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Off-Topic Discussion => General Discussion => Topic started by: Joey_21 on February 19, 2004, 11:51:15 pm

Title: Recreational Mathematicians...
Post by: Joey_21 on February 19, 2004, 11:51:15 pm
I've got a forum for all recreational mathematicians with a variety of subject coverage. Sign up if you're interested. :)

http://recreativemath.proboards27.com/
Title: Recreational Mathematicians...
Post by: Setekh on February 19, 2004, 11:56:59 pm
Wow... now that's what I call a niche board. ;) You know, I'll be taking a fair few maths subjects at uni, and I like to enjoy what I do, so I think I'll join you. :)
Title: Recreational Mathematicians...
Post by: CP5670 on February 20, 2004, 12:54:15 am
Make a special functions section for people like me. :D
Title: Recreational Mathematicians...
Post by: Joey_21 on February 20, 2004, 01:02:01 am
Quote
Originally posted by CP5670
Make a special functions section for people like me. :D


I believe the "Recreate" section would work quite nicely. :)
Title: Recreational Mathematicians...
Post by: CP5670 on February 20, 2004, 02:27:07 am
Since you seem to be into numerical evaluation computer programs, do you happen to know anything about Mathematica? This was my write-up for that ISTS thing: http://www.3dap.com/hlp/hosted/procyon/misc/AGFF.chm

I have been making a mathematica package to get numerical values (to get graphs and stuff) for this function but there are some things I don't really know how to do.
Title: Recreational Mathematicians...
Post by: Joey_21 on February 20, 2004, 07:23:31 am
Yum...  indeed, I have Mathematica 5.

I have a question about [2.02].
PF(n, x) = e^(HurwitzZeta'(-n, x)-Zeta'(-n))
= e^(HurwitzZeta'(-n, x)) / e^(Zeta'(-n))

When we refer to the derivative of the HurwitzZeta function and the regular Zeta function, we're referring to the derivative with respect to n, correct? If so then I will make a numerical computational derivative of PF(n, x) to get an idea of what it might look like.

d/dn Zeta(-n) ~ (Zeta(-(n+Dn))-Zeta(-(n-Dn)))/(2Dn)
d/dn HurwitzZeta(-n, x) ~ (HurwitzZeta(-(n+Dn), x)-HurwitzZeta(-(n-Dn), x))/(2Dn)

And then work from there...
Title: Recreational Mathematicians...
Post by: J3Vr6 on February 20, 2004, 08:10:47 am
[scoobydoo]

argh??


[/scoobydoo]




Mommy?
Title: Recreational Mathematicians...
Post by: Goober5000 on February 20, 2004, 08:24:00 am
CP, have you done any engineering math?  Convolution, for example?
Title: Recreational Mathematicians...
Post by: 01010 on February 20, 2004, 10:42:33 am
Recreational and mathematics are two words that normally only go together in my worst nightmares.
Title: Recreational Mathematicians...
Post by: Martinus on February 20, 2004, 01:32:34 pm
Quote
Originally posted by 01010
Recreational and mathematics are two words that normally only go together in my worst nightmares.

[color=66ff00]Yeah, it's like a dirty phrase I shudder everytime I think it.

*Recreational mathematics*

Gah!
[/color]
Title: Recreational Mathematicians...
Post by: aldo_14 on February 20, 2004, 01:38:28 pm
:shaking:
Title: Recreational Mathematicians...
Post by: Flipside on February 20, 2004, 01:40:30 pm
LOL Same here, though 'Recreational Mathmatics' sounds like 'Recreational Drugs' only with a worse headache the next morning ;)

Must be said though, Kudos to people who understand this stuff. Most boards I look at seem to be full of people who've only just figured out what the img tag does, yet HLP has actually got a very high percentage of intelligent and talented people :D

Just wish I was one of 'em ;)
Title: Recreational Mathematicians...
Post by: aldo_14 on February 20, 2004, 01:45:02 pm
I have nothing but respect for those who do maths.  I just can't understand why...........


:D
Title: Recreational Mathematicians...
Post by: ZylonBane on February 20, 2004, 02:02:23 pm
Sorry, I don't do recreational math. Math is a special thing, that should be reserved for the bonds of holy matrimony.
Title: Recreational Mathematicians...
Post by: CP5670 on February 20, 2004, 10:19:13 pm
Quote
Yum...  indeed, I have Mathematica 5.

I have a question about [2.02].
PF(n, x) = e^(HurwitzZeta'(-n, x)-Zeta'(-n))
= e^(HurwitzZeta'(-n, x)) / e^(Zeta'(-n))

When we refer to the derivative of the HurwitzZeta function and the regular Zeta function, we're referring to the derivative with respect to n, correct? If so then I will make a numerical computational derivative of PF(n, x) to get an idea of what it might look like.

d/dn Zeta(-n) ~ (Zeta(-(n+Dn))-Zeta(-(n-Dn)))/(2Dn)
d/dn HurwitzZeta(-n, x) ~ (HurwitzZeta(-(n+Dn), x)-HurwitzZeta(-(n-Dn), x))/(2Dn)

And then work from there...


Yeah, the derivative is on n; I figured that there is no ambiguity in writing z' because the derivative over x is basically another (normal) zeta function. The problem with doing it with Mathematica's built in zeta function is that the function is loaded with bugs (at least in 4.2, which is what I have; might have been fixed in 5), and with derivatives for non-real arguments, it's also slow. Instead, here is what I have done so far:
http://www.3dap.com/hlp/hosted/procyon/misc/PowerFactorial.m

This is a really messy WIP but it mostly works, although some things are definitely not being done in the most efficient way possible. The package can be loaded using << (Get) and contains the functions PowerFactorial[n,x] and LogPowerFactorial[n,x] (the second of these is analogous to LogGamma). For example, you can try Plot[Abs[PowerFactorial[1,x]],{x,-2,2}] or Plot3D[Re[PowerFactorial[3/2,x+I y]],{x,-2,2},{y,-2,2}] to see how it works.

The fastest method for numerical calculation is definitely the [7.01]/[7.02] asymptotic power series; an arbitrary accuracy goal cannot be specified for this but it still more than good enough for getting graphs of the function. [5.03] (the plana expansion integral) also seems to be fairly fast and can be made accurate to any required precision. These are the two methods I implemented into that package. I have not been able to get the function options to work correctly, so you will need to manually edit the default method in the package file to switch between the methods (look for the Options definition part and change it between PlanaIntegral and AsymptoticSeries).

The [2.03] difference equation needs to be used for both of these methods since the formulas are convergent/accurate only in parts of the x plane. The two formulas that are valid for all x, the [4.03] quasi-WFT product and the [7.03] limit, converge way too slowly to be of any use numerically, so this seems to be the only way to do it.

Quote
CP, have you done any engineering math? Convolution, for example?


You mean the convolution integral transform? I frequently run into integrals of that type but have only encountered the general transform once or twice. I've done all my stuff from a pure math, classical analysis-based perspective but am familiar with some physics and number theory applications.
Title: Recreational Mathematicians...
Post by: delta_7890 on February 20, 2004, 10:25:48 pm
..My head hurts..
Title: Recreational Mathematicians...
Post by: Carl on February 20, 2004, 10:31:27 pm
yeah, as soon as i saw this thread title i knew CP would be having fun.
Title: Recreational Mathematicians...
Post by: Setekh on February 21, 2004, 04:32:40 am
Quote
Originally posted by Flipside
Must be said though, Kudos to people who understand this stuff. Most boards I look at seem to be full of people who've only just figured out what the img tag does, yet HLP has actually got a very high percentage of intelligent and talented people :D

Just wish I was one of 'em ;)


Couldn't have said it better myself, Flipside... especially the "wish I was one of 'em" bit. :D

Btw, Joey, I love your motto. :D:yes:
Title: Recreational Mathematicians...
Post by: Joey_21 on February 21, 2004, 11:18:30 am
CP: I'll take a look at it. I have never programmed something for Mathematica so this is sort of new to me. I suppose the program is easy enough to get loaded so I'll attempt that now... Mainly I use Mathematica for numerical computations and for extra visualization of equations in my multivariable calculus class.

Quote
Originally posted by Setekh


Btw, Joey, I love your motto. :D:yes:


:D I had just realized where I heard something similar before... :nervous: