IIT JEE , AIEEE Forum - Mathematics , Algebra

edison

The "discovery" of the constant itself is credited to Jacob Bernoulli who attempted to find the value of the following expression (which is in fact e):

_n

_infinity (1+1/n)ⁿ

The mathematical constant e is the unique real number such that the value of the slope of the tangent of f(x) = e^x at the point x = 0 is exactly 1. The function e^x so defined is called the exponential function.

The value of e up to 20 digits of precision is

e=2.71828182845904523536.....

Representations of e:-

As a continued fraction-

$e= 2+\cfrac{1}{1+\cfrac{1}{2+\cfrac{2}{3+\cfrac{3}{4+\cfrac{4}{\ddots}}}}} \qquad e= 2+\cfrac{2}{2+\cfrac{3}{3+\cfrac{4}{4+\cfrac{5}{5+\cfrac{6}{\ddots\,}}}}}$

$e = 1+\cfrac{2}{1+\cfrac{1}{6+\cfrac{1}{10+\cfrac{1}{14+\cfrac{1}{\ddots\,}}}}}$

$e^x = 1+\cfrac{2x}{(2-x)+\cfrac{x^2}{6+\cfrac{x^2}{10+\cfrac{x^2}{14+\cfrac{x^2}{\ddots\,}}}}}$

$\frac{d}{dx}e^x = e^x.$

As a consequence, the exponential function with base e is particularly suited to doing calculus. Choosing e, as opposed to some other number, as the base of the exponential function makes calculations involving the derivative much simpler.

Considering the definition of the derivative of log_ax as the limit:

$\frac{d}{dx}\log_a x = \lim_{h\to 0}\frac{\log_a(x+h)-\log_a(x)}{h}=\frac{1}{x}\left(\lim_{u\to 0}\frac{1}{u}\log_a(1+u)\right).$

Once again, there is an undetermined limit which depends only on the base a, and if that base is e, the limit is one. So symbolically,

$\frac{d}{dx}\log_e x=\frac{1}{x}.$

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