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Subject: "Inverse to an eisenstein-related function"     Previous Topic | Next Topic
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Conferences The CTK Exchange College math Topic #534
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Gotti
guest
Oct-14-05, 08:41 AM (EST)
 
"Inverse to an eisenstein-related function"
 
   (since it'seems,that the plus-sign disappears in postings and I do not realize how to keep them I use the # sign in the following)

I am fiddling with some type of generalizing of sinh/cosh and sin/cos-function and am searching a way to find an inverse, like
the ln()-series as inverse of the exp()-series

Taking v as the 2'nd complex root of 1 (v=-1) the cosh/sinh-pair is defined as
cosh(x) = (exp(x) # exp(v*x))/2
sinh(x) = (exp(x) # v*exp(v*x))/2

Then, for instance
cosh(x) # sinh(x) = exp(x)
cosh(x) - sinh(x) = exp(-x)
cosh(x)' = sinh(x), sinh(x)' = cosh(x)
and much more.


Taking v as the 3'rd complex root of 1 (v=0.5(-1 # sqrt(3)i), which are he basis for the eisenstein-integers, I can construct:
f(x) = (exp(x) # exp(vx) # exp(vvx))/3
g(x) = (exp(x) # vv exp(vx) # v exp(vvx))/3
h(x) = (exp(x) # v exp(vx) # vv exp(vvx))/3

Since v+vv=-1 and v*vv=1 I get for instance:

f(x) # g(x) # h(x) = exp(x)

f(2x) = f(x)^2 # g(x)h(x)
g(2x) = g(x)^2 # h(x)f(x)
h(2x) = h(x)^2 # f(x)g(x)

f(3x) = f(x)^3 # g(x)^3 # g(x)^3 # 6 f(x)g(x)h(x)
g(3x) = 3( f(x)^2 g(x) # g(x)^2 h(x) # h(x)^2 f(x))
h(3x) = 3( g(x)^2 f(x) # h(x)^2 g(x) # f(x)^2 h(x))

f(x)' = h(x) g(x)'=f(x) h'(x) = g(x)

and some more interesting properties.
First I would like to find series expressions for the inverses of this functions to be able to play with other properties numerically.

A bit more complicated is with cos/sin-function; here we have
v=4'th primitive complex root of 1 = i , and w = vv
and the according representation of cos(x) and sin(x) is
more tedious.

But analoguously as for instance

cos(x) = 1 # vv x^2/2! # x^4/4! # vv x^6/6! ....

which we are used to with the alternating plus and minus sign,
I created the f(),g(),h()-functions, now using the primitive 9'th complex root of 1, thus, for instance

f(x) = 1 # vv x^3/3! # v x^6/6! # x^9/9! ...
g(x) = x/1! # vv x^4/4! # v x^7/7! # x^10/10! ...

which are functions with the oscillating property like cos/sin, but diverging with growing x.

I have not many properties of this group of functions; but to proceed I would like to find also a series for the inverse, so that I can try some properties at least by numerical approximation.

Gotti


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sfwc
Member since Jun-19-03
Oct-14-05, 10:40 AM (EST)
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1. "RE: Inverse to eisenstein-related function"
In response to message #0
 
   >Taking v as the 3'rd complex root of 1 (v=0.5(-1 #
>sqrt(3)i), which are he basis for the eisenstein-integers, I
>can construct:
> f(x) = (exp(x) # exp(vx) # exp(vvx))/3
> g(x) = (exp(x) # vv exp(vx) # v exp(vvx))/3
> h(x) = (exp(x) # v exp(vx) # vv exp(vvx))/3
More generally, let a_1 up to a_n be the roots of a polynomial p with real coefficients. We may define a function f_p: R -> C, x -> Sum(i = 1..n, exp(a_i x))/n. This function will be real-valued and it and its first n-1 derivatives will bear some interesting relationships to one another. You are considering f_p where p(X) = X^3 - 1 (or more generally X^n - 1) here.

>Since v+vv=-1 and v*vv=1 I get for instance:
>
> f(x) + g(x) + h(x) = exp(x)
>
> f(2x) = f(x)^2 + 2 g(x)h(x)
> g(2x) = g(x)^2 + 2 h(x)f(x)
> h(2x) = h(x)^2 + 2 f(x)g(x)
>
> f(3x) = f(x)^3 + g(x)^3 + h(x)^3 + 6 f(x)g(x)h(x)
> g(3x) = 3( f(x)^2 g(x) + g(x)^2 h(x) + h(x)^2 f(x))
> h(3x) = 3( g(x)^2 f(x) + h(x)^2 g(x) + f(x)^2 h(x))
>
> f(x)' = h(x) g(x)'=f(x) h'(x) = g(x)
I have corrected these, as they were very slightly wrong. The last property is tautologous in the scheme I mentioned above: It is the definition of g and of h.

>First I would like to find series expressions for the
>inverses of this functions to be able to play with other
>properties numerically.
I'm afraid I can't see how to do this in any obvious way. Good luck.

>A bit more complicated is with cos/sin-function; here we
>have
> v=4'th primitive complex root of 1 = i , and w = vv
>and the according representation of cos(x) and sin(x) is
>more tedious.
A way to represent cos and sin in the scheme I have presented is to note that cos(x) = f_p(x) where p(X) = X^2 + 1 is the fourth cyclotomic polynomial, and sin is the derivative of cos. The first, second and third cyclotomic polynomials give exp itself, exp(-x) and the function q(x) = 1/2(exp(vx) + exp(v^2 x)) where v is a cube root of unity. The natural generalisation of cos and sin would seem to be consideration of f_p for a general cyclotomic polynomial p.

>But analoguously as for instance
>
> cos(x) = 1 # vv x^2/2! # x^4/4! # vv x^6/6! ....
>
>which we are used to with the alternating plus and minus
>sign,
>I created the f(),g(),h()-functions, now using the primitive
>9'th complex root of 1, thus, for instance
>
> f(x) = 1 # vv x^3/3! # v x^6/6! # x^9/9! ...
> g(x) = x/1! # vv x^4/4! # v x^7/7! # x^10/10! ...
>
>which are functions with the oscillating property like
>cos/sin, but diverging with growing x.
The functions f and g which you construct are not real-valued on the reals, and there is not a clear analogy between the construction of f and g and that of the power series of cos and sin. But taking the real part of f gives f_p where p is the 9th cyclotomic polynomial, which is what I suggest you use as your generalisation.

Thankyou

sfwc
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