How did Euler come up with the value of e?

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Homework Statement: This is not a homework problem.
I am trying to imagine how Euler would have gone about getting the value of e, while he was trying to figure out the case of continuous compounding.
Homework Equations: I know how he reached up to the equation given below. I am not sure how that brilliant mind might have come to the conclusion that this looks like a constant.

$$ \lim_{n \rightarrow \infty} ~ ( 1 + \frac {1 }{N} )^N $$

Would he have calculated the output using different values of N and figured that it is progressing towards a limiting value as N increases?
 

CPW

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Math_QED

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I don't know if Euler did it like this, but

$$e=\sum_0^\infty \frac{1}{n!}$$

And you can estimate the error

$$R_N= e - \sum_{0}^N \frac{1}{n!}$$

Using this, you can estimate ##e## to arbitrary precision.
 
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Hello.
I do not now history how Euler get the formula so I write here my idea.
Say we set money deposit interest is double in one year, after one year the deposit of unit 1 will become
[tex]1+1=2[/tex].
If half year composite interest is applied, it is
[tex](1+1/2)^2=2.25[/tex]
If one third year composite interset is applied
[tex](1+1/3)^3=2.37...[/tex]
If one fourth year composite interest is applied
[tex](1+1/4)^4=2.44...[/tex]
...
If day by day composite interest is applied
[tex](1+1/365)^{365}=2.71...[/tex]

I do not think such shorter term composite interest way give us as much money as we would like so it should converge to the value called e for infinite N.
 
Last edited:

fresh_42

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Dieudonné notes that the elementary functions like ##e^x## have already be known in the 17th century and points to Cotes and de Moivre, i.e. to a time before Euler. They already knew
$$
(\cos x + i \sin x)^n = \cos nx + i \sin nx
$$
and that it behaves like an exponential function in ##n##. In a recession to my book, Juskevic claims, that already Cotes knew ##e^{ix}=\cos x + i \sin x##.

Dieudonné further says, that Euler mainly dealt with what we now call algebra of formal power series. He computed a lot of formulas in that algebra involving Bernoulli number as coefficients.

So once more, science has been a common effort and the fact that we name certain theorems and terms after certain scientists does not mean they have been first!
 
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$$\lim_{n \rightarrow \infty} ~ ( 1 + \frac {1 }{N} )^N = \lim_{N \rightarrow \infty} \sum_{k=0}^N N^{-k} {N \choose k} \\ = \lim_{N \rightarrow \infty} \sum_{k=0}^N N^{-k} \frac{k! (N-k)!}{N!} \\= \lim_{N \rightarrow \infty} \left(1+1+\frac {N-1}{2N} + \frac {(N-1)(N-2)}{3!N^2} + \dots\right)\\ = \dots = 1+1+\frac{1}{2!}+\frac{1}{3!}+ \dots =e$$

To show that the limit converges you can show that all terms are bounded by their limits (and they are all positive as well), that means you can switch the order of limit and sum.
 
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mathwonk

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You get one explanation, perhaps the best possible, by reading the works of Euler. Basically it simplifies the formulas for various exponential expressions to choose as base of the exponential functions, the base e.

I.e. in his Analysis of the Infinite, in paragraph 121, Euler observes that if one uses any positive base a for an exponential function, one has the formula log(1+x) = (1/k)(x/1 - x^2/2 + x^3/3 -.....), where log means log base a. Moreover the base itself has expression (paragraph 118): a = 1 + k/1 + k^2/1.2 + k^3/1.2.3 + .....

Hence it is natural to make this as simple as possible and take as base the number a such that k = 1, namely then a = e = 1 + 1/1.2 + 1/1.2.3 +......
 

WWGD

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And another good question is why it pops up in so many ( often unexpected) places.
 

HallsofIvy

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Like every one else here, I can't speak to what Euler did but this is how I would look at it: Given any function if the form [itex]f(x)= a^x[/itex], the derivative of the function is the limit of the difference quotient [itex]\frac{f(x+h)- f(x)}{h}= \frac{a^{x+h}- a^x}{h}= \frac{a^xa^h- a^h}{h}= a^x\frac{a^h- 1}{h}[/itex]. The "[itex]a^x[/itex]" is independent of h while "[itex]\frac{a^h-1}{h}[/itex]" has no x. Taking the limit as h goes to 0, [itex]\frac{a^h- 1}{h}= C_a[/itex], a number that depends upon a but not x. So the derivative of [itex]f(x)= a^x[/itex] is [itex]C_a a^x[/itex].

So what is [itex]C_a[/itex]? If a= 2, we can estimate [itex]C_2= \frac{2^h- 1}{h}[/itex] by taking h= 0.001. [itex]C_2= \frac{2^{0.001}- 1}{0.001}= \frac{1.00069- 1}{0.001}= \frac{0.0069}{0.001}= 0.69[/itex], less than 1.

If a= 3, we can estimate [itex]C_3= \frac{3^{0.001}- 1}{0.001}= \frac{1.00110- 1}{0.001}= \frac{0.00110}{0.001}= 1.1[/itex] which is larger than 1. There must be a value of a, between 2 and 3 such that [itex]C_a= 1[/itex] so that the derivative of [itex]f(x)= a^x[/itex] is again [itex]a^x[/itex]!

After tediously trying values of a between 2 and 3 (I am using a calculator. Euler didn't have a calculator but apparently had enormous patience!) I get to [itex]C_{2.7}[/itex] is approximately [itex]\frac{2.7^{0.001}- `1}{0.001}= \frac{1.00099- 1}{0.001}= \frac{0.00099}{0.001}= 0.99[/itex], less than 1, while [itex]C_{2.8}= \frac{2.8^{0.001}- 1}{0.001}= \frac{1.00103- 1}{0.001}= \frac{0.00103}{0.001}= 1.03[/itex], larger than 1.

Continuing like that we can zero in on a= e= 2.718... such that the derivative of [itex]e^x[/itex] is, again, [itex]e^x[/itex] the world's easiest derivative!
 

CPW

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The group of us on this thread should mark our calendars to commemorate this conversation on February 72, 2020. The nerdier holiday than March 14th.
 
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That date has some... trouble.

We can take the 27th of January, ideally in the year 8281, but maybe (20)82 works as well.
 

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