Euler's Formula From the Series for e to the ix

Write the Maclaurin series for e^y as the infinite sum y^n/n!. Substitute y = ix. You get e^(ix) = Σ (ix)^n/n!. The terms are complex numbers because i = √(−1). Now separate the sum into even n and odd n. For even n = 2k, (ix)^(2k) = i^(2k) x^(2k) = (i^2)^k x^(2k) = (−1)^k x^(2k). For odd n = 2k+1, (ix)^(2k+1) = i^(2k+1) x^(2k+1) = i * i^(2k) x^(2k+1) = i (−1)^k x^(2k+1). The even-index terms sum to cos x. The odd-index terms sum to i sin x. That is how the real and imaginary terms separate into cosine and sine series.

Take x = π/2. The series for e^(iπ/2) is Σ (iπ/2)^n/n!. Write the first four terms. n=0: 1. n=1: iπ/2 ≈ 1.5708i. n=2: (iπ/2)^2/2 = (−π^2/4)/2 = −π^2/8 ≈ −1.2337. n=3: (iπ/2)^3/6 = (−iπ^3/8)/6 = −iπ^3/48 ≈ −0.6455i. Group real terms: 1 − 1.2337 + … ≈ 0. The cosine of π/2 is 0. The series matches. Group imaginary terms: 1.5708 − 0.6455 + … ≈ 0.9253, approaching 1. The sine of π/2 is 1. The series converges to i.

Use a complex number calculator to sum more terms and watch the partial sums oscillate and settle. The separation works because the Maclaurin series for e^y converges for all real y, and absolute convergence lets you reorder the even and odd terms freely.

The pattern i^n cycles every four terms: i^0 = 1, i^1 = i, i^2 = −1, i^3 = −i, i^4 = 1. That cycle turns every even power of i into a real number (±1) and every odd power into an imaginary number (±i). The factorial denominator n! grows faster than any polynomial in x, so the ratio test limit is 0 for all x, and the series converges for all real x.

Absolute Convergence Allows Reordering

A conditionally convergent series cannot be rearranged without changing its sum. The series for e^(ix) converges absolutely for every complex x because the terms are bounded by |x|^n/n! and that sum converges. Absolute convergence lets you split the sum into even and odd indices and sum each group independently. Without that property, the separation into cos x and i sin x would be invalid.

What the Lagrange Remainder Tells You

If you truncate after N terms, the error is bounded by M |x|^(N+1)/(N+1)!, where M is the maximum of the (N+1)th derivative of e^(iy) on the interval from 0 to x. For real x, that derivative is i^(N+1) e^(ic) for some c between 0 and x, so M ≤ e^|x|. The factorial denominator guarantees the error shrinks to zero as N increases.

Start with the Maclaurin series for e^y: Σ y^n/n!. Substitute y = ix. Write e^(ix) = Σ (ix)^n/n!. Expand (ix)^n = i^n x^n. Separate the sum into n even and n odd: Σ_{k=0}∞ i^(2k) x^(2k)/(2k)! + Σ_{k=0}∞ i^(2k+1) x^(2k+1)/(2k+1)!. For even powers, i^(2k) = (−1)^k, giving Σ (−1)^k x^(2k)/(2k)!. That is the Maclaurin series for cos x. For odd powers, i^(2k+1) = i (−1)^k, giving i Σ (−1)^k x^(2k+1)/(2k+1)!. That is i sin x. The result is e^(ix) = cos x + i sin x.

Why This Is Not a Proof

The derivation assumes absolute convergence to reorder terms. Absolute convergence holds for all real x because the terms are bounded by |x|^n/n!, and the sum of |x|^n/n! converges to e^|x|. The derivation also assumes the Maclaurin series equals the function at every point, which is true for e^y because it is analytic on the whole complex plane. A rigorous proof requires showing that the remainder goes to zero, but the series method is the standard elementary derivation.

Failure Case: When x Is Complex

If x is not real, the separation into real and imaginary parts is not straightforward. For a complex number z = a + ib, e^z = e^a (cos b + i sin b). The series still converges for all complex z, but the real and imaginary parts of e^z are not simply cos of the real part and sin of the real part. The series for e^z mixes a and b in each term. You must treat the exponent as a single complex number.