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The Bridge Between Two Worlds: Euler’s formula is a profound mathematical identity that connects two entirely separate worlds: the mechanics of continuous growth represented by the constant \(e\), and the geometry of circles described by \(\sin\) and \(\cos\).

Euler’s formula, \(e^{ix}=\cos(x)+i\sin(x)\), provides the definitive bridge between these realms. By introducing the imaginary unit \(i\), we discover that exponential growth is not restricted to a straight line; instead, when the exponent is imaginary, that growth “curves.” It becomes a smooth, continuous rotation through the complex plane.

This identity allows us to describe periodic phenomena—from alternating currents to quantum mechanics—using a single, elegant algebraic expression. To understand why this is true, we look at the Taylor series.

Taylor Series: Polynomial Approximation

A Taylor series represents a function as an infinite sum of terms based on its derivatives at a single point. For a general function \(f(x)\): \[f(x)=\sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!}(x-a)^n\]

The Maclaurin series is a special case evaluated at \(a=0\): \[f(x)=\sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}x^n\]

Proof of Euler’s Formula

Step 1: Derivatives at \(x=0\)

The derivatives for \(e^x\) are all \(e^x\). At \(x=0\), \(e^0=1\).
The derivatives of \(\cos(x)\) follow a 4-step cycle: \(\{1, 0, -1, 0\}\).
The derivatives of \(\sin(x)\) follow a 4-step cycle: \(\{0, 1, 0, -1\}\).

Step 2: Maclaurin Expansions

\[e^x = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots = \sum_{n=0}^\infty \frac{x^n}{n!}\] \[\cos(x) = 1 – \frac{x^2}{2!} + \frac{x^4}{4!} – \cdots = \sum_{n=0}^\infty \frac{(-1)^nx^{2n}}{(2n)!}\] \[\sin(x) = x – \frac{x^3}{3!} + \frac{x^5}{5!} – \cdots = \sum_{n=0}^\infty \frac{(-1)^nx^{2n+1}}{(2n+1)!}\]

Step 3: Substitution and Separation

Substituting \(ix\) into the series for \(e^x\): \[e^{ix}=\sum_{n=0}^\infty \frac{(ix)^n}{n!} = \sum_{n=0}^\infty \frac{(ix)^{2n}}{(2n)!} + \sum_{n=0}^\infty \frac{(ix)^{2n+1}}{(2n+1)!}\]

Noting that \(i^{2n} = (-1)^n\) and \(i^{2n+1} = i(-1)^n\), we can factor out the \(i\): \[e^{ix} = \left[ \sum_{n=0}^\infty \frac{(-1)^nx^{2n}}{(2n)!} \right] + i \left[ \sum_{n=0}^\infty \frac{(-1)^nx^{2n+1}}{(2n+1)!} \right]\]

Substituting the series for \(\cos(x)\) and \(\sin(x)\) back in, we arrive at:

\[e^{ix} = \cos(x) + i\sin(x)\]

Alternative Proof: Differential Equations

This proof is adapted from Saff and Snider’s Fundamentals of Complex Analysis.

Let \(g(y) = e^{iy}\). Taking derivatives:

g′(y)=deiydy=ieiyg'(y) = \frac{d e^{iy}}{dy} = ie^{iy}
g′′(y)=d2eiydy2=i2eiy=−eiyg”(y) = \frac{d^2 e^{iy}}{dy^2} = i^2 e^{iy} = -e^{iy}

Therefore, \(g(y) := e^{iy}\) satisfies the differential equation:

d2gdy2=−g\frac{d^2 g}{dy^2} = -g

From the theory of differential equations, every solution of this equation has the form \(A\cos y + B\sin y\) for constants \(A\) and \(B\). Hence:

g(y)=Acos⁡y+Bsin⁡yg(y) = A\cos y + B\sin y

To find \(A\) and \(B\), we use initial conditions. At \(y = 0\):

g(0)=ei⋅0=1=Acos⁡0+Bsin⁡0=Ag(0) = e^{i \cdot 0} = 1 = A\cos 0 + B\sin 0 = A
g′(0)=iei⋅0=i=−Asin⁡0+Bcos⁡0=Bg'(0) = ie^{i \cdot 0} = i = -A\sin 0 + B\cos 0 = B

Thus \(A = 1\) and \(B = i\), giving us:

eiy=cos⁡y+isin⁡ye^{iy} = \cos y + i\sin y

Euler’s Identity

Evaluating the formula at \(x = \pi\): \[e^{i\pi} = \cos(\pi) + i\sin(\pi)\] \[e^{i\pi} = -1 + i(0)\] \[e^{i\pi} + 1 = 0\]

This identity links the five fundamental constants (\(0, 1, e, i, \pi\)) in a single, perfect expression. ■

References

Saff, Edward B., and Arthur David Snider. Fundamentals of Complex Analysis with Applications to Engineering and Science. 3rd ed., Pearson, 2003.

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