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

Problem 1

Given

$𝑭\mathit{(}𝒙\mathit{)}=\frac{{𝐞}^{𝐱}\mathbf{-}\mathrm{𝟏}}{𝐱}$

Find

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

Problem 9

Problem Statement

Pg. 7-1

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

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

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

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:

${𝑷}_{𝒏}\mathit{(}𝒙\mathit{)}\mathit{=}𝒇\mathit{(}{𝒙}_0\mathit{)}\mathit{+}\frac{(x-x_0)}{1!}{𝒇}^{\mathit{(}1\mathit{)}}\mathit{(}{𝒙}_0\mathit{)}\mathit{+}\mathit{.}\mathit{.}\mathit{.}\mathit{+}\frac{(x-x_0{)}^n}{n!}{𝒇}^{\mathit{(}𝒏\mathit{)}}\mathit{(}{𝒙}_0\mathit{)}$

[equation 4 p 2-2]

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

[equation 1 p 2-3]

Part 1

${𝑷}_{𝒏}\mathit{(}𝒙\mathit{)}\mathit{=}{𝒆}^{{𝒙}_0}\mathit{+}\frac{(x-x_0)}{1!}{𝒆}^{{𝒙}_0}\mathit{+}\mathit{.}\mathit{.}\mathit{.}\mathit{+}\frac{(x-x_0{)}^n}{n!}{𝒆}^{{𝒙}_0}$

for the case that $x_0=0$, we get,

${𝑷}_{𝒏}\mathit{(}𝒙\mathit{)}\mathit{=}1\mathit{+}\frac{(x)}{1!}\mathit{+}\mathit{.}\mathit{.}\mathit{.}\mathit{+}\frac{(x{)}^n}{n!}$

${𝑷}_{𝒏}\mathit{(}𝒙\mathit{)}$ =$\sum _{𝒋\mathit{=}0}^{\mathrm{\infty }}\frac{x^j}{j!}$

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

${𝑹}_{𝒏\mathit{+}1}\mathit{(}𝒙\mathit{)}\mathit{=}\frac{(x-x_0{)}^{n+1}}{(n+1)!}{𝒇}^{\mathit{(}𝒏\mathit{+}1\mathit{)}}\mathit{(}\mathit{\zeta }\mathit{(}𝒙\mathit{)}\mathit{)}$

for $x_0=0$, we get

${𝑹}_{𝒏\mathit{+}1}\mathit{(}𝒙\mathit{)}\mathit{=}\frac{(x{)}^{n+1}}{(n+1)!}{𝒇}^{\mathit{(}𝒏\mathit{+}1\mathit{)}}\mathit{(}\mathit{\zeta }\mathit{(}𝒙\mathit{)}\mathit{)}$

finally, $𝒇\mathit{(}𝒙\mathit{)}\mathit{=}{𝒆}^{𝒙}\mathit{=}\sum _{𝒋\mathit{=}1}^{\mathrm{\infty }}\frac{x^j}{j!}\mathit{+}\frac{(x{)}^{n+1}}{(n+1)!}{𝒇}^{\mathit{(}𝒏\mathit{+}1\mathit{)}}\mathit{(}\mathit{\zeta }\mathit{(}𝒙\mathit{)}\mathit{)}$

Part 2

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

${𝒆}^{𝒙}\mathit{=}\sum _{𝒋\mathit{=}0}^{\mathrm{\infty }}\frac{x^j}{j!}\mathit{=}1\mathit{+}\frac{x}{1!}\mathit{+}\frac{x^2}{2!}\mathit{+}\mathit{.}\mathit{.}\mathit{.}\mathit{+}\frac{x^n}{n!}$

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

dividing both sides by x we get,

$\frac{[e^x-1]}{x}\mathit{=}𝒇\mathit{(}𝒙\mathit{)}\mathit{=}\sum _{𝒋\mathit{=}1}^{\mathrm{\infty }}\frac{x^{j-1}}{j!}$

and remainder becomes

${𝑹}_{𝒏\mathit{+}1}\mathit{(}𝒙\mathit{)}\mathit{=}\frac{(x{)}^n}{(n+1)!}{𝒇}^{\mathit{(}𝒏\mathit{+}1\mathit{)}}\mathit{(}\mathit{\zeta }\mathit{(}𝒙\mathit{)}\mathit{)}$

since $x_0=0$, we have

${𝑹}_{𝒏\mathit{+}1}\mathit{(}𝒙\mathit{)}\mathit{=}\frac{(x{)}^n}{(n+1)!}{𝒇}^{\mathit{(}𝒏\mathit{+}1\mathit{)}}\mathit{(}\mathit{\zeta }\mathit{(}𝒙\mathit{)}$ where $\zeta (x))ϵ[0,x]$

Finally,

$𝒇\mathit{(}𝒙\mathit{)}\mathit{=}\frac{[e^x-1]}{x}\mathit{=}\sum _{𝒋\mathit{=}1}^{\mathrm{\infty }}\frac{x^{j-1}}{j!}\mathit{+}\frac{(x{)}^n}{(n+1)!}{𝒇}^{\mathit{(}𝒏\mathit{+}1\mathit{)}}\mathit{(}\mathit{\zeta }\mathit{(}𝒙\mathit{)}\mathit{)}$

problem 4

Problem Statement

Pg 5-1.

Prove the Integral Mean Value Theorem (IMVT) p. 2-3 for w(.) non-negative. i.e $w\ge 0$

Solution

We have the IMVT as

$\underset{𝒂}{\overset{𝒃}{\int }}𝒇\mathit{(}𝒙\mathit{)}𝒘\mathit{(}𝒙\mathit{)}𝒅𝒙\mathit{=}𝒇\mathit{(}\mathit{\xi }\mathit{)}\underset{𝒂}{\overset{𝒃}{\int }}𝒘\mathit{(}𝒙\mathit{)}𝒅𝒙$

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

Then we know that, $𝒎\mathit{\le }𝒇\mathit{(}𝒙\mathit{)}\mathit{\le }𝑴$

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

$\underset{𝒂}{\overset{𝒃}{\int }}𝒎\mathit{.}𝒘\mathit{(}𝒙\mathit{)}𝒅𝒙\mathit{=}\underset{𝒂}{\overset{𝒃}{\int }}𝒇\mathit{(}𝒙\mathit{)}\mathit{.}𝒘\mathit{(}𝒙\mathit{)}𝒅𝒙\mathit{=}\underset{𝒂}{\overset{𝒃}{\int }}𝑴\mathit{.}𝒘\mathit{(}𝒙\mathit{)}𝒅𝒙$

$𝒎\underset{𝒂}{\overset{𝒃}{\int }}𝒘\mathit{(}𝒙\mathit{)}𝒅𝒙\mathit{=}\underset{𝒂}{\overset{𝒃}{\int }}𝒇\mathit{(}𝒙\mathit{)}\mathit{.}𝒘\mathit{(}𝒙\mathit{)}𝒅𝒙\mathit{=}𝑴\underset{𝒂}{\overset{𝒃}{\int }}𝒘\mathit{(}𝒙\mathit{)}𝒅𝒙$

writing $\underset{𝒂}{\overset{𝒃}{\int }}𝒘\mathit{(}𝒙\mathit{)}\mathit{=}𝑰$, we get

$𝒎𝑰\mathit{\le }\underset{𝒂}{\overset{𝒃}{\int }}𝒇\mathit{(}𝒙\mathit{)}𝒘\mathit{(}𝒙\mathit{)}𝒅𝒙\mathit{\le }𝑴𝑰$

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

dividing throughout by $𝑰$

$𝒎\mathit{\le }\frac{1}{I}\underset{𝒂}{\overset{𝒃}{\int }}𝒇\mathit{(}𝒙\mathit{)}𝒘\mathit{(}𝒙\mathit{)}𝒅𝒙\mathit{\le }𝑴$

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

$𝒇\mathit{(}\mathit{\xi }\mathit{)}\mathit{=}\frac{1}{I}\underset{𝒂}{\overset{𝒃}{\int }}𝒇\mathit{(}𝒙\mathit{)}𝒘\mathit{(}𝒙\mathit{)}𝒅𝒙$

i.e

$𝒇\mathit{(}\mathit{\xi }\mathit{)}\underset{𝒂}{\overset{𝒃}{\int }}𝒘\mathit{(}𝒙\mathit{)}𝒅𝒙\mathit{=}\underset{𝒂}{\overset{𝒃}{\int }}𝒇\mathit{(}𝒙\mathit{)}𝒘\mathit{(}𝒙\mathit{)}𝒅𝒙$

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