Euler's Number is a constant that is commonly known by students and professors alike as \(e\). Believe it or not, there are many useful and striking properties that this number carries, and these will be discussed in the article. This includes a pattern in differential calculus and an application to Taylor Series.
Of course, since Euler's Number is a constant, its derivative is zero. However, the very common function
$$f(x) = e^x$$
has a more interesting derivative. To find it, remember the general derivative for functions of the form \(f(x) = a^x\):
$$\frac{d}{dx}(a^x) = \ln a \cdot a^x$$
Here, we can let \(a = e\) to get
$$\frac{d}{dx}(e^x) = \ln e \cdot e^x = 1 \cdot e^x = e^x$$
Therefore, this function's derivative is the original function.
$$\frac{d}{dx}{e^x} = e^x$$
In terms of differential equations, the function \(f(x) = e^x\) is a solution to the differential equation \(y' = y\).
This is rather straightforward; since it was proven that
$$\frac{d}{dx}{e^x} = e^x,$$
simply integrating this gives
$$\int{e^x} = e^x + C$$
Of course, it is possible for \(C = 0\), so you could integrate or differentiate \(e^x\) infinitely many times and continually get \(e^x\) as a result.
Earlier we stated that the function \(e^x\) has a derivative equal to itself, and that it is a solution to the differential equation
$$y = y'$$
as a result. We will solve this differential equation in the example below.
Example 1: Rewrite \(y'\) as \(\frac{dy}{dx}\):
$$y = \frac{dy}{dx}$$
Move all the terms with \(x\) to one side and all terms with \(y\) to the other--perform separation of variables:
$$dx = \frac{1}{y}dy$$
Now integrate both sides of this equation:
$$\int{dx} = \int{\frac{1}{y}dy} \Rightarrow$$ $$x + C = \ln(y)$$
We just put the constant on the left side for convenience; it really doesn't make a difference. Isolate \(y\) by applying the exponential function to both sides of the equation:
$$e^{x + C} = e^{\ln(y)} \Rightarrow$$ $$e^{x + C} = y$$
Remember, \(C\) can be any real constant.
Like many other functions, the function \(f(x) = e^x\) can be written as a Taylor Polynomial, a polynomial with infinitely many terms used as an alternative form of a non-polynomial function, where the \(n\)th term is the nth derivative of the function at \(x = a\) divided by \(n!\) and multiplied by \((x - a)^n\), where \(a\) is the center of the polynomial. Often, even before Calculus, you may see another form of \(e\):
$$e = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + ....$$
This is actually a Taylor Polynomial at a certain value. \(e^1 = e\), so this can be viewed as \(f(1)\) if we define \(f(x) = e^x\). The actual Taylor Polynomial looks similar:
$$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + .... \frac{x^n}{n!} + ....$$
As a quick check for validity, observe what happens if \(x = 0\) is plugged into the equation:
$$e^0 = 1 + 0 + \frac{0^2}{2!} + \frac{0^3}{3!} + .... \Rightarrow$$ $$e^0 = 1$$
This is absolutely true. If you differentiate the polynomial:
$$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ...$$
As expected, the same result occurs. There is one less term, but since this is a Taylor Polynomial, the number of terms is infinite anyway. Integrating the function also gives the same result provided you let the constant equal \(1\).
The powers of Euler's Number are very versatile, interesting, and beautiful. It can be seen how they branch out into different areas of mathematics.
Example 2: Use the Taylor Series of \(e^x\) to find the Taylor Series of \(e^{3x}\). Differentiate this series.
Solution: We begin with the Taylor Series for \(e^x\):
$$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ... + \frac{x^n}{n!} + ...$$
To get the Taylor Series for \(e^{3x}\), just substitute \(3x\) in wherever \(x\) appears (note that cubing the original series is perfectly valid algebraically, but the result is not a Taylor Series):
$$e^{3x} = 1 + 3x + \frac{(3x)^2}{2!} + \frac{(3x)^3}{3!} + ... + \frac{(3x)^n}{n!} + ...$$
Differentiate this just like a polynomial with a finite number of turns:
$$3e^{3x} = 3 + 3^2x + \frac{3^3x^2}{2!} + ... \frac{3^nx^{n - 1}}{(n - 1)!} + ...$$
Dividing this entire equation by \(3\) gives us the original Taylor Series for \(e^{3x}\), ensuring that we differentiated correctly.