User:Egm6341.s11.team3/sub1

= HW 1.1 =

Problem
Find the limit of function:


 * {| style="width:100%" border="0"

$$ \lim_{x\to 0} f(x) = \lim_{x\to 0} \frac {e^x - 1}{x}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle
 * }

Find

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$$ \lim_{x\to 0} f(x) = \lim_{x\to 0} \frac {e^x - 1}{x}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle (Eq. 1.1)
 * }

Solution
Take the limit of the function:


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$$ \lim_{x\to 0} f(x) = \frac {e^0 - 1}{0} = \frac {0}{0}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 * $$\displaystyle
 * }
 * }

Hence we can apply L-Hopital Rule:


 * {| style="width:100%" border="0"

$$ \lim_{x\to 0} f(n) = \frac {f'(n)}{g'(n)} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * $$\displaystyle (Eq. 1.2)
 * }
 * }

By above rule:


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$$ \lim_{x\to 0} f(x) = \lim_{x\to 0} \frac {e^x - 1}{x} =\lim_{x\to 0} \frac {f(x)}{g(x)} \Rightarrow
 * style="width:95%" |
 * style="width:95%" |

\begin{matrix} f'(x) = e^x -0 \\ g'(x) = 1 \end{matrix} $$

$$
 * $$\displaystyle
 * }
 * }


 * {| style="width:100%" border="0"

$$ \Rightarrow \lim_{x\to 0} f(0) = \frac {f'(0)}{g'(0)} = \frac {e^0 - 0}{1} = 1  $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * $$\displaystyle
 * }
 * }

$$ Ans.: \lim_{x\to 0} f(x) = \frac {e^x - 1}{x} = 1 $$ $$
 *  $$ \displaystyle (Eq. 1.3)

Extra Solution
From Taylor series :

$$ e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \frac{x^5}{120} + ... $$

Then the original fuction could be expressed as:

$$ f(x) = 1 + \frac{x}{2} + \frac{x^2}{6} + \frac{x^3}{24} + \frac{x^4}{120} + ... $$

$$ \Rightarrow \lim_{x\to 0} f(x) = 1 + \frac{0}{2} + \frac{0^2}{6} + \frac{0^3}{24} + \frac{0^4}{120} + ... =1 $$

The original function is plotted from 0 to 1. For comparison, serie-form of f(x) of order 2nd and 3rd are also shown in the following plot.



= HW 1.2 =

Problem
Given:


 * {| style="width:100%" border="0"

$$ a) \qquad f(x)= e^x $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle (Eq. 2.1)
 * }


 * {| style="width:100%" border="0"

$$ b) \qquad f(x)= \frac {e^x - 1}{x} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 2.2)
 * }

Find
<span id="(1)">
 * {| style="width:100%" border="0"

$$ P_n(x), R_{n+1}(x) \quad at \quad x_0 = 0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

Solution
<span id="(1)">
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$${P}_{n}(x) = f({x}_{0}) + \frac{(x-{x}_{0})}{1!}f'({x}_{0}) + .... + \frac{({x-{x}_{0}})^{n}}{n!}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 2.3)
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$${R}_{n+1}(x)= \frac{1}{n!}\int_{{x}_{0}}^{x}(x-t)^n f^{n+1}(t)\,dt = \frac{(x-x_0)^{n+1}}{(n+1)!}f^{n+1}(\xi), \quad \xi \in [x_0, x] $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 2.4)
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$$ f^{n+1}(x) = \frac{d^{n+1}}{dx^{n+1}}f(x) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

$$ a) \mathbf{\quad f(x) = e^x} $$

<span id="(1)">
 * {| style="width:100%" border="0"

$$
 * style="width:95%" |
 * style="width:95%" |

\begin{matrix} f(x) = e^x & f(0) = e^0 = 1 \\ f'(x) = e^x & f'(0) = e^0 = 1 \\ \vdots \\ f^n(x) = e^x & f(0) = e^0 = 1 \end{matrix} $$ $$
 * <p style="text-align:right"> $$ \displaystyle
 * }

From Eq. 2.3: <span id="(1)">
 * {| style="width:100%" border="0"

$${P}_{n}(x) = f({x}_{0}) + \frac{(x-{x}_{0})}{1!}f'({x}_{0}) + \frac{(x-{x}_{0})^2}{2!}f''({x}_{0})+.... + \frac{({x-{x}_{0}})^{n}}{1!}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$$ = \frac{x^0}{0!}(1) + \frac{x^1}{1!}(1) + \frac{x^2}{2!}(1) +....+ \frac{x^n}{n!}(1) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$$ = \sum_{n=0}^n (\frac{x^n}{n!}) \Rightarrow P_n(x) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 2.5)
 * }

From Eq. 2.4: <span id="(1)">
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$${R}_{n+1}(x)= \frac{1}{n!}\int_{{x}_{0}}^{x}(x-t)^n f^{n+1}(t)\,dt = \frac{(x-x_0)^{n+1}}{(n+1)!}f^{n+1}(\xi), \quad where \quad f^{n+1}(\xi) = \frac {d^{n+1}}{d \xi ^{n+1}}(e^\xi) = e^\xi $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$${R}_{n+1}(x)= \frac{x^{n+1}}{(n+1)!} e^\xi $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 2.6)
 * }

Therefore:

$$ f(x) = {P}_{n}(x) + {R}_{n+1}(x) = \sum_{n=0}^n (\frac{x^n}{n!}) + \frac{x^{n+1}}{(n+1)!} e^\xi $$ $$
 * <p style="text-align:right"> $$ \displaystyle (Eq. 2.7)

$$ b) \mathbf{ \quad f(x) = \frac {e^x - 1}{x} }$$

<span id="(1)">
 * {| style="width:100%" border="0"

$$ \begin{matrix} f(x) = {P}_{n}(x) + {R}_{n+1}(x) \end{matrix} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$$ \quad f(x) = \frac {e^x - 1}{x} \Rightarrow f(x_0) = f(0) = \frac {0}{0} = 1 \quad (from \quad equation \quad 1.3) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$$ \quad f'(x) = \frac{d}{dx}((e^x - 1)x^{-1}) = e^x(e^x - 1)^0x^{-1}+(-1(e^x - 1)x^{-2}) = e^xx^{-1} - (e^x - 1)x^{-2} = \frac {e^x(x-1)+1}{x^2}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$$ f'(x) = \frac {e^x(x-1)+1}{x^2} = \frac {f(x)}{g(x)}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$$ = \frac {f'(x)}{g'(x)} = \frac {e^x(x-1)+e^x}{2x^1} = \frac {xe^x}{2x^1} \Rightarrow \frac {f'(0)}{g'(0)}= \frac{0}{0} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$$ = \frac {f(x)}{g(x)} = \frac {e^x + xe^x}{2} \Rightarrow \frac {f(0)}{g(0)}= \frac{1}{2} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

In the same fashion we differentiated $$ f^{(2)}, f^{(3)},.... $$, then by superimposing multiple L'Hopital Rules for $$ x -> 0 $$ on existing derivatives we observed a number to approach a finite value.

Since passed $$ f''(x) $$ differentiation and L'Hoptal Rule application/differentiation became tedious, we generated Matlab code to differentiate $$ f(x) $$ for us and to superimpose multiple L'Hopital's rules in order to generate finite number as $$ x -> 0 $$

From Matlab code:

<span id="(1)">
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$$ \lim_{x\to 0} f(x) = 1$$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$$ \lim_{x\to 0} f^{(1)}(x) = \frac{1}{2}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$$ \lim_{x\to 0} f^{(2)}(x) = \frac{1}{3}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$$ \lim_{x\to 0} f^{(3)}(x) = \frac{1}{4}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$$ \lim_{x\to 0} f^{(4)}(x) = \frac{1}{5}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$$ \vdots $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$$ \lim_{x\to 0} f^{(n)}(x) = \frac{1}{n+1}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

From Eq. 2.3: <span id="(1)">
 * {| style="width:100%" border="0"

$${P}_{n}(x) = f({x}_{0}) + \frac{(x-{x}_{0})}{1!}f'({x}_{0}) + .... + \frac{({x-{x}_{0}})^{n}}{1!}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$${P}_{n}(x) = 1 + \frac{(x)}{1!}\frac{1}{2} + \frac{(x^2)}{2!}\frac{1}{3} + \frac{(x^3)}{3!}\frac{1}{4}+ .... + \frac{({x-{x}_{0}})^{n}}{n!} \frac{1}{n+1} \Rightarrow $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$${P}_{n}(x) = 1 + \sum_1^n \frac{(x^n)}{(n+1)!} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 2.8)
 * }

See the following extra solution for $$ R_n(x) $$

Extra Solution for HW1.2(b)
From Taylor series :

<span id="(1)">
 * {| style="width:100%" border="0"

$$ e^x = 1 + \sum_{i=1}^n \frac{x^i}{i!} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

Then the original fuction could be expressed as:

<span id="(1)">
 * {| style="width:100%" border="0"

$$ f(x) = \frac{e^{x}-1}{x} = \sum_{i=1}^n \frac{x^{i-1}}{i!} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$$ \Rightarrow P_n(x) = \sum_{i=0}^{n} \frac{x^{i}}{(i+1)!} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 2.12)
 * }

To find $$ R_n(x) $$, we need to solve for $$ f^{(n+1)}(x) $$ which is difficult. So, here we compute $$ R_n(x) $$ from the previous function $$ e^{x} $$. From (Eq. 2.7), we could write $$ e^{x} $$ as:

<span id="(1)">
 * {| style="width:100%" border="0"

$$ e^{x} = P(x) + R(x) = 1+ \sum_{i=1}^{n+1} \frac{x^i}{i!} + \frac{x^{n+2}}{(n+2)!} e^\xi = \sum_{i=0}^{n} \frac{x^i}{(i)!} +\frac{x^{n+1}}{(n+1)!} e^\xi, \quad \xi \in [0, x] $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle $$
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$$ \Rightarrow f(x) = \frac{e^{x}-1}{x} = \sum_{i=0}^{n} \frac{x^i}{(i+1)!} + \frac{x^{n+1}}{(n+2)!} e^\xi, \quad \xi \in [0, x] $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 2.13)

$$
 * }

Comparing (Eq. 2.12) and (Eq. 2.13) yields,

<span id="(1)">
 * {| style="width:100%" border="0"

$$ R_n(x) = \frac{x^{n+1}}{(n+2)!} e^\xi, \quad \xi \in [0, x] $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle

$$
 * }

$$ f(x) = {P}_{n}(x) + {R}_{n}(x) = \sum_{i=0}^{n} \frac{x^{i}}{(i+1)!} + \frac{x^{n+1}}{(n+2)!} e^\xi $$ $$
 * <p style="text-align:right"> $$ \displaystyle (Eq. 2.14)

= HW 1.3 =

Problem:
Given a smooth function that has maximum, minimum, and is solely all positive or all negative:

<span id="(1)">
 * {| style="width:100%" border="0"

$$ m\le f(x)\le M $$, $$ x\in [a,b] $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right;">$$\displaystyle (Eq. 3.1)
 * }
 * }

Find
Show at different weighting factors (positive, negative, and zero) that there exists such value in function over interval [a,b], such that area under the curve could be represented with a product of that value and the run of a function over abscissa axes.

Solution:
a) Consider a continuous function $$ f(x) $$ on $$ [a,b] $$, and its maximum and minimum value on this interval are respectively given by $$ M $$ and $$ m $$, we have

<span id="(1)">
 * {| style="width:100%" border="0"

$$ m\le f(x)\le M $$, $$ x\in [a,b] $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right;">$$\displaystyle (Eq. 3.1)
 * }
 * }

If the weight function $$ w(x)\ge 0 $$ on $$ [a,b] $$, then we have

<span id="(1)">
 * {| style="width:100%" border="0"

$$ w(x)m\le w(x)f(x)\le w(x)M $$, $$ x\in [a,b] $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right;">$$\displaystyle (Eq. 3.2)
 * }
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$$ \Rightarrow \int_a^b{w(x)m}dx\le\int_a^b{w(x)f(x)}dx\le\int_a^b{w(x)M}dx  $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$$ \Rightarrow m\int_a^b{w(x)}dx \le \int_a^b{w(x)f(x)}dx \le M\int_a^b{w(x)}dx $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }

i. if $$ \int_a^b{w(x)}dx\ne 0$$, this inequality can be rewritten as

<span id="(1)">
 * {| style="width:100%" border="0"

$$ m\le\dfrac{\int_a^b{w(x)f(x)}dx}{\int_a^b{w(x)}dx}\le M $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }

Hence, we can always find a $$ \xi \in [a,b] $$ that satisfies

<span id="(1)">
 * {| style="width:100%" border="0"

$$ f(\xi)=\dfrac{\int_a^b{w(x)f(x)}dx}{\int_a^b{w(x)}dx}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right;">$$\displaystyle (Eq. 3.3)
 * }
 * }

Therefore

$$ \int_a^b{w(x)f(x)}dx=f(\xi)\int_a^b{w(x)}dx,\ \ w(x)> 0 $$ $$
 * <p style="text-align:right;"> $$ \displaystyle (Eq. 3.4)

ii. if $$ \int_a^b{w(x)} = 0$$, then $$ w(x)\equiv 0 $$ on $$ [a,b] $$

<span id="(1)">
 * {| style="width:100%" border="0"

$$ \int_a^b{w(x)f(x)}dx\equiv 0$$, and $$ f(\xi)\int_a^b{w(x)}dx\equiv 0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }

Therefore

$$\int_a^b{w(x)f(x)}dx=f(\xi)\int_a^b{w(x)}dx,\ \ w(x) = 0 $$ $$
 * <p style="text-align:right;"> $$ \displaystyle (Eq. 3.6)

---

b) If the weight function $$ w(x)< 0 $$ on $$ [a,b] $$, we let $$ W(x)=-w(x)$$, then we know $$ W(x)> 0 $$ on $$ [a,b] $$. As been proven in $$ Eq. 3.5$$, we have

<span id="(1)">
 * {| style="width:100%" border="0"

$$\int_a^b{W(x)f(x)}dx=f(\xi)\int_a^b{W(x)}dx$$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }

which is

<span id="(1)">
 * {| style="width:100%" border="0"

$$ \int_a^b{-w(x)f(x)}dx=f(\xi)\int_a^b{-w(x)}dx $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }

Therefore

$$\int_a^b{w(x)f(x)}dx=f(\xi)\int_a^b{w(x)}dx,\ \ w(x)< 0 $$ $$\int_a^b{w(x)f(x)}dx=f(\xi)\int_a^b{w(x)}dx,\ \ w(x) = 0 $$ $$
 * <p style="text-align:right;"> $$ \displaystyle (Eq. 3.5)

Egm6341.s11.team3.ren 01:07, 22 January 2011 (UTC)

= HW 1.4 =

Given:
<span id="(1)">
 * {| style="width:100%" border="0"

$$f(x)=3x-2x^3\ $$ $$ <span id="(1)">
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 4.1)
 * }
 * {| style="width:100%" border="0"

$$g(x)=sinh(x)=\frac{1}{2}\left(e^{x}-e^{-x}\right)$$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 4.2)
 * }

Find
A) Plot $$ f(x) $$ and $$ g(x) $$

B) Find the following norms:

<span id="(1)">
 * {| style="width:100%" border="0"

$$||f(x)||_{\infty}, ||g(x)||_{\infty}, ||f(x) - g(x)||_{\infty}$$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Solution
A) Equations (4.1) and (4.2) are plotted over the domain of x=[0,1] as shown below:



The difference between these two functions will be called Y(x). It is defined below as:

<span id="(1)">
 * {| style="width:100%" border="0"

$$Y(x) = f(x)-g(x)=[3x-2x^3]-[sinh(x)]=3x-2x^3-\frac{1}{2}\left(e^{x}-e^{-x}\right)$$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 4.3)
 * }

The plot of equation (4.3) over the domain of x=[0,1] is shown below:



B) The infinity norms of these three equations using the restricted domain of x=[0,1] are found below:

The infinity norm of a function is simply the maximum value the function achieves over a specified domain. All three of the functions plotted above will approach infinity over the unrestricted domain of $$x=(-\infty,+\infty)$$.

However, since we are restricting the domain to x=[0,1], these norms must be calculated using techniques discussed below:

B.1) Upon observing the plot of equation (4.1), it is obvious that the maximum value of f(x) occurs at the vertex.  So, the infinity norm will be found by solving for the vertex.  This is accomplished by finding the point of the function where the derivative is zero, since f(x) is not changing at the vertex:

<span id="(1)">
 * {| style="width:100%" border="0"

$$f(x)=3x-2x^3\ $$ $$ <span id="(1)">
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 4.4)
 * }
 * {| style="width:100%" border="0"

$$ \frac{df(x)}{dx} = (1)3x^{(1-1)} - (3)2x^{(3-1)} = 3-6x^2$$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 4.5)
 * }

Solving equation (4.5) for x when the derivative of f(x) is zero yields the location of the vertex: <span id="(1)">
 * {| style="width:100%" border="0"

$$ \frac{df(x_{vertex})}{dx} = 3-6x_{vertex}^2=0$$ $$ <span id="(1)">
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 4.6)
 * }
 * {| style="width:100%" border="0"

$$ x_{vertex} = \sqrt{\frac{(0 - 3)}{-6}} = \frac{\sqrt{2}}{2}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

Substituting this value of x back into equation (4.1) will yield the value of f(x) at the vertex, which is also the infinity norm on x=[0,1]:

<span id="(1)">
 * {| style="width:100%" border="0"

$$ ||f(x)||_{\infty} = f(x_{vertex}) = 3\left(\frac{\sqrt{2}}{2}\right)-2\left(\frac{\sqrt{2}}{2}\right)^3 = \sqrt{2}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

B.2) Upon observing the plot of equation (4.2), it is obvious that the maximum value over the prescribed domain occurs at the right boundary of the domain, or when x=1.  So, the infinity norm may be found simply by substituting this value of x directly into the function g(x):

<span id="(1)">
 * {| style="width:100%" border="0"

$$g(x)=sinh(x)=\frac{1}{2}\left(e^{x}-e^{-x}\right)$$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 4.7)
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$$ ||g(x)||_{\infty} = g(1) = sinh(1) = \frac{1}{2}\left(e^{(1)}-e^{-(1)}\right) = 1.1752$$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

B.3) Upon observing the plot of Y(x), which is defined above by equation (4.3), it is once again obvious that the maximum value of Y(x) occurs at the vertex.  So, the infinity norm will again be found by solving for the vertex:

<span id="(1)">
 * {| style="width:100%" border="0"

$$Y(x)=3x-2x^3-\frac{1}{2}\left(e^{x}-e^{-x}\right)$$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 4.8)
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$$ \frac{dY(x)}{dx} = (1)3x^{(1-1)} - (3)2x^{(3-1)} - \frac{1}{2}\left((1)e^{x}-(-1)e^{-x}\right) = 3-6x^2 - \frac{1}{2}\left(e^{x}+e^{-x}\right)$$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 4.9)
 * }

Solving equation (4.9) for x when the derivative of Y(x) equals zero yields the location of the vertex: <span id="(1)">
 * {| style="width:100%" border="0"

$$ \frac{dY(x_{vertex})}{dx} = 3-6x_{vertex}^2 -\frac{1}{2}\left(e^{x_{vertex}}+e^{-x_{vertex}}\right)=0$$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 4.10)
 * }

This equation cannot be solved analytically. An iterative approach was used, and the solution verified by observation of the plot above:

<span id="(1)">
 * {| style="width:100%" border="0"

$$ x_{vertex} = 0.5541\ $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

Substituting this value of x back into equation (4.3) will yield the value of Y(x) at the vertex, which is also the infinity norm on x=[0,1]:

<span id="(1)">
 * {| style="width:100%" border="0"

$$ ||f(x)-g(x)||_{\infty}=||Y(x)||_{\infty} = Y(x_{vertex}) = 3(0.5541)-2(0.5541)^3-\frac{1}{2}\left(e^{(0.5541)}-e^{-(0.5541)}\right) = 0.7392$$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 4.11)
 * }

Thus, each norm has been found and the problem is complete:

When the domain is unrestricted, $$x=(-\infty,+\infty)$$ and the norms are: <span id="(1)">
 * {| style="width:100%" border="0"

$$||f(x)||_{\infty} = ||g(x)||_{\infty} = ||f(x) - g(x)||_{\infty} = \infty $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 4.12)
 * }

When the domain is restricted to $$x=[0,1]\ $$, the norms become:

Egm6341.s11.team3.russo 21:55, 21 January 2011 (UTC)

=Signatures=

Solved problem 1 -- Egm6341.s11.team3.rakesh 23:39, 26 January 2011 (UTC)

Solved problem 2 -- Egm6341.s11.team3.sungsikkim

Solved problem 3 -- Egm6341.s11.team3.ren 01:03, 22 January 2011 (UTC)

Solved problem 4 --Egm6341.s11.team3.russo 21:56, 21 January 2011 (UTC)

Reviewed problem 1 -- Egm6341.s11.team3.XiaEgm6341.s11.team3.elango 20:00, 22 January 2011 (UTC)

Reviewed problem 2 -- Egm6341.s11.team3.elangoEgm6341.s11.team3.Xia 20:00, 23 January 2011 (UTC)

Reviewed problem 3 -- Egm6341.s11.team3.Xia 20:10, 22 January 2011 (UTC)

Reviewed problem 4 --Egm6341.s11.team3.Xia 20:20, 22 January 2011 (UTC)

Wiki editing and reviewing problems: --Egm6341.s11.team3.JA 21:54, 24 January 2011 (UTC)

=Solution Assignments and Reviewers=

=References=