# Where does the natural logarithm constant "e" come

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#### bluewave

##### Junior Member level 1 natural logarithm hyperbola

I know pi=3.14 having a meaning in real world. It links the relationship between perimeter and radius of a circle. But in term of "e", I don't figure out how it comes out.
Could you be so kind to explain it in your words?
Thanks.

#### flatulent Re: Where does the natural logarithm constant "e"

It is a natural constant of mathematics. It pops out of calculus just like radians do.

#### the_penetrator

##### Full Member level 5 Re: Where does the natural logarithm constant "e"

If I can recall from high-school calculus it should be the limit of some expression:

lim(1 + (1/x))^x for x->∞

correct if'm wrong, it's been 10 years or so since last grade...

#### FoxyRick ### bluewave

Points: 2

#### electron_boy

##### Full Member level 3 Re: Where does the natural logarithm constant "e"

in many engineering aspects 'e' comes for example depth of penetration is 1/e
correct me if iam wrong

#### jimjim2k Hi

* -> t

tnx

#### checkmate Re: Where does the natural logarithm constant "e"

The importance in the natural log lies in its ability to define a standard model of decay, be it real damping decay or oscillating complex decay. Since decaying signals are quite prevalent in the real world, it has become necessary to be able to mathematically define decays.
If you consider the standard decay equation of e^-x, you'll find that the value decays by 1/e for every unit increase in x, a consistent definition regardless of where on the x-axis you are currently on. All other decay equations could simply be scaled from the standard decay equation as Ae^-kx, where k can be real or complex!

#### btbass Re: Where does the natural logarithm constant "e"

This number is trancendental-an irrational number that is not the root of any polynomial equation with integer coefficients. It was named e by Euler(1707-1783), a famous Swiss mathematician. He was the first to prove that e is the limit of (1 + 1/x)^x as x approaches infinity. Like all irrationals, the decimal expansion of e (2.718281828459045....) never repeats.

#### Fom Re: Where does the natural logarithm constant "e"

Wheb you will solve some physical or mathematical problems you will need to find function that equal to it's derivative. And you have to introduce "e", because this function is e^x.

#### nand_gates Re: Where does the natural logarithm constant "e"

You can refer to book
E Maor, e : the story of a number (Princeton, 1994).
Search google for "History of e"...
There is similar book for Pi "History of Pi"

#### the_penetrator

##### Full Member level 5 Re: Where does the natural logarithm constant "e"

Fom: obviously you have NEVER studied physics in a university or at least a decent one.

#### eda4you

##### Full Member level 5 Re: Where does the natural logarithm constant "e"

The "e" honours Euler for its work.

#### hwswboy

##### Member level 4 Re: Where does the natural logarithm constant "e"

e is the unique number with the property that the area of the region bounded by the hyperbola y=1/x , the x-axis, and the vertical lines x = 1 and x = e is 1. In other words:

/ e
|
| 1/x dx = 1 =Ln e
|
/ 1

Also but no original definition:

can be defined by the limit

Lim (1+1/x)^x=e
x->oo

or or by the infinite series

k=oo
SUM (1/k!)=e
k=0

Without mention of all world of definitions comed from complex analysis.... but the original math definition is first one

#### Fom Re: Where does the natural logarithm constant "e"

To the_penetrator

OK! May be it was not the best explanation. I tried to explain where this number come from.
I can quote explanation from the site pointed by jimjim2k:

"Does the number e have any real physical meaning, or is it just a mathematical convenience?
Yes, the number e does have physical meaning. It occurs naturally in any situation where a quantity increases at a rate proportional to its value, such as a bank account producing interest, or a population increasing as its members reproduce."

In other words "function is proportional to it's derivative or derivative is proportional to the function"

#### gm_navy

##### Newbie level 1 Suppose you borrow 1 Rupee from Uncle Scrooge, at an interest rate compounded annually of 100%. At the end of the year you have to pay Scrooge Rs 2. Money costs Money.
Suppose Scrooge, the scrooge that he is, starts to wonder …. why only 2 Rupees ? After all, the money owing to him should be accumulating as long as it is in the possession of the borrower. Why should he wait for a year to benefit from this ?
He now insists you should compute interest every month, and not at the end of the year. Thus, by the end of the first month, the borrowed capital should have grown to Rs (1+1/12), at the end of 2 months – to
〖Rs (1+1/12)〗^2 …. and at the end of the year Scrooge’s loaned amount should have grown to
〖Rs (1+1/12)〗^12. This works out to 〖(13/12)〗^12= which is the same as Rs 2.613035. (Try it out in Excel). That’s better. Now coming to think of it, why should Scrooge remain contented with just monthly computations ?
What if we compute interest every day , the year being reckoned at 365 days ? We would see that the capital grows by the end of the year to 〖(1+1/365)〗^365 which works out to Rs 2 .714567 … even better !
Why not compute the money every second ? There are 365x24x60x60 = 31536000 seconds in the year !
We get Rs 〖(1+1/31536000)〗^31536000 , which works out to 2.718282….
What if we find a way of Scrooge’s money keeping on growing every nanosecond, every instant … all the
time ? We would have to calculate 〖(1+1/n)〗^n where n grows to infinity in the limit ! The amount levels off to this grand, magic constant named by Euler as ‘e’ , namely, 2.718281828………… . Whew !
The way Euler computed this number was by expanding 〖(1+1/n)〗^n using the binomial theorem and getting an infinite series. We can see that this expansion is
〖nC^0×1〗^n+nC^1×1^(n-1)×1/n+nC^2×1^(n-2)×1/n^2 +nC^3×1^(n-3)×1/n^3 +⋯ =
n!/n!0!×1+n!/(n-1!1!)×1×1/n+n!/(n-2!2!)×1×1/n^2 +n!/(n-3!3!)×1×1/n^3 …. =
1/0!+1/1!+(n(n-1))/2!×1/n^2 +(n(n-1)(n-2))/3!×1/n^3 …. = 1/0!+1/1!+((1-1/n))/2!+((1-1/n)(1-2/n))/3!….
In the limit as n →∞ , this becomes
1/0!+1/1!+1/2!+1/3!+1/4!+⋯
(don’t forget, 0!=1)
This expression enables the computation of e in a far more rapid and easier manner, and we find the series sum approaches 2.718281828... Try it out on EXCEL.

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