o**a
发帖数: 76 | 1 the roots of cos(sqrt(z)) is (n*pi+pi/2)^2
and \sum 1/(n*pi+pi/2)^2 converges
so I set
cos(sqrt(z)) = g(z) * \prod (1-z/(n*pi+pi/2)^2)
and take logrithmic derivative on both side to solve for g(z)
but I finally got g(z) = 1/cos(sqrt(z))
What's the problem here? What's the genus of cos(sqrt(z)) indeed? |
w******o
发帖数: 442 | 2 the roots of cos(sqrt(z)) is (n*pi+pi/2)^2
so cos(sqrt(z)) = prod (1-z/(n*pi+pi/2)^2)
and cos(sqrt(z))^m = prod (1-z/(n*pi+pi/2)^2) m is positive integer.
so g(z) = cos(sqrt(z))^(1-m)
【在 o**a 的大作中提到】
: the roots of cos(sqrt(z)) is (n*pi+pi/2)^2
: and \sum 1/(n*pi+pi/2)^2 converges
: so I set
: cos(sqrt(z)) = g(z) * \prod (1-z/(n*pi+pi/2)^2)
: and take logrithmic derivative on both side to solve for g(z)
: but I finally got g(z) = 1/cos(sqrt(z))
: What's the problem here? What's the genus of cos(sqrt(z)) indeed?
|
o**a
发帖数: 76 | 3 I don't understand
【在 w******o 的大作中提到】
: the roots of cos(sqrt(z)) is (n*pi+pi/2)^2
: so cos(sqrt(z)) = prod (1-z/(n*pi+pi/2)^2)
: and cos(sqrt(z))^m = prod (1-z/(n*pi+pi/2)^2) m is positive integer.
: so g(z) = cos(sqrt(z))^(1-m)
|
w******o
发帖数: 442 | 4 first the roots of cos(sqrt(z)) is (n*pi+pi/2)^2 not (n*pi+pi/2)
second the roots of cos(sqrt(z))^m is (n*pi+pi/2)^2 too.
so prod(1-z/(n*pi+pi/2)^2) is the root expandation of cos(sqrt(z))^m
【在 o**a 的大作中提到】
: I don't understand
|
w******o
发帖数: 442 | 5 Please let me know the defination of 'genus'.
【在 o**a 的大作中提到】
: I don't understand
|
w******o
发帖数: 442 | 6 Sorry, I don't think I can fully understand your question based on my
mathematic background, and give you an answer.
【在 o**a 的大作中提到】
: I don't understand
|
o**a
发帖数: 76 | 7 我知道了，我的无穷乘积是从-\infty到\infty
实际上
cos(\sqrt(z))=\prod_{1}^{\infty} (1-z/(n*pi-pi/2)^2)
所以我得到了含根号的式子：
cos(\sqrt(z))=\sqrt[\prod_{-\infty}^{\infty} (1-z/(n*pi-pi/2)^2)]
【在 o**a 的大作中提到】
: the roots of cos(sqrt(z)) is (n*pi+pi/2)^2
: and \sum 1/(n*pi+pi/2)^2 converges
: so I set
: cos(sqrt(z)) = g(z) * \prod (1-z/(n*pi+pi/2)^2)
: and take logrithmic derivative on both side to solve for g(z)
: but I finally got g(z) = 1/cos(sqrt(z))
: What's the problem here? What's the genus of cos(sqrt(z)) indeed?
|
