What is the significance of the number e?

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In summary, the number e is special because it is the base of the natural logarithm and is a solution of y = y'. It is also equal to the limit of (1+1/n)^n as n approaches infinity, and can be represented as the infinite sum of 1/n!. Other interesting properties include its relationship to the area under the curve y = 1/x and its use in determining the derivative of exponential functions. The value of 0! is defined as 1, and e is just one of many mathematical constants with unique properties.
  • #1
okkvlt
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Why is 2.71828182846...etc special? Is there any reason why e should equal 2.71828182846 and not some other number? I read that e is "transcendental" (what the hell does that mean- some kind of hippie spiritualism thing?). There must be some reason for why e=2.71828182846.

i know that the function e^x=y is special in that y=y'. (which after reasoning through it i realize that there must be such a number, because for 2^x=y, y>y' for all x values, and for 3^x=y, y<y' for all x values, so there should be a number between 2 and 3 such that y=y')

But how is e=(1+1/∞)^∞? (i don't know how to use that latex thing for limit notation, but you get the idea. actually i kind of prefer this definition over the definition that uses limit notation)
Some interesting things i discovered while playing with my calculator and want to know why:
(1+1/∞)^-∞=e^-1 (which makes sense- no questions here.)
(1-1/∞)^∞=e^-1 (that's really, really weird. why is this?)
(1-1/∞)^-∞=e (apparently the negatives cancel out somehow, but why? subtraction and inverse exponentation are two completely different things!)

Some other questions about e:

Why is e = ∑(1/n!) where n goes from zero to infinity?

And why does e^x= ∑(x^n/n!) where n goes from zero to infinity?

Why does the natural logarithm function work when finding the derivative of a^x? I always think of the logarithm as just a way to solve for x in the situation a^x=y

Are there any other weird things about e that i should know?


Also, a kind of unrelated thing, why does 0!=1?
 
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  • #2
One can show that the derivative of the function f(x)= ax, for a any positive number, is of the form Caax where Ca is constant (with respect to x) but depends on a. One can also show that, for example, if a= 2, then C2< 1 while if a= 3, then C3> 1. "e" is defined as the number such that Ce is exactly equal to 1. For that reason, the derivative of ex is ex.

Yes, you can use a logarithm to solve ax= y precisely because "loga(x) is defined as the inverse function to ax so that loga(ax)= x. And, because of the property loga(xy)= y loga(x), you can use any base logarithm: If ax= y, then, taking the natural logarithm of both sides, ln(ax)= x ln(a)= ln(y) so x= ln(y)/ln(x). And, because loga(a^x)= x (from the definition above), loge(ex)= ln(ex)= x. From that
[tex]a^x= e^{ln(a^x)}= e^{x ln(a)}[/tex]
Now, the chain rule shows that
[tex]\frac{da^x}{dx}= e^{x ln(a)}\frac{d x ln(a)}{dx}= a^x ln(a)[/itex]

The fact that [itex]e^x= \sum x^n/n![/itex] follows from the fact that repeatedly differentiating ex always gives ex which is equal to 1 at x= 0, together with the formula for the Taylor's series.

And, of course, taking x= 1 in [itex]e^x= \sum x^n/n![/itex] immediately gives [itex]e= \sum 1/n![/itex].

But how is e=(1+1/∞)^∞? (i don't know how to use that latex thing for limit notation, but you get the idea. actually i kind of prefer this definition over the definition that uses limit notation)
Some interesting things i discovered while playing with my calculator and want to know why:
(1+1/∞)^-∞=e^-1 (which makes sense- no questions here.)
(1-1/∞)^∞=e^-1 (that's really, really weird. why is this?)
(1-1/∞)^-∞=e (apparently the negatives cancel out somehow, but why? subtraction and inverse exponentation are two completely different things!)
These are simply wrong. Those should all be limits, not numerical values.
 
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  • #3
A number is transcendental if it is not algebraic. If you're not sure about the meaning of either of those terms then a quick google will clarify them for you. It is nothing to do with 'hippy' anything - maths ain't like that.
 
  • #4
okkvlt said:
Why is 2.71828182846...etc special?

You listed several reasons that answer that question, eg.
e^x is a solution of y = y'
[tex]
lim[n\rightarrow\infty] \ (1+1/n)^n = e
[/tex]

Is there any reason why e should equal 2.71828182846 and not some other number?
Um, not really, other than it has to equal something.

Are there any other weird things about e that i should know?

The area under the curve y = 1/x is given by

[tex]
log_{e}(x)
[/tex]

Also, a kind of unrelated thing, why does 0!=1?

If you know that 3! is 6, then:
Divide 3! by 3 to get 2! = 6/3 = 2, then
Divide 2! by 2 to get 1! = 2/2 = 1.
Repeating this algorithm:
Divide 1! by 1 to get 0! = 1/1 = 1.

So 0! = 1 is consistent with properties of factorials for positive integers.

And moreover:
Try dividing 0! by 0 to get (-1)! is undefined, as are factorials for all negative integers.
 
  • #5
okkvlt said:
Also, a kind of unrelated thing, why does 0!=1?

Because it is useful and consistent to so declare it. Unless you say what you think the definition of n! is for n>0, we're guessing a little as to what you'd accept as justification.

A reasonable one is that n! for n>0 counts the number of strings made from n distinct symbols, with order important.

1!=1, since a is the unique string on one symbol, and 3!=6 since abc, acb, bac, bca, cab, cba are the 6 orderings.


0! should then count the strings on 0 symbols. And since there is a unique empty string, 0!=1 makes sense here.

It also fits into other arguments, and a nice recurrence relation.
 
  • #6
Redbelly98 said:
If you know that 3! is 6, then:
Divide 3! by 3 to get 2! = 6/3 = 2, then
Divide 2! by 2 to get 1! = 2/2 = 1.
Repeating this algorithm:
Divide 1! by 1 to get 0! = 1/1 = 1.

So 0! = 1 is consistent with properties of factorials for positive integers.

And moreover:
Try dividing 0! by 0 to get (-1)! is undefined, as are factorials for all negative integers.

What? 1!/1 is not 0!

To the poster, 0! is defined to be 1 because of the following:

n! = n x (n-1)!

dividing both sides by n,

[tex]\frac{n!}{n}[/tex] = (n-1)!

Choosing n=1,

[tex]\frac{1!}{1}[/tex] = 1 = (1-1)! = 0!



As for e? Who says its special. It just happens to be a constant with some nice properties. Just like pi is a constant with nice properties (one being that pi times diameter = circumference). As to where the value comes from? Well, you will have to read up on that.
 
  • #7
Howers said:
What? 1!/1 is not 0!

Howers said:
[tex]\frac{1!}{1}[/tex] = 1 = (1-1)! = 0!

Well, which is it then? 1!/1 either is or isn't 0!
 
  • #8
Redbelly98 said:
Well, which is it then? 1!/1 either is or isn't 0!

Since mathematicians use the incluse or, it is both!



(no not really. i just didn't find an algorithm as convincing as a general equality, even though in essence they were the same thing)
 
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  • #9
Since mathematicians use the incluse or, it is both!

Don't be ridiculous. Please.
 
  • #10
Search up Euler's number on wikipedia

en.wikipedia.org/wiki/E_(mathematical_constant)
 
  • #11
1!/1 = 1/1 = 0! also

0! is said to be equal to 1 as we can define n! as the number of ways in which n objects can be arranged (say, in a row), and since here you have 0 objects, there's only one way, and that is empty.

Because, for example in a permutation combination question, if you have used all the entities, you can multiply it with 0! any number of times (for zero objects remaining) , and the answer won't be affected.
 
  • #12
Ok then, in limit language:

limit of y for y=(1+1/x)^x as x approaches infinity=e
limit of y for y=(1-1/x)^x as x approaches infinity=e^-1

limit of y for y=(1-1/x)^-x as x approaches infinity=e

Why? how does a negative 1/x cause a multiplicative inverse of e, and a negative exponent x with a negative 1/x equals e?
 
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  • #13
okkvlt said:
Ok then, in limit language:

You've posted many requests on here for people to help you in your understanding of lots of concepts. One of the simplest things you can learn is to be precise, and to think things through properly and clearly: how are we supposed to know what is your ignorance and your completely nonstandard and unacceptable abuse of notation? A little gratitude wouldn't go amiss, but I doubt anyone expects that anymore.
 
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  • #14
The main reason for 0!=1 is that it fits many different situations, like the ones that everyone posted above. But actually, the factorial is a function that is defined as:

[tex]n! = \prod_{k=1}^{n}k, \forall n \in \mathbb{N}[/tex]

(in an informal language: n! = 1 x 2 x ... x n)

Since the domain of the function is [tex]\mathbb{N}[/tex], you don't have 0! defined here. That's why you need to define it. Since it's very useful for it to be 1,

[tex]n!=\begin{cases}
\prod_{k=1}^{n}k & \text{if } n \in \mathbb{N} \\
1 & \text{if } n=0[/tex]

It's all about being useful.
 

FAQ: What is the significance of the number e?

Why is the number e important in mathematics?

The number e, also known as Euler's number, is important in mathematics because it has many applications in different fields such as calculus, finance, and statistics. It is a fundamental mathematical constant that arises naturally in many mathematical equations and has many unique properties that make it special.

What is the significance of e in exponential growth?

The number e is the base of the natural logarithm and is closely related to exponential growth. It represents the rate of growth when continuously compounded. This means that the larger the value of e, the faster the growth. This is why it is commonly used in calculations involving population growth, radioactive decay, and compound interest.

How is e related to the concept of limits in calculus?

In calculus, the number e is often used to represent the limit of (1 + 1/n)^n as n approaches infinity. This limit is equal to e and has many applications in finding derivatives and integrals of exponential functions. The natural logarithm, ln(x), is also defined in terms of e, making it a useful tool in solving many calculus problems.

Can e be expressed as a finite decimal?

No, the value of e is an irrational number, which means it cannot be expressed as a finite decimal or a fraction. Its decimal representation is infinite and non-repeating, making it a unique and special number in mathematics. It is often approximated to 2.71828, but the exact value cannot be determined.

What is the origin of the number e?

The number e was first introduced by the Swiss mathematician Leonhard Euler in the 18th century while studying compound interest. However, its significance and applications were later discovered by many other mathematicians, including Jacob Bernoulli and Johann Bernoulli. The letter "e" was chosen to honor Euler's contributions to mathematics.

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