Kn = (R(1.5+n)+K0+...+Kn-1)/(R-1)

[TedChirvasiu]

Hello, guys! This is my first post on this forum so i apologize in advance if this is not in the right section or something like that.
As the title says, i have the following relation : K_n = (R(1.5+n)+K₀+K₁+...+K_n-1)/(R-1).
I know n, i know R. R is always a power of 2, if that helps.

My question is : how can i make it so that K_n will only depend on R and n? Can i do it without getting something very ugly? I can't quite get my head around this... For instance i know that K₀ is (R*1.5)/(R-1) and that K₁ is going to be (2.5*R²- R)/(R-1)² but from then on it gets pretty confusing for me.


This was pretty ugly in the first place. It has to do with stretching of some circles, but i won't get into further detail as i do not want to waste your time any more than i already do.





Any help or tips would be greatly appreciated. Thank you!

 

[tkhunny]

Why do you think it is generally possible?
Just what is it you are trying to do?

 

[Deleted member 4993]

Hello, guys! This is my first post on this forum so i apologize in advance if this is not in the right section or something like that.
As the title says, i have the following relation : K_n = (R(1.5+n)+K₀+K₁+...+K_n-1)/(R-1).
I know n, i know R. R is always a power of 2, if that helps.

My question is : how can i make it so that K_n will only depend on R and n? Can i do it without getting something very ugly? I can't quite get my head around this... For instance i know that K₀ is (R*1.5)/(R-1) and that K₁ is going to be (2.5*R²- R)/(R-1)² but from then on it gets pretty confusing for me.

This was pretty ugly in the first place. It has to do with stretching of some circles, but i won't get into further detail as i do not want to waste your time any more than i already do.


Any help or tips would be greatly appreciated. Thank you!

how can i make it so that K_n will only depend on R and n? - Can you please tell us why do you want to do that? If you are evaluating K_n in computer - the given expression would be fastest way to evaluate it.

 

[TedChirvasiu]

Hello, yes, well, i just wrote a code for it and it does take just a few miliseconds for 2²⁵.
I hoped there was more to do to it. I was thinking it would eventually simplify or there would be a way to write it without summing up all the K's. Like you would do for Gauss' sum.

Well, i guess this is the fastest way to compute it then. Fortunately it is fast enough for my project and i can also do some precaching.

Thanks, guys.

 

[Deleted member 4993]

I'm also confused...except evident that you're dealing with a geometric series.
Seems that you're trying to determine the value of nth term,
and then the sum of all terms including this nth term.

Let's use a simple example:
5, 15, 45, 135
a = 1st term = 5
m = multiplyer = 3
n = number of terms = 4

To calculate nth term:
a * m^(n - 1) : 5 * 3^(4 - 1) = 5 * 27 = 135

To calculate sum:
a * (m^n - 1) / (m - 1) = 5 * (3^4 - 1) / (3 - 1) = 5 * 80 / 2 = 200

With your problem, R seems to be the multiplyer: R-1 compared to m-1 above.
I presume R = radius?

Hope this helps...

\(\displaystyle \displaystyle \frac{K_2}{K_1} \ \ne \frac{K_3}{K_2}\) ................... right to the corner ..... 30 mts. .... if you protest it will increase by Geometric progression with r>1

 

[lookagain]

Let's use a simple example:
5, 15, 45, 135
a = 1st term = 5
m = > > multiplyer < < = 3
n = number of terms = 4

With your problem, R seems to be the > > multiplyer: < < R-1 compared to m-1 above.

In each of the online dictionaries I checked, they stated that "multiplyer" isn't in the dictionary.

 

[TedChirvasiu]

I'm also confused...except evident that you're dealing with a geometric series.
Seems that you're trying to determine the value of nth term,
and then the sum of all terms including this nth term.

Let's use a simple example:
5, 15, 45, 135
a = 1st term = 5
m = multiplyer = 3
n = number of terms = 4

To calculate nth term:
a * m^(n - 1) : 5 * 3^(4 - 1) = 5 * 27 = 135

To calculate sum:
a * (m^n - 1) / (m - 1) = 5 * (3^4 - 1) / (3 - 1) = 5 * 80 / 2 = 200

With your problem, R seems to be the multiplyer: R-1 compared to m-1 above.
I presume R = radius?

Hope this helps...

Hello, Denis, thanks a lot for your reply.
Yes, R is in fact the radius, which will be always a power of two.

Let me explain exacly what i am trying to do :
Imagine you have two circles. One has the diameter 2² and the other has the diameter 2³. Now, the circumference of the first circle will be 4*Pi and of the second one will be 2*4*Pi. So twice as big as the first. Let's say you shove 16 "squares" on the first circle and twice as many, so 32, on the second one.
Sort of like in this picture, but not as messy : http://blogs.discovermagazine.com/badastronomy/files/2009/08/squarecirclespiral_ann.jpg
The "squares" on the first circle will have a length of Circumference/16 = 4*Pi/16 = Pi/4 and the ones on the second circle C/32 = 8*Pi/32 = Pi/4. So, as you might have already noticed, the "squares" are going to have the same length on every 2^n diameter. Now in order for these squares to be somewhat square they will have to have a height aswell... Which will have to be somewhat equal with the length. So Pi/4.
Okay, keep in mind that between radius 4 and radius 8 there will be a gap of 4 units. Or will it? Well, actually the gap will be 4-2*Pi/4 because half of the squares on the lower circle and half of the squares on the upper circle will be falling inside of this gap. Okay. So, here's what i wanna do : I want to know how many circles i can put in this gap so that i get the best lateral stretch/vertical stretch ratio. It will never be 1, excepting for the 2^n diameter circles, but it will be very close.

Guys, i am very sorry, i know this is a pretty ambiguous explanation. I would need to make a drawing to explain it better and a way longer post. Please, there is no need to waste your time on this. I think what i have for now will do. I was hoping there would be a quick and easy way i haven't heard of for writing this. I am a 9th grade student so my math knowledge is super limited. I did feel from the very beginning that this is going to be either an arithmetic or a geometric series but this seems to be fast enough for my needs. Thank you all for your time and help!