# Little Big Math Challenge

#### MrAl

Joined Jun 17, 2014
8,479
Hi there,

This is kind of a small math challenge, but the application is interesting.
First, you dont have to be a genius to figure this out i dont think, but you have to be a genius if you can figure out the application for this progression if you have never seen it before or have no idea where it came from before you started to solve it today.

Here is the progression:
2/sqrt(3), 8/sqrt(15), 18/sqrt(35), 32/(3*sqrt(7)), 50/(3*sqrt(11)), 72/sqrt(143)

Those are all EXACT not approximate, and the first is for n=1 and the last is for n=6, so they go for n's:
1,2,3,4,5,6
and each n has one of those values in that order.
Maybe i should write this out ... for each 'n' from 1 to 6:

[1] 2/sqrt(3)
[2] 8/sqrt(15)
[3] 18/sqrt(35)
[4] 32/(3*sqrt(7))
[5] 50/(3*sqrt(11))
[6] 72/sqrt(143)

The first goal is to figure out the progression in terms of 'n' so it will be a function of 'n'.
The second goal, much more difficult unless you've seen this before, is to figure out what application it came out of.

The first solution is interesting, the second one may be surprising if you've never seen this before.

Have fun

#### jpanhalt

Joined Jan 18, 2008
11,088
Obviously, given nm, the answer is n/√(2n-1) m/√(2m-1)

Now what is "n""m" ?

n = principal quantum number. m = maximum number of electrons per principal energy level (e.g., first row, 1s =2 electrons, second row 2s +2p = 8 electrons and so forth)? I an just an organic chemist. Anything past a "d" shell , you had to ask the smart guys next door. See: https://chem.libretexts.org/Bookshe...chanical_Orbitals_and_Electron_Configurations.

Edit: Used "n" sloppily. Changed to "m". Added citation to spoiler.

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#### StayatHomeElectronics

Joined Sep 25, 2008
1,070
(2×n^2)/(sqrt((2×n)^2 - 1))

No idea...

#### MrAl

Joined Jun 17, 2014
8,479
Obviously, given n, the answer is n/√(2n-1)

Now what is "n" ?

Electron shells (e.g., first row, 1s =2 electrons, second row 2s +2p = 8 electrons and so forth)? I an just an organic chemist. Anything past a "d" shell , you had to ask the smart guys next door.
Hello,

The ordinate 'n' goes from 1 to 6.
The value given for n=6 is 72/sqrt(143)=6.020942472051054 approximately but
the value obtained from n/sqrt(2*n-1) for n=6 is 1.809068067466582 approximately so that cant be right the values are different by at least a factor of 3 or more.

#### MrAl

Joined Jun 17, 2014
8,479
(2×n^2)/(sqrt((2×n)^2 - 1))

No idea...
Hi,

Yes your #1 is correct. The answer for #2 is probably too hard to figure out without knowing something about the application maybe i should have given a hint.

#### StayatHomeElectronics

Joined Sep 25, 2008
1,070
I can see the trend but have not successfully applied it to anything yet. Thanks for the problem. I enjoyed the challenge of part one!

Hi,

Yes your #1 is correct. The answer for #2 is probably too hard to figure out without knowing something about the application maybe i should have given a hint.

#### ci139

Joined Jul 11, 2016
1,696
https://oeis.org/search?q=2,8,18,32,50,72&language=english&go=Search $$\rightarrow\ a_n=2·n^2\\ n^2=\sum_{j=1}^n\left(2·j-1\right)\\ k^2-1=\left(k-1\right)\left(k+1\right)$$
$\begin{array}{ccc} \left[{1}\right] & \frac{2}{\sqrt 3} & \frac{2}{\sqrt 3}\\ \left[{2}\right] & \frac{8}{\sqrt{15}} & \frac{8}{\sqrt{15}}\\ \left[{3}\right] & \frac{18}{\sqrt{35}} & \frac{18}{\sqrt{35}}\\ \left[{4}\right] & \frac{32}{3·\sqrt{7}} & \frac{32}{\sqrt{63}}\\ \left[{5}\right] & \frac{50}{3·\sqrt{11}} & \frac{50}{\sqrt{99}}\\ \left[{6}\right] & \frac{72}{\sqrt{143}} & \frac{72}{\sqrt{143}} \end{array}$

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#### MrAl

Joined Jun 17, 2014
8,479
I can see the trend but have not successfully applied it to anything yet. Thanks for the problem. I enjoyed the challenge of part one!
Hi,

You are welcome and thanks for trying this puzzle.
The second part is probably asking a lot because there are so many applications of math out there. I'll have to give a hint to at least reveal roughly where it comes from, what field of study.

BTW are you a prophet?
I ask because your screen name here is Stay-At-Home-Electronics and i see that you joined back in late 2008 long before the corona virus hit the world, and now just about every thing has to be "state at home" to do.

#### MrAl

Joined Jun 17, 2014
8,479
https://oeis.org/search?q=2,8,18,32,50,72&language=english&go=Search $$\rightarrow\ a_n=2·n^2\\ n^2=\sum_{j=1}^n\left(2·j-1\right)\\ k^2-1=\left(k-1\right)\left(k+1\right)$$
$\begin{array}{ccc} \left[{1}\right] & \frac{2}{\sqrt 3} & \frac{2}{\sqrt 3}\\ \left[{2}\right] & \frac{8}{\sqrt{15}} & \frac{8}{\sqrt{15}}\\ \left[{3}\right] & \frac{18}{\sqrt{35}} & \frac{18}{\sqrt{35}}\\ \left[{4}\right] & \frac{32}{3·\sqrt{7}} & \frac{32}{\sqrt{63}}\\ \left[{5}\right] & \frac{50}{3·\sqrt{11}} & \frac{50}{\sqrt{99}}\\ \left[{6}\right] & \frac{72}{\sqrt{143}} & \frac{72}{\sqrt{143}} \end{array}$
That's interesting because we usually dont think of the internet being around back in the 1960's. I dont think i knew about it back then but then i was quite young then. It gained momentum though around 1983 and then AOL started a couple years later, then sometime later it took off with everyone wanting to get "online". Now there are so many online it's incredible and with businesses coming online it changed the world forever.

#### MrAl

Joined Jun 17, 2014
8,479
Hello again,

For those interested, here is a hint.
The application for this sequence comes from the theory of 2nd order signal filters.
It may be well known already but i recently discovered it while trying to find a formula for something, and the formula came out very very complicated but after examining all of the possibilities the formula broke down into that simple sequence with the simple generator function you guys have found.
It is a very useful formula when it comes to 2nd order filters and would aid in the design.

#### StayatHomeElectronics

Joined Sep 25, 2008
1,070
Not a prophet, just another time in life when I got to spend a significant amount of time at home. Back then it was compliments of my kids instead of a pandemic.

Hi,

You are welcome and thanks for trying this puzzle.
The second part is probably asking a lot because there are so many applications of math out there. I'll have to give a hint to at least reveal roughly where it comes from, what field of study.

BTW are you a prophet?
I ask because your screen name here is Stay-At-Home-Electronics and i see that you joined back in late 2008 long before the corona virus hit the world, and now just about every thing has to be "state at home" to do.

#### LvW

Joined Jun 13, 2013
1,300
Hello again,

For those interested, here is a hint.
The application for this sequence comes from the theory of 2nd order signal filters.
It may be well known already but i recently discovered it while trying to find a formula for something, and the formula came out very very complicated but after examining all of the possibilities the formula broke down into that simple sequence with the simple generator function you guys have found.
It is a very useful formula when it comes to 2nd order filters and would aid in the design.
Hello MrAl, as you probably know, I am engaged in filter theory and systems since several years.....and the formula as given by StayatHomeElectronics looks rather familiar to me.
Am I allowed to reveal its meaning here?

#### MrAl

Joined Jun 17, 2014
8,479
Hello MrAl, as you probably know, I am engaged in filter theory and systems since several years.....and the formula as given by StayatHomeElectronics looks rather familiar to me.
Am I allowed to reveal its meaning here?
Hi,

Yeah i dont see why not, i think anyone who wants to try it has tried it by now.
I found it interesting because i started with a big formula and was able to reduce it a lot and come up with that fairly simple formula. I dont doubt that is has been mentioned before in some test book or something.

#### LvW

Joined Jun 13, 2013
1,300
Well, each 2nd-order lowpass (Q>0.707) shows a "peaking" effect at a frequency wx very close to the pole frequency wp, that means:
An amplitude peak somewhat larger than the amplitude at DC.

The frequency where such a peak occurs is: wx=wp*SQRT[1-(1/2Qp²)]

The value of such a gain peak is given by the formula: GP=2Qp²/SQRT[4Qp²-1].

For Qp=n this expression is identical to the formula which fulfills the sequency of numbers as given in your first post.

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#### MrAl

Joined Jun 17, 2014
8,479
Well, each 2nd-order lowpass (Q>0.707) shows a "peaking" effect at a frequency wx very close to the pole frequency wp, that means:
An amplitude peak somewhat larger than the amplitude at DC.

The frequency where such a peak occurs is: wx=wp*SQRT[1-(1/2Qp²)]

The value of such a gain peak is given by the formula: GP=2Qp²/SQRT[4Qp²-1].

For Qp=n this expression is identical to the formula which fulfills the sequency of numbers as given in your first post.
Yes that's it
It is also interesting to derive that formula from only the transfer function of a 2nd order low pass. If you feel like trying see what you can come up with.
H(s)=wo^2/(s^2+s*wo/Q+wo^2)

You can imagine i did this because we were talking about low pass filters in the other thread and i thought it would be interesting to find the peak for a given Q.

Also maybe a little interesting is that values from 0 to 0.5 result in only imaginary solutions, and at 0.5 we get a discontinuous point that could be infinity or minus infinity. Above 0.5 we get real solutions.

Also interesting is the behavior as the Q goes up. At around 2 we start to see the peak become nearly equal to the value of Q. At around 3.562 we see the result become within 1 percent of the Q, and after that the difference between Q and the peak gets smaller and smaller. At Q=10 the resulting peak comes out very close to to 10.0125 which is nearly 10.
With Q=100 we get very nearly 100.00125 which is really close to 100.
So around Q=2 and above the peak value (gain that is) becomes nearly equal to the Q itself.

#### LvW

Joined Jun 13, 2013
1,300
Yes that's it
It is also interesting to derive that formula from only the transfer function of a 2nd order low pass. If you feel like trying see what you can come up with.
H(s)=wo^2/(s^2+s*wo/Q+wo^2)

You can imagine i did this because we were talking about low pass filters in the other thread and i thought it would be interesting to find the peak for a given Q.

Also maybe a little interesting is that values from 0 to 0.5 result in only imaginary solutions, and at 0.5 we get a discontinuous point that could be infinity or minus infinity. Above 0.5 we get real solutions.

Also interesting is the behavior as the Q goes up. At around 2 we start to see the peak become nearly equal to the value of Q. At around 3.562 we see the result become within 1 percent of the Q, and after that the difference between Q and the peak gets smaller and smaller. At Q=10 the resulting peak comes out very close to to 10.0125 which is nearly 10.
With Q=100 we get very nearly 100.00125 which is really close to 100.
So around Q=2 and above the peak value (gain that is) becomes nearly equal to the Q itself.
Yes - I can agree to your observations.
The reason is as follows: From the normalized transfer function we can see that at the pole frequency wp (your notation: wo) the transfer function reduces to H(w=wp)=-jQ (that means: Magnitude=Q).
On the other hand, the frequency wx (peak frequency) and wp become closer and closer for rising Q values (see the formulas in my former post). This explains your observation.