muballitmitte
Nov3-09, 03:09 AM
I`ve encountered this limit in a book i`ve been reading.
\lim_{t_c \rightarrow 0} \frac{1}{t_c} \int_0^{\infty} \left[ \sum_{i=1}^M H(t-t_i)\exp(-\frac{t-t_i}{t_c}) - \sum_{j=1}^NH(t-t_j)\exp(-\frac{t-t_j}{t_c}) \right] ^2 \textrm{d}t
where H(x) is the heaviside step function : 1 when x >= 0 and 0 the other case. ( the t_i in the first sum indexes a different set than the t_j in the second so they are different ...but I used the same notation.
This limit is really strange. I tried to solve it by splitting it first into 3 limits after raising to the power of 2 inside the integral. So I got:
\lim_{t_c \rightarrow 0} \frac{1}{t_c} \int_0^{\infty} \left[ \sum_{i=1}^M H(t-t_i)\exp(-\frac{t-t_i}{t_c})\right] ^2 \textrm{d}t + \lim_{t_c \rightarrow 0} \frac{1}{t_c} \int_0^{\infty} \left[ \sum_{j=1}^NH(t-t_j)\exp(-\frac{t-t_j}{t_c}) \right] ^2 \textrm{d}t + + \lim_{t_c \rightarrow 0} \frac{1}{t_c} \int_0^{\infty} \left[ \sum_{i=1}^M H(t-t_i)\exp(-\frac{t-t_i}{t_c})* \sum_{j=1}^NH(t-t_j)\exp(-\frac{t-t_j}{t_c}) \right] \textrm{d}t
TO the first lim (call it A) i`ve applied the multinomial formula and and ended up with
A=\lim_{t_c \rightarrow 0} \frac{1}{t_c} \int_0^{\infty} \left[ \sum_{i=1}^M H^2(t-t_i)\exp(-2\frac{t-t_i}{t_c}) + \sum_{i \ne j}H(t-t_i)\exp(-\frac{t-t_i}{t_c}) H(t-t_j)\exp(-\frac{t-t_j}{t_c}) \right]\textrm{d}t
End the end I get that the first limit is M/2, the second is 0 and the first is N/2. Maybe this is correct. BUT i seriously doubt it. Also what really puzzles me is that when I compute te same limit with t_c \rightarrow \infty I get almost the same result. I must be missing something. Anyone has any ideas?
\lim_{t_c \rightarrow 0} \frac{1}{t_c} \int_0^{\infty} \left[ \sum_{i=1}^M H(t-t_i)\exp(-\frac{t-t_i}{t_c}) - \sum_{j=1}^NH(t-t_j)\exp(-\frac{t-t_j}{t_c}) \right] ^2 \textrm{d}t
where H(x) is the heaviside step function : 1 when x >= 0 and 0 the other case. ( the t_i in the first sum indexes a different set than the t_j in the second so they are different ...but I used the same notation.
This limit is really strange. I tried to solve it by splitting it first into 3 limits after raising to the power of 2 inside the integral. So I got:
\lim_{t_c \rightarrow 0} \frac{1}{t_c} \int_0^{\infty} \left[ \sum_{i=1}^M H(t-t_i)\exp(-\frac{t-t_i}{t_c})\right] ^2 \textrm{d}t + \lim_{t_c \rightarrow 0} \frac{1}{t_c} \int_0^{\infty} \left[ \sum_{j=1}^NH(t-t_j)\exp(-\frac{t-t_j}{t_c}) \right] ^2 \textrm{d}t + + \lim_{t_c \rightarrow 0} \frac{1}{t_c} \int_0^{\infty} \left[ \sum_{i=1}^M H(t-t_i)\exp(-\frac{t-t_i}{t_c})* \sum_{j=1}^NH(t-t_j)\exp(-\frac{t-t_j}{t_c}) \right] \textrm{d}t
TO the first lim (call it A) i`ve applied the multinomial formula and and ended up with
A=\lim_{t_c \rightarrow 0} \frac{1}{t_c} \int_0^{\infty} \left[ \sum_{i=1}^M H^2(t-t_i)\exp(-2\frac{t-t_i}{t_c}) + \sum_{i \ne j}H(t-t_i)\exp(-\frac{t-t_i}{t_c}) H(t-t_j)\exp(-\frac{t-t_j}{t_c}) \right]\textrm{d}t
End the end I get that the first limit is M/2, the second is 0 and the first is N/2. Maybe this is correct. BUT i seriously doubt it. Also what really puzzles me is that when I compute te same limit with t_c \rightarrow \infty I get almost the same result. I must be missing something. Anyone has any ideas?