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I am attempting to prove the following: prove that there exists a function $\exp:\mathbb{Q}\rightarrow\mathbb{R}$ such that $\exp(1)=\lim_{n\rightarrow\infty}\sum_{k=0}^{n}{\frac{1}{k!}}$. It seems to be a bit circular in logic to assume that it exists and to show that the properties hold. So I started with a defined function $\exp(x)=\lim_{n\rightarrow\infty}\sum_{k=0}^{n}{\frac{x^k}{k!}}$.

I took $x\in\mathbb{Q}$ letting $x=\frac{p}{q}$ where $p,q\in\mathbb{Z}$ and had for $x\in\mathbb{Q}$ that $\exp(x)=\sum_{k=0}^{\infty}{\frac{(\frac{p}{q})^k}{k!}}=\sum_{k=0}^{\infty}{\frac{(p)^k}{(q)^kk!}}$ and since it follows that each term being summed will be a rational number since the numerator and denominator portion will each be integers. Then it follows that for $x\in\mathbb{Q}$ it holds that $\exp(x)\in\mathbb{R}$ and so definging $\exp$ as above there exists a function $\exp:\mathbb{Q}\rightarrow\mathbb{R}$ such that$\exp(1)=\lim_{n\rightarrow\infty}\sum_{k=0}^{n}{\frac{1}{k!}}$

Is this sound reasoning? I am sure that there are other ways of proving this.

I have encountered this proof while learning real analysis and am open to proof that use real analysis.

Kenta S
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lambdaepsilon
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    I think you skipped a huge and rather important step: you must prove the series defining your $;exp(x);$ converges for all the real numbers $;x;$ that you want... – DonAntonio Jul 30 '20 at 21:24
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    Acutally, in the form stated, there exist infinitely many such functions (just consider $f(1) = e$, and define $f(x)$ arbitrarily if $x\neq 1$). Assuming you know that the limit $\lim_{n\to \infty}\sum_{k=0}^{n}\frac{1}{k!}$ exists in $\Bbb{R}$, you can appeal to an existence and uniqueness theorem: for any $a\in \Bbb{R}$, there exists a unique $F_a:\Bbb{Q} \to \Bbb{R}$ such that $F_a(1) = a$ and for all $x,y\in \Bbb{Q}$, $F_a(x+y) = F_a(x) F_a(y)$. I recently wrote an answer which touches on this matter among other things. – peek-a-boo Jul 30 '20 at 21:25
  • @DonAntonio yes you are correct, for completeness this must be included. I have proved that before and will include. – lambdaepsilon Jul 30 '20 at 21:31
  • @peek-a-boo in this instance then showing that the limit exists and showing that $F_{a}(1)=a$ and for all$ x,y\in\mathbb{Q}, F_{a}(x+y)=F_{a}(x)F_{a}(y) $it can be assumed from the theorem that there exist a unique continous function as defined from the rationals to the reals? I am wondering if I can begin with showing the properties hold first then alluding to the theorem to prove it. Further is the existence of the limit sufficient to show continuity of the function in this case? – lambdaepsilon Jul 30 '20 at 21:55
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    Let's denote $e:= \lim_{n\to \infty}\sum_{k=0}^n\frac{1}{k!}$ (I assume you've somehow shown this limit exists say for example using ratio test). If you only care about the domain being $\Bbb{Q}$, then this is a relatively simple theorem to prove the existence and uniqueness of $F_e$ (and if you only care about $\Bbb{Q}$, continuity is not needed), and if you look at the proof given in the answer, it shows you exactly how to define $F_e$. The alternative is to directly define $\exp(x)=\sum_{k=0}^{\infty}\frac{x^k}{k!}$, show the the series converges everywhere, and satisfies all the properties – peek-a-boo Jul 30 '20 at 22:02
  • @peek-a-boo why is showing continuity unnecessary for $\mathbb{Q}$? What would be different if we were trying to prove say the existence of $exp:\mathbb{R}\rightarrow\mathbb{R}$ without imposing the condition of $\exp(1)=\lim_{n\rightarrow\infty}\sum_{k=0}^{n}{\frac{1}{k!}}$ – lambdaepsilon Jul 30 '20 at 22:54
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    did you take a look at the proof provided there? Anyway, the function will turn out to be continuous (because it is the restriction of a continuous function on $\Bbb{R}$ to $\Bbb{Q}$), but this fact is not at all important when showing existence and uniqueness. – peek-a-boo Jul 30 '20 at 23:02
  • @peek-a-boo Yes I did look at the proof. Thank you. Further what is this theorem called exactly? I am trying to wrap my head around the theorem.This seems to imply differentiability, which is what is used in the existence and uniqueness theorem I am familiar with, is this correct? Also thanks for all the clarification youve been a great help. – lambdaepsilon Jul 30 '20 at 23:05
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    I'm not sure of what this theorem is called. Regarding differentiability... I'm not sure about differentiability would even work on $\Bbb{Q}$? But on $\Bbb{R}$, of course the function is differentiable, infinitely differentiable, analytic, everything :) – peek-a-boo Jul 30 '20 at 23:10
  • okay thanks. Im going to try to show this for complex to complex now. – lambdaepsilon Jul 30 '20 at 23:34

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