Register to reply 
Number eby Hepic
Tags: number 
Share this thread: 
#19
Mar3114, 12:59 PM

Sci Advisor
HW Helper
P: 11,915

Actually a more interesting discussion would be the universality of e:
[tex] \sum_{k=0}^{\infty} \frac{1}{k!} = \lim_{n\rightarrow\infty} \left(1+\frac{1}{n}\right)^n [/tex] 


#20
Mar3114, 01:06 PM

Sci Advisor
HW Helper
PF Gold
P: 12,016

Reminds me of my student days, when I and a couple of costudents became determined to prove that identity directly. We were proud of ourselves when we managed to do so, not the least because we found it rather troublesome to achieve. 


#21
Mar3114, 01:07 PM

P: 1,055




#22
Mar3114, 01:09 PM

Sci Advisor
HW Helper
PF Gold
P: 12,016

That ends my involvement in this thread. 


#23
Mar3114, 01:09 PM

Mentor
P: 18,240

[tex]\int_1^e \frac{1}{x} = 1[/tex] then I would think "alright, cool". But I wouldn't see at all the importance of what ##e## is. Why would we care about this new number ##e## in the first place? This definition doesn't answer this question. It's a bit like Rudin does the following, he defines [tex]\sin(x) = x  \frac{x^3}{3!} + \frac{x^5}{5!}  ...[/tex] and then he defines ##\pi## as the smallest positive zero. This definition is correct, it is important and it makes it very easy to derive all the analysis properties. But it doesn't show why I would care about ##\pi##. So in that sense, I don't like the definition. 


#24
Mar3114, 01:14 PM

P: 1,055




#25
Mar3114, 01:17 PM

Mentor
P: 18,240




#26
Mar3114, 01:33 PM

Mentor
P: 15,153

Let's expand upon Integral's integral a bit. Define ##g(x)=\int_1^x \frac {dt}{t}##. Here's the key question: For what value of x is this g(x) equal to one? For lack of a better name, let's call the solution to g(x)=1 e. In other words, e is the unique value that satisfies ##\int_1^e \frac {dt}{t} = 1## (i.e., Integral's integral.) To find this value e it will help to find the inverse function of g(x). Let's call this inverse function f(x), defined by f(g(x))=x. That f'(x)=f(x) pops right out of this definition. Since g(1)=0, f(0)=1. These two results immediately lead to ##f(x) = \sum_{n=0}^{\infty} \frac {x^n}{n!}##, from whence ##e = \sum_{n=0}^{\infty} \frac {1}{n!}##. 


#27
Mar3114, 02:18 PM

P: 1,295

Of course it actually has everything to do with the functional antiderivative when we know such things.
The point is that the relationship described by the integral does not depend on understanding the relationship between area and antiderivative. This is something that aldrino and I can agree on, and if you are claiming otherwise then any argument you had for the definition being fundamental goes right out the window. And yes, obviously we can derive other definitions and properties about e from an expression that uniquely describes e. The point is that the integral definition of e, no matter how important of a question it answers, or how cool it is, does not give any intuitive understanding of why e is important or where it will appear in mathematics. It is the area of this region, OK, so what? Who cares about this region? If someone asked you what e was and why it was portant, would you really tell them that it was the area of this region? Or even mention that fact? 


#28
Mar3114, 03:12 PM

Engineering
Sci Advisor
HW Helper
Thanks
P: 7,115

First, define the idea of derivatives (for example using the epsilondelta definition) and antiderivatives. Then you can easily prove that the derivative of ##x^n## is ##nx^{n1}##, for all ##n \ne 0##. That leaves an unanswered question: what is the antiderivative ##A(x)## of ##1/x##? You can prove that ##A(x)## has the properties of a logarithm, to some (unknown) base  for example ##A(xy) = A(x) + A(y)##  direct from the definition of the antiderivative. So you can write ##A(x) = \log_e x##, and Integral's integral defines the value of ##e##. It then follows that the derivative of the inverse function, ##e^x##, is ##e^x##, and hence we get the power series for ##e^x##. And after jumping through those hoops to motivate the definition, it's much simpler just to define ##e^x## as a power series (for all complex values of ##x## not just real values), and define ##\ln x## as the inverse function of ##e^x##. 


#29
Mar3114, 06:43 PM

Mentor
P: 7,318

Wow! That's a lot of discussion for such a simple statement. Obviously area is a more fundamental concept then rate of change. History shows that area has been a useful concept since the very first writings of mankind, rate of change is a much later and subtler concept.
Expressing e as a upper limit in a integral is not the most computable way of going about it. But for me tying it to a simple geometry is much more meaningful then expressing it as an infinite series or any other more esoteric definitions. This integral is a large portion of the reason e shows up in nature as much as it does. 


#30
Mar3114, 07:07 PM

Engineering
Sci Advisor
HW Helper
Thanks
P: 7,115

I guess it depends which bits of nature you look at. You could argue that ##e^{iz} = \cos z + i \sin z## shows up just as much as the integral.
Personally I find the connection with ##\lim_{n \rightarrow \infty}(1 + \frac 1 n) ^n## the least motivational approach. But mathematicians are fairly agnostic about the "right way" to define things. Back in my day, if a math exam paper asked you to prove three results about X, it was perfectly acceptable to define X three different ways for the three proofs, without bothering to show that the definitions were equivalent! 


#31
Apr114, 04:34 AM

Sci Advisor
HW Helper
PF Gold
P: 12,016




#32
Apr114, 05:15 AM

P: 542

Wish I could remember where I read about the early origin of e, long before it was known.
What I read stated that farmers that grow seed crops need to put a certain amount of the seed crop aside to plant the next season... but a portion of that put aside will actually be accounting for the seed crop to be put aside for the season after that, and so on ad infinitum, converging to what would later be found as e. So staying above e was crucial to surviving... anyone ever hear about this? 


#33
Apr114, 03:26 PM

Sci Advisor
HW Helper
PF Gold
P: 3,288

How much will this grow after one year? It depends on how often the interest is paid. If paid annually, we will simply have ##C(1+r)##. More generally, if it is paid ##n## times per year, then we will have ##C(1+r/n)^n##, which is larger than ##C(1+r)## because we were able to earn interest on each interest payment in addition to our original sum. The best case would be if interest was paid continuously, in which case we get ##Ce^r##, where we have defined ##e = \lim_{n\rightarrow \infty}(1+1/n)^n##. Of course, we have to prove that this limit exists and perform the simple change of variables to evaluate ##\lim_{n\rightarrow \infty}(1+r/n)^n##. I certainly found this more motivating than any of the competing definitions. But I like money. 


Register to reply 
Related Discussions  
Determining the atomic number, mass number, and chemical name during beta decay  Introductory Physics Homework  5  
[C++] Efficiently counting the number of '1' bits inside blocks of a binary number  Programming & Computer Science  4  
Maximal number of bases for which composite number is Fermat pseudoprime  Linear & Abstract Algebra  2  
[number theory] find number in certain domain with two prime factorizations  Calculus & Beyond Homework  3  
Number of ways to make change for dollar with even number of coins  Calculus & Beyond Homework  1 