NoteNextra-origin/content/Math4121/Math4121_L5.md

Math4121 Lecture 5

Continue on differentiation

L'Hôpital's Rule

Suppose f and g are real differentiable on (a,b) and g'(x)\neq 0 for all x\in (a,b).

Suppose \frac{f'(x)}{g'(x)}\to A as x\to a,

If f(x)\to 0 and g(x)\to 0 as x\to a,

or g(x)\to \infty as x\to a,

then \frac{f(x)}{g(x)}\to A as x\to a.

Proof:

Main step: Let -\infty<a<b<\infty.for any q>A, there exists c\in (a,b) such that \frac{f(x)}{g(x)}<q for all x\in (a,c).

Topological definition of limit:

h(x)\to A as x\to a if \forall \epsilon>0, \exists \delta>0 such that |x-a|<\delta implies |h(x)-A|<\epsilon.

In other words, if for any open neighborhood V of A, there exists an open neighborhood U of a such that h(U)\subseteq V.

Case 1: A=-\infty, for any q>A, there exists \delta>0 such that x\in (a,a+\delta) implies \frac{f(x)}{g(x)}<q.

Case 2: A=\infty, we change the function F(x)=-f(x) and apply the case 1.

Case 3: A\in \mathbb{R}, Let \epsilon>0 and take q=A+\epsilon. \exists c_1\in (a,b) such that \forall x\in (a,c_1), \frac{f(x)}{g(x)}<q.

Set F(x)=-f(x). and q=-A+\epsilon>-A. Apply main step, \exists c_2\in (a,b) such that \forall x\in (a,c_2), \frac{F(x)}{g(x)}<-A+\epsilon. so \forall x\in (a,c_2), \frac{f(x)}{g(x)}>A-\epsilon.

We take c=\min(c_1,c_2). Then \forall x\in (a,c), \frac{f(x)}{g(x)}<q.

QED

Higher Order Derivatives

Definition 5.14

If f is differentiable on (a,b), then we define f'(x)=\lim_{t\to x}\frac{f(t)-f(x)}{t-x}.

If f' is differentiable on (a,b), then we define f''(x)=(f')'(x).

If f^{(k)} is differentiable on (a,b), then we define f^{(k+1)}(x)=(f^{(k)})'(x).

Theorem 5.15 Taylor's Theorem

Let f:[a,b]\to \mathbb{R}, and n be a positive integer.

Let f^{(n-1)} be continuous on [a,b], and differentiable on (a,b).

For \alpha\in [a,b], define the Taylor polynomial of order n-1 at \alpha by

P(t)=\sum_{k=0}^{n-1}\frac{f^{(k)}(\alpha)}{k!}(t-\alpha)^k

Example:

When n=1, P(t)=f(\alpha).

When n=2, P(t)=f(\alpha)+f'(\alpha)(t-\alpha).

When n=3, P(t)=f(\alpha)+f'(\alpha)(t-\alpha)+\frac{f''(\alpha)}{2}(t-\alpha)^2.

Key property:

P^{(k)}(\alpha)=f^{(k)}(\alpha)\quad \forall k=0,1,\cdots,n-1

For each \beta\in [a,b], there exists x between \alpha and \beta such that

f(\beta)=P(\beta)+\frac{f^{(n)}(x)}{n!}(\alpha-\beta)^n

On rudin, it is

f(\beta)=P(\beta)+\frac{f^{(n)}(x)}{n!}(\beta-\alpha)^n

Proof:

Let M=\frac{f(\beta)-P(\beta)}{(\beta-\alpha)^n}.

So that f(\beta)=P(\beta)+M(\beta-\alpha)^n.

Need to show that n!M=f^{(n)}(x). for some x\in (\alpha,\beta). Defined g(t)=f(t)-P(t)-M(t-\alpha)^n.

By our choice of M, g(\alpha)=g(\beta)=0.

g(t)=f(t)-P(t)-M(n(n-1)(n-2)\cdots(n-k+1)(t-\alpha)^{n-k})

for k=1,2,\cdots,n-1. And when k=n, g^{(n)}(t)=f^{(n)}(t)-0-M(n!).

Need to show that \exists x\in (\alpha,\beta) such that g^{(n)}(x)=0.

By Mean Value Theorem, \exists x_1\in (\alpha,\beta) such that g'(x_1)=0.

By Mean Value Theorem, \exists x_2\in (\alpha,x_1) such that g''(x_2)=0.

By Mean Value Theorem, \exists x_3\in (\alpha,x_2) such that g^{(3)}(x_3)=0.

\cdots

By Mean Value Theorem, \exists x_n\in (\alpha,x_{n-1}) such that g^{(n)}(x_n)=0.

Since g^{(n)}(\alpha)=0 for k=0,1,2,\cdots,n-1, we can find x_n\in (\alpha,x_{n-1}) such that g^{(n)}(x_n)=0.

QED

Chapter 6: Riemann-Stieltjes Integral