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

Given

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

Find

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

Solution

Problem Statement

Pg. 7-1

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

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

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

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:

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

[equation 4 p 2-2]

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

[equation 1 p 2-3]

Part 1

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

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

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

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

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

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

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

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

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

Part 2

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

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

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

dividing both sides by x we get,

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

and remainder becomes

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

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

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

Finally,

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

Problem Statement

Pg 5-1.

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

Solution

We have the IMVT as

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

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

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

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

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

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

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

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

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

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

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

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

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

i.e

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

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