0

Here, take $[x]$ to be the smallest integer function.

I like this more than $\lim_{n\to \infty} (1+\frac{x}{n})^n$ because it seems to make the property $\exp(x+y)=\exp(x)\exp(y)$ obvious.

This is because either $[\frac{x+y}{h}]=[\frac{x}{h}]+[\frac{y}{h}]$ or $[\frac{x+y}{h}]=[\frac{x}{h}]+[\frac{y}{h}]-1$

If the former equality holds for some small $h$, then $\exp(x+y)\approx (1+h)^{[\frac{x+y}{h}]}=(1+h)^{[\frac{x}{h}]+[\frac{y}{h}]}=(1+h)^{[\frac{x}{h}]}(1+h)^{[\frac{y}{h}]}\approx \exp(x)\exp(y)$

Similarly, if the latter equality holds for some small $h$, then

$\exp(x+y)\approx (1+h)^{[\frac{x+y}{h}]}= (1+h)^{[\frac{x}{h}]+[\frac{y}{h}]-1}=\frac{(1+h)^{[\frac{x}{h}]}(1+h)^{[\frac{y}{h}]}}{1+h}$

$\approx \frac{\exp(x)\exp(y)}{1+h}$

But in this case, the denominator $1+h$ tends to $1$ as $h$ tends to zero, which again implies $\exp(x+y)=\exp(x)\exp(y)$ in the limit.

The smallest integer function ensures that the 'repeated multiplication definition' of exponents can be used.

But would this definition really give the same exponential function as $\lim_{n\rightarrow \infty} (1+\frac{x}{n})^n$? If there are some flaws, then any suggestions to correct then?

Xiangxiang Xu
  • 1,815
Ryder Rude
  • 1,417

1 Answers1

2

Your question clearly arises from curiosity in alternative approaches, so I do not think it deserves to be downvoted. While your explanation was not rigorous, it can be expressed rigorously as follows: $ \def\lfrac#1#2{{\large\frac{#1}{#2}}} $

If we have defined integer exponentiation on the reals, and proven that $x^{a+b} = x^a·x^b$ for any real $x≠0$ and integers $a,b$, and we have proven that $\lim_{h→0} (1+h)^{\lfloor x/h \rfloor}$ exists for every real $x$, then we can define $f(x) = \lim_{h→0} (1+h)^{\lfloor x/h \rfloor}$ for every real $x$, and can prove the property you wanted as follows:

As $h→0$, we have $f(x)·f(y) ≈ (1+h)^{\lfloor x/h \rfloor} · (1+h)^{\lfloor y/h \rfloor}$ $= (1+h)^{\lfloor x/h \rfloor+\lfloor y/h \rfloor}$ $∈ (1+h)^{\lfloor (x+y)/h \rfloor+\{0,1\}}$ $= (1+h)^{\lfloor (x+y)/h \rfloor} · \{1,1+h\}$ $≈ f(x+y)·\{1\}$. Therefore $f(x)·f(y) = f(x+y)$.

Note that the above proof uses pointwise operations on sets of reals, and at your level it is probably better if you write down the ε-δ proof, because I am not sure whether you made a mistake in your reasoning or not.

However, there is a fundamental flaw in your 'approach': it is actually less convenient! Compare with this post, for example, which sketches a quick path to the complex exponential function using well-motivated yet entirely elementary techniques. The point is never merely about whether you can define something, but whether you can prove all the desired properties or not.

Specifically, first you would have to prove that your stated limit exists. Can you? If you were to ask me to do it, I would in fact use my preferred approach to define $\exp$ first, and then use asymptotic analysis to trivially compute your limit! In more detail, when we have $\exp,\ln$ then we could define real exponentiation by $a^b = \exp(b·\ln(a))$ for $a>0$. Then for every real $x$ we easily obtain that as $h→0$ we have $f(x) ≈ (1+h)^{\lfloor x/h \rfloor}$ $= \exp(\lfloor x/h \rfloor·\ln(1+h))$ $∈ \exp((x/h+O(1))·(h+o(h)))$ $⊆ \exp(x+O(h))$ $= \exp(x)·\exp(O(h))$ $⊆ \exp(x)·(1+O(h))$ $→ \exp(x)$, and therefore conclude that $f(x) = \exp(x)$.

Furthermore, there is no obvious reason that $f$ is differentiable, nor that $f' = f$ on the reals. By the property we have just discussed, you can obtain $\lfrac{f(x+t)-f(x)}{t} = f(x)·\lfrac{f(t)-1}{t}$ for every reals $x,t$ with $t≠0$, and hence you would still have to prove $\lim_{t→0} \lfrac{f(t)-1}{t} = 1$. Can you?

Furthermore, your definition works only for the real exponential function. If you want to get to the complex exponential function eventually, then it is obviously better to start with the Taylor series definition instead.

Community
  • 1
user21820
  • 57,693
  • 9
  • 98
  • 256
  • 1
    Very nice answer. +1 for highlighting the two issues in this approach: existence of limit and extension to complex exponents. Hope things are fine with you in these difficult times. – Paramanand Singh May 02 '20 at 02:24
  • @ParamanandSingh: Thanks a lot for your comment and your concern! I am indeed fine, and I hope you too have been doing fine so far! Take care! =) – user21820 May 02 '20 at 02:32
  • Actually, it can be used to define $e^{ix}$. There you just approach $h\rightarrow 0$ from the imaginary axis, or in other words $(1+ih)^{[\frac{x}{h}]}$ has real $h$ tends to 0. And this approach also makes complex exponentiation obvious as the argument of $1+ih$ has $h$ tends to 0 is $h$ radians. So we end up rotating by $h$ radians $\frac{x}{h}$ times, which would be $x$ radians. – Ryder Rude May 03 '20 at 14:59
  • @RyderRude: You can do a hundred and one things to get from the real to the complex exponential function. But none of them are well motivated unless you look at the Taylor series or you believe there is a complex-differentiable function satisfying the canonical differential equation. For example you say you can use $(1+ih)^{x/h}$ as $h→0$ to define $e^{ix}$. The question is, why should you? And the only possible answer is, so that you can get certain properties. But that is exactly why I said that if you do everything rigorously the best way is still via Taylor series. – user21820 May 03 '20 at 15:04
  • Do not forget that you still want to prove $\exp(ix) = \cos(x)+i·\sin(x)$ at some point, and this holds for any complex $x$. So ultimately, if you want to develop all the standard mathematics involving the exponential function, the Taylor series is the best route. And I notice you didn't answer the questions regarding whether you can actually prove that your limit exists. I do not see any way that is simpler than using the Taylor series definition! – user21820 May 03 '20 at 15:07
  • By the way, the last sentence of your above comment is wrong; you cannot anyhow take limits like that as it is not rigorous and can lead to nonsense. – user21820 May 03 '20 at 15:10
  • Isn't that Taylor series just the binomial expansion of $\lim_{n\rightarrow \infty} (1+\frac{1}{n})^n)$? If both definitions are this closely connected, then I'd prefer the limit one because its very intuitive to me (it also makes Euler's formula obvious, while the Taylor series proof of Euler's formula seems like a hack). And then my approach modifies the limit definition a bit to make $exp(x+y)=exp(x)exp(y)$ obvious. But, as you said, it makes some other properties like differentiability less obvious. I wish there was a way around to that as I really like my approach. – Ryder Rude May 03 '20 at 15:20
  • I think there should be some easy way we are able to prove things like differentiability in my approach without relying on the other definitions of the exponential function. – Ryder Rude May 03 '20 at 15:21
  • @RyderRude: No!! Firstly, binomial expansion has nothing to do with limits! Secondly, I don't know why on earth you say the limit definition makes Euler's formula obvious when I already explained why your earlier comments are totally wrong. Thirdly, if you feel that the proof of Euler's formula via Taylor series seems like a hack, then it implies that you do not truly understand $\exp$. Look at the differential equations for $\exp,\cos,\sin$ as described in the linked post and you can see why they must have those Taylor series. – user21820 May 03 '20 at 15:24
  • Given your responses, I strongly advise you to pick up Spivak's Calculus textbook and work properly through it, as it would provide a proper rigorous foundation for real analysis. It does not matter what definition Spivak uses, but it would be rigorous. Then you can see for yourself what is true mathematical rigour, and after that when you come back to this question you will appreciate what I have been saying. – user21820 May 03 '20 at 15:26
  • "By the way, the last sentence of your above comment is wrong; you cannot anyhow take limits like that as it is not rigorous and can lead to nonsense". I know it's not a rigorous proof of $e^{ix}=cos(x)+isin(x)$. I was just giving the underlying idea there. I'm sure it can be made rigorous because the idea is there (Just like my proof of $e^{x+y}=e^xe^y$ wasn't rigorous in my post, but you made it rigorous) – Ryder Rude May 03 '20 at 15:26
  • @RyderRude: No. I know this stuff very well. The idea in your question that I made rigorous was nearly a trivial thing, which is why it is easy to make rigorous. The parts of your idea that you would need to make rigorous to get your idea to work (existence of limit and differentiability) are incredibly difficult to do so, and even I took a long time to convince myself that I could do it without using Taylor series... – user21820 May 03 '20 at 15:29
  • Your idea for defining $\exp$ makes the proof of $\exp(x+y) = \exp(x)·\exp(y)$ easy, but makes everything else much much harder. As I said, do Spivak's Calculus. I like your curiosity and creativity (I upvoted your question), but you must really put in a lot of hard work to be able to understand real analysis properly. – user21820 May 03 '20 at 15:31
  • Thanks for your responses. It means a lot. One last thing, what is wrong with expanding $(1+\frac{x}{n})^n$ using binomial theorem and then applying $\lim_{n\rightarrow \infty}$ to arrive at the Taylor series? I always saw that method as the link between the limit definition and the Taylor series. If that method is fine, then we can use it to easily arrive at the Taylor series from my definition, and then use the Taylor series to prove the existence of the limit in my definition. – Ryder Rude May 03 '20 at 15:48
  • @RyderRude: Firstly, you cannot swap infinite summation and limit in general. There are some powerful theorems that can be used to handle this specific case, but the proofs of these theorems themselves rely on a significant amount of real analysis. Secondly, $\lim_{n→∞} (1+x/n)^n$ is very different from $\lim_{n→∞} (1+1/n)^{\lfloor x·n \rfloor}$; I do not see any easy way to get from one to the other. – user21820 May 03 '20 at 16:02
  • https://en.m.wikipedia.org/wiki/Characterizations_of_the_exponential_function Here, in the equivalence of characterisations section, they've established the equivalence of the limit and series definitions by only taking $\limsup$ and $\liminf$ of the binomial expansion. And if we substitute $k=[xn]$ in my definition, then $\frac{k-1}{x}< n\leq \frac{k}{x}$. This means my limit will lie between $\lim_{k\rightarrow \infty} (1+\frac{x}{k})^k$ and $\lim_{k\rightarrow \infty} (1+\frac{x}{k-1})^k$. We can show that both of these are the same limit using the substitution $m=k-1$ in the latter limit. – Ryder Rude May 04 '20 at 01:32
  • @RyderRude: The proof you cited only established equivalent of series and limit definition for positive real exponent. Also, you should have observed that they used the series definition to get $\exp' = \exp$... so as I had said so many times already, you are not making it easier. As for your second half of your comment, you made a mistake. If $x≥0$ and $k=[xn]$ then $k/x≤n<(k+1)/x$. If $x<0$ then it's different. Your later argument is not quite correct (it fails if $x=0$), but if you learn to be rigorous you can patch it up. – user21820 May 04 '20 at 04:26
  • Anyway, I don't think I want to spend much more time on this since in my opinion there is little appeal in your proposed definition due to its narrow scope; unlike the standard limit definition it cannot extend to complex exponents at all, and it is easier to prove all the desired properties of $\exp$ in other approaches. Why don't you just take my advice and learn at least one rigorous approach before you come back to this? (And, no, I don't count Wikipedia as rigorous...) – user21820 May 04 '20 at 04:30
  • Thanks I'll read Spivak's calculus then. My calculus level is high school right now. Does it cover things like measure theory? – Ryder Rude May 04 '20 at 05:19
  • @RyderRude: It doesn't cover measure theory, but it is just right for your current level and interests, so I think you will love it! =) By the way, once you get a good foundation in basic logic and real analysis, measure theory will not be that hard. If you have any questions about the stuff in Spivak's calculus, feel free to ask me in chat. – user21820 May 04 '20 at 05:29