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Libniz
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\[\sum _{k=\infty }^{\infty } \theta (nk)\theta (k1)a^k\]
 one year ago
 one year ago
Libniz Group Title
\[\sum _{k=\infty }^{\infty } \theta (nk)\theta (k1)a^k\]
 one year ago
 one year ago

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Libniz Group TitleBest ResponseYou've already chosen the best response.0
\[\theta\] is step function
 one year ago

Libniz Group TitleBest ResponseYou've already chosen the best response.0
I can show you the work , it would be great if you can explain it
 one year ago

Libniz Group TitleBest ResponseYou've already chosen the best response.0
I get the n<=1 part but not the n>1
 one year ago

psi9epsilon Group TitleBest ResponseYou've already chosen the best response.0
this is convultion
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
I know u(k1) is 1 for k= inf to 1 u(x) is 0 for x<0, and 1 for x≥0 u(nk) is 1 for k= inf to +n
 one year ago

psi9epsilon Group TitleBest ResponseYou've already chosen the best response.0
use convolution property
 one year ago

psi9epsilon Group TitleBest ResponseYou've already chosen the best response.0
you can break up the sequence from infinity to 0 and from 0 to infinity for 0 to infinity we know what are the results of a step function
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
when you multiply the 2 step functions the one that "steps down" (because we are doing u(k) ) first sets the upper limit n wins for n< 1, and the other wins for n> 1
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
dw:1347579756548:dw
 one year ago

psi9epsilon Group TitleBest ResponseYou've already chosen the best response.0
you can split up the basic sequence too, its way easier to do it
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
when n>1 u(nk) times u(k1) is zero for k≥ 1
 one year ago

Libniz Group TitleBest ResponseYou've already chosen the best response.0
lol, I am reading all your posts over , will ask questions later
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
The first confusion is u(k) is 0 up to k=0 , then it jumps to 1 but u(k) is 1 and drops down to 0 at k=0 that is because when we plug in (for example) k=2, we get u((2))= u(2) =1 (2>0 means we jumped up) Both of these steps are backwards, are at 1 and drop to zero at some point
 one year ago

Libniz Group TitleBest ResponseYou've already chosen the best response.0
I get it how u(t) is reflected across y axis
 one year ago

Libniz Group TitleBest ResponseYou've already chosen the best response.0
trying to parse this "when you multiply the 2 step functions the one that "steps down" (because we are doing u(k) ) first sets the upper limit n wins for n< 1, and the other wins for n> 1"
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
u(k1) never changes. It is 1 from inf to 1, then drops to 0 we multiply this step times u(nk) if n is bigger than 1 this second step drops to zero after 1 but it does not do anything, because we are multiplying the 2 steps and everything past 1 is 0 (from the first step function). if n is less than 1, it drops to zero before the first step does, and it sets the upper limit where the product of the 2 step functions is 1.
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
the picture is trying to show the two cases. If it helps, think logic, both steps have to be 1
 one year ago

Libniz Group TitleBest ResponseYou've already chosen the best response.0
oh, so u(k1) is one constant step function whereas u(nk) is many different function based on differing value of n so say if n=2 u(2k) is it drop to zero before first step does
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
yes, u(2k) drops to zero when 2k=0 or k=2 when you multiply against the first step, you get 1 only up to k=2 (that is, k= n)
 one year ago

Libniz Group TitleBest ResponseYou've already chosen the best response.0
I guess my main confusion is \[\sum _{k=\infty }^{1} a^k\] n>1 < why do we use this even though n is not used in summation itself
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
n is controlling the upper limit. n bigger than 1 cannot make the limit bigger than 1 , but n< 1 can change the upper limit it n (n being smaller than 1)
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
still puzzling over it?
 one year ago

Libniz Group TitleBest ResponseYou've already chosen the best response.0
yeah, a lot of mental gymnastic, but thanks though
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
maybe this helps the u(k1) is 1 from inf to 1 and zero afterwards so the sum \[\sum_{k= \infty}^{+\infty}u(k1)a^k= \sum_{k= \infty}^{1}a^k\] the step function does not "do" anything except zero out all k's bigger than 1 on the other hand, the step (nk) affects the sum \[\sum_{k= \infty}^{+\infty}u(nk)a^k= \sum_{k= \infty}^{n}a^k\] by tossing all entries where k>n so the step function is setting the upper limit. when we use both (multiply both together), both have to be true (1). you get 2 cases: n<1 (where the upper limit is n) and n> 1 (where the upper limit is 1)
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
the next trick if figuring out how to do the summations. For a>1, you can change the limits to positive values and invert a: a^value= (1/a)^+value now it is a geometric series with a closed form solution.
 one year ago

Libniz Group TitleBest ResponseYou've already chosen the best response.0
^that one did it, much more clear now
 one year ago
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