What is the significance of the number e in calculus?

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SUMMARY

The number e, approximately 2.718281, is defined mathematically by the limit \lim_{h \rightarrow 0} \frac{e^h-1}{h}=1 and is significant in calculus due to its unique property where the derivative of the function f(x) = e^x is itself, e^x. Other definitions include e = \lim_{h \rightarrow \infty} (1 + 1/h)^h and e = \sum_{i=0}^{\infty} \frac{1}{i!}. The integral \int_1^e \frac{dx}{x} = 1 illustrates the area under the curve of the natural logarithm, further emphasizing the importance of e in mathematical analysis.

PREREQUISITES
  • Understanding of limits in calculus
  • Familiarity with derivatives and integrals
  • Basic knowledge of exponential functions
  • Introduction to series expansions
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  • Study the properties of the exponential function and its derivatives
  • Explore Euler's identity: e^{i\pi} + 1 = 0
  • Learn about the significance of the natural logarithm and its relationship with e
  • Investigate the convergence of series and the concept of irrational numbers in exponentiation
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Students in calculus courses, mathematics educators, and anyone interested in the foundational concepts of exponential functions and their applications in calculus.

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I just started calc 2. And my book defined e as the number such that: \lim_{h \rightarrow 0} \frac{e^h-1}{h}=1 I’m having trouble picturing what e is. Is there another definition of e? Is a actual physical ratio, (like pi or the golden ratio) or is it just some random number?
 
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JonF said:
I just started calc 2. And my book defined e as the number such that: \lim_{h \rightarrow 0} \frac{e^h-1}{h}=1 I’m having trouble picturing what e is. Is there another definition of e? Is a actual physical ratio, (like pi or the golden ratio) or is it just some random number?
For one, the exponential function f(x) = e^x is the same as all of its derivatives (and anti-derivatives if you ignore the constant). There are other definitions for e, perhaps look it up at wikipedia and mathworld.
 
e = lim as h->infinity of (1+1/h)^h
 
Another popular version is:
e=\sum_{i=0}^{\infty} \frac{1}{i!}
 
Thanks nate that was exactly what I was looking for, I can picture that one
 
But that's (the series expansion) only saying how big e is (2.718281...). It doesn't tell you anything more about why e is useful or interesting.
 
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Other than the limits, and the mclaurin series, all i can think of at the moment is euler's (identity?)

e^(i*Pi) - 1 = 0

For a calc 2 perspective, you'll probably use all three.

Here's the easiest way to think of e.

d e^x / dx = e^x

The function is its own derivative (...and integral)
 
Last edited:
Gokul43201 said:
But that's (the series expansion) only saying how big e is (2.718281...). It doesn't tell you anything more about why e is useful or interesting.

If I'm not mistaken, I believe that the function Exp(x) defined by:
Exp(x)=\sum_{i=0}^{\infty}\frac{x^{i}}{i!}
is necessary in order to rigourously define the exponentiation process in general (for example, to introduce the concept of an irrational number raised to an irrational exponent (I think this is the rational way to do it..)).

Besides, the sequence of finite series approximations to Exp(1) converges quite fast.
So, Exp(1) is perhaps a form of e worth mentioning.
 
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The magic of e lies in this expression

\int ^e _1 \frac {dx} x =1

so the area under the inverse curve between 1 and e is exactly 1 square unit.
 
  • #10
The point of \lim_{h \rightarrow 0} \frac{e^h-1}{h}=1 is that it is easy to show that the derivative of ax is
(\lim_{h \rightarrow 0} \frac{a^h-1}{h})a^x.

Since \lim_{h \rightarrow 0} \frac{e^h-1}{h}=1,
ex has the nice property that its derivative is just ex again.

Think of it this way: the derivative of ax is C ax where C is a constant (i.e. does not depend on x) the does depend on C. e is defined as the number for which that C is 1.
 

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