# Week 6 - Taylor Series

## Brief

• This week talks about how to transfer a function $f(x)$ to a power series.

## Better Than Linear Approximation

• In calculus one, we’ve learned $f(x) \approx f(a) + f'(a) (x - a)$ to approximately get value of $f(x)$.
• This time we use another function $g(x) = f(a) + f'(a) \cdot (x - a) + \frac{f''(a)}{2} \cdot (x -a)^2 + \frac{f'''(a)}{6} \cdot (x - a)^3$
• With some calculation, we can get:
• $g(a) = f(a)$
• $g'(a) = f'(a)$
• $g''(a) = f''(a) + f''(a) (x - a)$
• $g'''(a) = f'''(a) + f'''(a) (x - a)$
• So, $g(x)$ is a better approximation for $f(x)$.
• Example: $f(x) = \sin(x)$
• $f(x) = \sin(x), f(0) = 0$
• $f(x) = \cos(x), f'(0) = 1$
• $f(x) = - \sin(x), f''(0) = 0$
• $f(x) = - \cos(x), f'''(0) = -1$
• $g(x) = 0 + 1 \cdot x + \frac{0}{2} \cdot (x)^2 + \frac{-1}{6}(x)^3 = x - \frac{x^3}{6}$
• $f(\frac{1}{2}) = \sin(\frac{1}{2}) \approx 0.4794$
• $g(\frac{1}{2}) = \frac{1}{2} - \frac{(1/2)^3}{6} \approx 0.4792$

## Taylor Theorem

### the Taylor Series for f Around Zero(Maclaurin Series)

• Suppose $f(x) = \sum_{n=0}^{\infty} a_n x^n,\ |x| < R$

• $= a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \cdots$
• Then, $f(0) = a_0 + a_1 \cdot 0 + a_2 \cdot 0 + a_3 \cdot 0 + \cdots = a_0$

• $f'(x) = \sum_{n=1}^{\infty} a_n \cdot n \cdot x^{n-1}, |x| < R$

• $= a_1 + a_2 \cdot 2 \cdot x + \cdots$
• => $f'(0) = a_1$
• $f''(x) = \sum_{n=2}^{\infty} a_n \cdot n \cdot (n-1) \cdot x^{n-2}, |x| < R$

• $= a_2 \cdot 2 \cdot 1 \cdot 1 + a_3 \cdot 3 \cdot 2 \cdot x + \cdots$
• => $f''(0) = a_2 \cdot 2$
• => $a_2 = \frac{f''(0)}{2}$
• $a_3 = \frac{f'''(0)}{3!}$

• $\cdots$

• So, Assume $f(x) = \sum_{n=0}^{\infty} a_n x^n,\ |x| < R$

• Then $a_n = \frac{f^{n}(0)}{n!}$

#### Definition

• If $\displaystyle f(x) = \sum_{n=0}^{\infty} a_n x^n,\ \text{where}\ |x| < R$
• Then $\displaystyle f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} \cdot x^n$
• This series is called Taylor series of the function f at 0 or Maclaurin series.

### The Taylor Series for f Centered Around a

• $\displaystyle f(x) = \sum_{n=0}^{\infty} c_n (x-a)^n$
• $c_0 + c_1 (x - a) + c_2 (x - a)^2 + \cdots$
• $f(a) = c_0$
• $f'(a) = c_1$
• To summary this:
• The Taylor Series for f Centered Around a is $$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} \cdot (x-a)^n$$
• Example $f(x) = \sin(x)$
• The Taylor Series for $\sin$ around 0 is $$f(x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} \cdot x^{2n+1}$$
• Quiz 6: Let $f(x) = \cos(x^5)$. By considering the Taylor series for $f$ around 0, compute $f^{(90)}(0)$.
• The Taylor Series for $\cos$ around 0 is $$f(x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!} \cdot x^{2n}$$
• $f(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots + \frac{x^{16} }{16!} - \frac{x^{18} }{18!} + \cdots$
• $f(x^5) = 1 - \frac{x^{10} }{2!} + \frac{x^{20} }{4!} - \frac{x^{30} }{6!} + \cdots + \frac{x^{80} }{16!} - \frac{x^{90} }{18!} + \cdots$
• So we just need to differentiate term $\frac{d^{90} }{dx^{90} } (- \frac{x^{90} }{18!})$ which $= - \frac{90! \cdot x^0}{18!} = - \frac{90!}{18!}$

### Taylor’s Theorem

• Suppose $f: \mathbb{R} \to \mathbb{R}$ is infinitely differentiable.
• f(x) = (\sum_{n=0}^{N} \frac{f^{(n)}(0)}{n!} \cdot x^n) + R_N{x}$$. **R** stands for remainder. #### Usage • Use Taylor’s Theorem to prove$$\sin{x} = \sum_{n=0}{\infty}\frac{(-1)n}{(2n+1)!} \cdot x^{2n+1}$$• $f(x) = \sin x$ • With Taylor’s Theorem,$$\begin{aligned}
\sin{x} &- \sum_{n=0}^{N} \frac{f^{(n)}(0)}{n!} \cdot x^n = R_N(x) \
R_N(x) &= \frac{f^{(N+1)}(z)}{(N+1)!} x^{N+1}
\sum_{n=0}^{\infty} \frac{(ix)^n}{n!} = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n} }{2n!} + i \cdot \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1} }{(2n+1)!}$$. Set $x = \pi$:$$e^{i \pi} = \cos \pi + i \cdot \sin \pi = -1$$• Taylor series are the first step into the theory of complex analysis(复分析理论). #### Mean Value Theorem • $f: [a, b] \to \mathbb{R} \text{ continuous}$ • and, on $(a, b)$ differentiable, • Then, there is a point $c \in (a, b)$, so that$$f’© = \frac{f(b) - f(a)}{b - a}$$• replace c and b with z and x, we get: • $f'(z) = \frac{f(x) - f(a)}{x - a}$ • $f(x) = f(a) + R_0(x), \text{where}\ R_0(x) = \frac{f'(z)}{1!} \cdot (x-a)^1$, and z is between x and a. • It is the Taylor Theorem when $N = 0$ • Example $f(t) = \text{your position at time t sec}$ • $f(0) = 0$, $f'(0) = 0$, $f''(t) \le 250 \frac{m}{s^2}$ • Question: How big can $f(60)$ be? • Use Taylor Theorem, we get: • $f(t) = f(0) + f'(0) \cdot (t - 0) + R_1(t)$, $R_1(t) = \frac{f''(z)}{2!} \cdot (t - 0)^2$ • $|R_1(t)| \le \frac{250}{2!} \cdot t^2$ • So $f(t) \le f(0) + f'(0) \cdot t + \frac{250}{2!} \cdot t^2 = 125 t^2$ • $f(60) \le 125 \cdot 60^2 = 450000 \text{m} = 450 \text{km}$ ## Practice ### Uniform Convergence #### Approximate cos x when x is near zero • Goal: Find polynomial $p(x)$ so that $|p(x) - \cos x| < \frac{1}{100}$, where $x \in [-1, 1]$ • $\displaystyle \cos x = \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!} \cdot x^{2n}$ • $\displaystyle \cos x - \sum_{n=0}^{N}\frac{f^{(n)}(0)}{n!} \cdot x^{n} = R_N(x)$ • $\displaystyle R_N(x) = \frac{f^{(N+1)}(z)}{(N+1)!} \cdot x^{N+1}$, $z \in (0, x)$ • $x \in [-1, 1]$, then $|x^{N+1}| \le 1$ • => $R_N(x) \le \frac{f^{(N+1)}(z)}{(N+1)!}$ • $f^{(N+1)}(z) = \pm \sin z\ \text{or}\ \pm \cos z \le 1$ • => $R_N(x) \le \frac{1}{(N+1)!}$ • We want $\frac{1}{(N+1)!} < \frac{1}{100}$ and $\frac{1}{5!} = \frac{1}{120}$ • => $N = 4$ • $\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} -\frac{x^6}{720} + \cdots$ • There is no x to the 5th term, so we can make $N = 5$ • $\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} -R_5$ • => $R_5(x) \le \frac{1}{(5+1)!} = \frac{1}{6!} = \frac{1}{720} < \frac{1}{100}$ • So $p(x) = 1 - \frac{x^2}{2} + \frac{x^4}{24} \approx \cos x$ ### Limits • $\displaystyle \lim_{x \to 0} \frac{\sin x}{x}$ • use Taylor Series, we know: • $\sin x = x - \frac{x^3}{6} + \frac{x^5}{5!} - \cdots$ • $\sin x \approx x +\ \text{higher order terms}$ • then $\displaystyle \lim_{x \to 0} \frac{\sin x}{x} = \lim_{x \to 0} \frac{x + \text{higher order terms} }{x} = 1 + \lim_{x \to 0} \frac{\text{higher order terms} }{x}$ ### Real Analytic Function • Theorem: • The function $f$ is real analytic at a if there is some $R > 0$, • So that$$f(x) = \sum_{n = 0}^{\infty} \frac{f{(n)}(a)}{n!}(x-a)n \ \text{when } |x-a|<R
• Real Analytic Functions ∈ Infinitely Differentiable Functions(Smooth Functions($C^{\infty}$)) ∈ Differentiable Functions ∈ Continuous Functions ∈ All Functions

#### Holograms

• A little bit of a hologram records everything in the function.
• For example $\sin x$.
• Only look the points near 0, we know:
• $f(0) = 0$
• $f'(0) = 1$
• $f''(0) = 0$
• $f'''(0) = -1$
• We can calculate all of the derivatives of the function by just looking the area near 0.
• And use Taylor Series, we can get $\displaystyle f(x) = \sum_{n = 0}^{\infty} \frac{f{(n)}(0)}{n!} x^n$

## Important Taylor Series and Their Radii of Convergence

• $\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n = 1 + x + x^2 + x^3 + \cdots, R = 1$

• $e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots, R = \infty$

• $\sin x = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1} }{(2n+1)!} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots, R = \infty$

• $\cos x = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n} }{(2n)!} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots, R = \infty$

• $\tan^{-1} x = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1} }{2n+1} = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \cdots, R = 1$

• $\ln (1 + x) = \sum_{n=0}^{\infty} \frac{(-1)^{n-1} x^{n} }{n} = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots, R = 1$

• $(1 - x)^k = \sum_{n=0}^{\infty} (n {k})x^{n} = 1 + kx + \frac{k(k-1)}{2!}x^2 + \frac{k(k-1)(k-2)}{3!}x^3 + \cdots, R = 1$