What is e?

J7 · Dec16-04, 03:44 PM

Hi, can anyone explain "e" to me? It's used all the time in calc and I don't understand what it represents or it's value. Thanks

**Muzza** · Dec16-04, 03:51 PM

e is a number, approximately equal to 2.71828183. It's irrational and transcendental. The function f: R -> R, f(x) = e^x satisfies f'(x) = f(x).

**dextercioby** · Dec16-04, 03:55 PM

Originally Posted by J7

Hi, can anyone explain "e" to me? It's used all the time in calc and I don't understand what it represents or it's value. Thanks

Along with $LaTeX Code: \\pi$ ,

and

,it is the most important number in mathematics,hence in science.
It is a real number,mathematicians call it irrational and transcendent.It has an infinite number of decimals,the first fifteen (hopefully i haven't forgotten them

) are:

and it is defined as the limit of the sequence:
$LaTeX Code: e=:lim_{n\\rightarrow +\\infty} (1+\\frac{1}{n})^{n}$

It has an interesting history and some mathematicians call it "Euler's number",hence the letter "e".

Daniel.

**matt grime** · Dec16-04, 04:07 PM

It is also

$LaTeX Code: \\sum_{r=0}^{\\infty}\\frac{1}{r!}$

(can we ignore the "infinite number of decimals" thing - that's neither here nor there, as well as incorrect English 1/3 has also not a finitely long decimal expansion in base 10, and? And there were no rational transcendental numbers. A number is transcendental if it is not a root of any polynomial with integer (whole number) coefficients.The complementary notion is 'algebraic'; all integers are algebraic, sqrt(2) is algebraic, pi isn't.)

I, e, t arises as the (unique) solution to f'=f, which tells you that it is important in "the real world" since we model that with differential equations, and importantly, once we've defined e, we can solve f'=kf, as well as a whole load of other differential equations without needing to define anything else.

**HallsofIvy** · Dec16-04, 08:09 PM

It's just this number, y'know? Any function of the form f(x)= a^x has the property that its derivative (rate of change) is proportional to a^x itself.
e (which, as Muzza said, is "approximately equal to 2.71828183.") has the nice property that the constant of proportionality is 1- that is, the rate of change of the function e^x is precisely e^x.

~~QuantumTheory~~ · Dec16-04, 11:32 PM

Originally Posted by HallsofIvy

It's just this number, y'know? Any function of the form f(x)= a^x has the property that its derivative (rate of change) is proportional to a^x itself.
e (which, as Muzza said, is "approximately equal to 2.71828183.") has the nice property that the constant of proportionality is 1- that is, the rate of change of the function e^x is precisely e^x.

Great explanation! I know some calculus, mostly all elementary. And I was thinking about what you said, it makes sense. I'm trying to figure out partial derivatives.

Could you explain some real-life explainations of

? Like an example, then solve it? That'd be real helpful to me, if you could think of one.

Before you posted this, I thought e was actually just related to the logaritam.

Since its proportion is 1:1, would this mean that..

$LaTeX Code: \\int \\frac{dy}{dx} e^2 dx = [tex]1/3^3e + C$

**Chrono** · Dec17-04, 02:39 AM

What about this? I remember this from a class.

$LaTeX Code: e = (1 + \\frac{1}{x})^x$

**Integral** · Dec17-04, 03:01 AM

$LaTeX Code: 1 = \\int_1^e \\frac {dx} x$

I believe that it this relationship which gives the number physical significance.

**shmoe** · Dec17-04, 03:13 AM

Originally Posted by Chrono

What about this? I remember this from a class.

$LaTeX Code: e = (1 + \\frac{1}{x})^x$

You probably mean:

$LaTeX Code: e=\\lim_{x\\rightarrow\\infty}(1 + \\frac{1}{x})^x$

Or more generally:

$LaTeX Code: e^a=\\lim_{x\\rightarrow\\infty}(1 + \\frac{a}{x})^x$

**matt grime** · Dec17-04, 03:20 AM

Originally Posted by QuantumTheory

I thought e was actually just related to the logaritam.

Since its proportion is 1:1, would this mean that..

$LaTeX Code: \\int \\frac{dy}{dx} e^2 dx = [tex]1/3^3e + C$

nb.your backslashes and forward slashes are wonky

It is related to the logarithm, and the examples ought to tell you you've underestimated the importance of log.

THe second bit is wrong. e^2 is just a number so you're asking for

$LaTeX Code: e^2 \\int \\frac{dy}{dx}dx$

which is just ye^2

**HallsofIvy** · Dec17-04, 08:08 AM

There are many applications in which the "rate of change" of some quantity is proportional to the quantity itself: population growth, radioactivity, etc. That is:
dy/dx= C y so that y is necessarily an exponential. For example, suppose you have a radioactive element that has a "half-life" of 1000 years. It's easy to see that, if you start with M₀ grams, the amount left after T years is M₀ (1/2)[sup]T/1000[/sub]. The "T/1000" just counts "the number of times you multiply by 1/2".
A more formal way of deriving that would be to write dM/dt= kM (k is the unknown constant of proportionality) so that (1/M)dM= kdt. Integrating both sides: ln(M)= kt+ C so that M= e^kt+C= e^Ce^kt= C' e^kt (where C'= e^C). Setting t= 0 we get M(0)= C' so that the coefficient is in fact M₀, the initial amount. Knowing that the half-life is 1000, tells us that M(1000)= M₀e^1000k= (1/2)M₀. We solve for k by taking the logarithm of both sides: 1000k= ln(1/2) so k= ln(1/2)/1000.

That is: M(t)= M₀e[sup]ln(1/2)t/1000[/sub]. Of course if you are clever, you will recognize that ln(1/2)t/1000 is the same as ln((1/2)^t/1000 so that e^{ln(1/2)t/1000}= (1/2)^t/1000.

**HallsofIvy** · Dec17-04, 08:11 AM

There are many applications in which the "rate of change" of some quantity is proportional to the quantity itself: population growth, radioactivity, etc. That is:
dy/dx= C y so that y is necessarily an exponential. For example, suppose you have a radioactive element that has a "half-life" of 1000 years. It's easy to see that, if you start with M₀ grams, the amount left after T years is M₀ (1/2)^T/1000. The "T/1000" just counts "the number of times you multiply by 1/2".
A more formal way of deriving that would be to write dM/dt= kM (k is the unknown constant of proportionality) so that (1/M)dM= kdt. Integrating both sides: ln(M)= kt+ C so that M= e^kt+C= e^Ce^kt= C' e^kt (where C'= e^C). Setting t= 0 we get M(0)= C' so that the coefficient is in fact M₀, the initial amount. Knowing that the half-life is 1000, tells us that M(1000)= M₀e^1000k= (1/2)M₀. We solve for k by taking the logarithm of both sides: 1000k= ln(1/2) so k= ln(1/2)/1000.

That is: M(t)= M₀e[sup]ln(1/2)t/1000[/sub]. Of course if you are clever, you will recognize that ln(1/2)t/1000 is the same as ln((1/2)^t/1000 so that e^{ln(1/2)t/1000}= (1/2)^t/1000.

**Hessam** · Dec23-04, 01:08 PM

isnt there also a relation between pi and e?

its something like this....

(pi^5 + pi^5)^(1/6)

or something of the sort.... please feel free to correct me

**arildno** · Dec23-04, 01:35 PM

The arguably most famous relation is:
$LaTeX Code: e^{i\\pi}+1=0$

**matt grime** · Dec23-04, 01:40 PM

Originally Posted by Hessam

isnt there also a relation between pi and e?

its something like this....

(pi^5 + pi^5)^(1/6)

or something of the sort.... please feel free to correct me

It is, at least I think so at any rate, an open question as to whether any such algebraic relation exists betweene them

**HallsofIvy** · Dec23-04, 02:03 PM

Well, there is certainly no question as to whether THAT relation is true:

(pi^5+ pi^5)^(1/6)= (2pi^5)^(1/6)= approximately 2.9138 which is not particularly close to e!