d*n
发帖数: 137 | 1 Exp(a v D_v)Exp(-v*v) (1)
D_v is a differenctial operator with respect to v.
If use Taylor expansion, we need to this calculation
(v D_v)^nExp(-v*v) (2)
which is equal to
(V D_V)^nExp(-V). (3)
For small n, it's easy to calculate (3), the problem is how to get
the general expression of (3). | s***e
发帖数: 3 | 2 How about expand Exp(-v*v) also. Try this:
Exp(a v D_v) = Sum(a v D_v)^n
Exp(-v*v) = Sum(-v*v)^m
Then time the two series together. If n>m, the element is 0; else you can
connect the upper limit of summation on n with m.
Not sure what you are really expecting. But, that might lead you to some luck.
Good luck.
【在 d*n 的大作中提到】
: Exp(a v D_v)Exp(-v*v) (1)
: D_v is a differenctial operator with respect to v.
: If use Taylor expansion, we need to this calculation
: (v D_v)^nExp(-v*v) (2)
: which is equal to
: (V D_V)^nExp(-V). (3)
: For small n, it's easy to calculate (3), the problem is how to get
: the general expression of (3).
| y***u
发帖数: 25 | 3 可以精确解出吧，如下
可以验证：Exp{a*D_x}f(x)=f(x+a), where f(x) is an arbitrary differentiable
function. 其实D_x就是x的平移算子。
你的问题:
Exp{a*x*D_x}Exp{-x^2}
since x*D_x=D_{lnx}
set y=lnx
so
Exp{a*x*D_x}Exp{-x^2}=Exp{a*D_y}Exp{-Exp{2y}}
=Exp{-Exp{2(y+a)}}=Exp{-Exp{2a}*x^2}
如果你不放心可以这么办， 设原来的函数为
f(a)=Exp{a*x*D_x}Exp{-x^2}
so,
f'(a)=x*D_x f(a) and f(0)=Exp{-x^2}
把求得的结果带进来正好满足这个微分方程，所以肯定是对的。
【在 d*n 的大作中提到】
: Exp(a v D_v)Exp(-v*v) (1)
: D_v is a differenctial operator with respect to v.
: If use Taylor expansion, we need to this calculation
: (v D_v)^nExp(-v*v) (2)
: which is equal to
: (V D_V)^nExp(-V). (3)
: For small n, it's easy to calculate (3), the problem is how to get
: the general expression of (3).
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