Is e^x zero in some extended number system?
Dave L. Renfro
and I'm curious if any related work has been done
along these lines.
Paul Dienes, "The exponential function in linear algebras",
Quarterly Journal of Mathematics (Oxford) 1 (1930), 300-309.
http://www.emis.de/cgi-bin/JFM-item?56.0151.02
The text of the first two paragraphs follows.
"The introduction of complex numbers was chiefly
suggested by the problem of determining the zeros
of polynomials. Some integral functions, such as
e^x, have no zero in the field of complex numbers.
This fact suggests the following question. Can we
generalize the idea of number to such an extent
that the exponential function may have a zero in
the extended field?"
"We shall prove in this Note that the exponential
function has no zero in the linear associative
algebra to a finite base, and that it has no
zero in finite non-associative linear algebras.
[Dienes assumes, of course, that the algebra
product has a multiplicative identity.] This
result extends to a large class of algebras to
an infinite base. In particular, the exponential
function has no zero in the tensor algebra of
relativity theory and it misses only singular
tensor values which do not divide some tensor.
Moreover, in Hilbert's [*] algebra of infinite
bounded matrices, so important in atom mechanics,
the exponential function has no absolutely
bounded matrix zero."
[*] "D. Hilbert, 'Grundzüge einer allgemeinen
Theorie der linearen Integralgleichungen',
Leipzig (1912), pp. 128-9."
Dave L. Renfro
David W. Cantrell
To answer the question in your title: Sure.
In [-oo, +oo], the two-point extension of the reals, we have e^(-oo) = 0.
But I certainly suppose you already knew that.
David Cantrell
> [*] "D. Hilbert, 'Grundz=FCge einer allgemeinen
Dave L. Renfro
> To answer the question in your title: Sure.
> In [-oo, +oo], the two-point extension of the
> reals, we have e^(-oo) = 0. But I certainly
> suppose you already knew that.
Well, in retrospect I know that, but I didn't
think about this possibility of looking at the
issue I raised!
Dave L. Renfro
Robert Israel
What is he using as definition of the exponential
function? Maybe I'm missing something, but all
definitions of "exponential function" that I've seen
(e.g. on Lie algebras or Banach algebras) are either
explicitly mapping something into a group or make it
easy to prove that exp(x) has an inverse, namely
exp(-x).
Robert Israel isr...@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
Dave L. Renfro
> What is he using as definition of the exponential
> function? Maybe I'm missing something, but all
> definitions of "exponential function" that I've seen
> (e.g. on Lie algebras or Banach algebras) are either
> explicitly mapping something into a group or make it
> easy to prove that exp(x) has an inverse, namely
> exp(-x).
Continuing quoting from his paper . . .
"Since we have to define the symbol
1 + sum(n=1 to oo) of (x^n)/(n!)
for numbers x of a general linear algebra, it is
necessary to suppose that 1 is a number of the
algebra, i.e. that the algebra has a modulus e_0
such that if x is any number of the algebra,
x(e_0) = (e_0)x = x. Obviously the algebra must
also contain a zero element."
"2. Construction of an algebra to an infinite base
The numbers of an algebra to the base e_0 = 1,
e_1, ..., e_n, ... are, by definition, the forms
x = sum(i=0 to oo) (x_i)(e_i), where the components
x_i are real or complex numbers. Addition is defined
of the corresponding components. Multiplication is
supposed to be distributive with respect to addition
and is defined (a) by the multiplication table
(e_i)(e_j) = sum(k=0 to oo) (m_ijk)(e_k),
where m_ijk are arbitrarily given real or complex
numbers, i.e. by
xy = sum(k=0 to oo) { sum(i,j = 0 to oo) (x_i)(y_j)(m_ijk) } e_k,
and (b) by the convention that if c is any real
or complex number,
cx = xc = sum(i=0 to oo) (c)(x_i)(e_i).
Zero is the number of the algebra whose components
are all zero. We see that 0x = x0 = 0."
"We also notice that multiplication does not lead
to a number, unless the infinity of double series
sum(i,j = 0 to oo) (x_i)(y_j)(m_ijk) (k = 0, 1, ...)
are all convergent, so that, in every algebra to an
infinite base, there are numbers whose product is
meaningless, unless, for every k and for i > N,
j > N, we have m_ijk = 0. In fact, we can choose
x_i, y_i such that (x_i)(y_j)(m_ijk) = 1 for an
infinity of suffixes i and j."
"The necessary and sufficient conditions that
the multiplication of units should be associative
is that [(e_i)(e_j)](e_k) = (e_i)[(e_j)(e_k)], i.e.
sum(p=0 to oo) (m_ijp)(m_pkl)
= sum(p=0 to oo) (m_ipl)(m_jkp) (4)
for every i, j, k, l. In an algebra to an infinite
base these conditions are not sufficient to make
multiplication associative in the whole algebra.
In fact, the detailed form of the equation
(xy)z = x(yz) is
sum(p,k) sum(i,j) (x_i)(y_j)(z_k)(m_ijp)(m_pkl)
= sum(p,i) sum(j,k) (x_i)(y_j)(z_k)(m_jkp)(m_ipl), (5)
and unless the order of summation can be inverted,
(5) is not necessarily a consequence of (4)."
"For example, in the infinite matrix algebra to a
double array of units e_ij with the multiplication
table
(e_ij)(e_kl) = 0 if j not equal to k
= e_il if j = k,
multiplication of units is associative, but for
suitably chosen coefficients, the equation
[ messy stuff omitted ]
is not satisfied. This remark seems to be necessary,
for in papers on quantum theory the associative
property is taken for granted also for non-bounded
infinite matrices."
Dave L. Renfro