Taylor's Theorem

The Taylor's theorem lets one to write down a function in terms of its derivatives and gives a direct way to compute limits in a few cases. For example, it is useful in computing limits of integrals as will be shown later in the post with the help of an example. Some of the basic theorems are stated/ proved first before giving a statement of the Taylor's theorem. The basic ideas are from Apostol. (All the integrals are Riemann).

Mean Value Theorem (Differentiation) : If $f:[a,b]\to \mathbb{R}$ is continuous on $[a,b]$ and differentiable on $(a,b)$ then $f(b)-f(a)=f^\prime(c)(b-a)$ for some $c\in (a,b)$.

Mean Value Theorem (Integration): Assume that $\int_a^b f(x)dx$ exists. Let $M = \sup_{x \in [a,b]} f(x), m = \inf_{x \in [a,b]} f(x)$. Then there exists a real number $c$ such that $m \leq c \leq M$ and $$\int_a^b f(x)dx = c(b-a)$$ If $f$ is continuous on $[a,b]$ then, $c=f(x_0)$ for some $x_0\in [a,b]$

See Apostol for a Proof.

First Fundamental Theorem of Calculus: Assume that $\int_a^b f(x)dx$ exists. Let $F(x) = \int_a^x f(x)dx$. Then $F^\prime(x)$ exists at each point $x \in [a,b]$ where $f$ is continuous.

For $y\in (a,b)$ we have $F(y+h) - F(y) = \int_y^{y+h} f(x)dx = c(y,h)h$ where $\inf_{x \in [y,y+h]} f(x) \leq c(y,h) \leq \sup_{x \in [y,y+h]} f(x)$. Therefore, $\frac{F(y+h) - F(y)}{h} = c(y,h)$. If $f$ is continuous at $y$, then both $\inf_{x \in [y,y+h]} f(x)$ and $\sup_{x \in [y,y+h]} f(x)$ converge to $f(y)$ as $h \to 0$. The same reasoning will give that $\frac{F(y)-F(y-h)}{h}$ also converges to $f(y)$. Hence $F^\prime(y) = f(y)$. For the cases $y=a,y=b$, if we define one side limits of $F$, then clearly the above reasoning applies and we get $F^\prime(a)=f(a)$ and $F^\prime(b)=f(b)$ if $f$ is continuous at $a$ and $b$ respectively. (Ofcourse, continuity is also one-sided here).

Second Fundamental Theorem of Calculus: Assume that $\int_a^b f(x)dx$ exists. Let $g : [a,b] \to \mathbb{R}$ so that $g^\prime(x) = f(x)$ for all $(a,b)$ and $g(a) - g(a+) = g(b) - g(b-)$ (note that $g(a+),g(b-)$ should exist). Then $$\int_a^b f(x)dx = \int_a^b g^\prime(x)dx = g(b) - g(a)$$

Taylor's Theorm : Let $f:[a,b] \to \mathbb{R}$ be such that $f^{(n)}$ is finite on $(a,b)$ and $f^{(n-1)}$ is continuous on $[a,b]$. Assume that $c\in [a,b]$. Then for every $x \in [a,b], x\neq c$ there exits $x_1$ (which depends on $x$) interior to the interval joining $x$ and $c$ such that $$f(x) = f(c) + \sum_{k=1}^{n-1}\frac{f^{(k)}(c)}{k!}(x-c)^k + \frac{f^{(n)}(x_1)}{n!}(x-c)^n$$

If it is known that $f\in C^n[a,b]$ then Taylor's theorem gives a simple bounds as $f^{(n)}$ is bounded on $[a,b]$. A simple application of Taylor's theorem is given below. Suppose $f\in C^2_b(\mathbb{R})$ and we want to estimate the limit $$\lim_{t \to 0} \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty \frac{f(x+y\sqrt{t})-f(x)}{t}\exp{(-\frac{1}{2}y^2)}dy$$ Using Taylor's theorem, we can write $f(x+y\sqrt{t}) = f(x) + f^{(1)}(x)y\sqrt{t} +\frac{f^{(2)}(x+\theta(y)\sqrt{t})}{2}ty^2$. Therefore, the limit becomes $$\lim_{t \to 0} \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty \frac{f^{(1)}(x)y\sqrt{t} +\frac{f^{(2)}(x+\theta(y)\sqrt{t})}{2}ty^2}{t}\exp{(-\frac{1}{2}y^2)}dy$$ $$\lim_{t \to 0} \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty \frac{f^{(2)}(x+\theta(y)\sqrt{t})}{2}y^2\exp{(-\frac{1}{2}y^2)}dy$$ Now, using dominated convergence (based on the fact that $f^{(2)}(x)$ is bounded and continuous), the limit can be pushed inside the integral. Finally we get that the original expression equals $$\frac{1}{2}f^{(2)}(x)$$