Finding Finite Series for Euler's Number (e)

[eddybob123]

well, i kno that e is derived from (1+(1/n))^n, as n goes to infinity, but i want to find a finite series . if i plot a curve like this on a graph, then take its derivative, i can tske a value of e.

 

[disregardthat]

well, i kno that e is derived from (1+(1/n))^n, as n goes to infinity, but i want to find a finite series . if i plot a curve like this on a graph, then take its derivative, i can tske a value of e.

A "finite series"? What do you mean? If you meant an infinite series a common one is $e = \frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+...$. In fact e^x is sometimes defined as
$$\frac{1}{0!}+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+...$$

 

[TylerH]

Another useful definition of e, especially for calc I derivation of the e^x rule, is $$lim_{n \to 0} (1+n)^{1/n}$$.

$$\frac{d}{dx}e^x=lim_{h \to 0} \frac{e^{x+h}-e^x}{h}$$
$$\frac{d}{dx}e^x=e^xlim_{h \to 0} \frac{e^h-1}{h}$$
$$\frac{d}{dx}e^x=e^xlim_{h \to 0} \frac{\left(\left(1+h\right)^{1/h}\right)^h-1}{h}$$
$$\frac{d}{dx}e^x=e^xlim_{h \to 0} \frac{1+h-1}{h}$$
$$\frac{d}{dx}e^x=e^x$$

 

[HallsofIvy]

well, i kno that e is derived from (1+(1/n))^n, as n goes to infinity, but i want to find a finite series . if i plot a curve like this on a graph, then take its derivative, i can tske a value of e.

Any finite series, involving only integers and fractions of integers, must give a rational number. e is not rational so there can be no such finite series. In fact, because e is transcendental, any such series would have to involve transcendental numbers as coefficients which would be as difficult to work with as e itself.

 

[CellCoree]

I would like to first clarify that the value of Euler's number, e, is a mathematical constant that represents the base of the natural logarithm. It is approximately equal to 2.71828 and is an important number in many mathematical and scientific applications.

Now, to address the content of finding a finite series for e, I would like to mention that there are several ways to approximate the value of e using finite series. One common method is to use the Maclaurin series expansion, which is a special case of the Taylor series expansion. This series is given by:

e^x = 1 + x + (x^2)/2! + (x^3)/3! + (x^4)/4! + ...

By setting x=1, we get the series for e:

e = 1 + 1 + (1^2)/2! + (1^3)/3! + (1^4)/4! + ...

= 1 + 1 + (1/2) + (1/6) + (1/24) + ...

= 2.71828

This series converges to the value of e as we add more terms, but it is not a finite series as it has infinite terms.

Another way to approximate e using a finite series is by using the continued fraction representation of e:

e = 2 + 1/(1+1/(1+1/(1+...)))

By truncating this continued fraction at a certain point, we can get a finite series that will approximate the value of e. However, this method may not give a very accurate result as it depends on where we choose to truncate the continued fraction.

In conclusion, there are ways to approximate the value of e using finite series, but they will always be approximations as the true value of e is an irrational number with infinite decimal places. The most accurate way to find the value of e is by using its definition as the limit of (1+(1/n))^n as n approaches infinity.