# University of Florida/Egm6341/s10.team2.niki/HW1
## Problem 1
### 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 $\mathbf \left[0,1\right]$

### Solution
## Problem 9
##### 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 4
### Problem Statement

Pg 5-1.
Prove the Integral Mean Value Theorem (IMVT) p. 2-3 for w(.) non-negative. i.e $\ w\geq 0$

### 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)