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Euler's number

by derek181
Tags: euler, number
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derek181
#1
Apr19-14, 01:12 PM
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Can anyone give me a good definition of Euler's number and its significance. I see it everywhere, it's prolific in science and engineering.
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micromass
#2
Apr19-14, 01:31 PM
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http://en.wikipedia.org/wiki/E_(mathematical_constant)
homeomorphic
#3
Apr19-14, 01:49 PM
P: 1,268
There are two standard definitions:

[itex] e = lim_{ n \to \infty} (1+\frac{1}{n})^n[/itex]

[itex] e = 1 + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \ldots [/itex]

The first one is the result of constructing a function, [itex]y = exp(t)[/itex] that solves the differential equation

[itex]y = y' [/itex]

with the initial condition

[itex]y(0) = 1[/itex]

using Euler's method with step size 1/n and taking the limit as n goes to infinity.

Euler's method is glorified name for following a slope field (or vector field if the dimension is greater than 1) along to approximate a solution.

http://en.wikipedia.org/wiki/Euler_method

You could call this solution [itex]y = exp(t)[/itex]. It then turns out that [itex]exp(a+b) = exp(a)exp(b)[/itex]. This gives us a lot of information about the function. For example, [itex]exp(5) = 5exp(1)[/itex] and [itex]1 = exp(1-1) = exp(1)exp(-1)[/itex], so [itex]exp(-1) = 1/exp(1)[/itex]. So, this is looking a lot like a function [itex]a^t[/itex]. If you argue further along these lines, you see that that is indeed the case. So, we define [itex]e = exp(1)[/itex]. It then follows that [itex]e^t = exp(t)[/itex], so this function, [itex]exp(t)[/itex] that solves the differential equation turns out to be some number, which we call e, raised to the power t.

You can also interpret the limit using compound interest (or any form of growth with constant relative rate, like population growth). Khan Academy explains it well from this point of view, for example.

The 2nd formula for e solves the same differential equation, using power series, rather than Euler's method. The differential equation with initial condition determines a power series for [itex]e^x[/itex] and when you plug in x = 1, you get the formula for e.

Matterwave
#4
Apr19-14, 02:12 PM
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Euler's number

Quote Quote by homeomorphic View Post
For example, [itex]exp(5) = 5exp(1)[/itex]
This can't be right...but I'm also not sure what you were going for with this equality...
homeomorphic
#5
Apr19-14, 02:48 PM
P: 1,268
Oops, I meant exp(5) = exp(1)^5.
Curious3141
#6
Apr20-14, 07:59 AM
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Quote Quote by homeomorphic View Post
Oops, I meant exp(5) = exp(1)^5.

Ambiguous. You should write exp(5) = [exp(1)]^5

But there's nothing special about that since you're just saying x^5 = (x)^5
micromass
#7
Apr20-14, 08:05 AM
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Quote Quote by Curious3141 View Post
Ambiguous. You should write exp(5) = [exp(1)]^5
The equality posted by homeomorphic is perfectly clear and unambiguous. I'm not sure how you would interpret it in any other way.

But there's nothing special about that since you're just saying x^5 = (x)^5
You're missing his point. He did not define the exponential as ##\textrm{exp}(x) = e^x##. He defined the exponential as the unique function ##y## such that ##y^\prime = y## and ##y(0) = 1##. As such, saying that ##\textrm{exp}(5) = \textrm{exp}(1)^5## is not as trivial and actually serves to proving that the exponential function is of the form ##e^x## for some ##e##.
Curious3141
#8
Apr21-14, 09:16 AM
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Quote Quote by micromass View Post
The equality posted by homeomorphic is perfectly clear and unambiguous. I'm not sure how you would interpret it in any other way.



You're missing his point. He did not define the exponential as ##\textrm{exp}(x) = e^x##. He defined the exponential as the unique function ##y## such that ##y^\prime = y## and ##y(0) = 1##. As such, saying that ##\textrm{exp}(5) = \textrm{exp}(1)^5## is not as trivial and actually serves to proving that the exponential function is of the form ##e^x## for some ##e##.
Yes, I see his point now.


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