The Main Idea Behind Taylor’s Theorem
A Proof Sketch via Iterated Integration by Parts
1 Overview
The core insight behind Taylor’s theorem is that iterated integration by parts accumulates successive derivative terms at the base point while packaging the cumulative approximation error into an exact integral remainder.
2 Base Step: Fundamental Theorem of Calculus
Let \(f \in C^{n+1}([a, x])\). By the Fundamental Theorem of Calculus:
\[f(x) = f(a) + \int_{a}^{x} f'(t) \, dt\]
3 Step 1: First Integration by Parts
To extract the linear derivative term, apply integration by parts to \(\int_{a}^{x} f'(t) \, dt\).
A clever choice of antiderivative for \(1 \cdot dt\) makes the boundary term vanish at the upper limit \(t = x\):
- Let \(u = f'(t) \implies du = f''(t) \, dt\)
- Let \(dv = dt \implies v = -(x - t)\)
Evaluating the integral:
\[\int_{a}^{x} f'(t) \, dt = \Big[ -f'(t)(x - t) \Big]_{a}^{x} - \int_{a}^{x} -(x - t) f''(t) \, dt\]
Because the upper evaluation at \(t = x\) is zero:
\[\int_{a}^{x} f'(t) \, dt = f'(a)(x - a) + \int_{a}^{x} (x - t) f''(t) \, dt\]
Substituting back into the base equation yields:
\[f(x) = f(a) + f'(a)(x - a) + \int_{a}^{x} (x - t) f''(t) \, dt\]
4 Step 2: Inductive Iteration
Repeat the integration by parts on the remainder integral \(\int_{a}^{x} (x - t) f''(t) \, dt\):
- Let \(u = f''(t) \implies du = f'''(t) \, dt\)
- Let \(dv = (x - t) \, dt \implies v = -\frac{(x - t)^2}{2}\)
Evaluating:
\[\int_{a}^{x} (x - t) f''(t) \, dt = \left[ -f''(t) \frac{(x - t)^2}{2} \right]_{a}^{x} + \int_{a}^{x} \frac{(x - t)^2}{2} f'''(t) \, dt\]
\[\int_{a}^{x} (x - t) f''(t) \, dt = \frac{f''(a)}{2}(x - a)^2 + \int_{a}^{x} \frac{(x - t)^2}{2} f'''(t) \, dt\]
5 General Form: Taylor’s Formula with Integral Remainder
Continuing this procedure \(n\) times yields:
\[f(x) = \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!}(x - a)^k + R_n(x)\]
where the exact remainder is given by the Cauchy integral form:
\[R_n(x) = \frac{1}{n!} \int_{a}^{x} (x - t)^n f^{(n+1)}(t) \, dt\]
6 Deriving the Lagrange Form of the Remainder
Because \((x - t)^n\) does not change sign for \(t\) between \(a\) and \(x\), we apply the Mean Value Theorem for Integrals:
\[R_n(x) = \frac{f^{(n+1)}(\xi)}{n!} \int_{a}^{x} (x - t)^n \, dt \quad \text{for some } \xi \in (a, x)\]
Evaluating the remaining polynomial integral:
\[\int_{a}^{x} (x - t)^n \, dt = \left[ -\frac{(x - t)^{n+1}}{n+1} \right]_{a}^{x} = \frac{(x - a)^{n+1}}{n+1}\]
Substituting this back gives Lagrange’s form of the remainder:
\[R_n(x) = \frac{f^{(n+1)}(\xi)}{(n+1)!}(x - a)^{n+1}\]
When \(f \in C^\infty\) and \(\lim_{n \to \infty} R_n(x) = 0\), the partial sums converge to the infinite Taylor series.