Hi all!

Below is a discussion on the natural exponent and its application. If things do not copy (this is a great hindrance in the yahoo browser, especially since its recent 'upgrading'; then I advise the interested reader to visit this post at my website (freeyellow below), where it is duplicated.

I am posting this, as the factor used here (1+1/n) has an even greater significance for modern physics as hitherto realised. Its inverse as (n/(n+1)) is used in Quantum Relativity as the expansion parameter in Einstein's Field Equations applied to modern cosmology.

The thread below serves however to show some well known intricacies of the exponetial and logarithmic functions for the mathematical novice.

Tony B.

Consider the following function:

f(n)=(1+1/n)ⁿ.

Evaluate for increasing counts n:

f(1)=(1+1/1)^1=2

f(2)=(1+1/2)^2=9/4=2.25

f(3)=(1+1/3)^3=64/27=2.370370...

...

f(10)=(1.1)^10=2.59374246........

...

f(100)=(1.01)^100=2.704813829...

...

f(Infinity)=℮=2.718281828.... This function is limited by a number 'e' for increasing n.

Definition: ℮=lim{1+1/n}ⁿ for n->Infinity

But this function approches infinity as n approaches -1 from -1- (from -2 say).

This function is limited by 1 as n approaches 0 from 0+ from above (>1) is UNDEFINED for n=0 and is limited by 1 from below (<1) as n approaches 0 from 0-.

The function is again UNDEFINED for n=-1 and behaves 'complexified irrationally' in the interval from n=-1 to n=0. Say the function assumes the complex value 1/i=i/ii=-i for n=-1/2.

As n decreases from -1 to negative infinity, it decreases from Infinity at n=-1 to e as n approaches large negative values.

f(1/100)=(101)^1/100=1.04723.

f(1/2)=(1+2)^1/2=√3

f(-100)=(1-1/100)^-100=2.732.

f(-10)=(1-1/10)^-10=2.867972.

f(-3)=(1-1/3)^-3=27/8=3.375

f(-2)=(1-1/2)^-2=4

f(-1)=(1-1)^-1=1/0

f(-1/2)=(1-2)^-1/2=1/√-1=1/i=-i

f(0)=(1+1/0)^0=(1+Infinity)^0=Undefined

The natural log at e, Ln(e), is equal to 1

These basic considerations in regard to this (transcendental) number e lead to an infinite series expansion for this number and where factorial function n!=n(n+1)..2.1, relates to a recurrence relation, known as Gamma Function: Gamma(x+1)=x.Gamma(x)=n! as a special case. The Gamma Function is similarly undefined for n=0 and n=-1.

{The Gamma Function is defined in a Euler Integral below and has the value 1 for n=0. The integral evaluates then as a limit for -1/e^x for the interval from Infinity to 0, which is (0+1)=1.

But now we have Gamma(x+1)=0!=x.Gamma(x) and by the recurrence of the integral as shown below, for n=1, we also have Gamma(x+1)=1!=1=x.Gamma(x).

Without evaluating the integral, we find the integrals as relating to each other in the Factorial Function.

I1=1.I0 this is 1.0!=1!

I2=2.I1=2.1.I0 this is 2.1!=2!

I3=3.I2=3.2.1.I0 this is 3.2!=3!

I4=4.I3=4.3.2.1.I0 this is 4.3!=4!

In=n.In-1=n! generalising as n.(n-1)!=n!

For the Integration by parts simply use d(uv)/dx =(udv/dx+vdu/dx).

Let u=x^n and v=-e^-x. Then du/dx=n.x^n-1 and dv/dx=e^-x.

Then since Integral (d(uv)/dx)dx=uv; the procedure below follows in writing: uv=Integral{d(uv)/dx}dx=Integral{udv/dx}dx + Integral{vdu/dx}dx.

The integral equation is then: Integral{-udv}=In=uv-Integral{vdu}=uv-Integral{-e^-x.n.x^n-1}dx.

This is an interesting application of integration by parts. Let's have a look at the integral

I_n = $\displaystyle \int_{0}^{\infty}$ xⁿe^-x dx

where n is some non-negative whole number.

This seems to be a candidate for integration by parts. Differentiate the xⁿ and integrate the e^-x. This will give us

I_n = $\displaystyle \left[\vphantom{-x^ne^{-x}}\right.$ - xⁿe^-x $\displaystyle \left.\vphantom{-x^ne^{-x}}\right]_{0}^{\infty}$ + n $\displaystyle \int_{0}^{\infty}$ x^{n - 1}e^-x dx

It is an important property of the exponential function that, whatever the value of n, xⁿe^-x $\to$ 0 as x $\to$ $\infty$ (e^x grows far faster than any power of x). So, if n > 0 the first term on the RHS vanishes at both ends. The other term on the RHS involves an integral which is just the same as our original one with n replaced by n - 1. So we have

I_n = n.I_{n - 1} for n $\displaystyle \ge$ 1

You grumble that this has not evaluated the integral--it has just turned it into another similar integral. True, but that does actually make it easy to evaluate the original integral. The point is that the above formula is valid for any positive integer n. So, climbing up the ladder, we have

I₁	= 1.I₀
I₂	= 2.I₁ = 2.1.I₀
I₃	= 3.I₂ = 3.2.1.I₀
I₄	= 4.I₃ = 4.3.2.1.I₀

--and so on. You will probably believe me if I say that the overall result is that I_n = n!.I₀.

This has reduced everything to the single problem of evaluating

I₀ = $\displaystyle \int_{0}^{\infty}$ x⁰e^-x dx = 1

(as you should check).

So we have obtained the rather nice result that

$\displaystyle \int_{0}^{\infty}$ xⁿe^-x dx = n!

Now for a generalisation. We had a positive integer n in the integrand as a power. There was actually nothing in our integration procedure (the integration by parts) that relied on the fact that n was a whole number. We could just as well look at the integral

f (a) = $\displaystyle \int_{0}^{\infty}$ x^ae^-x dx

where a is any positive value. Doing integration by parts on this will give

f (a) = a.f (a - 1) if a > 1

as before. If a is a whole number then f (a) = a!, but f (a) makes perfectly good sense for any positive value a. So we have produced a kind of generalisation of the factorial function. We can now talk about the factorial of ${\frac{1}{2}}$ or $\pi$ . This may sound a rather silly thing to do, but it does actually have wide application.

Traditionally, we do not work with the function f (a) but with the function $\Gamma$ (a) = f (a - 1). This is known as the Gamma Function,

$\displaystyle \Gamma$ (a) = $\displaystyle \int_{0}^{\infty}$ x^{a - 1}e^-x dx, $\displaystyle \Gamma$ (a + 1) = a. $\displaystyle \Gamma$ (a)

for a > 0. Of course, we have $\Gamma$ (1) = $\Gamma$ (2) = 1.

To satisfy idle curiosity I can tell you that $\Gamma$ ( ${\frac{1}{2}}$ ) = $\sqrt{\pi}$ }

$e = \sum_{n = 0}^\infty \frac{1}{n!} = \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \cdots$

An important consequence then follows for a similar expansion for the exponential function.

$e^{x} = 1 + {x \over 1!} + {x^{2} \over 2!} + {x^{3} \over 3!} + \cdots$

Fig. 6.1

Graph of y = e^x is mirror image of that of y = ln x in line
y = x.

Note the following:

Approximations Of e

Setting x = 1 we get:

This formula can be used to compute approximations of e, as shown in Fig. 9.1.

Fig. 9.1

Computation Of Approximations Of e.

The approximation gets better and better as n > 0 gets larger and
larger.

For any integer n > 0 let n! = 1 x 2 x 3 x ... x n. The notation n! is read “n factorial”, for the reason that it's a product of
positive integer factors from 1 to n. For example, 3! = 1 x 2 x 3 = 6. Define 0! = 1. It can be shown that a better way to
compute approximations of e is to utilize the infinite series (sum of infinitely many terms):

which is usually encountered in a second calculus course or in a first analysis course.

e in calculus

Image

The natural log at e, Ln(e), is equal to 1

The principal motivation for introducing the number e, particularly in calculus, is to perform differential and integral calculus with exponential functions and logarithms.^[5] A general exponential function y=a^x has derivative given as the limit:

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

The limit on the right-hand side is independent of the variable x: it depends only on the base a. When the base is e, this limit is equal to one, and so e is symbolically defined by the equation:

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

Consequently, 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.
Another motivation comes from considering the base-a logarithm.^[6] 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,

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

The logarithm in this special base is called the natural logarithm (often represented as "ln"), and it also behaves well under differentiation since there is no undetermined limit to carry through the calculations.
There are thus two ways in which to select a special number a=e. One way is to set the derivative of the exponential function a^x to a^x. The other way is to set the derivative of the base a logarithm to 1/x. In each case, one arrives at a convenient choice of base for doing calculus. In fact, these two bases are actually the same, the number e.

[edit] Alternative characterizations

See also: Representations of e

Other characterizations of e are also possible: one is as the limit of a sequence, another is as the sum of an infinite series, and still others rely on integral calculus. So far, the following two (equivalent) properties have been introduced:
1. The number e is the unique positive real number such that

$\frac{d}{dt}e^t = e^t.$

2. The number e is the unique positive real number such that

$\frac{d}{dt} \log_e t = \frac{1}{t}.$

The following three characterizations can be proven equivalent:
3. The number e is the limit

$e = \lim_{n\to\infty} \left( 1 + \frac{1}{n} \right)^n$

Image

The area under the graph y = 1/x is equal to 1 over the interval 1 ≤ x ≤ e.

4. The number e is the sum of the infinite series

$e = \sum_{n = 0}^\infty \frac{1}{n!} = \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \cdots$

where n! is the factorial of n.
5. The number e is the unique positive real number such that

$\int_{1}^{e} \frac{1}{t} \, dt = {1}$ .

[edit] Properties

[edit] Calculus

As in the motivation, the exponential function f(x) = e^x is important in part because it is the unique nontrivial function (up to multiplication by a constant) which is its own derivative

Fig. 6.1

Graph of y = ex is mirror image of that of y = ln x in line y = x.

Fig. 9.1

Computation Of Approximations Of e.

Image

[edit] Alternative characterizations

Image

[edit] Properties

[edit] Calculus

Graph of y = e^x is mirror image of that of y = ln x in line
y = x.