The Taylor series of f centered at a is the infinite power series sum from n=0 to infinity of f^(n)(a)/n! * (x - a)^n. If this series converges to f(x), then f has a power series representation. The Taylor series extends the Taylor polynomial to infinite degree, providing an exact representation (not just an approximation) within the radius of convergence. Not all functions equal their Taylor series (the remainder must go to zero).
Derive Taylor series for e^x, sin(x), cos(x), and 1/(1 - x) from the definition. Verify convergence using the ratio test. Show that the remainder term goes to zero (at least for the standard functions). Practice manipulating known Taylor series (substitution, differentiation, integration) to find new ones.
You have already worked with Taylor polynomials, which approximate a smooth function near a point by matching the function's value and derivatives up to some finite degree. A degree-3 Taylor polynomial for sin(x) near 0 gives x − x³/6, which is excellent near x = 0 but drifts away from sin(x) as x grows. The natural question is: what if we never stop adding terms? The Taylor series is the answer — it extends the Taylor polynomial to an infinite sum, and when it converges to the function, it gives an *exact* representation rather than an approximation.
The Taylor series of f centered at a is the infinite sum: f(a) + f'(a)(x−a) + f''(a)(x−a)²/2! + f'''(a)(x−a)³/3! + ··· You write this compactly as Σ (f⁽ⁿ⁾(a)/n!) (x−a)ⁿ from n = 0 to ∞. You have already computed these coefficients for Taylor polynomials; the Taylor series just keeps going. The coefficients are determined entirely by the derivatives of f at the single point a — the remarkable claim is that, for well-behaved functions, all the information about f near a is encoded in those derivatives.
But here is the critical caveat: not every infinitely differentiable function equals its Taylor series. The series always converges to *something*, but that something may not equal f(x). To confirm that f(x) equals its Taylor series on an interval, you must show that the remainder term Rₙ(x) — the error between f and the nth partial sum — goes to zero as n → ∞. For e^x, sin(x), cos(x), and 1/(1−x) this can be verified directly, which is why these are the standard examples. A pathological function like e^(−1/x²), however, has all zero derivatives at x = 0, so its Taylor series is identically 0 — clearly not equal to the function for x ≠ 0.
The radius of convergence from your power series work remains central here. The Taylor series for 1/(1−x) = 1 + x + x² + ··· converges only for |x| < 1, even though the function itself is defined for all x ≠ 1. Within the radius, the series equals the function; outside it, the series diverges. For e^x the radius is infinite — the series converges everywhere. Understanding where equality holds is as important as knowing the series itself.
In practice, the most powerful technique is often *not* recomputing from the definition but instead manipulating known series. If you know the series for e^x, you can substitute −x² to get the series for e^(−x²) without computing a single new derivative. You can differentiate or integrate term-by-term inside the radius of convergence. This toolkit — derive the four or five standard series once, then transform them — is what makes Taylor series genuinely useful in applied mathematics, physics, and numerical analysis.