Explain me, please.

- Loser66

Explain me, please.

- katieb

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- dan815

for a decreasing sequence bn that summation converges

- dan815

is it dirichlet?

- Loser66

Those are what my prof did in class.

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## More answers

- Loser66

The red parts are my explanation on the lecture so that I remember what is going on when reviewing for test. :)

- dan815

it is still downloading, its going very slow

- Loser66

dan!!

- dan815

ok hmm i see

- dan815

lets take cos(ntheta)= Real part of {e^(itheta)}

- dan815

what they are using as their argument is that hey look just like how an alternating sequence of (-1)^n multiplied by a decreasing series goes to 0, this also does the same thing

- dan815

because for any thetea u pick, for varying n, you keep getting some number of positive and some number of negatives , both equal to each other, tho the addition of this decreasing bn sequences works slightly differently

- dan815

but you can still definately show that you will be subtracting greater part than what u are adding in the next set

- dan815

here is a picture to help you see

- dan815

|dw:1441645035166:dw|

- dan815

varying theta is just like varying the period you sample your dots

- Loser66

next?

- dan815

|dw:1441645239534:dw|

- dan815

so in fact this is not much different from alternating series at all, its just put in a different way

- dan815

let f(n) be some decreasing function
for the first 3 values of n let cosntheta be positive
for the next 3 values of n let cos n*theta be negative
can you see why now, this has to be convergin

- dan815

liket lets take a random decreasing function

- dan815

|dw:1441645504019:dw|

- dan815

|dw:1441645527838:dw|

- Loser66

|dw:1441645615856:dw|

- dan815

thats wht u are doing with chos (ntheta) you are taking samples in each part of that curve adding and subtracting

- Loser66

dan, slow down on let f(n) be some decreasing function for the first 3 values of n let cosntheta be positive for the next 3 values of n let cos n*theta be negative can you see why now, this has to be convergin

- anonymous

wassup dan

- dan815

|dw:1441645682902:dw|

- Loser66

f(n) = cos (ntheta) 3 values of n let cos n theta > 0

- dan815

these could be your sambled points when you multiply by cos n theta

- dan815

yeah like

- dan815

|dw:1441645810574:dw|

- Loser66

|dw:1441645965886:dw|

- Loser66

They are the value of cos (n theta) when n = 1, 2, 3... right?

- dan815

|dw:1441646040644:dw|

- Loser66

yes

- dan815

yeah

- dan815

now you say the sum of all these samples values in between is bounded by that formula they gave u up at the top

- Loser66

suppose 0

- Loser66

now what?

- Loser66

sum cos (n theta) = cos (0theta) + cos (1 theta) + cos (2 theta) + ..... cos (ntheta) , right? then?

- Loser66

is it not that it is =0?

- dan815

|dw:1441646218632:dw|

- dan815

now u can use that bound

- anonymous

- dan815

notice how if theta -->0 its going to finity as u will be sampling an infinite points in each of the interval

- Loser66

then?

- dan815

@empty how do u show this series converges

- Loser66

Is it not that if it is bounded and continuous, it converges?

- dan815

its kind of weird to think about actually

- dan815

like say you take an finite number of points on the highest parto of your decreasing function

- dan815

|dw:1441646971233:dw|

- dan815

now lets say u subtract an infinite set of points but lesser value.. the difference between these 2 is still infinity is it notq

- dan815

|dw:1441647101921:dw|

- dan815

but then u subtraced with some other difference an infinite set of points when compared to the next block of addition

- Loser66

I gotta go. You guys PPPPPPLLLLLEASE, freely discuss. I' ll take it later. Thanks in advance.

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