l'hopital's rule (which would give the right answer), as it relies on differentiating an exponential - our thing to be proven.
So, basically, I would really, really appreciate it if somebody could attempt to prove my limit is true, or comment that they could not (so I know the ability levels it's going to take).
Thank you, and remember: No circular reasoning! No proving that e^x differentiates to itself by assuming it in the first place!
Michael.blackburn 20:56, 12 September 2006 (UTC)
I found an easy solution:
Using definition
so iff
This would seem to require L'Hôpital's rule, but what do you all think?
M.manary
21:11, 12 September 2006 (UTC)
I also believe that TeaDrinker's solution will rely on a formula already using d/dx e^x, as Taylor series can ONLY be derived from that notion (try it yourself and see), so that solution is no-go. M.manary 21:15, 12 September 2006 (UTC)
M.manary, I think using L'Hôpital's rule is okay here, as long as we don't use it with exponentials. It can be proven from fairly basic principles. Michael.blackburn 21:17, 12 September 2006 (UTC)
I just found a proof that you can read at:
http://www.ltcconline.net/greenl/courses/106/ApproxOther/lhop.htm
so: Q.E.D.
M.manary
21:19, 12 September 2006 (UTC)
Okay, thank you (Very) much for the awnsers to the previous question. This one is probably a lot more obvious, yet through searching Wikipedia and the internet I've still failed to find a solution. My mathematics teachers were also uncertain.
Basically, the question is "Why radians?"
From thinking about it, it's obvious that the Calculus can only work with a single angle measure, and obviously it's the radian. But the best explaination I've recieved is that solving Taylor series of Trig. functions works in radians only... but those Taylor series surely required Radians in the first place.
Can -anybody- offer a sensible 'proof' or reasoning that you have to use 2pi 'units' per circle for the Calculus to work, without starting in radians in the first place?
Thank you again!
Oh, so.. if you used degrees for example, and arclength is.. kθ, so sin(θ) ~ kθ in degrees (where k isn't one), and so the results go icky. Is that it? Michael.blackburn 21:44, 12 September 2006 (UTC)
So the definition of a radian is that in a circle of radius r, a central angle of one radian will define an arclength of the circle that is exactly r long. If the radius of a circle is r, then there are (by circumference) 2π radii in the arclength of the circle. SO when we say an angle = 2 radians, it inscribes an arclength of 2r out of its circle.
The numbers actually don't work out that well, as you may notice, because we always have to say an angle is π/3 or 2π/7 radians, which really aren't that great numbers... M.manary 23:12, 12 September 2006 (UTC)
We are learning determinants and inverses in matrices in math. My teacher said that if you have a determinant of "0' then there is no inverse. I went on to say that in some remote field of mathematics, there is probably a way to fond the inverse of matrix with d=0. She said, "Maybe, but I doubt it. Why don't you look that up and tell us tomorrow." I took look at abstract algebra, and it was kind of confusing. I did some relevant Google searches, but to no avail. My question is: am I right? Is there, in some field of mathematics, a way to find the inverse of a matrix that has a determinant of 0? schyler 23:37, 12 September 2006 (UTC)
Please tell me if I've got this right, or is there a better formula I could use?
I have ten spot heights scattered irregularly over a retangular map. I intend to estimate the height of every point on this map so that I can create a contour map.
I am going to estimate the height of each point by the weighted average of all the spot heights. The weight I am going to use is the inverse of the distance squared.
In fact I am going to use a further refinement - instead of just using the square, I am going to estimate the exact power to use by disregarding each of the ten spot heights in turn, and finding what the power is that best predicts the disregarded spot height from the nine known spot heights, and then calculate the arithmetic average of these ten powers.
So my weights will be 1/d^n where d is distance and n is the power. My questions are-
a) is there any better (ie more accurate estimator) formula to use for the weights than 1/d^n?
b) should I use an arithmetic average of the ten powers, or some other kind of average?
c) is there any better approach I could use, even though the scheme described above is easy to program? Thanks 62.253.52.8 23:55, 12 September 2006 (UTC)
Thanks. Although Delaunay triangulation is attractive, I think it would be far too difficult to program - it would require several days I expect.
I did wonder if there would be any advantage in using weights based on formulas such as:
where x is a constant. I think the magnitude of x would determine to what extent the estimated local height was based on the average of all heights, rather than those of the nearer spot heights. Are there any better formulas I could use?
I don't actually know how I would estimate x and n by regression or otherwise - could any help me please? Thanks.