University of Florida/Egm6341/s10.team2.niki/HW1

Given
$$\boldsymbol{F(x)}=\frac{\mathbf{e^x -1}}{\mathbf{x}}$$

Find
Determine the limit of the given function $$\boldsymbol{F(x)}$$ and plot it in the interval $$\mathbf{} \left [ 0,1 \right ]$$

Problem Statement
Pg. 7-1

$$\frac{\mathbf{[e^x -1]}}{\boldsymbol{x}} = \boldsymbol{\frac{1}{x}[e^x-1]} = \boldsymbol{f(x)}$$

1) Expand $$\exp^{x}$$ in Taylor Series w/ remainder:

$$ \boldsymbol{R(x) = \frac{(x-0)^{n+1}}{(n+1)!}exp \left[ \zeta(x) \right] }$$

2) Find Taylor Series Expansion and Remainder of f(x). eq. 4 of p 6-3.--Egm6341.s10.team2.niki 02:26, 26 January 2010 (UTC)

Solution
Given:

$$\boldsymbol{P_{n}(x)=f(x_{0})+\frac{(x-x_{0})}{1!}f^{(1)}(x_{0})+...+\frac{(x-x_{0})^{n}}{n!}f^{(n)}(x_{0})}$$[equation 4 p 2-2]

$$ \boldsymbol{R(x) = \frac{(x-0)^{n+1}}{(n+1)!}exp \left[ \zeta(x) \right] }$$ [equation 1 p 2-3]

Part 1
$$\boldsymbol{P_{n}(x)=e^{x_{0}}+\frac{(x-x_{0})}{1!}e^{x_{0}}+...+\frac{(x-x_{0})^{n}}{n!}e^{x_{0}}}$$

for the case that $$x_{0}=0$$, we get,

$$\boldsymbol{P_{n}(x)=1+\frac{(x)}{1!}+...+\frac{(x)^{n}}{n!}}$$

$$\boldsymbol{P_{n}(x)}$$ =$$\boldsymbol{\sum_{j=0}^{\infty }\frac{x^{j}}{j!}}$$

Using equation 1 p 2-3, we get the remainder as

$$ \boldsymbol{R_{n+1}(x) = \frac{(x-x_{0})^{n+1}}{(n+1)!}f^{(n+1)} (\zeta(x))}$$

for $$x_{0}=0$$, we get

$$\boldsymbol{R_{n+1}(x) = \frac{(x)^{n+1}}{(n+1)!}f^{(n+1)} (\zeta(x))}$$

finally, $$\boldsymbol{f(x)= e^x = \sum_{j=1}^{\infty } \frac{x^{j}}{j!} + \frac{(x)^{n+1}}{(n+1)!}f^{(n+1)} (\zeta(x))}$$

Part 2
$$\boldsymbol{f(x)=\frac{1}{x}[e^{x}-1]}$$

$$\boldsymbol{e^{x}=\sum_{j=0}^{\infty }\frac{x^j}{j!} = 1+ \frac{x}{1!}+\frac{x^2}{2!}+...+\frac{x^n}{n!}}$$

$$\therefore \boldsymbol{[e^x -1]= \frac{x}{1!}+\frac{x^2}{2!}+...+\frac{x^n}{n!}}$$

dividing both sides by x we get,

$$\boldsymbol{\frac{[e^x - 1]}{x} = f(x) = \sum_{j=1}^{\infty } \frac{x^{j-1}}{j!}}$$

and remainder becomes

$$\boldsymbol{R_{n+1}(x) = \frac{(x)^{n}}{(n+1)!}f^{(n+1)} (\zeta(x))}$$

since $$x_{0}= 0$$, we have

$$\boldsymbol{R_{n+1}(x) = \frac{(x)^{n}}{(n+1)!}f^{(n+1)} (\zeta(x)}$$ where $$\zeta (x) )\epsilon [0,x]$$

Finally,

$$\boldsymbol{f(x)= \frac{[e^x -1 ]}{x} = \sum_{j=1}^{\infty } \frac{x^{j-1}}{j!} + \frac{(x)^{n}}{(n+1)!}f^{(n+1)} (\zeta(x))}$$

Problem Statement
Pg 5-1.

Prove the Integral Mean Value Theorem (IMVT) p. 2-3 for w(.) non-negative. i.e $$\ w\geq0 $$

Solution
We have the IMVT as

$$\boldsymbol{ \int_{a}^{b} f(x)w(x)dx = f(\xi )\int_{a}^{b} w(x)dx}$$

For a given function $$\boldsymbol{f(x)}$$ Let m be the minimum of the function and M be the maximum of the same function

Then we know that, $$\boldsymbol{m \leq f(x)\leq M}$$

multiplying the inequality throughout by $$\boldsymbol{w(x) \geq 0}$$ and integrating between $$\boldsymbol{[a,b]}$$ we get

$$\boldsymbol{\int_{a}^{b}m. w(x)dx = \int_{a}^{b} f(x).w(x)dx = \int_{a}^{b}M.w(x)dx}$$

$$\boldsymbol{m\int_{a}^{b} w(x)dx = \int_{a}^{b} f(x).w(x)dx = M\int_{a}^{b}w(x)dx}$$

writing $$\boldsymbol{\int_{a}^{b}w(x) = I}$$, we get

$$\boldsymbol{mI\leq \int_{a}^{b} f(x)w(x)dx \leq MI}$$

It is seen that when w(x) = 0, the result is valid. Consider the case when w(x) > 0

dividing throughout by $$\boldsymbol{I}$$

$$\boldsymbol{m\leq \frac{1}{I}\int_{a}^{b} f(x)w(x)dx \leq M}$$

From the Intermediate Value Theorem, we know that there exists $$\boldsymbol{\xi \epsilon[a,b] }$$ such that

$$\boldsymbol{f(\xi )= \frac{1}{I}\int_{a}^{b} f(x)w(x)dx}$$

i.e

$$\boldsymbol{f(\xi )\int_{a}^{b} w(x)dx= \int_{a}^{b} f(x)w(x)dx}$$

Hence Proved --Egm6341.s10.team2.niki 02:28, 27 January 2010 (UTC)