📂Calculus

Conditions for a function and its Taylor series to be equal

Theorem¹

A function $f$ that is infinitely differentiable around point $a$, a necessary and sufficient condition for $\displaystyle f(x) = \sum_{n=0}^{\infty} {{f^{(n)} (a)}\over{n!}} {(x-a)}^n$ is that for some $\xi \in \mathscr{H} \left\{ x , a \right\}$

$$ \lim_{n \to \infty} {{f^{(n)} (\xi)}\over{n!}} {(x-a)}^n = 0 $$

where $\xi \in \mathscr{H} \left\{ x , a \right\}$ means that $\xi$ is in either $(x,a)$ or $(a,x)$.

Explanation

The Taylor theorem often represents a function that can be infinitely differentiated as an infinite series. This is called a Taylor series, and specifically, when $a=0$, it is called a Maclaurin series. The Taylor series is also commonly referred to as Taylor Formula, Taylor Expansion.

Proof

Taylor theorem

If a function $f(x)$ is continuous at $[a,b]$ and differentiable $n$ times at $(a,b)$, then for $x_{0} \in (a,b)$

$$ f(x) = \sum_{k=0}^{n-1} {{( x - x_{0} )^{k}\over{ k! }}{f^{(k)}( x_{0} )}} + {(x - x_{0} )^{n}\over{ n! }}{f^{(n)}(\xi)} $$

there exists a $\xi \in (a,b)$ that satisfies the above.

By the Taylor theorem,

$$ f(x) = \sum_{k=0}^{n-1} {{( x - a )^{k}\over{ k! }}{f^{(k)}( a )}} + {(x - a )^{n}\over{ n! }}{f^{(n)}(\xi)} $$

there exists at least one $\xi$ between $x$ and $a$ that satisfies the above. Since the function $f$ is infinitely differentiable,

$$ f(x) =\lim_{n \to \infty} \left[ \sum_{k=0}^{n-1} {{f^{(k)} (a)}\over{k!}} {(x-a)}^k + {{f^{(n)} (a)}\over{n!}} {(x-a)}^n \right] $$

If $\displaystyle \lim_{n \to \infty} {{f^{(n)} (a)}\over{n!}} {(x-a)}^n = 0$ then,

$$ f(x) =\lim_{n \to \infty} \sum_{k=0}^{n-1} {{f^{(k)} (a)}\over{k!}} {(x-a)}^k = \sum_{n=0}^{\infty} {{f^{(n)} (a)}\over{n!}} {(x-a)}^n $$

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