f*****k
发帖数: 353 | 1 1.Given a diagonalizable square matrix A and a noise matrix E, find
conditions on A and E such that for each eigenvalue L of A, there is an
eigenvalue M of (A+E) such that abs(L-M) < K for a given K. (this is
obviously an open ended question, try to tighten the bounds to the best of
your ability)
开始假设E是diagonlizable的，这样，可以得到一个界，接下来怎么搞？
2.求解：-4 t x'' - x' + 4 t^3 x =0
这个开始找不到代换，直接用幂级数展开，好像只能得到零解 | S*********g
发帖数: 5298 | 2 this is from walleye, right?
For the 2nd one, introduce y=a*t^2.
find a such that the x' term cancels out.
【在 f*****k 的大作中提到】
: 1.Given a diagonalizable square matrix A and a noise matrix E, find
: conditions on A and E such that for each eigenvalue L of A, there is an
: eigenvalue M of (A+E) such that abs(L-M) < K for a given K. (this is
: obviously an open ended question, try to tighten the bounds to the best of
: your ability)
: 开始假设E是diagonlizable的，这样，可以得到一个界，接下来怎么搞？
: 2.求解：-4 t x'' - x' + 4 t^3 x =0
: 这个开始找不到代换，直接用幂级数展开，好像只能得到零解
| f*****k
发帖数: 353 | 3 got it，真是傻了，光想到替换x了,thanks
是walleye，给您发信了，请查收
【在 S*********g 的大作中提到】
: this is from walleye, right?
: For the 2nd one, introduce y=a*t^2.
: find a such that the x' term cancels out.
| p*****k
发帖数: 318 | 4
seems to me it's related to Bessel functions. see e.g.,
http://mathworld.wolfram.com/BesselDifferentialEquation.html
Eq.(6)
with alpha=3/8, beta=i/2, gamma=2, and n=3/16
so the general solution is:
x(t)=t^(3/8)*[C1*I_{3/16}(t^2/2) + C2*I_{-3/16}(t^2/2)],
where I is the modified Bessel function
【在 f*****k 的大作中提到】
: got it，真是傻了，光想到替换x了,thanks
: 是walleye，给您发信了，请查收
| f*****k
发帖数: 353 | 5 Thanks a lot! You two BIG COW do know everything!
【在 p*****k 的大作中提到】
:
: seems to me it's related to Bessel functions. see e.g.,
: http://mathworld.wolfram.com/BesselDifferentialEquation.html
: Eq.(6)
: with alpha=3/8, beta=i/2, gamma=2, and n=3/16
: so the general solution is:
: x(t)=t^(3/8)*[C1*I_{3/16}(t^2/2) + C2*I_{-3/16}(t^2/2)],
: where I is the modified Bessel function
| c**********e
发帖数: 2007 | 6 这个 a为什么cancel不掉？令y=a*t^2得到
x'' + 5/8y x' - x/(4a*a)=0.
【在 S*********g 的大作中提到】
: this is from walleye, right?
: For the 2nd one, introduce y=a*t^2.
: find a such that the x' term cancels out.
| x*****i
发帖数: 287 | | L**********u
发帖数: 194 | 8 这个好像也不对。
用y作为新的变量，将x看作是y的函数。
呵呵。
这两个题目都没有多少意思，
很明显是一个做计算pde的人出的。
用自己的专业考非专业的人士。
【在 x*****i 的大作中提到】
: 同问 怎么用 y=a*t^2. Thanks!
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