Quick Math Question

[RedArmy]

Our teacher didn't go into much detail about this and our book seems to be lacking, and the internetz is currently failing me so here I am.

I know how to do a Dirac Delta problem with a bounded integral, but in that case as long as it stays inside the bounds, you just kinda do it.

When it comes to summations, you don't really have bounds, so when I get my value (say like delta[n-5]=0 so 5] and I have say,some value to the n for the summation, what do I do?.

Function I'm trying to describe: Summation from something to something of : delta[n-5]*(3^n)

Ugh, I hate describing math online.

Thanks for the help ATOT, I :heart: you all (not really)

 

[Jschmuck2]

 

[JohnCU]

erhuhuhdhdsahu humm.. is this for a discrete laplace transform or something? i vaguely remember this.

 

[chuckywang]

There's a discrete delta function, it's 1 when n=0, 0 otherwise.

 

[RedArmy]

Originally posted by: JohnCU
erhuhuhdhdsahu humm.. is this for a discrete laplace transformer or something? i vaguely remember this.

I honestly have no idea, it's just a problem we have to do, I know for the integrals it's called the 'sifting' property.

 

[RedArmy]

Originally posted by: chuckywang
There's a discrete delta function, it's 1 when n=0, 0 otherwise.

So like if my summation is from say 2 to 8, then the answer is 0? This is sounding vaguely familiar.

 

[JohnCU]

Originally posted by: RedArmy

Originally posted by: chuckywang
There's a discrete delta function, it's 1 when n=0, 0 otherwise.

So like if my summation is from say 2 to 8, then the answer is 0? This is sounding vaguely familiar.

wouldn't it be 3^5?

because delta[n-5] = 1 @ n = 5

 

[RedArmy]

Originally posted by: JohnCU

Originally posted by: RedArmy

Originally posted by: chuckywang
There's a discrete delta function, it's 1 when n=0, 0 otherwise.

So like if my summation is from say 2 to 8, then the answer is 0? This is sounding vaguely familiar.

wouldn't it be 3^5?

because delta[n-5] = 1 @ n = 5

Oh man, I read what he said completely wrong. I thought what he meant as the final answer, not just the delta function itself. Where's does the 2 to 8 come into play then for the summation?

Edit: Also, he said that it's 1 when n=0, but n is never 0 in this case, sooooo, yeah, I'm confused.

 

[JohnCU]

Originally posted by: RedArmy

Originally posted by: JohnCU

Originally posted by: RedArmy

Originally posted by: chuckywang
There's a discrete delta function, it's 1 when n=0, 0 otherwise.

So like if my summation is from say 2 to 8, then the answer is 0? This is sounding vaguely familiar.

wouldn't it be 3^5?

because delta[n-5] = 1 @ n = 5

Oh man, I read what he said completely wrong. I thought what he meant as the final answer, not just the delta function itself. Where's does the 2 to 8 come into play then for the summation?

it's just a bunch of 0's until you get to n=5, at which delta[n-5] becomes delta[0] which is 1 and then you multiply 1 by 3^n and since n=5, you get 0 + 0 + 0 + 0 + 3^5 + 0 + 0 + 0 = 3^5

 

[RedArmy]

Originally posted by: JohnCU

Originally posted by: RedArmy

Originally posted by: JohnCU

Originally posted by: RedArmy

Originally posted by: chuckywang
There's a discrete delta function, it's 1 when n=0, 0 otherwise.

So like if my summation is from say 2 to 8, then the answer is 0? This is sounding vaguely familiar.

wouldn't it be 3^5?

because delta[n-5] = 1 @ n = 5

Oh man, I read what he said completely wrong. I thought what he meant as the final answer, not just the delta function itself. Where's does the 2 to 8 come into play then for the summation?

it's just a bunch of 0's until you get to n=5, at which delta[n-5] becomes delta[0] which is 1 and then you multiply 1 by 3^n and since n=5, you get 0 + 0 + 0 + 0 + 3^5 + 0 + 0 + 0 = 3^5

ha ha! I get it now. That's what I was thinking, but I have low self confidence. So basically you can ignore everything except where the function itself becomes 0 and subsequently get one as an answer.

 

[JohnCU]

Originally posted by: RedArmy

Originally posted by: JohnCU

Originally posted by: RedArmy

Originally posted by: JohnCU

Originally posted by: RedArmy

Originally posted by: chuckywang
There's a discrete delta function, it's 1 when n=0, 0 otherwise.

So like if my summation is from say 2 to 8, then the answer is 0? This is sounding vaguely familiar.

wouldn't it be 3^5?

because delta[n-5] = 1 @ n = 5

Oh man, I read what he said completely wrong. I thought what he meant as the final answer, not just the delta function itself. Where's does the 2 to 8 come into play then for the summation?

it's just a bunch of 0's until you get to n=5, at which delta[n-5] becomes delta[0] which is 1 and then you multiply 1 by 3^n and since n=5, you get 0 + 0 + 0 + 0 + 3^5 + 0 + 0 + 0 = 3^5

ha ha! I get it now. That's what I was thinking, but I have low self confidence. So basically you can ignore everything except where the function itself becomes 0 and subsequently get one as an answer.

well the function doesn't become zero but the input becomes zero (delta[0] = 1).

it's a weird thing. infinite height and zero width and an area of exactly 1.

 

[RedArmy]

Originally posted by: JohnCU

Originally posted by: RedArmy

Originally posted by: JohnCU

Originally posted by: RedArmy

Originally posted by: JohnCU

Originally posted by: RedArmy

Originally posted by: chuckywang
There's a discrete delta function, it's 1 when n=0, 0 otherwise.

So like if my summation is from say 2 to 8, then the answer is 0? This is sounding vaguely familiar.

wouldn't it be 3^5?

because delta[n-5] = 1 @ n = 5

Oh man, I read what he said completely wrong. I thought what he meant as the final answer, not just the delta function itself. Where's does the 2 to 8 come into play then for the summation?

it's just a bunch of 0's until you get to n=5, at which delta[n-5] becomes delta[0] which is 1 and then you multiply 1 by 3^n and since n=5, you get 0 + 0 + 0 + 0 + 3^5 + 0 + 0 + 0 = 3^5

ha ha! I get it now. That's what I was thinking, but I have low self confidence. So basically you can ignore everything except where the function itself becomes 0 and subsequently get one as an answer.

well the function doesn't become zero but the input becomes zero (delta[0] = 1).

it's a weird thing. infinite height and zero width and an area of exactly 1.

Yeah, that's what I meant, I was gonna edit that, but as long as I understand it in my own demented ways it doesn't really matter. I remember our teacher talking about that infinite height and zero width stuff but I don't really see how it applied to anything...until now! Thanks for the help!!