text:
User:Egm6341.s11.team3/sub2

= HW 2.1 =

Given
Given a generic function $$ f(x) \quad $$

Find
a) Integrate by parts to obtain $$ {f(x)=f(x_0)+\frac {x-x_0}{1!} f(x_0)+\frac {(x-x_0)^2}{2} f''(x_0)+ \frac {(x-x_0)^3}{3!} f(t)+\frac {(x-x_0)^4}{4!}f'(x_0) } $$

b) show the remainder in to obtain in terms of $$ f^5(\xi) \xi for (x,x_0) \quad $$

c)Then assume Eq. (3) and Eq. (4) Lec. 3-3 are true and obtain $$ R_{n+2} \quad $$ d)Then obtain Eq.(5)from Eq. (4) [http://en.wikiversity.org/w/index.php?title=File%3ANm1.s11.mtg3.djvu&page=3 Lec. 3-3] are true and do integration by parts once more.

Solution
a) Given That
 * {| style="width:100%" border="0"

$$ {f(x)=f(x_0)+\frac {x-x_0}{1!} f(x_0)+\int\limits_{x}^{x_0}(x-t)f''(t)dt } $$|| $$ \displaystyle $$
 * style="width:95%" |
 * style="width:95%" |
 * }

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

$$ \int\limits_{x}^{x_0}(x-t)f(t)dt = [-\frac {(x-t)^2}{2}]_{t=x_0}^{t=x} f(t)-\int\limits_{x}^{x_0}\frac{-(x-t)^2}{2}f(t)dt $$| $$ \displaystyle $$
 * style="width:95%" |
 * style="width:95%" |
 * }

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

$$ \int\limits_{x}^{x_0}(x-t)f(t)dt = \frac {(x-x_0)^2}{2} f(x_0)+\int\limits_{x}^{x_0}\frac {(x-t)^2}{2}f(t)dt $$| $$ \displaystyle $$
 * style="width:95%" |
 * style="width:95%" |
 * }

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

$$ \int\limits_{x}^{x_0}\frac {(x-t)^2}{2}f(t)dt = -[\frac {(x-t)^3}{2*3}]_{t=x_0}^{t=x} f(t)-\int\limits_{x}^{x_0}\frac {-(x-t)^3}{2*3}f'(t)dt $$| $$ \displaystyle $$
 * style="width:95%" |
 * style="width:95%" |
 * }

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

$$ \int\limits_{x}^{x_0}\frac {(x-t)^2}{2}f(t)dt = \frac {(x-x_0)^3}{3!} f(t)+\int\limits_{x}^{x_0}\frac {(x-t)^3}{3!}f'(t)dt $$| $$ \displaystyle $$
 * style="width:95%" |
 * style="width:95%" |
 * }

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

$$ \int\limits_{x}^{x_0}\frac {(x-t)^3}{3}f'(t)dt = -[\frac {(x-t)^4}{4*3!}]_{t=x_0}^{t=x} f'(t)-\int\limits_{x}^{x_0}-\frac {(x-t)^4}{4!}f''(t)dt $$| $$ \displaystyle $$
 * style="width:95%" |
 * style="width:95%" |
 * }

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

$$ \int\limits_{x}^{x_0}\frac {(x-t)^3}{3}f'(t)dt = \frac {(x-x_0)^4}{4!} f'(t)+\int\limits_{x}^{x_0}\frac {(x-t)^4}{4!}f'(t)dt $$| $$ \displaystyle $$
 * style="width:95%" |
 * style="width:95%" |
 * }

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

$$ {f(x)=f(x_0)+\frac {x-x_0}{1!} f(x_0)+\frac {(x-x_0)^2}{2} f''(x_0)+ \frac {(x-x_0)^3}{3!} f(t)+\frac {(x-x_0)^4}{4!} f'(x_0)} $$|| $$ \displaystyle $$
 * style="width:95%" |
 * style="width:95%" |
 * }

b) using IMVT THEOREM
 * {| style="width:100%" border="0"

$$ \int\limits_{x}^{x_0}\frac {(x-t)^4}{4}f^(4)(t)dt =  f^5(\xi)\frac {(x-x_0)^5}{5!} $$, $$   \xi  \in (x,x_0) $$||
 $$ \displaystyle $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

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

 * style="width:95%" |Integrate by parts of $$R_{n+1}\left ( x \right )$$
 * $$\frac{1}{n!}\int_{x_{0}}^{x}\left ( x-t \right )^{n}f^{(n+1)}\left ( t \right )dt=\frac{1}{n!}\left \{ \left [ -\frac{(x-t)^{n}}{(n+1)}f^{(n+1)}\left ( t \right ) \right ]_{x_{0}}^{x} - \int_{x_{0}}^{x}\frac{-(x-t)^{n+1}}{(n+1)}f^{(n+1)}\left ( t \right )dt\right \}$$||
$$\displaystyle
 * $$\frac{1}{n!}\int_{x_{0}}^{x}\left ( x-t \right )^{n}f^{(n+1)}\left ( t \right )dt=\frac{1}{n!}\left \{ \left [ -\frac{(x-t)^{n}}{(n+1)}f^{(n+1)}\left ( t \right ) \right ]_{x_{0}}^{x} - \int_{x_{0}}^{x}\frac{-(x-t)^{n+1}}{(n+1)}f^{(n+1)}\left ( t \right )dt\right \}$$||
$$\displaystyle

$$
 * }
 * }

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

 * style="width:95%" |
 * $$ \frac{(x-x_{0})^{n+1}}{(n+1)!}f^{(n+1)}\left ( x_{0} \right )+\int_{x_{0}}^{x}\frac{(x-t)^{n+1}}{(n+1)!}f^{(n+2)}dt$$
$$\displaystyle
 * $$ \frac{(x-x_{0})^{n+1}}{(n+1)!}f^{(n+1)}\left ( x_{0} \right )+\int_{x_{0}}^{x}\frac{(x-t)^{n+1}}{(n+1)!}f^{(n+2)}dt$$
$$\displaystyle

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

 * style="width:95%" |
 * $$f\left ( x \right )=P_{n+1}\left ( x \right )+R_{n+2}\left ( x \right )$$||
$$\displaystyle
 * $$f\left ( x \right )=P_{n+1}\left ( x \right )+R_{n+2}\left ( x \right )$$||
$$\displaystyle

$$
 * }
 * }

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

$$ P_{n+1}\left ( x \right )=f\left ( x_{0} \right )+\frac{(x-x_{0})}{1!}f^{(1)}\left ( x_{0} \right )+...+\frac{(x-x_{0})^{n+1}}{(n+1)!}f^{(n+1)}\left ( x_{0} \right ) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
$$\displaystyle
 * }
 * }

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

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

R_{n+2}\left ( x \right )=\frac{1}{(n+1)!}\int_{x_{0}}^{x}(x-t)^{n+1}f^{(n+2)}dt $$ $$
 *
$$\displaystyle
 * }
 * }

d) APPLYING IMVT TO EQUATION

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

$$\int\limits_{x}^{x_0}\frac{(x-t)^n}{n!}f^{n+1}(t)dt = f^{n+1}(\xi)\frac {(x-x_0)^{n+1}}{n+1!} \xi \in (x,x_0)$$
 $$ \displaystyle $$ Egm6341.s11.team3.rakesh 20:00 02 February 2011 (UTC)
 * style="width:95%" |
 * style="width:95%" |
 * }

= HW 2.2 =

Given
a) Given That
 * {| style="width:100%" border="0"

$$f(x) = \sin x, x\in(0,\pi) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ f(x) = \sin x,  \quad x\in(0,\pi)$$    |$$ x_0=\frac{3\pi}{4}$$ $$
 * style="width:95%"  |
 * style="width:95%"  |
 *
 $$ \displaystyle
 * }

Find
Construct the Taylor series of $$f(x) \,$$ around $$x_0=\frac{\pi}{4},\frac{3\pi}{4}$$ for $$n=0,1,...,10 \,$$.

Then estimate the maximum of $$R(x) \,$$ at $$x=\frac{\pi}{2},\frac{3\pi}{4}$$

Solution
a) Given That
 * {| style="width:100%" border="0"

$$f(x) = \sin x, x\in(0,\pi) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ f(x) = \sin x  \quad $$ $$ x\in(0,\pi)$$   |$$ x_0=\frac{\pi}{4}$$ $$
 * style="width:95%"  |
 * style="width:95%"  |
 *
 $$ \displaystyle
 * }

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

$$
 * style="width:95%" |$$  f'(x) = \cos x \quad$$
 *
$$ \displaystyle
 *
$$ \displaystyle
 * }

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

$$
 * style="width:95%" |$$  f''(x) = -\sin x \quad $$
 *
 $$ \displaystyle
 *
 $$ \displaystyle
 * }

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

$$
 * style="width:95%" |$$ f(x) = -\cos x \quad $$
 *
 $$ \displaystyle
 *
 $$ \displaystyle
 * }

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

$$
 * $$ f'(x) =  \sin x \quad $$
 *
 $$ \displaystyle
 *
 $$ \displaystyle
 * }

we know from taylor series
 * {| style="width:100%" border="0"

$$ f(x) = P_n + R_{n+1} \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
$$ \displaystyle
 * }

where as

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

$$P_n =f(x_0)+\frac {x-x_0}{1!} f(x_0)+\frac {(x-x_0)^2}{2} f''(x_0)+\frac {(x-x_0)^3}{3!} f(t)+.....+\frac {(x-x_0)^n}{n!} f^n(t) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
$$ \displaystyle
 * }

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

$$R_{n+1} = \frac {(x-x_0)^{n+1}}{n+1!} f^{n+1}(\S) $$ $$ when n=0
 * style="width:95%" |
 * style="width:95%" |
 *
$$ \displaystyle
 * }

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

$$
 * style="width:95%" |$$  f(x) = P_0+ R_1 \quad $$
 *
$$ \displaystyle
 *
$$ \displaystyle
 * }

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

$$P_0 = f(x_0) = \sin\frac{\pi}{4} = \frac{1}{\sqrt{2}}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
$$ \displaystyle
 * }

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

$$R_1= (x-\frac{\pi}{4})$$   $$ \,f(\S)$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ f(x_0) = \frac{1}{\sqrt{2}}+(x-\frac{\pi}{4}) \,f(\S) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
$$ \displaystyle
 * }

when n=1
 * {| style="width:100%" border="0"

$$ f(x) = P_1+ R_2 \quad$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
$$ \displaystyle
 * }

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

$$P_0 = f(x_0) = \sin\frac{\pi}{4} = \frac{1}{\sqrt{2}}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
$$ \displaystyle
 * }

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

$$P_1 = f(x_1) = \frac{x-\frac{\pi}{4}}{1!}\cos\frac{\pi}{4}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
$$ \displaystyle
 * }

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

$$R_2= \frac{(x-\frac{\pi}{4})^2}{2!} \,f^2(\S)$$$$ \displaystyle $$ $$P_1 + R_2= \frac{1}{\sqrt{2}}+\frac{x-\frac{\pi}{4}}{1!}\frac{1}{\sqrt{2}}+\frac{(x-\frac{\pi}{4})^2}{2!} \,f^2(\S) $$ $$ when n=2
 * style="width:95%" |
 * style="width:95%" |
 *
$$ \displaystyle
 * }

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

$$ f(x) = P_2+ R_3 \quad$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
$$ \displaystyle
 * }

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

$$ =\frac{1}{\sqrt{2}}+\frac{x-\frac{\pi}{4}}{1!}\frac{1}{\sqrt{2}}-\frac{({x-\frac{\pi}{4})}^2}{2!}\frac{1}{\sqrt{2}}+\frac{(x-\frac{\pi}{4})^3}{3!} \,f^3(\S)  $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
$$ \displaystyle
 * }

when n=3

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

$$ f(x) = P_3+ R_4 \quad$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
$$ \displaystyle
 * }

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

$$ =\frac{1}{\sqrt{2}}+\frac{x-\frac{\pi}{4}}{1!}\frac{1}{\sqrt{2}}-\frac{({x-\frac{\pi}{4})}^2}{2!}\frac{1}{\sqrt{2}}-\frac{({x-\frac{\pi}{4})}^3}{3!}\frac{1}{\sqrt{2}}+\frac{(x-\frac{\pi}{4})^4}{4!} \,f^4(\S)  $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
$$ \displaystyle
 * }

when n=4

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

$$ f(x) = P_4+ R_5 \quad$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
$$ \displaystyle
 * }

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

$$ =\frac{1}{\sqrt{2}}+\frac{x-\frac{\pi}{4}}{1!}\frac{1}{\sqrt{2}}-\frac{({x-\frac{\pi}{4})}^2}{2!}\frac{1}{\sqrt{2}}-\frac{({x-\frac{\pi}{4})}^3}{3!}\frac{1}{\sqrt{2}}+\frac{({x-\frac{\pi}{4})}^4}{4!}\frac{1}{\sqrt{2}}+\frac{(x-\frac{\pi}{4})^5}{5!} \,f^5(\S)$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
$$ \displaystyle
 * }

so on and when n=10
 * {| style="width:100%" border="0"

$$ f(x) = P_{10}+ R_{11}\quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ =\frac{1}{\sqrt{2}}+\frac{x-\frac{\pi}{4}}{1!}\frac{1}{\sqrt{2}}-\frac{({x-\frac{\pi}{4})}^2}{2!}\frac{1}{\sqrt{2}}-\frac{({x-\frac{\pi}{4})}^3}{3!}\frac{1}{\sqrt{2}}+\frac{({x-\frac{\pi}{4})}^4}{4!}\frac{1}{\sqrt{2}}+.......-\frac{({x-\frac{\pi}{4})}^{10}}{10!}\frac{1}{\sqrt{2}}+\frac{(x-\frac{\pi}{4})^{11}}{11!} \,f^{11}(\S)  $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

finally the maximum can be computed by

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

$$R_{n+1} = \frac {(x-x_0)^{n+1}}{n+1!} f^{n+1}(\S)$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$R_{n+1}\le\frac {(x-x_0)^{n+1}}{n+1!} |maxf^{n+1}(\S)|\le\frac {(x-x_0)^{n+1}}{n+1!}=\frac {(\pi/4)^{n+1}}{n+1!}\le(\pi/4)$$ $$ b)
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ f(x) = \sin x  \quad $$ $$ x\in(0,\pi)$$   |$$ x_0=\frac{3\pi}{4}$$ $$
 * style="width:95%"  |
 * style="width:95%"  |
 *
 $$ \displaystyle
 * }

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

$$
 * style="width:95%" |$$  f'(x) = \cos x \quad$$
 *
$$ \displaystyle
 *
$$ \displaystyle
 * }

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

$$
 * style="width:95%" |$$  f''(x) = -\sin x \quad $$
 *
 $$ \displaystyle
 *
 $$ \displaystyle
 * }

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

$$
 * style="width:95%" |$$ f(x) = -\cos x \quad $$
 *
 $$ \displaystyle
 *
 $$ \displaystyle
 * }

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

$$
 * $$ f'(x) =  \sin x \quad $$
 *
 $$ \displaystyle
 *
 $$ \displaystyle
 * }

we know from taylor series
 * {| style="width:100%" border="0"

$$ f(x) = P_n + R_n+1 \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
$$ \displaystyle
 * }

where as

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

$$P_n =f(x_0)+\frac {x-x_0}{1!} f(x_0)+\frac {(x-x_0)^2}{2} f''(x_0)+\frac {(x-x_0)^3}{3!} f(t)+.....+\frac {(x-x_0)^n}{n!} f^n(t) \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
$$ \displaystyle
 * }

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

$$R_n+1 = \frac {(x-x_0)^n+1}{n+1!} f^{n+1}(\S) $$ $$ when n =0
 * style="width:95%" |
 * style="width:95%" |
 *
$$ \displaystyle
 * }

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

$$P_0 = f(x_0) = \sin\frac{3\pi}{4} = \frac{1}{\sqrt{2}}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
$$ \displaystyle
 * }

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

$$R_1= (x-\frac{3\pi}{4})$$   $$ \,f(\S)$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ f(x_0) = \frac{1}{\sqrt{2}}+(x-\frac{3\pi}{4}) \,f(\S) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
$$ \displaystyle
 * }

when n=1
 * {| style="width:100%" border="0"

$$ f(x) = P_1+ R_2 \quad$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
$$ \displaystyle
 * }

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

$$R_2= \frac{(x-\frac{3\pi}{4})^2}{2!} \,f^2(\S)$$$$ \displaystyle $$ $$f(x)= \frac{1}{\sqrt{2}}-\frac{x-\frac{3\pi}{4}}{1!}\frac{1}{\sqrt{2}}+\frac{(x-\frac{3\pi}{4})^2}{2!} \,f^2(\S) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
$$ \displaystyle
 * }

when n=2

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

$$ f(x) = P_2+ R_3 \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
$$ \displaystyle
 * }

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

$$ =\frac{1}{\sqrt{2}}-\frac{x-\frac{3\pi}{4}}{1!}\frac{1}{\sqrt{2}}+\frac{({x-\frac{3\pi}{4})}^2}{2!}\frac{1}{\sqrt{2}}+\frac{(x-\frac{3\pi}{4})^3}{3!} \,f^3(\S)  $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
$$ \displaystyle
 * }

when n=3

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

$$ f(x) = P_3+ R_4 \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
$$ \displaystyle
 * }

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

$$ =\frac{1}{\sqrt{2}}-\frac{x-\frac{3\pi}{4}}{1!}\frac{1}{\sqrt{2}}+\frac{({x-\frac{3\pi}{4})}^2}{2!}\frac{1}{\sqrt{2}}+\frac{({x-\frac{3\pi}{4})}^3}{3!}\frac{1}{\sqrt{2}}+\frac{(x-\frac{3\pi}{4})^4}{4!} \,f^4(\S)  $$ $$ ...so on when n=10
 * style="width:95%" |
 * style="width:95%" |
 *
$$ \displaystyle
 * }
 * {| style="width:100%" border="0"

$$ f(x) = P_{10}+ R_{11} \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ =\frac{1}{\sqrt{2}}-\frac{x-\frac{3\pi}{4}}{1!}\frac{1}{\sqrt{2}}+\frac{({x-\frac{3\pi}{4})}^2}{2!}\frac{1}{\sqrt{2}}+\frac{({x-\frac{3\pi}{4})}^3}{3!}\frac{1}{\sqrt{2}}++.......-\frac{({x-\frac{3\pi}{4})}^{10}}{10!}\frac{1}{\sqrt{2}}+\frac{(x-\frac{\pi}{4})^{11}}{11!} \,f^{11}(\S)  $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

finally the maximum can be computed by

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

$$R_{n+1} = \frac {(x-x_0)^{n+1}}{n+1!} f^{n+1}(\S)$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$R_{n+1}\le\frac {(x-x_0)^{n+1}}{n+1!} |maxf^{n+1}(\S)|\le\frac {(x-x_0)^{n+1}}{n+1!}=\frac {(\pi/2)^{n+1}}{n+1!}\le(\pi/2)$$ $$ Egm6341.s11.team3.rakesh 20:21, 2 February 2011 (UTC)
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

= HW 2.3 =

Given

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

$$   f(x) = \frac {e^x - 1}{x} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 3.1)
 * }

Find
1. Expand $$ e^x \quad $$ in TS (Taylor Series) up to order $$ n \quad $$ with remainder $$ R_{n+1}[e^x; x] \quad $$

2. Find TS of $$ f(x) \quad $$ up to order $$ n \quad $$ with reminder $$ R_{n+1}[f(x); x] \quad $$ and with remainder $$ R_{n+1}[e^x,x] \quad $$

Solution

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

$${P}_{n}(x) = f({x}_{0}) + \frac{(x-{x}_{0})}{1!}f'({x}_{0}) + .... + \frac{({x-{x}_{0}})^{n}}{n!}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 3.2)
 * }

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

$${R}_{n+1}(x)= \frac{1}{n!}\int_{{x}_{0}}^{x}(x-t)^n f^{n+1}(t)\,dt = \frac{(x-x_0)^{n+1}}{(n+1)!}f^{n+1}(\xi), \quad \xi \in [x_0, x] $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 3.3)
 * }

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

$$ f^{n+1}(x) = \frac{d^{n+1}}{dx^{n+1}}f(x) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

 * {| 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} $$ $$
 *
 $$ \displaystyle
 * }

From Eq. 3.3:
 * {| style="width:100%" border="0"

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

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

$$ = \frac{x^0}{0!}(1) + \frac{x^1}{1!}(1) + \frac{x^2}{2!}(1) +....+ \frac{x^n}{n!}(1) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ = \sum_{n=0}^n (\frac{x^n}{n!}) \Rightarrow P_n(x) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 3.4)
 * }

From Eq. 3.4:
 * {| style="width:100%" border="0"

$${R}_{n+1}(x)= \frac{1}{n!}\int_{{x}_{0}}^{x}(x-t)^n f^{n+1}(t)\,dt = \frac{(x-x_0)^{n+1}}{(n+1)!}f^{n+1}(\xi), \quad where \quad f^{n+1}(\xi) = \frac {d^{n+1}}{d \xi ^{n+1}}(e^\xi) = e^\xi $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$${R}_{n+1}(x)= {R}_{n+1}[e^x;x]= \frac{x^{n+1}}{(n+1)!} e^\xi $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 3.5)
 * }

Therefore:

$$ f(x) = {P}_{n}(x) + {R}_{n+1}(x) = \sum_{i=0}^n (\frac{x^i}{i!}) + \frac{x^{i+1}}{(i+1)!} e^\xi = \sum_{i=0}^n (\frac{x^i}{i!}) + {R}_{n+1}[e^x;x] $$ $$
 *
 $$ \displaystyle (Eq. 3.6)

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

From Taylor series of $$ \quad e^x $$ (Eq. 3.6), we further expand $$ \quad e^x $$ to $$ \quad (n+1)^{th} $$ order:

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

$$ e^{x} = P(x) + R(x) = 1+ \sum_{i=1}^{n+1} \frac{x^i}{i!} + \frac{x^{n+2}}{(n+2)!} e^\xi \quad \xi \in [0, x] $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 3.7)
 * }

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

$$ \Rightarrow f(x) = \frac{e^{x}-1}{x} = \sum_{i=0}^{n} \frac{x^i}{(i+1)!} + \frac{x^{n+1}}{(n+2)!} e^\xi, \quad \xi \in [0, x] $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 3.8)
 * }

Therefore, $$ \quad R_{n+1}[f(x);x] $$ could be expressed as:
 * {| style="width:100%" border="0"

$$ R_{n+1}(x) = R_{n+1}[f(x);x] = \frac{x^{n+1}}{(n+2)!} e^\xi, \quad \xi \in [0, x] $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 3.9)
 * }

Then $$ f(x) \quad $$ is expanded to order $$ n \quad $$ with remainder $$ R_{n+1}[f(x); x] \quad $$:

$$ f(x) = \sum_{i=0}^{n} \frac{x^{i}}{(i+1)!} + R_{n+1}[f(x);x] = \sum_{i=0}^{n} \frac{x^{i}}{(i+1)!} + \frac{x^{n+1}}{(n+2)!} e^\xi, \quad \xi \in [0, x] $$ $$
 *
 $$ \displaystyle (Eq. 3.10)

Combining equation 3.5 and 3.10

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

$$ R_{n+1}(x) = R_{n+1}[f(x);x] = \frac{R_{n+1}[e^x;x]}{n+2} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 3.11)
 * }

So $$ f(x) \quad $$ is expanded to order $$ n \quad $$ with remainder $$ R_{n+1}[e^x; x] \quad $$ as:

$$ f(x) = \sum_{i=0}^{n} \frac{x^{i}}{(i+1)!} + \frac{R_{n+1}[e^x;x]}{n+2} = \sum_{i=0}^{n} \frac{x^{i}}{(i+1)!} + \frac{1}{n+2} \frac{x^{n+1}}{(n+1)!} e^\xi, \quad \xi \in [0, x] $$ $$
 *
 $$ \displaystyle (Eq. 3.12)

--Egm6341.s11.team3.JA 20:55, 2 February 2011 (UTC)

= HW 2.4 =

Given

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

$$ I := \int_{-1}^{+1} \frac {e^x - 1}{x} dx $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 4.1)
 * }

Find
Plot $$ f(x) \quad $$ and integrate $$f(x) \quad $$ in $$[-1, +1] \quad $$ (find $$ I_n \quad $$) by using:

 * Taylor Series
 * Composite Trapezoid Rule
 * Composite Simpson's Rule
 * Gauss-Legendre Quadrature

for n = 2, 4, 8,.... until the error is of order $$ 10^{-6} \quad $$

2.4(a)
Plot of $$ \quad f(x) $$:

Solution
Noting from equation 3.10

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

$$ f_n(x) = {P}_{n}(x) = \sum_{i=0}^{n} \frac{x^{i}}{(i+1)!} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 4.1)
 * }

So the integration could be derived as

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

$$ I_n = \int_{-1}^{1} f_n(x)dx = \int_{-1}^{1} \sum_{i=0}^{n} \frac{x^{i}}{(i+1)!}dx = \sum_{i=0}^{n} \int_{-1}^{1}\frac{x^{i}}{(i+1)!}dx $$ $$ substitute $$ \quad n $$ with $$ \quad 2k $$ yields,
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 4.2)
 * }

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

$$ I_n = \sum_{i=0}^{k} \frac{2}{(2i+1)(2i+1)!} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 4.3)
 * }

As $$ \quad n = 2,4,8,16,... $$, use matlab to solve for the integration and the code is attached below:

The results are shown in the following table:

HW2.4(b) Solution
From Composite Trapezoidal rule :

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

$$ I_n(x) = \frac {b-a}{n} [\frac{f_a+f_b}{2} +\sum_{i=1}^{n-1} f_(a+\frac{k*(b-a)}{n})] $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

Maple code is written to solve the problem and attached below

for n from 2 by 2 to 16 do

evalf(2*(((exp(-1)-1)/(-1)+exp(1)-1)*(1/2)+sum(exp(-1+2*i/n), i = 1 .. n-1))/n) end do

Similarly, for varying values of n the Integral value was calculated

Maple output

HW2.4(c) Solution
From Simpson's rule :

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

$$ I_n(x) = \frac {h}{3} [f_0 + 4\sum_{i=1}^{n/2} f_{2i-1} + 2\sum_{i=1}^{n/2-1} f_{2i} + f_n] $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

Matlab code is written to solve the problem and attached below:

The solution and iteration history is shown in following table:

Egm6341.s11.team3.Xia 19:50, 31 January 2011 (UTC)

HW2.4(d) Solution
For Gauss-Legendre quad, the integration could be written as :

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

$$ I_n(x) = \sum_{i=1}^{n} w_if(x_i) $$ for $$ \quad i=1,2,...,n $$ and $$ \quad x_i $$ are the roots for (Eq) $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

The weight functions are given by :

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

$$ w_i = \frac {-2}{(n+1)P_{n}^{(1)}(x_i)P_{n+1}(x_i)} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ P_{n}(x) = \sum_{i=0}^{[n/2]} (-1)^i\frac{(2n-2i)!x^{n-2i}}{2^ni!(n-i)!(n-2i)!} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

Combining the above equations, we can obtain the roots and weight function values for n=2 and n=4 as presented in lecture notes. The following matlab code uses those values directly and compute the integrations for n=2 and n=4.

The results are shown in the following table:

Obviously from the above table, the error for n=4 already smaller than the order of 0.000001, so there is no need of further computaions.

Egm6341.s11.team3.Xia 22:00, 31 January 2011 (UTC)

= HW 2.5 =

Given
A Legendre Polynomial $$ f(x) \quad $$ at integration limits ranging from $$ -1 \quad $$ to $$ +1 \quad $$

Find
1. Verify weight functions for Legendre Polynomials 2. Values of n and points x to evaluate W

Solution
For the integration problem stated above, the associated polynomials are Legendre polynomials, Pn(x). With the nth polynomial normalized to give Pn(1) = 1, the ith Gauss node, xi, is the ith root of Pn; its weight is given by
 * $$ w_i = \frac{2}{\left( 1-x_i^2 \right) [P'_n(x_i)]^2} \,\!$$

Egm6341.s11.team3.elango 20:08, 22 January 2011 (UTC)

= HW 2.6 =

Given
To verify P[n] for n = 0..7

Given
Expression for P[n]

Solution
MAPLE CODE

for n from 0 to 4 do P[n] := sum((-1)^i*factorial(2*n-2*i)*x^(n-2*i)/(2^n*factorial(i)*factorial(n-i)*factorial(n-2*i)), i = 0 .. (1/2)*n) end do;

MAPLE OUTPUT

$${P}_{0}(x)= 1 \quad $$ $${P}_{1}(x)= x$$ $${P}_{2}(x)= \frac{1}{2}(3x^2-1) $$

$${P}_{3}(x)= \frac{1}{2}(5x^3-3x) $$

$${P}_{4}(x)= \frac{1}{8}(35x^4-30x^2+3) $$

$${P}_{5}(x)= \frac{1}{8}(63x^5-70x^3+15x) $$

Egm6341.s11.team3.elango 26:00, 22 January 2011 (UTC)

= HW 2.7 =

Given
$${{P}_{i}(x), \quad i = 0, 1, 2, ...}$$ is orthogonal functions for $$n=0,1,2,3,4,5 \quad$$

Find
Verify that $${{P}_{i}(x), i = 0, 1, 2, ...}$$ is orthogonal functions for $$n=0,1,2,3,4,5$$

Construct Gram Matrix

Show Diag of Gram Matrix

Show Det of Gram

Solution
$${P}_{0}(x)= 1 \quad $$

$${P}_{1}(x)= x \quad $$

$${P}_{2}(x)= \frac{1}{2}(3x^2-1) $$

$${P}_{3}(x)= \frac{1}{2}(5x^3-3x) $$

$${P}_{4}(x)= \frac{1}{8}(35x^4-30x^2+3) $$

$${P}_{5}(x)= \frac{1}{8}(63x^5-70x^3+15x) $$

Gram Matrix $$ {r}_{ij}=(<{p}_{i},{p}_{j}>) \quad $$

where $$<{p}_{i},{p}_{j}>= \int{p}_{i}{p}_{j}$$

$${p}_{00}=\int_{-1}^{1}1dx =2 $$

$${p}_{11}=\int_{-1}^{1}x^2dx =\frac{2}{3} $$

$${p}_{22}=\int_{-1}^{1}(\frac{1}{2}(3x^2-1))^2dx =\frac{2}{5} $$

$${p}_{33}=\int_{-1}^{1}(\frac{1}{2}(5x^3-3x))^2dx =\frac{2}{7} $$

$${p}_{44}=\int_{-1}^{1}(\frac{1}{8}(35x^4-30x^2+3))^2dx =\frac{2}{9} $$

$${p}_{55}=\int_{-1}^{1}(\frac{1}{8}(63x^5-70x^3+15x^2))^2dx =\frac{1}{11} $$

$${p}_{ij}={p}_{ji} \quad$$

where

$$\int_{-a}^{a} (Even Function) = const $$

$$\int_{-a}^{a} (Odd Function) = 0 $$

Even x Odd = Odd

Even x Even = Even Odd x Odd = Even

Therefore

when $$i/= j,{p}_{ij} \quad $$ is Odd Function, $${p}_{ij}=0 \quad $$

Egm6341.s11.team3.sungsikkimEgm6341.s11.team3.sungsikkim 04:04, 2 February 2011 (UTC)

= HW 2.8 =

Given
Generic function $$ f(x) \quad $$

Find
Show that

$$I={I}_{1}(f)=\int_{a}^{b}{f}_{1}(x)dx$$ $$=[\int_{a}^{b}{l}_{0}xdx]f({x}_{0})+[\int_{a}^{b}{l}_{1}(x)dx]f({x}_{1})$$

using simple Trap. Rule

Solution
Trap. Rule

$${I}_{1}=\frac{b-a}{2}[f(a)+f(b)]$$

$${l}_{0}=\frac{x-{x}_{1}}{{x}_{0}-{x}_{1}}$$

$${l}_{1}=\frac{x-{x}_{0}}{{x}_{1}-{x}_{0}}$$

Let $$ a={x}_{0},b={x}_{1} \quad $$

$${l}_{0}(x)=\frac{x-b}{a-b}$$

$${l}_{1}(x)=\frac{x-a}{b-a}$$

$$[\int_{1}^{2}\frac{x-b}{a-b}dx]f(a)=[\frac{1}{b-a}(\frac{{b}^{2}}{2}+\frac{{a}^{2}}{2}-ab)]f(a)$$

$$[\int_{1}^{2}\frac{x-a}{b-a}dx]f(b)=[\frac{1}{b-a}(\frac{{b}^{2}}{2}+\frac{{a}^{2}}{2}-ab)]f(b)$$

$$[\int_{1}^{2}{l}_{0}(x)dx]f(a) + [\int_{1}^{2}{l}_{1}(x)dx]f(b)=\frac{b-a}{2}(f(a)+f(b))$$

Examples

Let $$f(x)=x, a=1={x}_{0},b=2={x}_{1} \quad $$

From Trap. Rule

$${I}_{1}=\frac{2-1}{2}[1+2]=\frac{3}{2}$$

Proof

$${l}_{0}(x)=\frac{x-2}{1-2}=2-x, f({x}_{0})=1$$

$${l}_{1}(x)=\frac{x-1}{2-1}=x-1, f({x}_{1})=2$$

$$I={I}_{1}(f)=\int_{1}^{2}xdx$$ $$=[\int_{1}^{2}(2-x)dx](1)+[\int_{1}^{2}(x-1)dx](2)=\frac{1}{2}+1=\frac{3}{2}$$

The results are the same as $$\frac{3}{2} \quad $$

Let $$f(x)={x}^{2}, a=1={x}_{0},b=2={x}_{1} \quad $$

From Trap. Rule

$${I}_{1}=\frac{2-1}{2}[1+4]=\frac{5}{2}$$

Proof

$${l}_{0}(x)=\frac{x-2}{1-2}=2-x, f({x}_{0})=1$$

$${l}_{1}(x)=\frac{x-1}{2-1}=x-1, f({x}_{1})=4$$

$$I={I}_{1}(f)=\int_{1}^{2}xdx$$ $$=[\int_{1}^{2}(2-x)dx](1)+[\int_{1}^{2}(x-1)dx](4)=\frac{1}{2}+2=\frac{5}{2}$$

The results are the same as $$\frac{5}{2}$$

Egm6341.s11.team3.sungsikkim 04:05, 2 February 2011 (UTC)

= HW 2.9 =

Given
Legendre polynomial general form for $$ P_n(x) \quad $$

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

$$ P_n(x) = \sum_{i=0}^{n/2} (-1)^i \frac{(2n-2i)! x^{(n-2i)}}{2^ni!(n-i)!(n-2i)!} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 9.1)
 * }

where $$ [\frac{n}{2}] $$ = integer part of $$ \frac{n}{2} $$

Find
By using Legendre polynomial generate $$ P_5(x) \quad $$ and matlab command "roots" to compute roots for $$ P_5(x) \quad $$.

Plot roots on $$ [-1,+1] \quad $$ interval. Repeat the above for $$ P_{10}(x) \quad $$.

Observe the location of hte roots near the end points $$ -1 \quad $$ and $$ +1 \quad $$ and discuss.

Solution
Given from Professor's Vu Quac notes provided table, the values for $$ P_{1-5}(x) $$ , and original source the values do match.

By using the above matlab code, for $$ P_5(x) \quad $$ and $$ P_{10}(x) \quad $$ the governing equation and roots came out to be:

>> sub2ex9(5,10)

Equation for $$ P_5(x) \quad $$ :

(63*x^5)/8 - (35*x^3)/4 + (15*x)/8 $$
 *
 $$ \displaystyle (Eq. 1.3)

Symbolic roots for $$ P_5(x) \quad $$ :

0

-(5/9 - (2*2^(1/2)*35^(1/2))/63)^(1/2)

-((2*2^(1/2)*35^(1/2))/63 + 5/9)^(1/2)

(5/9 - (2*2^(1/2)*35^(1/2))/63)^(1/2)

((2*2^(1/2)*35^(1/2))/63 + 5/9)^(1/2) $$
 *
 $$ \displaystyle

Numeric roots for $$ P_5(x) \quad $$ :

0

-0.9062

-0.5385

0.9062

0.5385 $$
 *
 $$ \displaystyle

Equation for $$ P_{10}(x) \quad $$ :

(46189*x^10)/256 - (109395*x^8)/256 + (45045*x^6)/128 - (15015*x^4)/128 + (3465*x^2)/256 - 63/256 $$
 *
 $$ \displaystyle (Eq. 1.3)

Symbolic roots for $$ P_{10}(x) \quad $$ :

solvelib::Union({[-z4^(1/2), 1], [z4^(1/2), 1]}, z4, RootOf(X42^5 - (45*X42^4)/19 + (630*X42^3)/323 - (210*X42^2)/323 + (315*X42)/4199 - 63/46189, X42))**** $$
 *
 $$ \displaystyle

Numeric roots for $$ P_{10}(x) \quad $$ :

-0.9739

-0.8651

-0.6794

0.9739

0.8651

0.6794

-0.4334

0.4334

-0.1489

0.1489 $$
 *
 $$ \displaystyle

that there might be imaginary part of an equation as well. Therefore, the answer was converted to "double numerical" format and "roots" for polynomial were used to get the answer.
 * As you may see, matlab could not solve symbolically for roots for given $$ P_{10}(x) \quad $$ equation, mainly because it assumed

Also, the plot was generated for $$ P_5(x)\quad $$ and $$ P_{10}(x) \quad $$ cases. Additionally, superimposed 2 more plots to show the places $$ P_5(x) \quad $$ and $$ P_{10}(x) \quad $$ hit $$ f(x) = 0 \quad $$ values.

It could be observed that as degree of Legendre polynomials increases the more and more oscillations appear near the ends of the functions. This is known as Runge's phenomenon

--Egm6341.s11.team3.JA 21:13, 2 February 2011 (UTC)

= HW 2.10 =

Given
Given general $$ 2^{nd} \quad $$ order polynomial interpolation expression

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

$$ f_2(x) = p_2(x) = c_2x^2 + c_1x + c_0 \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 10.1)
 * }

Find
At the set nodes which correspond to set function values, express unknown constants $$ c_2, c_1, c_0 = {c_i} \quad $$ in terms of known node and function values $$ (x_i, f(x_i)) \quad$$

Solution
From given problem set up, we express plug in each of the given/known nodes into the polynomial expression:

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

$$ p_2(x_0) = f(x_0) = c_2x_0^2 + c_1x_0 + c_0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ p_2(x_1) = f(x_1) = c_2x_1^2 + c_1x_1 + c_0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ p_2(x_2) = f(x_2) = c_2x_2^2 + c_1x_2 + c_0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

Bring eq. 2-4 right hand side members to the other side leaving only $$ c_0 \quad $$ on the right hand side.

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

$$ f(x_0) - (c_2x_0^2 + c_1x_0) = c_0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 10.2a)
 * }

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

$$ f(x_1) - (c_2x_1^2 + c_1x_1) = c_0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 10.2b)
 * }

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

$$ f(x_2) - (c_2x_2^2 + c_1x_2) = c_0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 10.2c)
 * }

Then substitute for $$ c_0 \quad $$ eq. 2 in eq. 3 & 4

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

$$ f(x_0) - (c_2x_0^2 + c_1x_0) = f(x_1) - (c_2x_1^2 + c_1x_1) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ f(x_0) - (c_2x_0^2 + c_1x_0) = f(x_2) - (c_2x_2^2 + c_1x_2) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

Next rearranging terms and isolating $$ c_1 \quad $$ on the right hand side:

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

$$ [f(x_0) - f(x_1)] + [c_2x_1^2 - c_2x_0^2] = c_1x_0 - c_1x_1 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ [f(x_0) - f(x_2)] + [c_2x_2^2 - c_2x_0^2] = c_1x_0 - c_1x_2 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

Cleaning up and grouping

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

$$ [f(x_0) - f(x_1)] + [c_2(x_1^2 - x_0^2)] = c_1(x_0 - x_1) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ [f(x_0) - f(x_2)] + [c_2(x_2^2 - x_0^2)] = c_1(x_0 - x_2) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

Isolating $$ c_1 \quad $$

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

$$ \frac {[f(x_0) - f(x_1)] + [c_2(x_1^2 - x_0^2)]}{x_0 - x_1} = c_1 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 10.3a)
 * }

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

$$ \frac {[f(x_0) - f(x_2)] + [c_2(x_2^2 - x_0^2)]}{x_0 - x_2} = c_1 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 10.3b)
 * }

Equating:

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

$$ \frac {[f(x_0) - f(x_1)] + [c_2(x_1^2 - x_0^2)]}{x_0 - x_1} = \frac {[f(x_0) - f(x_2)] + [c_2(x_2^2 - x_0^2)]}{x_0 - x_2} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

Cross multiplying and bringing $$ c_2 \quad $$ to another side

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

$$ (x_0 - x_2)[f(x_0) - f(x_1)] - (x_0 - x_1)[f(x_0) - f(x_2)] = - c_2(x_0 - x_2)(x_1^2 - x_0^2) + c_2(x_0 - x_1)(x_2^2 - x_0^2) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

$$ c_2 = \frac {(x_0 - x_2)[f(x_0) - f(x_1)] - (x_0 - x_1)[f(x_0) - f(x_2)]}{(x_0 - x_1)(x_2^2 - x_0^2) - (x_0 - x_2)(x_1^2 - x_0^2)} $$ $$
 *
 $$ \displaystyle (Eq. 10.4)

Substituting $$ c_2 \quad $$ into $$ c_1 \quad $$ expression:

$$ c_1 = \frac {[f(x_0) - f(x_2)] + [\frac {(x_0 - x_2)[f(x_0) - f(x_1)] - (x_0 - x_1)[f(x_0) - f(x_2)]}{(x_0 - x_1)(x_2^2 - x_0^2) - (x_0 - x_2)(x_1^2 - x_0^2)}(x_2^2 - x_0^2)]}{x_0 - x_2} $$ $$
 *
 $$ \displaystyle (Eq. 10.5)

and $$ c_0 \quad $$

$$ c_0  = f(x_0) - (\frac {(x_0 - x_2)[f(x_0) - f(x_1)] - (x_0 - x_1)[f(x_0) - f(x_2)]}{(x_0 - x_1)(x_2^2 - x_0^2) - (x_0 - x_2)(x_1^2 - x_0^2)} x_0^2 + $$ $$
 *
 $$ \displaystyle

$$ \qquad + \frac {[f(x_0) - f(x_2)] + [\frac {(x_0 - x_2)[f(x_0) - f(x_1)] - (x_0 - x_1)[f(x_0) - f(x_2)]}{(x_0 - x_1)(x_2^2 - x_0^2) - (x_0 - x_2)(x_1^2 - x_0^2)}(x_2^2 - x_0^2)]}{x_0 - x_2}x_0)$$ $$
 *
 $$ \displaystyle (Eq. 10.6)

--Egm6341.s11.team3.JA 21:14, 2 February 2011 (UTC)

= HW 2.11 =

Given
Given from lecture notes:

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

$$ p_2(x)=\sum_{i=0}^{n=2} l_{i,2}(x)f(x_i) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 11.1)
 * }
 * {| style="width:100%" border="0"

$$ l_{i,n}=\prod_{j=0,j \neq i}^{n} \frac{x-x_j}{x_i-x_j} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 11.2)
 * }

Find
Simple simpson's rule.

Solution
From definition:

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

$$ I_2(f) = \int_{a}^{b} f_2(x)dx = \sum_{i=0}^{2} \int_{a}^{b} l_{i,2}(x)dx f(x_i) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 11.3)
 * }

With

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

$$ x_0=a;x_1=\frac {a+b}{2};x_2=b. $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

from (Eq. 11.2), we can obtain:

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

$$ l_{0,2}(x) = \frac {(x-x_1)(x-x_2)}{(x_0-x_1)(x_0-x_2)} = \frac {2}{(a-b)^2} [x^2-\frac{a+3b}{2} x +\frac{(a+b)b}{2}] $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 11.4)
 * }

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

$$ l_{1,2}(x) = \frac {(x-x_0)(x-x_2)}{(x_1-x_0)(x_1-x_2)} = -\frac {4}{(a-b)^2} [x^2-(a+b) x +ab] $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 11.5)
 * }

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

$$ l_{2,2}(x) = \frac {(x-x_0)(x-x_1)}{(x_2-x_0)(x_2-x_1)} = \frac {2}{(b-a)^2} [x^2-\frac{3a+b}{2} x +\frac{(a+b)a}{2}] $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 11.6)
 * }

Integrate (Eq. 11.4)(Eq. 11.5)(Eq. 11.6), we can solve

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

$$ \int_{a}^{b}l_{0,2}(x) = \frac {2}{(b-a)^2} [-\frac{1}{12} (a-b)^3] = \frac{1}{6}(b-a) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 11.7)
 * }

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

$$ \int_{a}^{b}l_{1,2}(x) = \frac {4}{(a-b)^2} [-\frac{1}{6} (a-b)^3] = \frac{4}{6}(b-a) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 11.8)
 * }

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

$$ \int_{a}^{b}l_{2,2}(x) = \frac {2}{(b-a)^2} [-\frac{1}{12} (a-b)^3] = \frac{1}{6}(b-a) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 11.9)
 * }

With

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

$$ h=\frac {b-a}{2} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

Plug (Eq. 11.7)(Eq. 11.8)(Eq. 11.9) into (Eq. 11.3), yields

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

$$ I_2(f) = \int_{a}^{b} l_{0,2}(x)dx f(x_0) + \int_{a}^{b} l_{1,2}(x)dx f(x_1) + \int_{a}^{b} l_{2,2}(x)dx f(x_2) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 11.10)
 * }

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

$$ \Rightarrow I_2(f) = \frac {h}{3} (f(x_0) + 4f(x_1) + f(x_2)) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

Which is right the simple simpson's rule.

Egm6341.s11.team3.Xia 05:30, 31 January 2011 (UTC)

= HW 2.12 =

Given

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

$$ f(x)=\frac {e^x-1}{x} on [-1,1], x_0=-1,x_n=1 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 12.1)
 * }

Find
1) Construct $$ \quad f_n(x) $$ by lagrange interpolation functions for $$ \quad n=1,2,4,8,16 $$ and then plot $$ \quad f(x) $$ and $$ \quad f_n(x) $$. 2) Compute weight functions for $$ \quad n=1,2,4,8,16 $$ and then form tables. 3) Compute $$ \quad I_n(x) $$ for $$ \quad n=1,2,4,8 $$ and compare to $$ \quad I $$ that is solved using WolframAlpha with more digits. 4) For $$ \quad n=5 $$, plot $$ \quad l_0,l_1,l_2 $$ and then predict the shapes of $$ \quad l_3,l_4,l_5 $$ based on those.

Solution
1) According to the problem, we construct the fuctions from following equations:

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

$$ f_n(x)=p_n(x)=\sum_{i=0}^{n} l_{i,n}(x)f(x_i) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 12.2)
 * }
 * {| style="width:100%" border="0"

$$ l_{i,n}=\prod_{j=0,j \neq i}^{n} \frac{x-x_j}{x_i-x_j} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 12.3)
 * }

For n=1,

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

$$ \quad x_0=-1,x_1=1 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ l_{0,1}(x) = \frac {x-x_1}{x_0-x_1} = -\frac{1}{2}(x-1) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 12.4)
 * }

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

$$ l_{1,1}(x) = \frac {x-x_0}{x_1-x_0} = \frac{1}{2}(x+1) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 12.5)
 * }

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

$$ f(x_0) = \frac {e^{x_0}-1}{x_0} = 1-e^{-1} \displaystyle $$ $$ f(x_1) = \frac {e^{x_1}-1}{x_1} = e-1 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ \Rightarrow f_1(x) = l_{0,1}(x)f(x_0) + l_{1,1}(x)f(x_1) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 12.6)
 * }

For n=2,

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

$$ \quad x_0=-1,x_1=0,x_2=1 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ l_{0,2}(x) = \frac {(x-x_1)(x-x_2)}{(x_0-x_1)(x_0-x_2)} = \frac{1}{2}x(x-1) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 12.7)
 * }

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

$$ l_{1,2}(x) = \frac {(x-x_0)(x-x_2)}{(x_1-x_0)(x_1-x_2)} = -(x+1)(x-1) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 12.8)
 * }

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

$$ l_{1,2}(x) = \frac {(x-x_0)(x-x_1)}{(x_2-x_0)(x_2-x_1)} = \frac{1}{2}(x+1)x $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 12.9)
 * }

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

$$ f(x_0) = \frac {e^{x_0}-1}{x_0} = 1-e^{-1} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ f(x_1) = \frac {e^{x_1}-1}{x_1} = 1 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ f(x_2) = \frac {e^{x_2}-1}{x_2} = e-1 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ \Rightarrow f_2(x) = l_{0,2}(x)f(x_0) + l_{1,2}(x)f(x_1) +  l_{2,2}(x)f(x_2)$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 12.10)
 * }

For n=4,8,16, we use matlab for plotting with (Eq. 12.2) and (Eq. 12.3). The code is attached below:

The plots are shown as following:

2) For n=1, integrate previous (Eq. 12.4) and (Eq. 12.5) from [-1,1], we get,

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

$$ w_{0,1} = \int_{-1}^{1} l_{0,1}(x)dx = 1 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ w_{1,1} = \int_{-1}^{1} l_{1,1}(x)dx = 1 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

For n=2, integrate previous (Eq. 12.7), (Eq. 12.8) and (Eq. 12.9) from [-1,1], we get,

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

$$ w_{0,2} = \int_{-1}^{1} l_{0,2}(x)dx = \frac{1}{3} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ w_{1,2} = \int_{-1}^{1} l_{1,2}(x)dx = \frac{4}{3} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ w_{2,2} = \int_{-1}^{1} l_{2,2}(x)dx = \frac{1}{3} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

For n=4,8,16, by apply LIET, we obtain the linear system equations:

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

$$ \sum_{i=0}^{n}w_{i,n}(x_i)^j = \frac{b^{j+1}-a^{j+1}}{j+1} $$ when $$ \quad j=0,1,2,...,n $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 12.11)
 * }

Solving (Eq. 12.11) for n=4,8,16 by matlab and the code is attached below:

The results are shown in the following tables:

3) Integrate using Newton-Cotes Method, The integration could be expressed as:

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

$$ I_n(f) = \int_{a}^{b} f_n(x)dx = \sum_{i=0}^{n} w_{i,n}f(x_i) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 12.12)
 * }

With previously calculated value of weight functions, The results of the integrations are shown in the following table and compared with that obtained from WolframAlpha.

4) For n=5, the lagrange functions are computed from (Eq. 12.3) using matlab, the code is attached below:

The first three fuctions are plotted as follows:

Now, we will analyze the shapes of the lagrange functions. First it is obvious to identify:

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

$$ \quad x_i = -x_{n-i} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

From (Eq. 12.3), we could derive

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

$$ l_{n-i,n}(x) = \prod_{j=0,j \neq n-i}^{n} \frac{x-x_j}{x_{n-i}-x_j} = \prod_{j=0,j \neq n-i}^{n} \frac{x+x_{n-j}}{x_{n-i}+x_{n-j}} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ = \prod_{j=0,n-j \neq i}^{n} \frac{-x-x_{n-j}}{x_{i}-x_{n-j}} = \prod_{k=0,k \neq i}^{n} \frac{-x-x_{k}}{x_{i}-x_{k}} = l_{i,n}(-x) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 12.13)
 * }

So from (Eq. 12.13), we can predict that $$ \quad l_3 $$ and $$ \quad l_2 $$, $$ \quad l_4 $$ and $$ \quad l_1 $$, $$ \quad l_5 $$ and $$ \quad l_0 $$ are axis-symmetric function pairs (With respect to y axis). We further plot $$ \quad l_3 $$, $$ \quad l_4 $$, $$ \quad l_5 $$ in the following figure and the prediction is verified.

Egm6341.s11.team3.Xia 05:31, 31 January 2011 (UTC)

= HW 2.13 =

Given
(a) Simple Trapezoid rule rule:

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

$$I_1=\dfrac{b-a}{2}\left[f(a)+f(b)\right] $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 13.1)
 * }

(b) Simple Simptson's rule:

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

$$I_2=\dfrac{h}{3}[f(x_0)+4f(x_1)+f(x_2)]$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 13.2)
 * }

Find
(a) Derive the composite Trapezoidal rule.

(b) Derive the Simpson's composite rule.

Solution
(a) Trapezoidal rule

Simple rule:

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

$$I_1=\dfrac{b-a}{2}\left[f(a)+f(b)\right] $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

Composite rule:

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

$$I_n=\dfrac{x_1-x_0}{2}[f(x_0)+f(x_1)]+\dfrac{x_2-x_1}{2}[f(x_1)+f(x_2)]+\ \cdot\cdot\cdot\ +\dfrac{x_n-x_{n-1}}{2}[f(x_{n-1})+f(x_n)]$$
 * style="width:95%" |
 * style="width:95%" |

$$=\sum^n_{j=0}\dfrac{x_j-x_{j-1}}{2}[f(x_{j-1})+f(x_j)]$$ $$
 *
 $$ \displaystyle
 * }

Using uniform intervals, $$h=x_j-x_{j-1} (j=0,1,2,\cdot\cdot\cdot,n)$$, the integral becomes

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

$$I_n=\frac{h}{2}\sum^n_{j=0}[f(x_{j-1})+f(x_j)]=\frac{h}{2}[f(x_0)+2f(x_1)+2f(x_2)+\ \cdot\cdot\cdot\ +2f(x_{n-1})+f(x_n)]$$
 * style="width:95%" |
 * style="width:95%" |

$$=h[\frac{1}{2}f(x_1)+f(x_2)+\ \cdot\cdot\cdot\ +f(x_{n-1})+\frac{1}{2}f(x_n)]$$ $$
 *
 $$ \displaystyle
 * }

(b) Simpson's rule

Simple rule:

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

$$I_2=\dfrac{h}{3}[f(x_0)+4f(x_1)+f(x_2)]$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

Composite rule:

Using uniform intervals, $$h=(b-a)/n\ (n=2k, k==0,1,2,\cdot\cdot\cdot)$$, the integral can be written as

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

$$I_n=\frac{h}{3}\sum^{n-1}_{j=1,3,5,\cdot\cdot\cdot}[f(x_{j-1})+4f(x_j)+f(x_{j+1})]$$
 * style="width:95%" |
 * style="width:95%" |

$$=\dfrac{h}{3}[f(x_0)+4f(x_1)+f(x_2)]+\dfrac{h}{3}[f(x_2)+4f(x_3)+f(x_4)]+\ \cdot\cdot\cdot\ +\dfrac{h}{3}[f(x_{n-2})+4f(x_n)+f(x_n)]$$

$$=\dfrac{h}{3}[f(x_0)+4f(x_1)+2f(x_2)+4f(x_3)+2f(x_4)+ \cdot\cdot\cdot\ +2f(x_{n-2})+4f(x_n)+f(x_n)]$$ $$
 *
 $$ \displaystyle
 * }

Egm6341.s11.team3.ren 02:35, 31 January 2011 (UTC)

= HW 2.14 =

Given
Given expression $$\int^{b}_{a}f(x)dx=\int^{1}_{-1}\bar f(y)dy$$

Find
Show the detail of $$\int^{b}_{a}f(x)dx=\int^{1}_{-1}\bar f(y)dy$$

Solution
Assume y is related to x in a simple form $$y=g(x)=C_1x+C_2 \quad $$. We want y=-1 as x=a, and y=1 as x=b, which is
 * {| style="width:100%" border="0"

$$g(a)=aC_1+C_2=-1 \quad $$
 * style="width:95%" |
 * style="width:95%" |

$$ g(b)=bC_1+C_2=1 \quad $$ $$
 *
 $$ \displaystyle
 * }

Solving for $$C_1 \quad $$ and $$C_2 \quad $$, we have
 * {| style="width:100%" border="0"

$$C_1=\frac{2}{b-a}$$
 * style="width:95%" |
 * style="width:95%" |

$$C_2=\frac{a+b}{a-b} \quad $$

$$\Rightarrow\ y=\frac{2}{b-a}x+\frac{a+b}{a-b}\ \Rightarrow\ x=\frac{b-a}{2}y-\frac{1}{2}\ \Rightarrow\ dx=\frac{b-a}{2}dy$$

$$\Rightarrow\ \int^{b}_{a}f(x)dx=\int^{1}_{-1}f(\frac{b-a}{2}y-\frac{1}{2})\frac{b-a}{2}dy=\int^{1}_{-1}\bar f(y)dy$$ $$ Therefore,
 *
 $$ \displaystyle
 * }
 * {| style="width:100%" border="0"

$$\bar f(y)=\frac{b-a}{2}f(\frac{b-a}{2}y-\frac{1}{2})$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

Egm6341.s11.team3.ren 03:18, 31 January 2011 (UTC)

=References =

= Contirubtors =

Egm6341.s11.team3.rakesh 21:18, 02 February 2011 (UTC)

Egm6341.s11.team3.sungsikkim 21:12, 02 February 2011 (UTC)

Egm6341.s11.team3.ren 21:22, 02 February 2011 (UTC)

Egm6341.s11.team3.russo 21:05, 02 February 2011 (UTC)

Egm6341.s11.team3.Xia 20:35, 02 February 2011 (UTC)

Egm6341.s11.team3.elango 20:00, 22 January 2011 (UTC)

Egm6341.s11.team3.JA 21:26, 2 February 2011 (UTC)

title: User:Egm6341.s11.team3/sub2


text:
Solution Architecture/Elaborating

Determining the Solution Architecture

 * 1) The Assessment Framework
 * 2) Understanding Security

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

title: Solution Architecture/Elaborating


text:
Solution Architecture/Elaborating/Assessment Framework

The Assessment Framework
Template An Assessment Framework template can be found in google docs;
 * https://spreadsheets.google.com/ccc?key=0Ak8Mcp1f93B1dExnSUtkU2hCZVRZNHJtaUg3eUZoZmc&hl=en

Sample

Product Features
Does the product have the features that fulfill the business case requirements? List the high level features to be assessed against the requirements.

Ability to Deliver
Does the vendor / solution provider / consulting firm have the proven ability to deliver the solution? Reference existing relationships with vendors, client references or online evidence to assess candidates’ ability to deliver.

Cost + TCO
When a solution is evaluated it should also consider three cost attributes; Solutions can vary greatly within these three cost attributes. Ongoing costs should also include all HR related costs of running the solution from an operational perspective.
 * 1) The initial cost of software, consulting and required infrastructure.
 * 2) The expected life of this product.
 * 3) The ongoing costs of maintaining the software and infrastructure.

Sustainability
This is a care and feeding question. Does the solution require a great deal of care or can it self-heal? The sustainability also includes the solution’s ability to stay current to the business and strategic needs. This is more an effort question rather than a cost. Sustainability can become a distraction for an organization or IT Team.

Alignment
Alignment includes many attributes, does the solution align with current technology infrastructure, the organizations current internal skills and knowledge, current roadmap and current strategy. Intellectual Property related issues also need to be aligned.

Risk
What risks do you assume by adopting the technology? What risks do you assume if you do not adopt the technology?