Talk:Characterizations of the exponential function

message: Give me a sec, I think the "intuitive" proof actually links defs 1 and 3, not 1 and 2, and might actually be able to replace the more clunky Havil proof I have here. But, I'm pooped and will return later. Revolver 01:21, 24 Jul 2004 (UTC)
replies: text: I don't really care, to be honest. Call it what you want. Revolver 22:42, 24 Jul 2004 (UTC)

text: The main problem that pops into my mind is that it's not immediately obvious why such a function exists, i.e. why there is y s.t. y' = y and y(0) = 1. Without falling back on one of the others defs, I mean. If anyone knows of a nice way to prove this existence independent of the other 3 defs, I'd be happy to include it. Revolver 08:35, 20 Sep 2004 (UTC)
thread_title: Revamp

message: The proof of the equivelance of charactaristics 1 and 3 isn't actually a proof, it's a circular argument and I've no idea why it's there. It claims that: 
\lim_{n\to\infty}\ln\left(1+\frac{x}{n}\right)^n=\lim_{n\to\infty}n\ln \left(1+\frac{x}{n}\right).
which should only hold surely, if we were dealing with a logarithm, but that's exactly what we're trying to proove, that ln x is the same thing as loge(x)
* That's easy to prove; integral of dt/t from 1 to x^n, by the substitution u^n=t, becomes the integral of du n u^(n-1)/u^n = n u/du for u = 1 to x. Any good calculus book should have this proof. User:Ben Standeven as 70.249.214.16 03:05, 14 October 2005 (UTC)
thread_title:  What did I miss? 

message: Hi,
The point raised on the discussion page about the equivalence of 1 and 3 is worthy of being addressed in the main article, I believe, as a perceptive reader will see that one is using a result whose prior establishment is not clear.  (Indeed, I thought there was a problem and, having satisfied myself that there wasn't, inserted an edit to clear it up, before I had read the point on the discussion page. [Sorry for not checking first]) I hope the edit is still deemed worthy of retention.
Hugh McManus 14:56, 26 February 2007 (UTC)
thread_title: Equivalence of 1 and 3

message: The proof given only works for x \geq 0.  In line 6, you are arguing (1-1/n)(1-2/n)\leq 1 implies that (x^3/3!)(1-1/n)(1-2/n)\leq x^3/3! and this requires non-negative x.
Indeed, the statement in Characterization 1 that a_n=(1+x/n)^n is an increasing sequence may not be  valid if x<0. Fathead99 (talk) 14:24, 30 January 2008 (UTC)
thread_title:  Equivalence of characterizations 1 and 2 

message: From the limit (char-1) follows the differential equation (char-4) by differentiation, and then the taylor series (char-2). 
It is unclear whether the article talks about the real or complex exponential function. The reference to positive values of x indicate real exponentials, but the results are important for complex exponentials. 
The formulation "in several cases can be extended to any Banach algebra" is confusing; in which cases can it be extended, and in which cases can it not be extended? Is differentiation defined for any Banach algebra? The Banach algebra article didn't tell. 
Bo Jacoby 07:29, 23 July 2007 (UTC)
thread_title:  Taylor series. 

message: For some of the characterizations given here (e.g. power series) one may take the domain to be the whole complex plane, but for others (monotonicity plus the functional equation) one must assume the domain is the real line.  (There do exist continuous functions on the complex plane that satisfy the functional equation and the initial condition but are not equal to the natural exponential function; they are nowhere (complex-)differentiable.  For example, observe that
: 
::  g(x + iy) = e^x(\cos(2y) + i\sin(2y))\, 
: 
for x and y real, is continuous everywhere in the complex plane and satisfies the functional equation
: 
::  g(z + w) = g(z)g(w)\, 
: 
for z and w complex, and the initial condition
: 
::  g(1) = e\,
: 
but not the Cauchy-Riemann equations, and is therefore nowhere complex-differentiable.) Michael Hardy 13:16, 26 August 2007 (UTC)
thread_title:  Real versus complex 

message: I think at least definitions 1., 2., and 5. may seem to a beginning student like rabbits pulled out of a hat. Among all possible limits, why study \lim_{n\to\infty} \left(1+\frac{x}{n}\right)^n? Among all possible power series, why study \sum_{n=0}^\infty {x^n \over n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots? Among all measurable functions f satisfing f(x+y)=f(x)f(y), why study the one satisfying f(1)=e? I will try to answer these quations in what follows.
Even definitions 3. and 4. (which are really two sides of the same coin, the relation dy=ydx first when x is considered a function of y and then when y is considered a function of x) may need some motivation. In 3., why start the integral at 1? In 4., why chose y(0)=1? The answer is that otherwise we do not get the properties log(xy)=log(x)+log(y) and e^{(x+y)}=e^xe^y.
The differential equation y'=y is sufficiently well-behaved to have a power series solution. From 4. we thus get 2. Of course, 3. also leads to a power series: log(1+x)=x-x^2/2+x^3/3-x^4/4+.... Perhaps this power series should also be included among the definitions.
Although it is commonplace, I find the choice of the variable name "n" in 1. unfortunate. It suggests an integer variable when any complex number is perfectly permissible. I also prefer to use a variable that approaches 0 to one that approaches infinity, so rather than motivating 1., I will motivate: e^x = \lim_{h \to 0} {(1+xh)}^{1/h}.
If you try to differentiate a^x with respect to x in the most straightforward of ways, you get \lim_{h\to0}\frac{a^h-1}{h} a^x. Since we know that the derivative should be log(a)a^x, and since a can be any positive real number, we must have come across an expression for the function log. Since the exponential function is just the inverse of log, it is natural to wonder whether you can turn the expression for log into one for the exponential function. You can, and what you get is exactly e^x = \lim_{h \to 0} (1+xh)^{1/h}. (And once we have this expression it is a good idea to use the binomial series to expand it; this constitutes an alternative path to the power series 2.)
As for 5., it is really incomplete as a definition since it presupposes that you already have a definition of e. It seems to me that what we do have is an interesting characterisation of the function f(x)=a^x. It is determined once you know:
I. that f(x+y)={f(x)}{f(y)},
II. that f is measurable at one point, and
III. the value of f(x) for any value of x except 0. Mattias Wikstrom (talk) 18:02, 8 February 2008 (UTC)
thread_title:  Motivating the definitions? 

message: Looking through this, I notice that there is a proof that 5 implies 1, but not that 1 (or one of the others) implies 5.  Should be easy to fix (a proof should be in Rudin or any other standard textbook). — Steven G. Johnson (talk) 18:09, 11 December 2011 (UTC)
thread_title:  Missing proof 

message: Am I missing something or does the article never link characterization 4 to anything else? Dmcq (talk) 04:49, 8 February 2012 (UTC)
replies: text: Plugging the Taylor series (characterization 2) into characterization 4 quickly shows the relation.  About characterization 5, even though it is a nice characterization, it is far less common than the other four characterizations. It should probably be not placed on the same footing.  Perhaps the main list should have only 4 characterizations (we can keep the 5'th, writing something like:  Besides these, there are other characterizations as well, for example: .. #5).  71.229.28.197 (talk) 17:38, 22 February 2014 (UTC)
replies: text: Perhaps Dmcq's point is that we should have a subsection Equivalence of characterizations 2 and 4, rather than leaving it to the reader? —Quondum 18:13, 22 February 2014 (UTC)
thread_title:  Characterization 4 again 

message: This article contains the errorneous series for exponent
: e^x = \sum_{n=0}^\infty {x^n \over n!}
But it's clear that for x=0 the first item {0^0 \over 0!} is nonsense because 0^0 isn't correct mathematical expression, see Indeterminate form.
I modified the text to correct syntax:
: e^x = 1 + \sum_{n=1}^\infty {x^n \over n!}
but  undid my modification with comment: «Unsourced claim of error in edit summary, and atypical presentation». No source is needed for the obvious failure. LGB (talk) 10:40, 14 February 2014 (UTC)
replies: text: You're confusing the indeterminate form 00 with the expression 00. Talk:Exponentiation and several archives on the topic should make it clear that it is not so simple. —Quondum 06:29, 15 February 2014 (UTC)
replies: text:  (1) The section «the expression 00» has no reference to sources. (2) The same Exponentiation section warns honestly: «Not all sources define 0^0 to be 1». (3) You have no basis  to suppose that all Wikipedia readers know and support such strange convention — even without any commentary and reference to Exponentiation section. My modified formula is clearly better for a reader and I hope you'll not insist upon equivocal and confusing formula. Otherwise let Wikipedia talk:WikiProject Mathematics will discuss the subject. LGB (talk) 11:22, 15 February 2014 (UTC)
replies: text: You're welcome to do discuss the matter there, but be aware that this topic has been discussed at length on Wikipedia, at times quite heatedly. It would make sense to first acquaint yourself with the existing consensus (or lack thereof!) by finding and examining the extensive past discussions on the topic, such as the example to which I linked, before reopening the discussion, as many will have tired of it already. If the rationale behind your edit were to be accepted, it would imply a similar revision to many other articles, including Exponentiation, Exponential function, Taylor series, Laurent series, Binomial theorem, Derivative and at a guess no less than hundreds of others. There is no shortage of sources that use the expression  to mean 1 in exactly the way that you object to. To find a source that carefully avoids its use with this meaning might be a challenge, though. —Quondum 16:29, 15 February 2014 (UTC)
replies: text: I agree with LGB that it is wrong to say in one page that 0^0 is undefined, while assuming in another page that 0^0 will be evaluated to 1.  Unfortunately, that is precisely what many calculus textbooks do.  The mathematical solution to this problem is very easy (accept the empty product rule, which makes 0! and 0^0 equal to 1).  The goal in wikipedia is to say what the textbooks say, so if the textbooks contradict themselves, then wikipedia will do likewise.  This means that the only way to fix this contradiction on wikipedia is to convince authors of calculus books that 0^0 is 1.  MvH (talk) 18:51, 21 February 2014 (UTC)
replies: text: This article says nothing of the sort. Perhaps you're confusing an expression with a limiting form? The shorthand 00 for the limit \lim_{x\to 0^+,y\to 0}x^y looking like the expression might be confusing, but that is all it is: a shorthand.  —Quondum 02:38, 22 February 2014 (UTC)
replies: text: Quondum, yes, I agree that one can do this correctly (0^0 = 1, 0/0 = undefined, while "0^0" and "0/0" refer to limits that require more than direct substitution).  Widespread confusion arises because there are many calculus textbooks that leave 0^0 undefined, even though the same textbooks expect the reader to evaluate 0^0 to 1 in many formulas.  It's not the readers fault when they are confused. In the past, 1 was considered a prime number, until it was found that this was inconvenient in an increasingly large number of situations.  The same will happen with the view that 0^0 is undefined, and for the same reason.  The option "undefined" is inconvenient in an increasingly large number of places, so eventually people will settle on "1". In research papers, 0^0=1 is already the status quo. Calculus textbooks will likely be the last ones to adopt this.  71.229.28.197 (talk) 16:03, 22 February 2014 (UTC)
replies: text: I agree with you about problems relating to any (lack of) definition of 00 and the consequent confusion, but that was not my point, so I guess I was unclear. I'm saying that since "00" and "the indeterminate form 00" refers to distinct concepts, one must take care to use the full phrase in the latter. Conflating the two distinct concepts seems to have been underlying LGB's assertion. The eventual resolution may occur differently from your prediction. I would like to see notation arising that distinguishes between the two fundamentally different types of exponentiation, which will allow 00=1 without those in analysis objecting (other than to the change of notation). —Quondum 17:50, 22 February 2014 (UTC)
thread_title:  Error in series for exponent 

message: One characterization I've seen used in calc textbooks (eg Briggs, Cochran, Gillett): let $e$ be the number such that $\lim e^h-1/h=1.$ Then $e^x$ is the exponential function with that base. This should also be shown equivalent. -lethe talk [ +] 04:29, 21 October 2015 (UTC)
thread_title:  Missing characterization 

message: The generalisation to negative values of x is done in an ad-hoc way. It raises the question of how to do the same for the complex numbers.
I think that using the Dominated convergence theorem allows for a generalisation to all unital Banach algebras.
--Svennik (talk) 21:50, 30 September 2019 (UTC)
thread_title:  More general proof that 1 implies 2 using the Dominated Convergence Theorem