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Suppose I have a ring extension of Z, say Zw].
mike40033 ( talk) 01:05, 13 May 2008 (UTC)
Okay, so in my numerical analysis class, studying for the finals, I came across the Gerschgorin criterion. One of the exercises is the proof of the criterion which is split up into three parts. I have proved the first two parts but the third one (which I learned later is called the Taussky's theorem), I can't get. Here is everything I have done:
Let be an arbitrary singular complex matrix.
Then there exists a nonzero such that .
Choose such that .
This maximal element is of course, nonzero.
By considering the th row of , I have already proven that
.
Next, taking a dxd matrix B and letting be an arbitrary eigenvalue of B.
Substituting, , I have also shown that is in a Gerschgorin disc where
And since, was an arbitrary eigenvalue of B, the entire spectrum of B is contained the union of all the Gerschgorin discs.
Now here is the problem which I can't get. Let us suppose that the matrix B is irreducible and the previous inequality holds as an equality, i.e.
for some between 1 and d. I need to show that this equality implies that
for all k between 1 and d.
This would imply that if an eigenvalue is on the boundary of one Gerschgorin disc, then that eigenvalue will be on the boundary of all Gerschgorin discs. Any ideas on how to prove it? I have looked in several books but everyone just states it and calls it a famous result but no one proves it.
A Real Kaiser (
talk) 05:51, 13 May 2008 (UTC)
On a completely different note, I am trying to prove some properties for this function.
Let be defined as
where [x] is the truncation function (i.e. [x]=the integer part of x). Notice that at each unit fraction, the function has a jump discontinuity and in between, the function is a constant. We have already had some discussion about this function but now I am trying to prove that this function is Riemann integrable. Here are some facts that I proved:
is a non-decreasing function
Proof:
Let
.
is continuous at from the right.
Proof:Let be given and then let .
(for sufficiently small)
is Riemann integrable. This will be shown by proving that for every , we can find a partition p of [0,1] such that the upper sum U(f,p) minus the lower sum L(f,p) using p, is smaller than .
Proof: For a given , let , and let N=the largest integer less than or equal to 1/. This gives us that 1/N will be the next unit fraction. Hence x=1/N will be the next point (on the right) after where will have a jump discontinuity. Let and let the partition be
This is a perfectly valid partition, because the division by 2N ensures that none of the points go out of order. The intervals will not overlap. The point of the first fact was to make the Riemann evaluation easier. Since the function is nondecreasing, in order evaluate the area of a Riemann rectangle, for the upper sum, we simply take the value of the function on the right endpoint, and for the lower sum, we simply take the value of the function at the left endpoint. Both facts also show (for my teacher) that the function does NOT become unbounded as x goes to zero. In fact, if we extend the function to the interval [-1,1], the limit from the left as x goes to zero also exists so I can show that the function is actually continuous at x=0. But, that is just something extra.
Now when the difference between the upper sum and the lower sum is taken, the rectangles which fall on the flat region, with the base being of the form to cancel out because the upper sum is the same as the lower sum. The same thing happens with the last rectangle. The only rectangles that remain are the columns of width which have the jump discontinuity in the middle.
So,
Since height of each rectangle is less than or equal to one, and the number of terms with is precisely N-1, we have that
.
Therefore, the function is Riemann integrable on the interval [0,1]. The reason, I put all of this up is that I just want to run it by a couple of people here to make sure that I have not made a stupid mistake and to see if this is "rigorous" enough. I will be presenting this to him (and maybe we can finally settle this argument once and for all about this function being Riemann integrable) and he is very particular about rigor. So, do you guys think that am I missing out some particular detail which I should include or is this good enough for a rigorous Mathematical argument? Any comments and suggestion will be welcome about the proof of all three facts. The third one is the most important one, of course. In addition, in the proof of the second fact, in line three, it is obvious that the line is true for a sufficiently small epsilon but how can I write that better? I don't really like it the way I have it but I need it there. Thanks everyone! A Real Kaiser ( talk) 09:33, 13 May 2008 (UTC)
Algebraist, you are absolutely right. I just said the opposite of what is true. So here is the revised proof.
Proof:Let be given and then let .
because we have that
Here is another thing I thought of. I will try to prove that for all .
Proof:
First, let {x}=x-[x] be the fractional part of a real number x.
Now we have that for all and such that and .
Now let us consider all such that .
We now have that .
for all
and using the fact that for all
and then using the squeeze theorem, I can show that the limit of f(x) from the right as x goes to zero is indeed zero and this function is most definitely not unbounded in [0,1]. I think that after these four facts combined, there is no question that f(x) is bounded and Riemann integrable in [0,1].
In fact, we can also find the exact value of the integral as follows:
Since, the function as a jump discontinuity at every unit fraction, with 1/2 being the first one from the left as x goes to zero, and in between two consecutive unit fractions, the function is constant
(i.e. ), we can just find the area underneath the curve by drawing rectangles
with length with height and then add up the areas of all such rectangles.
The first sum is just from the Riemann Zeta function (the values are known) and the second sum is a telescoping series. We can split up the series because it is absolutely convergent. So, what do you guys think? Should I change something or is this enough? Any more stupid mistakes in what I wrote or my reasoning? Come on guys, don't be shy. I just need your input and some constructive criticism. A Real Kaiser ( talk) 23:23, 13 May 2008 (UTC)
"Let ABCDEFGH be an eight digit number where A,B,C,D,E,F,G,H are the digits 1,2,3,4,5,6,7,8 in some order. For example, A=4,B=1,C=3,D=2,E=5,F=6,G=7,H=8 gives 41325678. FInd other eight digit numbers in which the three digit number ABC is divisible by 7, the three-digit number BCD is divisible by 6, CDE is divisible by 5, DEF is divisible by 4, EFG is divisible by 3 and FGH is divisible by 2."
I know that E is 5 but don't konw the others. Can someone pls give me some hints as the maths exam is round the corner(to be exact, tomorrow)? tks. —Preceding unsigned comment added by Invisiblebug590 ( talk • contribs) 10:26, 13 May 2008 (UTC)
Unless I've missed the point, 73485612 is invalid as the 3-digit number 734 isn't divisible by 7.… 81.159.11.144 ( talk) 11:50, 14 May 2008 (UTC)
Please provide symbol/math tags in TeX/HTML for logic gates like NOT, AND, OR, XOR,... Currently, users are uploading image files to depict circuit diagrams. A XML version would be immediately useful beyond Wiki. Anwar ( talk) 14:35, 13 May 2008 (UTC)
Hi, I am an academic student currently researching on China's demographic. I am having a difficult time locating the most current information concerning "the homeless percentage in China overall and by region or province and cities vs. rural." I would really appreciate it if anyone can provide me with the most up to date data or point me to the right direction on where to search.
Thank you for your time and help. Your prompt reply will be greatly appreciated.
Sincerely, username: darjeelinguru
Using the Mathcad statistical functions how would I convert percentile to a z-score and vice versa? 71.100.14.205 ( talk) 16:32, 13 May 2008 (UTC)
When was pi first used in England how was it measured? —Preceding unsigned comment added by 82.71.27.169 ( talk) 18:26, 13 May 2008 (UTC)
An episode of Numb3rs involved a (false) proof for the Riemann hypothesis that was used as a ransom with the intent of breaking encryption. Did they get it mixed up with the P = NP problem, or are they that closely related? — DanielLC 23:33, 13 May 2008 (UTC)
Mathematics desk | ||
---|---|---|
< May 12 | << Apr | May | Jun >> | May 14 > |
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. |
Suppose I have a ring extension of Z, say Zw].
mike40033 ( talk) 01:05, 13 May 2008 (UTC)
Okay, so in my numerical analysis class, studying for the finals, I came across the Gerschgorin criterion. One of the exercises is the proof of the criterion which is split up into three parts. I have proved the first two parts but the third one (which I learned later is called the Taussky's theorem), I can't get. Here is everything I have done:
Let be an arbitrary singular complex matrix.
Then there exists a nonzero such that .
Choose such that .
This maximal element is of course, nonzero.
By considering the th row of , I have already proven that
.
Next, taking a dxd matrix B and letting be an arbitrary eigenvalue of B.
Substituting, , I have also shown that is in a Gerschgorin disc where
And since, was an arbitrary eigenvalue of B, the entire spectrum of B is contained the union of all the Gerschgorin discs.
Now here is the problem which I can't get. Let us suppose that the matrix B is irreducible and the previous inequality holds as an equality, i.e.
for some between 1 and d. I need to show that this equality implies that
for all k between 1 and d.
This would imply that if an eigenvalue is on the boundary of one Gerschgorin disc, then that eigenvalue will be on the boundary of all Gerschgorin discs. Any ideas on how to prove it? I have looked in several books but everyone just states it and calls it a famous result but no one proves it.
A Real Kaiser (
talk) 05:51, 13 May 2008 (UTC)
On a completely different note, I am trying to prove some properties for this function.
Let be defined as
where [x] is the truncation function (i.e. [x]=the integer part of x). Notice that at each unit fraction, the function has a jump discontinuity and in between, the function is a constant. We have already had some discussion about this function but now I am trying to prove that this function is Riemann integrable. Here are some facts that I proved:
is a non-decreasing function
Proof:
Let
.
is continuous at from the right.
Proof:Let be given and then let .
(for sufficiently small)
is Riemann integrable. This will be shown by proving that for every , we can find a partition p of [0,1] such that the upper sum U(f,p) minus the lower sum L(f,p) using p, is smaller than .
Proof: For a given , let , and let N=the largest integer less than or equal to 1/. This gives us that 1/N will be the next unit fraction. Hence x=1/N will be the next point (on the right) after where will have a jump discontinuity. Let and let the partition be
This is a perfectly valid partition, because the division by 2N ensures that none of the points go out of order. The intervals will not overlap. The point of the first fact was to make the Riemann evaluation easier. Since the function is nondecreasing, in order evaluate the area of a Riemann rectangle, for the upper sum, we simply take the value of the function on the right endpoint, and for the lower sum, we simply take the value of the function at the left endpoint. Both facts also show (for my teacher) that the function does NOT become unbounded as x goes to zero. In fact, if we extend the function to the interval [-1,1], the limit from the left as x goes to zero also exists so I can show that the function is actually continuous at x=0. But, that is just something extra.
Now when the difference between the upper sum and the lower sum is taken, the rectangles which fall on the flat region, with the base being of the form to cancel out because the upper sum is the same as the lower sum. The same thing happens with the last rectangle. The only rectangles that remain are the columns of width which have the jump discontinuity in the middle.
So,
Since height of each rectangle is less than or equal to one, and the number of terms with is precisely N-1, we have that
.
Therefore, the function is Riemann integrable on the interval [0,1]. The reason, I put all of this up is that I just want to run it by a couple of people here to make sure that I have not made a stupid mistake and to see if this is "rigorous" enough. I will be presenting this to him (and maybe we can finally settle this argument once and for all about this function being Riemann integrable) and he is very particular about rigor. So, do you guys think that am I missing out some particular detail which I should include or is this good enough for a rigorous Mathematical argument? Any comments and suggestion will be welcome about the proof of all three facts. The third one is the most important one, of course. In addition, in the proof of the second fact, in line three, it is obvious that the line is true for a sufficiently small epsilon but how can I write that better? I don't really like it the way I have it but I need it there. Thanks everyone! A Real Kaiser ( talk) 09:33, 13 May 2008 (UTC)
Algebraist, you are absolutely right. I just said the opposite of what is true. So here is the revised proof.
Proof:Let be given and then let .
because we have that
Here is another thing I thought of. I will try to prove that for all .
Proof:
First, let {x}=x-[x] be the fractional part of a real number x.
Now we have that for all and such that and .
Now let us consider all such that .
We now have that .
for all
and using the fact that for all
and then using the squeeze theorem, I can show that the limit of f(x) from the right as x goes to zero is indeed zero and this function is most definitely not unbounded in [0,1]. I think that after these four facts combined, there is no question that f(x) is bounded and Riemann integrable in [0,1].
In fact, we can also find the exact value of the integral as follows:
Since, the function as a jump discontinuity at every unit fraction, with 1/2 being the first one from the left as x goes to zero, and in between two consecutive unit fractions, the function is constant
(i.e. ), we can just find the area underneath the curve by drawing rectangles
with length with height and then add up the areas of all such rectangles.
The first sum is just from the Riemann Zeta function (the values are known) and the second sum is a telescoping series. We can split up the series because it is absolutely convergent. So, what do you guys think? Should I change something or is this enough? Any more stupid mistakes in what I wrote or my reasoning? Come on guys, don't be shy. I just need your input and some constructive criticism. A Real Kaiser ( talk) 23:23, 13 May 2008 (UTC)
"Let ABCDEFGH be an eight digit number where A,B,C,D,E,F,G,H are the digits 1,2,3,4,5,6,7,8 in some order. For example, A=4,B=1,C=3,D=2,E=5,F=6,G=7,H=8 gives 41325678. FInd other eight digit numbers in which the three digit number ABC is divisible by 7, the three-digit number BCD is divisible by 6, CDE is divisible by 5, DEF is divisible by 4, EFG is divisible by 3 and FGH is divisible by 2."
I know that E is 5 but don't konw the others. Can someone pls give me some hints as the maths exam is round the corner(to be exact, tomorrow)? tks. —Preceding unsigned comment added by Invisiblebug590 ( talk • contribs) 10:26, 13 May 2008 (UTC)
Unless I've missed the point, 73485612 is invalid as the 3-digit number 734 isn't divisible by 7.… 81.159.11.144 ( talk) 11:50, 14 May 2008 (UTC)
Please provide symbol/math tags in TeX/HTML for logic gates like NOT, AND, OR, XOR,... Currently, users are uploading image files to depict circuit diagrams. A XML version would be immediately useful beyond Wiki. Anwar ( talk) 14:35, 13 May 2008 (UTC)
Hi, I am an academic student currently researching on China's demographic. I am having a difficult time locating the most current information concerning "the homeless percentage in China overall and by region or province and cities vs. rural." I would really appreciate it if anyone can provide me with the most up to date data or point me to the right direction on where to search.
Thank you for your time and help. Your prompt reply will be greatly appreciated.
Sincerely, username: darjeelinguru
Using the Mathcad statistical functions how would I convert percentile to a z-score and vice versa? 71.100.14.205 ( talk) 16:32, 13 May 2008 (UTC)
When was pi first used in England how was it measured? —Preceding unsigned comment added by 82.71.27.169 ( talk) 18:26, 13 May 2008 (UTC)
An episode of Numb3rs involved a (false) proof for the Riemann hypothesis that was used as a ransom with the intent of breaking encryption. Did they get it mixed up with the P = NP problem, or are they that closely related? — DanielLC 23:33, 13 May 2008 (UTC)