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Young physicist said:Well, with my proof, that n can be replaced by 4,5 or any natural number.
What about this, as a variation of your proof?
##1 = \frac{a^0}{a^0} = a^{0-0} = a^0##
What do you think of that?
Young physicist said:Well, with my proof, that n can be replaced by 4,5 or any natural number.
Ibix said:On the argument about definition versus proof, is part of the problem that there's only one sensible extension of ##a^n## from shorthand for "##n## ##a##s multiplied together" to the case of ##n\leq 0##? Several ways have been shown of doing it in this thread, but they all lead to the same result. Would a counter-example be trying to extend division (which already gives us a way into non-integer numbers starting from the integers) to cover the case of 0/0? You could argue it should be 1 because it's something divided by itself, or 0 because it's zero times something, or infinite because it's something divided by zero. There's no single definition consistent with all the rules, so we leave it undefined (note "undefined" not "unprovable" or something of that nature) and try to take limits on a case-by-case basis?
Too many question marks there, but am conscious that I'm outside my field here.
Ibix said:Too many question marks there, but am conscious that I'm outside my field here.
Try this...jbriggs444 said:You seem to have skipped a few steps. The definition in question does not involve multiplication at all. But let us ignore that.
Let a = x/2.
Let b = x/2
By definition f(x)=f(a+b)=f(a)f(b)f(x)=f(a+b)=f(a)f(b)f(x) = f(a+b) = \frac{f(a)}{f(b)}
But a=b so f(a) = f(b) and f(a)f(b)=1f(a)f(b)=1\frac{f(a)}{f(b)}=1
bhobba said:What I am about to say is not at the B level. So please B level people ignore it - it involves calculus.
For those still with me please read the Harvey Mudd notes I posted. Its basic simple calculus. There is no mystery like 0/0 or anything like that. Its simply you can rigorously define e^x for all real x. Its very basic really.
Normally this stuff is done before calculus so you must resort to definitions that make sense - but once you have the tool of calculus it falls out straight away. Pedagogically I prefer the calculus way - but either way is logically valid - its just I remember when I came across this stuff in the classroom I had the feeling of the first post - 'for some reason this does make sense to me but I have a feeling the result is not satisfying enough.'. That's all there is to it - some thinking students find the usual way not satisfying - me amongst them. I was fortunate in knowing calculus from self study - so it wasn't that 'unsatisfying' - but many do not have that tool yet. I am also suggesting maybe they should do them at the same time - hand-wavy calculus is not hard and much more satisfying than definitions. Just me - others it may not worry.
Thanks
Bill
Well, except zero.PeroK said:What about this, as a variation of your proof?
##1 = \frac{a^0}{a^0} = a^{0-0} = a^0##
What do you think of that?
Young physicist said:Well, except zero.
Well, we are trying to prove a0 = 1 ,right?PeroK said:Why?
Young physicist said:Well, we are trying to prove a0 = 1 ,right?
You can't use the result of a proof in the proof!
Just like you can't use the vocabulary itself in it's own definition.
PeroK said:But, in the sense of mathematical development, where does calculus come from?
bhobba said:A rigorous development? That's analysis and the motivation is the issues that can arise in calculus if you are not careful. By that stage you will know a^0 = 1 for sure. But a hand-wavy development doesn't require it eg:
https://en.wikipedia.org/wiki/Calculus_Made_Easy
Its based on intuitive ideas like very small numbers dx you can for all practical purposes ignore and most certainly you can ignore dx^2. You can interweave it into a pre-calculus or algebra 2 + trig course they call it in the US before a student tackles calculus proper with ideas of limits etc made clearer. In Aus calculus and pre-calculus is taught in an integrated way and you can most certainly do it in such a course. I have to say however I don't think they do it like that here in Aus - I think they do it like the link given previously to the UNSW paper on it. Why - well I do not think most students are like the OP and feel dissatisfied with the definition route - only people like me that like to think about it feel a bit uneasy - it's in the back of your mind - why is this based on definitions - there must be something more going on - and indeed there is.
Thanks
Bill
Yes. Recall that we are operating under the proposed property that ##f(a+b)=\frac{f(a)}{f(b)}##.Rada Demorn said:Try this...
Let a = x/3
Let b = x/3
Let c = x/3
What do you get? Is it still equal to 1?
No.jbriggs444 said:Yes. Recall that we are operating under the proposed property that ##f(a+b)=\frac{f(a)}{f(b)}##.
We have already established that if f(x) is defined at all (and is non-zero) then it must be equal to 1 everywhere. You ask about f(a+b+c) with a=b=c.
##f(a+b+c) = f((a+b)+c)= \frac{f(a+b)}{f(c)} = \frac{f(a)/f(b)}{f(c)}= \frac {1/1} {1} = \frac{1}{1} = 1##
##f(a+b+c) = f(a+(b+c)) = \frac{f(a)}{f(b+c)} = \frac{f(a)}{f(b)/f(c)} = \frac {1}{1/1} = \frac{1}{1} = 1##
As I had pointed out previously, division is associative if you only ever divide by one.
You seem to have a fundamental misunderstanding.Rada Demorn said:you are taking f(c) = 1 having only noticed this before and only for the case f (c/2+c/2) = 1. But what if c = c/3 + c/3 + c/3?
jbriggs444 said:If we demonstrate that f(c) must be equal to 1, then f(c) must be equal to 1. If we are able to demonstrate this without assuming anything about c then it follows that f(x) = 1 for all x. We have, in fact provided such a demonstration.
No need to assume. Both(*) are obviously true and easily provable.Rada Demorn said:You are assuming that c= c/2 + c/2 or x = x/2 + x/2.
PeroK said:##e^x = \Sigma_{n = 0}^{\infty} \frac{x^n}{n!}##
PeroK said:This is interesting. I've never questioned the need to define certain things in mathematics. I suppose having studied pure maths it just is such a part of mathematics.
bhobba said:There is no logical issue - its done all the time. But in this case what you are doing by these definitions is extending, in a reasonable way, via the property you would like, namely x^(a+b) = x^a*x^b, what a^x is. Note - you can only go as far as the rationals via this. It just screams something more elegant should exist - that's the feeling I got from the final sentence of the original post. And it does - and it even exists for the reals - not just rationals. IMHO that more elegant way is better. But to be fair I don't think most students really care - only a few like the OP see surely there is something better than just defining things - and of those that do even less want to pursue it - even though if they did they will learn a lot about some more advanced math - namely calculus which will be to their credit.
Thanks
Bill
PeroK said:Well, it's all a matter of taste, I guess, and your definition of elegance.
PeroK said:My guess, from this thread, is that many students would prefer 2). As a 16-year-old I would have been very unhappy with 3). Not to say baffled by it!
Rada Demorn said:You better leave f(x) undetermined and admit that:
nuuskur said:Uh, I remember I started real analysis I on my first semester and it was considered to be hand-wavy (with many results revised and given proofs for in analysis III). Even the hand-wavy course put rocks in my head![]()
nuuskur said:It is shown that all elementary functions are continuous in their domain, therefore due to continuity of ##x\mapsto a^x ## we have ##a^{q_n} \to a^0 ##.
bhobba said:Still for those interested in starting to learn calculus, Calculus Made Easy is a good choice.
Ahh..see? Already got stuck in the circle :(. Well, then the whole argument is null.Stephen Tashi said:How do we show the elementary function ##f(x) = a^x## is continuous at ##x= 0## without having a definition of ##f(0)##?
PeroK said:The fundamental issue is that when you use some mathematical symbols, you must define what you mean by that arrangement of symbols. Until you know what you mean by those symbols, you cannot start to do mathematics using them. In this case, for example, you might write:
##2^0##
But, what does that mean? There's no immediate way to "multiply 2 by itself 0 times". Unlike ##2^1, 2^2, 2^3 \dots ##, which have a simple, clear definition.
My recommended approach is to define ##2^0 = 1## before you go any further. Then you know what those symbols mean.
Now, of course, you need to be careful that a definition is consistent with other definitions, and you need to understand the implications of a certain definition.
In this case, the only other candidate might be to define ##2^0 = 0##. But, when you look at the way powers work, you see that defining ##2^0 =1## is logical and consistent.
mathwonk said:note also that then f(1/n)...f(1/n), (n times), = f(1/n+...+1/n) = f(1) = a, so f(1/n) = nth root of a = a^(1/n). Thus also f(m/n) = f(1/n)...f(1/n), (m times), = a^(1/n)...a^(1/n), (m times).
Thus f is determined on all rational numbers, and hence by continuity also on all real numbers.
mathwonk said:then this function f(x) is a good candidate for a^x.
Ok, we can define ##a^x## in terms of ##e^x## for ##a > 0##. Then we have the problem of the case ##a \le 0##.At least it agrees with a^x when x is a positive integer, and it has the right additive and multiplicative property.
Now sir, I think that it is actually a mistake to consider the Functional Equation: f(x+y) = f(x).f(y) over the Reals. If we take this over the Natural numbers it has solution ## a^x ## and I think this is obvious.mathwonk said:pardon me rada, i did not explain clearly what i was proving. there are two issues here, one is how to define a^x for all real x, and you are right i have not done this. the other issue is whether any definition at all, will yield a^0 = 1. Therse are called the existence and the uniqueness aspects of the exponential function. I have done only the uniqueness aspect. I.e. I have not shown how to define f(x) = a^x for all x, but rather i have proved that any definition at all, if one is possible satisfying the rule f(x+y) = f(x).f(y), must then obey the rule that a^0 = 1.
PeroK said:Anyway, with that definition of the exponential function, let's evaluate ##e^0##:
##e^0 = \Sigma_{n = 0}^{\infty} \frac{0^n}{n!} = \frac{0^0}{0!} + \frac{0^1}{1!} + \dots = \frac{0^0}{0!} ##
Hmm. Maybe that doesn't resolve all your misgivings after all!
bhobba said:but the power series expansion of e(x) is (not using the compact summation formula)
e(x) = 1 + x + x^2/2! + x^3/3! + ...
##f(0)=f(0)^2## has two solutions. The other solution is alluded to by @PeroK in #23.Rada Demorn said:Now we need only the restriction that f(0) is not equal to zero and let x=y=0 so that f(0+0)=f(0)=f(0).f(0) which gives immediately f(0)=1 or ## a^0=1 ## as required.
Rada Demorn said:How do you like this for a definition?
a is a Natural Number.
a^3 is a volume, a cube.
a^2 is an area, a square.
a^1 is a length, a line.
What else a^0 could be but a point, an individual, a 1?
PeroK said:If we have a cube of side 2, it's volume is 8, Or, if we have a cube of bricks, with each side 2 bricks long, we have 8 bricks.
For a square of side 2, it's area is 4. Or, a square of bricks needs 4 bricks.
For a line of length 2, it's length is 2. Or, a line of bricks needs 2 bricks.
For a point, it's size is 0. Or, if we have a zero-dimensional array of bricks, then there are no bricks.
By this intuitive reasoning, we should have ##2^0 = 0##.