Couple of questions for you Math guys. Please Help!
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From: Hacienda Heights/Irvine
yeah...I talked to my TA about it, and he told me that I couldn't prove it that way. He said probably no on is going to get it. I am having trouble following your explanation on #2. I kind of get how you get your answer, but I don't know how to prove it with work. Thanks again for your help
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From: Yorba Linda, CA
Originally posted by frycow
The series will converge, because the upper bound of cos n/n from 1 to infinity is 1 and the lower bound is -1.
The upper limit (n->infinity) is 0.
The lower limit (n->1) is 0 as well.
That should get you started.
The series will converge, because the upper bound of cos n/n from 1 to infinity is 1 and the lower bound is -1.
The upper limit (n->infinity) is 0.
The lower limit (n->1) is 0 as well.
That should get you started.
I haven't worked with infinite series for quite a while. I'll have to ponder this one a bit.
Originally posted by frycow
To prove that the series is not absolutely convergent, you must prove that the sum of |cos n/n| (absolute value) is divergent.
Hint: A series with a sum that is always greater than or equal to than another, divergent series, is divergent.
Hint: The series 1/n from 1 to inifinity is divergent.......
That should get you started.
To prove that the series is not absolutely convergent, you must prove that the sum of |cos n/n| (absolute value) is divergent.
Hint: A series with a sum that is always greater than or equal to than another, divergent series, is divergent.
Hint: The series 1/n from 1 to inifinity is divergent.......
That should get you started.
Originally posted by frycow
Since A^2 = A, you know that A*A = A. Therefore, A^-1 = the identity matrix. Since A has in inverse, A is nonsingular. A = I^-1. Therefore, A = I.
Since A^2 = A, you know that A*A = A. Therefore, A^-1 = the identity matrix. Since A has in inverse, A is nonsingular. A = I^-1. Therefore, A = I.
As an example:
1 0
0 0
Originally posted by frycow
When finding eigenvalues, you want to find lambda such that A*x = lambda*x.
That can be rewritten as (A - lambda * the identity matrix of A)*x = theta, where x does not equal theta.
Skipping ahead, the only possible eigenvalue of A is 1. I'll let you prove it.
When finding eigenvalues, you want to find lambda such that A*x = lambda*x.
That can be rewritten as (A - lambda * the identity matrix of A)*x = theta, where x does not equal theta.
Skipping ahead, the only possible eigenvalue of A is 1. I'll let you prove it.
A*x = kx, where k is an eigenvalue of A.
(A*A)*x = A*(A*x) = A*kx = (k*k)x, where k is still an eigenvalue of A.
But, A*A = A, so kx = (k*k)x for all eigenvectors x. Solve this equation; it has solutions other than k = 1.
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From: Yorba Linda, CA
Originally posted by CyrusA
what math is this? i could possibly do a) and b) but i wouldnt know how to set up the diff eq for c
what math is this? i could possibly do a) and b) but i wouldnt know how to set up the diff eq for c
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From: Yorba Linda, CA
Originally posted by EvoVII
PM magician if you have any other problems, if he's online he'll help, I think he's a math teacher or something
PM magician if you have any other problems, if he's online he'll help, I think he's a math teacher or something
I teach math at CSUF and may be teaching risk management at UCI this Fall.
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From: Yorba Linda, CA
Originally posted by CyrusA
oh. i havent taken lin alg, just used eigenvalues/vectors in diff eq
oh. i havent taken lin alg, just used eigenvalues/vectors in diff eq
(I've taught a course in linear algebra and differential equations several times.)
