What does e represent in calculus and why is it used?

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The number "e," approximately equal to 2.71828183, is a fundamental constant in calculus, known for its unique property where the derivative of the function f(x) = e^x is equal to the function itself. It is classified as an irrational and transcendental number, meaning it cannot be expressed as a fraction or the root of any polynomial with integer coefficients. "e" arises in various real-world applications, such as modeling population growth and radioactive decay, where the rate of change is proportional to the quantity itself. The limit definition of "e" is given by e = lim (n→∞) (1 + 1/n)^n, highlighting its mathematical significance. Understanding "e" is crucial for solving differential equations and its relationship with logarithms, particularly the natural logarithm, ln(x).
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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
 
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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).
 
J7 said:
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 \pi,1 and i,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 :-p ) are:
2.718281828459045... and it is defined as the limit of the sequence:
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.
 
It is also

\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.
 
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It's just this number, y'know? Any function of the form f(x)= ax has the property that its derivative (rate of change) is proportional to ax 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 ex is precisely ex.
 
HallsofIvy said:
It's just this number, y'know? Any function of the form f(x)= ax has the property that its derivative (rate of change) is proportional to ax 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 ex is precisely ex.


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 explanations of e? 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..

\int \frac{dy}{dx} e^2 dx = 1/3^3e + C
 
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What about this? I remember this from a class.

e = (1 + \frac{1}{x})^x
 
1 = \int_1^e \frac {dx} x

I believe that it this relationship which gives the number physical significance.
 
Chrono said:
What about this? I remember this from a class.

e = (1 + \frac{1}{x})^x

You probably mean:

e=\lim_{x\rightarrow\infty}(1 + \frac{1}{x})^x

Or more generally:

e^a=\lim_{x\rightarrow\infty}(1 + \frac{a}{x})^x
 
  • #10
QuantumTheory said:
I thought e was actually just related to the logaritam.

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

\int \frac{dy}{dx} e^2 dx = 1/3^3e + C
<br /> <br /> <br /> nb.your backslashes and forward slashes are wonky<br /> <br /> It is related to the logarithm, and the examples ought to tell you you&#039;ve underestimated the importance of log.<br /> <br /> THe second bit is wrong. e^2 is just a number so you&#039;re asking for <br /> <br /> e^2 \int \frac{dy}{dx}dx<br /> <br /> which is just ye^2
 
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  • #11
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[/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= ekt+C= eCekt= C' ekt (where C'= eC). 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)= M0e1000k= (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)= M0eln(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 eln(1/2)t/1000= (1/2)t/1000.
 
  • #12
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= ekt+C= eCekt= C' ekt (where C'= eC). 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)= M0e1000k= (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)= M0eln(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 eln(1/2)t/1000= (1/2)t/1000.
 
  • #13
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
 
  • #14
The arguably most famous relation is:
e^{i\pi}+1=0
 
  • #15
Hessam said:
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
 
  • #16
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!
 
  • #17
HallsofIvy said:
(pi^5+ pi^5)^(1/6)= (2pi^5)^(1/6)= approximately 2.9138 which is not particularly close to e!

The correct one is (pi^4 + pi^5)^(1/6).

Forgive me for not using latex, I just don't want to go through all that trouble with it now.
 
  • #18
Chrono said:
The correct one is (pi^4 + pi^5)^(1/6).

Forgive me for not using latex, I just don't want to go through all that trouble with it now.
Is this another piece of Ramanurjan magic?
 
  • #19
well, log of that number, is according to my not very accurate caculator 0.999 something

so close, but probably no cigar (seriously, it is an open question, and that ain't the solution)
 
  • #20
Hessam said:
isnt there also a relation between pi and e?
its something like this...
(pi^4 + pi^5)^(1/6)
or something of the sort... please feel free to correct me

e\simeq \sqrt[6] {\pi^{4}+\pi^{5}}
,the fourth decimal is different.

Matt,it sounds like the Goldbach conjecture... :-p As for that,i still trust mathematicians would solve it,in the next millenium. :wink: But for an algebraic relation beween the most famous transcendental numbers,i frankly don't...Feel free to contradict me. :-p It would mean proving the existence of this relation,a great benefit to mathematics and generally science+humanity+get u a Medal.I would have said Nobel prize,but that dude hated your kind... :-p

Daniel.
 
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  • #21
Actually Nobel didn't hate mathematicians. An old wives tale is taht he didnt' create one for maths because his wife had an affair with one of us. As he never married I think we can dismiss that.
 
  • #22
arildno said:
Is this another piece of Ramanurjan magic?

Never heard of the guy.U probably meant this dude
http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Ramanujan.html

No offense,but if u were a mathematician,i'd feel sorry for you... :-p

Daniel.

PS.U may want to check other pages from that site.Who knows how many names u got wrong... :wink:
PPS.Yeah,Matt.He never got married.So does a mathematician know the reason why a brilliant chemist and inventor didn't consider mathematics to as important as other disciplines??
http://nobelprize.org/nobel/alfred-nobel/biographical/life-work/gradeschool.html
 
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  • #23
Forgive me if I'm missing the point, but surely a much easier way to explain what "e" is all about would be to say:

e is the only number which when you make f(x) = logaX with the base a = e, that the gradient at any point will equal 1/x. ie: the derivative of logeX is 1/x. It can be derived from first principles when trying to work out how to find the gradint of a curve y = logaX, then later placing a = e. I can post this if you would like? That might clarify for you exactly where e comes from. :smile: NB - the usual way of writing logeX is lnX

All e basically is is just another constant like pi, etc, though it is used a lot more, since we use 1/x so often in maths and science and the integral of 1/x is lnX. ( since the derivative of lnX is 1/x, etc)

Any clearer J7? :smile:
 
  • #24
dextercioby said:
e\simeq \sqrt[6] {\pi^{4}+\pi^{5}}
,the fourth decimal is different.

Wasn't that what I said it was?

And when I checked it it was accurate to seven decimal places.
 
  • #25
Your implication was that they are the same, by saying there is a relation between them.
Noting that one may apply some arithmetic operations to get a number close to the other arguably isn't "related", that#s all.

After all, pi is just a little bit bigger than 3, and e just a little bit smaller than 3 so we're going to be able to manufacture something like that.
 
  • #26
Actually, with enough operations, I would think one use any number to make an estimation of another
 
  • #27
http://www.maa.org/devlin/devlin_04_04.html


here (about 1/2 way down) is a reasonable summary of what Nobel may have been thinking, and a list of mathematicians who have won Nobel PRizes
 
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  • #28
The vast majority of the people on that list are not mathematicians.
 
  • #29
Deadwolfe,
hmm give me one name on that list which is not listed here
http://www-groups.dcs.st-and.ac.uk/~history/Indexes/Full_Alph.html
as a mathematician.

Just because their contribution to the field of mathematics was little does not make them any less of a mathematician.

On the Nobel Prize :
Nobel made his wealth from explosives. Now if i were to believe the historians, then its said he was deeply disturbed by the amount of disaster his explosives are making. Hence he decided to dedicate his wealth to fields which would improve the state of the society as a whole. Mathematics as such has no direct implications on the welfare of the society, which is why i think, he never offered the prize for mathematicians. (Even if this is not true, any person who is narrow minded as to be obsessed with his wife's affair wouldn't be as broad minded to gift his wealth away IMHO).

-- AI
 
  • #30
(Even if this is not true, any person who is narrow minded as to be obsessed with his wife's affair wouldn't be as broad minded to gift his wealth away IMHO).

It would particularly freaky in Nobel's case, seeing as he didn't even have a wife. ;)
 

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