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1.2

Recall 1:

We say \(f\) is equal to \(g\) at \(a\) up to degree \(N\) if \(\lim_{x \to a} \frac{f(x) - g(x)}{(x-a)^N} = 0\)

Let \(h(x) = f(x) - g(x)\), then \(h\) is equal to zero up to degree \(N\) at \(a\) if \(\lim_{x \to a} \frac{h(x)}{(x-a)^N} = 0\)

Note that:
If \(\lim_{x \to a} \frac{h(x)}{(x-a)^N} = 0\), then \(\lim_{x\to a}\frac{h(x)}{(x-a)^{N}}\cdot(x-a)^{k}=0\quad\forall k\geq0\)

  • For \(k = N\): \(\lim_{x \to a} h(x) = 0\)

  • For \(k = N-1\): \(\lim_{x\to a}\frac{h(x)}{(x-a)}=0\Rightarrow h^{\prime}(a)=0\)

  • ....

  • For \(k = 1\): \(\lim_{x \to a} \frac{h(x)}{(x-a)^{N-1}} = 0\)

  • For \(k = 0\): \(\lim_{x \to a} \frac{h(x)}{(x-a)^N} = 0\)​

Thus \(h^{(k)}(a)=0,\forall k=0,...,N\)

Suppose in addition that \(f\) and \(g\) are \(N\)-times differentiable at \((a-\delta, a+\delta)\).
\(h\) is \(N\)-times differentiable at \(a\). (And \(f, f', f'', \dots, f^{(N-1)}, f^{(N)}\), and \(g\) are continuous.)

We obtain that: \(f(a) = g(a), \quad f'(a) = g'(a), \quad \dots, \quad f^{(N)}(a) = g^{(N)}(a)\)

Recall 2

If \(P\) and \(Q\) are polynomials of degree \(\leq N\) and \(P\) and \(Q\) have \(N+1\) "linear coincidences",

\(\implies P = Q.\)"Linear coincidences": \(P(x_0) = Q(x_0), \dots, P(x_N) = Q(x_N)\)(\(N+1\) coincidences)

For \(N = 1\): Two lines coinciding in 2 points are the same.

For \(N = 2\): Two parabolas with 3 coincidences are the same parabola.

For a general case: \(P(a) = Q(a), \quad P'(a) = Q'(a), \quad \dots, \quad P^{(N)}(a) = Q^{(N)}(a)\)​\(\implies P = Q.\)

Recall 3

Theorem: If \(f : (a-\delta, a+\delta) \rightarrow \mathbb{R}\) and it is \(N\)-times differentiable at \(a\) \(\Rightarrow \lim_{x \to a} \frac{f(x) - T_{N,a}(x)}{(x - a)^N} = 0.\)

Proof: Apply \(N-1\) times L'Hôpital's Rule. \(T_{N,a}(x) = \sum_{n=0}^N \frac{f^{(n)}(a)}{n!} (x-a)^n\)

Strong Taylor's Theorem

Let \(f : (a-\delta, a+\delta) \to \mathbb{R}\) be \(N+1\) times differentiable in \((a-\delta, a+\delta)\). Then

  1. \(f(x) - T_{N,a}(x) = \frac{f^{(N+1)}(c)}{N!} (x-a) (x-c)^N \quad \text{for some } c \text{ between } a \text{ and } x\)

    (Error of using \(T_{N,a}\) instead of \(f\).)

  2. \(f(x) - T_{N,a}(x) = \frac{f^{(N+1)}(c)}{(N+1)!} (x-a)^{N+1} \quad \text{for some } c \text{ between } a \text{ and } x\)

    Notation: \(E_{N,a}=R_{N,a}=\frac{f^{(N+1)}(c)}{(N+1)!}(x-a)^{N+1}\quad\)Remainder term

Example

\(f(x) = \sin(x)\) \(T_{5,0}(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!}\)

image​

\(a=0\)

\(\left|\sin(x)-\left(\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}+\ldots\right)=\begin{cases}0\quad\text{if }N\text{ is even}\\ \frac{x^{N}}{N!}\quad\text{if }N\equiv1\left(4\right)\\ -\frac{x^{N}}{N!}\quad\text{if }N\equiv3\left(4\right)\end{cases}\right)\right|=|\sin^{(N+1)}(c)|\frac{|x|^{N+1}}{(N+1)!}\leq\frac{|x|^{N+1}}{(N+1)!}\to0\quad\text{as }N\to\infty\text{ for fixed }x.\)

We conclude that for any \(x\in\mathbb{R},\quad\lim_{N\to\infty}T_{N,a}(x)=\sin(x)\)​

In other words, the infinite series \(\sum_{k=0}^{\infty}\frac{(-1)^{k+1}x^{2k+1}}{(2k+1)!}=\sin(x)\text{ for all }x\in\mathbb{R},n=2k+1\)

Example

\(f(x)=\sqrt x\), want to know \(\sqrt{60} = f(60) = ?\) Answer with 5 correct digits.

What is \(a\)?

If \(a = 60\)

\(T_{2,60}(x) = f(a) + \frac{f'(a)}{1!}(x-a) + \frac{f''(a)}{2!}(x-a)^2\), then \(T_{2,60}(60) = f(60)\). But we don't know \(\sqrt{60} = ?\)

Thus we need try some \(a\) that square root can be calculated

\(a = 64\) ✅
​\(a = 49\) ✅

\(f(64)\)​ \(f'(64)\)​ \(f''(64)\)​ \(f'''(64)\)​ \(f''''(64)\)​
\(x\)​ \(x^{1/2}\)​ \(\frac{1}{2}x^{-1/2}\)​ \(-\frac{1}{4}x^{-3/2}\)​ \(+\frac{3}{8}x^{-5/2}\)​ \(-\frac{15}{16}x^{-7/2}\)​
\(a = 64\)​ \(8\)​ \(\frac 12\cdot \frac 18\)​ \(-\frac14\cdot \frac{1}{8^3}\)​ \(\frac38\cdot\frac{1}{8^5}\)​ \(-\frac{15}{16}\cdot\frac{1}{8^7}\)​

\(T_{3,64}(x) = 8 + \frac{1}{16} \cdot \frac{(x - 64)}{1!} - \frac{1}{4 \cdot 8^3} \cdot \frac{(x - 64)^2}{2!} + \frac{3}{8 \cdot 8^5} \cdot \frac{(x - 64)^3}{3!}\)

Then \(T_{3,64}(60) = 8 + \frac{1}{16} \cdot 4 - \frac{1}{4 \cdot 8^3} \cdot \frac{4^2}{2} + \frac{3}{8 \cdot 8^5} \cdot \frac{4^3}{6}=7.745971\)

And \(\sqrt{60}=7.745967\)

Then error: \(\left| \sqrt{60} - T_{3,64}(60) \right| = \frac{15}{16} \cdot \frac{1}{\sqrt{C^7}} \cdot \frac{4^4}{4!}\)

Since \(49 < C \in (60, 64)\), then \(< \frac{15}{16} \cdot \frac{1}{7^7} \cdot \frac{4^4}{4!}\)

Proof

Let's assume \(a\) fixed and \(x\in (a-\delta,a+\delta)\) also fixed.

Let's assume \(x>a\) and think on the fixed domain \([a,x]\)

\(E\left(t\right)=f(x)-T_{N,t}\left(x\right)\quad t\in\left\lbrack a,x\right\rbrack\)​

We want to apply MVT to \(E(t)\), that is \(\exists c\in[a,x]:\frac{E\left(x\right)-E\left(a\right)}{x-a}=E^{\prime}\left(c\right)\)

Note that \(E(t)=f(x)-T_{N,t}(x)=f(x)-\sum_{n=0}^{N}\frac{f^{(n)}(t)}{n!}(x-t)^{n}\)​

\(= f(x) - \left( f(t) + f'(t)(x-t) + \frac{f''(t)}{2!}(x-t)^2 + \ldots + \frac{f^{(N)}(t)}{N!}(x-t)^N \right)\)

Since \(E\) is cont on \([a, x]\) and diff \(\forall t \in (a, x)\), then by MVT \(\text{LHS} = \frac{E(x) - E(a)}{x - a} = \frac{0 - (f(x) - T_{N,a}(x))}{x - a}\)

\(\text{RHS} = ?\)

\(E^{\prime}(t)=0-\left(f^{\prime}(t)+f''(t)(x-t)-f^{\prime}(t)\cdot1+\frac{f'''(t)}{2!}(x-t)^2-2\frac{f''(t)}{2}(x-t)+\frac{f^{(4)}(t)}{3!}(x-t)^3-3\frac{f'''(t)}{3}(x-t)^2+\ldots+\frac{f^{(N+1)}(t)}{N!}(x-t)^{N}\right.-N\frac{f^{(N)}(t)}{N!}\left(x-t)^{N-1}\right)\)​

Thus \(E'(t) = -\frac{f^{(N+1)}(t)(x-t)^N}{N!}\), then \(E'(c) = -\frac{f^{(N+1)}(c)(x-c)^N}{N!} \quad \text{RHS}\)

Therefore \(f(x)-T_{N,a}(x)=\frac{f^{(N+1)}(c)}{N!}(x-a)(x-c)^{N}\)

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