Uncovering the Truth about e: Proofs and Derivatives

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In summary: Second one is pretty easy to show using the definition of the exponential function and the chain ruleIn summary, the conversation discusses two results related to the number e and its properties. The first result is the universality of e, which states that the sum of the infinite series represented by e is equal to e. The second result is the derivative of e^x, which is equal to e^x for all real values of x. The proofs for these results are not provided, but it is mentioned that they can be found using the geometric definition of natural log and its derivative, the binomial expansion of e, and the definition of the exponential function.
  • #1
dextercioby
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I'm looking for proofs to the 2 following results.

Let

[tex] \displaystyle{e=: \lim_{n\rightarrow +\infty} \left(1+\frac{1}{n}\right)^n} [/tex]

Show that:

1. Universality of e.

[tex] \sum_{k=0}^{\infty} \frac{1}{k!} = e [/tex]

2. Derivative of [itex]e^x [/itex].

[tex] (e^x)' = e^x, ~ \forall x\in\mathbb{R} [/tex]

Searching google didn't get me satisfactory results.

Could you, please, post or link to proofs ? Thank you!
 
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  • #2
The classic definition is based on the fact that d(2^x)/dx ~ 0.693147 and d(3^x)/dx~1.098612 at x=0. Then you guess that there must be a number such that f'(0)=1 where f=a^x (it turns out a =e and is unique).

===
g(x)=b^x

g'(x) =lim h->0 b^(x+h)-b^x/h
=lim h->0 (a^x)(a^h -1)/h
=(a^x) lim h->0 (a^h -1)/h
and the derivative is where at x=0 we get
lim h->0 (a^h -1)/h = f'(0)

f ' (0) = (2^x)(a^h -1)/h ~ 0.693147
f ' (0) = (3^x)(a^h -1)/h ~1.098612
So it seems like there should be a number that converges to 1.
But I don't know the proof that you mentioned. Hope that helps somewhat.
 
  • #3
Also, if you assume the geometric definition of natural log and its derivative then the derivative of e^x can be computed from the fact that

d(ln(exp(x))/dx = (d/dx (e^x))*(1/e^x) = 1 >>> d/dx (e^x) = e^x
probably not the most satisfactory answer since it pushes off the proof to something contingent on the natural log. But its perfectly accurate.

Of course you can also try the taylor expansion but that's not really "first principals"
 
  • #4
dextercioby said:
[tex] \sum_{k=0}^{\infty} \frac{1}{k!} = e [/tex]
Lemma:
[tex]\lim_{n\to\infty}\frac{n!}{n^k (n-k)!} = 1[/tex]
Use the above in conjunction with the binomial expansion of
[tex]e=\lim_{n\to\infty}\left(1+\frac 1 n\right)^n[/tex]
[tex] (e^x)' = e^x, ~ \forall x\in\mathbb{R} [/tex]
You need a definition of exp(x) for this. One definition of exp(x) is that it is the function that is equal to its own derivative such that exp(0)=1. That makes the proof a bit too easy. Try using
[tex]\exp(x)=\lim_{n\to\infty}\left(1+\frac x n\right)^n[/tex]
 
  • #5
First one follows from second if you use taylor series.
 

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