what r u looking for?
it seems to me you are calculating

\Sig_k E^k

instead of

\Sig_n n^k

引用:

原帖由 andrew 于 2007-4-3 11:13 PM 发表
X = E^(a)+ E^(2a)+...+E^(na)
X E^(a) = E^(2a)+ E^(3a)+.... + E^( (n+1) a)
2 式減1式
X (E^a -1) = E^( (n+1) a) - E^a

X = (E^( (n+1) a) - E^a)/ (E^a -1)

E^a + E^(2a) +... E^(n a)
differentiate with respect to a twice....
1^2 E^a + 2^2 E^(2a) + ... n^2 E^(n a)
take limit a->0
1^2 + 2^2 + ... + n^2

since the first equation equals X, doing the same operation to X gives the answer to the last sum

E^(ka)
1st derivative = aE^(ka-k)
2nd derivative = E^(ka-k)+a(a-1)E^(ka-2k)
let a->0, we have E^(-k)
let c=E^(-1), we have c^k
so essentially, you are calculating
\Sig_k c^k
that is, summation over the power, instead of the base

the E here is exponential ..... ^ means power.....
1st d w.r.t. a = E^( k a) = k E^(k a )

oh, ic. ur E is exp , right? i thought it is a constant, then its set
this is essentially a Fourier transform

引用:

原帖由 andrew 于 2007-4-3 11:41 PM 发表
the E here is exponential ..... ^ means power.....
1st d w.r.t. a = E^( k a) = k E^(k a )

[ 本帖最后由 aeolus 于 2007-4-3 11:54 PM 编辑 ]

引用:

原帖由 aeolus 于 2007-4-3 11:47 PM 发表
btw, this is essentially a Fouriere Transform

yes, I somehow follow the same idea of characteristic function in probability ..... except that F.T. is E^(i k x) instead

引用:

原帖由 andrew 于 2007-4-3 10:29 PM 发表
我的方法是先算 X(a, n)=E^(a)+E^(2a)+...+E^(n a)
有 close form的
然後把答案對a微分２次，再 take limit a->0 (這步最煩)
基本上任何次方都可以算出來
但前題是要用 mathematica take limit....

So X(a,n)=[Exp(a)-Exp((n+1)a)]/[Exp(a)-1]

I suggest you Taylor expand this to series of powers of a

then the coefficient of a^2 is just 1^2+2^2+...+n^2
of course coefficient of a^3 is 1^3+2^3+3^3+...+n^3

and so on..........

en, good method.
with 2nd derivative, you multiple your constant signal by t^2, that is why the sum is \Sig_t t^2
i wonder signal processing guys can answer this problem in no time

引用:

原帖由 andrew 于 2007-4-3 11:50 PM 发表


yes, I somehow follow the same idea of characteristic function in probability ..... except that F.T. is E^(i k x) instead

nice!

引用:

原帖由 dongdongfish 于 2007-4-3 11:55 PM 发表


So X(a,n)=/

I suggest you Taylor expand this to series of powers of a

then the coefficient of a^2 is just 1^2+2^2+...+n^2
of course coefficient of a^3 is 1^3+2^3+3^3+...+n^3

and so on ...

coef of taylor is just take derivatives, the same Andrew did above, isnt it?

引用:

原帖由 dongdongfish 于 2007-4-3 11:55 PM 发表


So X(a,n)=/

I suggest you Taylor expand this to series of powers of a

then the coefficient of a^2 is just 1^2+2^2+...+n^2
of course coefficient of a^3 is 1^3+2^3+3^3+...+n^3

and so on ...