p**l
发帖数: 102 | 1 0
what is the limit of the below formula?
1-(1-alpha^1)*(1-alpha^2)*(1-alpha^3)*.....*(1-alpha^n)
as n -> positive infinity
Thanks a lot! | x******g
发帖数: 318 | 2 ∏[n=1...](1-a^n)=∑[n=0,1,-1,2,-2..](-1)^n*a^(n(3n+1)/2)
【在 p**l 的大作中提到】
: 0: what is the limit of the below formula?
: 1-(1-alpha^1)*(1-alpha^2)*(1-alpha^3)*.....*(1-alpha^n)
: as n -> positive infinity
: Thanks a lot!
| p**l
发帖数: 102 | 3 Sorry but I did not follow your answer.
Can you tell me how to calculate the right part in your formula? For example,
when n=1, the left part is 1-a, but the right part is
(-1)^0*a^(0)+(-1)^1*a^2=1-a^2, is this right?
【在 x******g 的大作中提到】
: ∏[n=1...](1-a^n)=∑[n=0,1,-1,2,-2..](-1)^n*a^(n(3n+1)/2)
| x******g
发帖数: 318 | 4 我说的是极限形式的
你看到那个点点点了吧
【在 p**l 的大作中提到】
: Sorry but I did not follow your answer.
: Can you tell me how to calculate the right part in your formula? For example,
: when n=1, the left part is 1-a, but the right part is
: (-1)^0*a^(0)+(-1)^1*a^2=1-a^2, is this right?
| y**t
发帖数: 50 | 5 This is almost a Dedekind eta function which is classic.
【在 p**l 的大作中提到】
: Sorry but I did not follow your answer.
: Can you tell me how to calculate the right part in your formula? For example,
: when n=1, the left part is 1-a, but the right part is
: (-1)^0*a^(0)+(-1)^1*a^2=1-a^2, is this right?
|
|
