When Does the Exponential Function Reach Its Maximum Value?

[JeremyEbert]

why is it that the largest value of n^( (((n-k)*(k-1)/(2k)) + (k-1))/(n-1) ) always seems to be when k=36?

 

[SteamKing]

What is n?

 

[JeremyEbert]

What is n?

integers
0<n<LIM

 

[Dr. Seafood]

^ What's "LIM"? Do you mean n is any positive integer?

I suppose the job to be done here is to find which value of k maximizes the expression in question. Let Q_n(k) be that expression. The problem is to find k_0 such that, for each n, Q_n(k_0) \geq Q_n(k) for all k -- and subsequently, to show that apparently k_0 = 36. Is this what you're asking?

 

[JeremyEbert]

^ What's "LIM"? Do you mean n is any positive integer?

I suppose the job to be done here is to find which value of k maximizes the expression in question. Let Q_n(k) be that expression. The problem is to find k_0 such that, for each n, Q_n(k_0) \geq Q_n(k) for all k -- and subsequently, to show that apparently k_0 = 36. Is this what you're asking?

Yes, you are correct. Thanks for stating it in a better way.

 

[JeremyEbert]

Yes, you are correct. Thanks for stating it in a better way.

Well, I feel real dumb... Nevermind this thread, I had an issue with the application I was using to compute the results. Sorry to all who spent any time on this.

 

[JeremyEbert]

why is it that the largest value of n^( (((n-k)*(k-1)/(2k)) + (k-1))/(n-1) ) always seems to be when k=36?

Interesting function none the less.

t=(((n-k)*(k-1)/(2k)) + (k-1))/(n-1)

n^(t)

when k=1 then t=0 and n^(t)=k

when k=n^(1/2) then t=0.5 and n^(t)=k

when k=n then t=1 and n^(t)=k

 

[JeremyEbert]

WolframAlpha shows some interesting results at well. The series expansion shows terms involving double factorial numbers.

http://www.wolframalpha.com/widgets/view.jsp?id=daf29fc2857c2b71d6be58dcc6e7ef49

http://www.wolframalpha.com/widgets/view.jsp?id=b8ba4c95900e275211a98d8bd1b0a53c

 

[JeremyEbert]

Interesting function none the less.

t=(((n-k)*(k-1)/(2k)) + (k-1))/(n-1)

n^(t)

when k=1 then t=0 and n^(t)=k

when k=n^(1/2) then t=0.5 and n^(t)=k

when k=n then t=1 and n^(t)=k

Looking at its deviation from k is very interesting:

k - n^( (((n-k)*(k-1)/(2k)) + (k-1))/(n-1) )


http://dl.dropbox.com/u/13155084/nt-k-4.png
http://dl.dropbox.com/u/13155084/nt-k-9.png
http://dl.dropbox.com/u/13155084/nt-k-16.png
http://dl.dropbox.com/u/13155084/nt-k-25.png
http://dl.dropbox.com/u/13155084/nt-k-36.png
http://dl.dropbox.com/u/13155084/nt-k-49.png

 

[JeremyEbert]

Looking at its deviation from k is very interesting:

k - n^( (((n-k)*(k-1)/(2k)) + (k-1))/(n-1) )


http://dl.dropbox.com/u/13155084/nt-k-4.png
http://dl.dropbox.com/u/13155084/nt-k-9.png
http://dl.dropbox.com/u/13155084/nt-k-16.png
http://dl.dropbox.com/u/13155084/nt-k-25.png
http://dl.dropbox.com/u/13155084/nt-k-36.png
http://dl.dropbox.com/u/13155084/nt-k-49.png

so basically the roots of the function:

log(n,k) - ((((n-k)*(k-1)/(2k)) + (k-1))/(n-1) )

are

k=1
k=n^(1/2)
k=n

Is this a correct statement?

 

[JeremyEbert]

more information

"Another link into Eulers Generalized Pentagonal Numbers and the divisor function d(n):



For our Divisor summatory function we have:

D(n) = SUM(d(n)) :

for k = 0 --> floor [sqrt n]
SUM (d(n)) = SUM ((2*floor[(n - k^2)/k]) + 1)


The notable difference in the equation from the published version is the:

(n - k^2)/k (congruence of squares)

which is derived from the

z = (n - k^2)/2k + i n^(1/2)

forming a parabolic coordinate system.


The function (n - k^2)/2k forms a divisor symmetry centered on the square-root of n.

Example:

k = divisors of n {1,2,3,4,6,9,12,18,36}
n = 36


+17.5, +8.0, +4.5, +2.5, 00.0, -2.5, -4.5, -8.0, -17.5




key results:
sqrt(n) = 0
Sum Terms = 0

Offsetting by -((n-1)/2) = -17.5 and taking the absolute values gives us:


0, 9.5, 13, 15, 17.5, 20, 22, 25.5, 35

Key results:
sqrt(n) = (n-1)/2;




Another way to generate these terms is:

((n-k)*(k-1)/2k) + (k-1)

The key ratio here being the (k-1)/2k function.

reducing this ratio sequence we get:

01/04, 01/03, 03/08, 02/05, 05/12, 03/07, 07/06, 04/09, 09/20, 05/11, 11/24, 06/13, 13/28, 07/15, 15/32, 08/17, 17/36

or

01 01 03 02 05 03 07 04 09 05 11 06 13 07 15 08 17
04 03 08 05 12 07 16 09 20 11 24 13 28 15 32 17 36


Showing a direct connection to Eulers Generalized Pentagonal Numbers and the divisor function d(n)



**

A026741 ( n if n odd, n/2 if n even. ) = xx, 00, 01, 01, 03, 02, 05, 03, 07, 04, 09, 05, 11, 06, 13, 07, 15, 08, 17
A022998 ( If n is odd then n else 2*n. ) = 00, 01, 04, 03, 08, 05, 12, 07, 16, 09, 20, 11, 24, 13, 28, 15, 32, 17, 36

A026741 = Partial sums give Generalized Pentagonal Numbers A001318 = 00, 01, 02, 05, 07, 12, 15, 22, 26, 35, 40, 51, 57
A022998 = Partial sums give Generalized Octagonal Numbers A001082 = 00, 01, 05, 08, 16, 21, 33, 40, 56, 65, 85, 96, 120 "

 

[JeremyEbert]

More WolframAlpha:

cos( k - (((n-k)*(k-1)/(2k)) + (k-1))/(n-1))

http://www.wolframalpha.com/input/?i=cos(+k+-+(((n-k)*(k-1)/(2k))+++(k-1))/(n-1))

 

[JeremyEbert]

so basically the roots of the function:

log(n,k) - ((((n-k)*(k-1)/(2k)) + (k-1))/(n-1) )

are

k=1
k=n^(1/2)
k=n

Is this a correct statement?

The divisor symmetry still shows up nicely. For example 36:

(ln(x)/ln(36)) - (((36-x)*(x-1)/(2*x)) + (x-1))/(36-1))

http://dl.dropbox.com/u/13155084/36.png

 

[JeremyEbert]

for the function

f(n,k) = ( (ln(k)/ln(n)) - ((((n-k)*(k-1)/(2k)) + (k-1))/(n-1) )

the

local minimum = (((n-1)/2)-sqrt(((n-1)/2)^2-(n*ln^2(sqrt(n)))))/log(sqrt(n))
local maximum = (((n-1)/2)+sqrt(((n-1)/2)^2-(n*ln^2(sqrt(n)))))/log(sqrt(n))

ex: n=49

49 minimum = (24-sqrt(576-49 log^2(7)))/(log(7))
49 maximum = (24+sqrt(576-49 log^2(7)))/(log(7))


and min * max = n

 

[JeremyEbert]

more information

"Another link into Eulers Generalized Pentagonal Numbers and the divisor function d(n):



For our Divisor summatory function we have:

D(n) = SUM(d(n)) :

for k = 0 --> floor [sqrt n]
SUM (d(n)) = SUM ((2*floor[(n - k^2)/k]) + 1)


The notable difference in the equation from the published version is the:

(n - k^2)/k (congruence of squares)

which is derived from the

z = (n - k^2)/2k + i n^(1/2)

forming a parabolic coordinate system.


The function (n - k^2)/2k forms a divisor symmetry centered on the square-root of n.

Example:

k = divisors of n {1,2,3,4,6,9,12,18,36}
n = 36


+17.5, +8.0, +4.5, +2.5, 00.0, -2.5, -4.5, -8.0, -17.5




key results:
sqrt(n) = 0
Sum Terms = 0

Offsetting by -((n-1)/2) = -17.5 and taking the absolute values gives us:


0, 9.5, 13, 15, 17.5, 20, 22, 25.5, 35

Key results:
sqrt(n) = (n-1)/2;




Another way to generate these terms is:

((n-k)*(k-1)/2k) + (k-1)

The key ratio here being the (k-1)/2k function.

reducing this ratio sequence we get:

01/04, 01/03, 03/08, 02/05, 05/12, 03/07, 07/06, 04/09, 09/20, 05/11, 11/24, 06/13, 13/28, 07/15, 15/32, 08/17, 17/36

or

01 01 03 02 05 03 07 04 09 05 11 06 13 07 15 08 17
04 03 08 05 12 07 16 09 20 11 24 13 28 15 32 17 36


Showing a direct connection to Eulers Generalized Pentagonal Numbers and the divisor function d(n)



**

A026741 ( n if n odd, n/2 if n even. ) = xx, 00, 01, 01, 03, 02, 05, 03, 07, 04, 09, 05, 11, 06, 13, 07, 15, 08, 17
A022998 ( If n is odd then n else 2*n. ) = 00, 01, 04, 03, 08, 05, 12, 07, 16, 09, 20, 11, 24, 13, 28, 15, 32, 17, 36

A026741 = Partial sums give Generalized Pentagonal Numbers A001318 = 00, 01, 02, 05, 07, 12, 15, 22, 26, 35, 40, 51, 57
A022998 = Partial sums give Generalized Octagonal Numbers A001082 = 00, 01, 05, 08, 16, 21, 33, 40, 56, 65, 85, 96, 120 "

the contour plot shows the divisor function very nicely:

(n-k^2)/2k mod .5

http://www.wolframalpha.com/input/?i=ContourPlot[Mod[(-k^2+++n)/(2+k),+0.5],+{k,+-2,+2},+{n,+-4,+4}]

 

[JeremyEbert]

the contour plot shows the divisor function very nicely:

(n-k^2)/2k mod .5

http://www.wolframalpha.com/input/?i=ContourPlot[Mod[(-k^2+++n)/(2+k),+0.5],+{k,+-2,+2},+{n,+-4,+4}]

for those who have problems with wolframalpha.com

http://dl.dropbox.com/u/13155084/mod%20point%205.png

 

[JeremyEbert]

for the function

f(n,k) = ( (ln(k)/ln(n)) - ((((n-k)*(k-1)/(2k)) + (k-1))/(n-1) )

the

local minimum = (((n-1)/2)-sqrt(((n-1)/2)^2-(n*ln^2(sqrt(n)))))/log(sqrt(n))
local maximum = (((n-1)/2)+sqrt(((n-1)/2)^2-(n*ln^2(sqrt(n)))))/log(sqrt(n))

ex: n=49

49 minimum = (24-sqrt(576-49 log^2(7)))/(log(7))
49 maximum = (24+sqrt(576-49 log^2(7)))/(log(7))


and min * max = n

A better way to plot this:

e^(pi i ( ((n-k^2)/(2k)) / ((n-1)/2) ) ) for k=1 to 36, n=36

http://dl.dropbox.com/u/13155084/unit%20circle.gif