When Does the Exponential Function Reach Its Maximum Value?

In summary: 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)=kwhen k=n^(1/2) then t=0.5 and n^(t)=kwhen k=n then t=1 and n^(t)=kLooking 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/
  • #1
JeremyEbert
204
0
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?
 
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  • #2
What is n?
 
  • #3
SteamKing said:
What is n?
integers
0<n<LIM
 
  • #4
^ 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 [itex]Q_n(k)[/itex] be that expression. The problem is to find [itex]k_0[/itex] such that, for each n, [itex]Q_n(k_0) \geq Q_n(k)[/itex] for all k -- and subsequently, to show that apparently [itex]k_0 = 36[/itex]. Is this what you're asking?
 
  • #5
Dr. Seafood said:
^ 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 [itex]Q_n(k)[/itex] be that expression. The problem is to find [itex]k_0[/itex] such that, for each n, [itex]Q_n(k_0) \geq Q_n(k)[/itex] for all k -- and subsequently, to show that apparently [itex]k_0 = 36[/itex]. Is this what you're asking?

Yes, you are correct. Thanks for stating it in a better way.
 
  • #6
JeremyEbert said:
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.
 
  • #7
JeremyEbert said:
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
 
  • #8
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 [Broken]

http://www.wolframalpha.com/widgets/view.jsp?id=b8ba4c95900e275211a98d8bd1b0a53c [Broken]
 
Last edited by a moderator:
  • #9
JeremyEbert said:
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 [Broken]
http://dl.dropbox.com/u/13155084/nt-k-9.png [Broken]
http://dl.dropbox.com/u/13155084/nt-k-16.png [Broken]
http://dl.dropbox.com/u/13155084/nt-k-25.png [Broken]
http://dl.dropbox.com/u/13155084/nt-k-36.png [Broken]
http://dl.dropbox.com/u/13155084/nt-k-49.png [Broken]
 
Last edited by a moderator:
  • #10
JeremyEbert said:
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 [Broken]
http://dl.dropbox.com/u/13155084/nt-k-9.png [Broken]
http://dl.dropbox.com/u/13155084/nt-k-16.png [Broken]
http://dl.dropbox.com/u/13155084/nt-k-25.png [Broken]
http://dl.dropbox.com/u/13155084/nt-k-36.png [Broken]
http://dl.dropbox.com/u/13155084/nt-k-49.png [Broken]

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?
 
Last edited by a moderator:
  • #11
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 "
 
Last edited:
  • #13
JeremyEbert said:
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 [Broken]
 
Last edited by a moderator:
  • #14
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
 
  • #15
JeremyEbert said:
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}]
 
  • #17
JeremyEbert said:
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 [Broken]
 
Last edited by a moderator:

1. What is an exponential function?

An exponential function is a mathematical function in the form of f(x) = ab^x, where a and b are constants and x is a variable. It is a type of non-linear function in which the input variable appears in the exponent. The graph of an exponential function is a curve that increases or decreases rapidly.

2. What is the difference between an exponential function and a linear function?

The main difference between an exponential function and a linear function is that the variable in an exponential function appears in the exponent, while in a linear function it appears in the base. This results in a curved graph for an exponential function and a straight line for a linear function.

3. How is an exponential function used in real life?

An exponential function is used to model situations where the rate of change is proportional to the current value. It is commonly used in finance, population growth, radioactive decay, and other natural phenomena. It is also used in computer science and engineering to analyze algorithms and growth of data.

4. How do you solve an exponential function?

To solve an exponential function, you can use logarithms, which are the inverse of exponential functions. By taking the logarithm of both sides of the equation, you can isolate the variable and solve for it. Another method is to use a graphing calculator or software to find the intersection of the exponential function and a straight line.

5. What are the properties of exponential functions?

Some of the properties of exponential functions include:
1. The domain is all real numbers.
2. The range is positive real numbers if the base is greater than 1, or negative real numbers if the base is between 0 and 1.
3. The function is continuous and differentiable for all real numbers.
4. The graph is always increasing or decreasing, depending on the value of the base.
5. The x-intercept is always 0, and the y-intercept is 1.
6. The function has a horizontal asymptote at y = 0.

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