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

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

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


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

$$
 * }
 * }

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


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

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


 * style="width:95%" |
 * $$f\left ( x \right )=P_{n+1}\left ( x \right )+R_{n+2}\left ( x \right )$$|| <p style="text-align:right;">$$\displaystyle
 * $$f\left ( x \right )=P_{n+1}\left ( x \right )+R_{n+2}\left ( x \right )$$|| <p style="text-align:right;">$$\displaystyle

$$
 * }
 * }

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

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

<span id="(1)">
 * {| 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 $$ $$
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }

d) APPLYING IMVT TO EQUATION

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

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

= HW 2.2 =

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

$$f(x) = \sin x, x\in(0,\pi) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

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

$$ f(x) = \sin x,  \quad x\in(0,\pi)$$    |$$ x_0=\frac{3\pi}{4}$$ $$
 * style="width:95%"  |
 * style="width:95%"  |
 * <p style="text-align:right"> $$ \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 <span id="(1)">
 * {| style="width:100%" border="0"

$$f(x) = \sin x, x\in(0,\pi) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

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

$$ f(x) = \sin x  \quad $$ $$ x\in(0,\pi)$$   |$$ x_0=\frac{\pi}{4}$$ $$
 * style="width:95%"  |
 * style="width:95%"  |
 * <p style="text-align:right"> $$ \displaystyle
 * }

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

$$
 * style="width:95%" |$$  f'(x) = \cos x \quad$$
 * <p style="text-align:right">$$ \displaystyle
 * <p style="text-align:right">$$ \displaystyle
 * }

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

$$
 * style="width:95%" |$$  f''(x) = -\sin x \quad $$
 * <p style="text-align:right"> $$ \displaystyle
 * <p style="text-align:right"> $$ \displaystyle
 * }

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

$$
 * style="width:95%" |$$ f(x) = -\cos x \quad $$
 * <p style="text-align:right"> $$ \displaystyle
 * <p style="text-align:right"> $$ \displaystyle
 * }

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

$$
 * $$ f'(x) =  \sin x \quad $$
 * <p style="text-align:right"> $$ \displaystyle
 * <p style="text-align:right"> $$ \displaystyle
 * }

we know from taylor series <span id="(1)">
 * {| style="width:100%" border="0"

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

where as

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

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

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

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

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

$$
 * style="width:95%" |$$  f(x) = P_0+ R_1 \quad $$
 * <p style="text-align:right">$$ \displaystyle
 * <p style="text-align:right">$$ \displaystyle
 * }

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

$$P_0 = f(x_0) = \sin\frac{\pi}{4} = \frac{1}{\sqrt{2}}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">$$ \displaystyle
 * }

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

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

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

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

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

$$ f(x) = P_1+ R_2 \quad$$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">$$ \displaystyle
 * }

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

$$P_0 = f(x_0) = \sin\frac{\pi}{4} = \frac{1}{\sqrt{2}}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">$$ \displaystyle
 * }

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

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

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

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

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

$$ f(x) = P_2+ R_3 \quad$$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">$$ \displaystyle
 * }

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

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

when n=3

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

$$ f(x) = P_3+ R_4 \quad$$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">$$ \displaystyle
 * }

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

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

when n=4

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

$$ f(x) = P_4+ R_5 \quad$$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">$$ \displaystyle
 * }

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

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

so on and when n=10 <span id="(1)">
 * {| style="width:100%" border="0"

$$ f(x) = P_{10}+ R_{11}\quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

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

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

finally the maximum can be computed by

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

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

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

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

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

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

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

$$
 * style="width:95%" |$$  f'(x) = \cos x \quad$$
 * <p style="text-align:right">$$ \displaystyle
 * <p style="text-align:right">$$ \displaystyle
 * }

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

$$
 * style="width:95%" |$$  f''(x) = -\sin x \quad $$
 * <p style="text-align:right"> $$ \displaystyle
 * <p style="text-align:right"> $$ \displaystyle
 * }

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

$$
 * style="width:95%" |$$ f(x) = -\cos x \quad $$
 * <p style="text-align:right"> $$ \displaystyle
 * <p style="text-align:right"> $$ \displaystyle
 * }

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

$$
 * $$ f'(x) =  \sin x \quad $$
 * <p style="text-align:right"> $$ \displaystyle
 * <p style="text-align:right"> $$ \displaystyle
 * }

we know from taylor series <span id="(1)">
 * {| style="width:100%" border="0"

$$ f(x) = P_n + R_n+1 \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">$$ \displaystyle
 * }

where as

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

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

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

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

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

$$P_0 = f(x_0) = \sin\frac{3\pi}{4} = \frac{1}{\sqrt{2}}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">$$ \displaystyle
 * }

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

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

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

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

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

$$ f(x) = P_1+ R_2 \quad$$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">$$ \displaystyle
 * }

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

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

when n=2

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

$$ f(x) = P_2+ R_3 \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">$$ \displaystyle
 * }

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

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

when n=3

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

$$ f(x) = P_3+ R_4 \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">$$ \displaystyle
 * }

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

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

$$ f(x) = P_{10}+ R_{11} \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

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

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

finally the maximum can be computed by

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

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

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

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

= HW 2.3 =

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

$$   f(x) = \frac {e^x - 1}{x} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \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
<span id="(1)">
 * {| style="width:100%" border="0"

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

<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. 3.3)
 * }

<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. 3.3: <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. 3.4)
 * }

From Eq. 3.4: <span id="(1)">
 * {| style="width:100%" border="0"

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

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

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

Therefore:

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

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

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

<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. 3.8)
 * }

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

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

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

$$ f(x) = \sum_{i=0}^{n} \frac{x^{i}}{(i+1)!} + R_{n+1}[f(x);x] = \sum_{i=0}^{n} \frac{x^{i}}{(i+1)!} + \frac{x^{n+1}}{(n+2)!} e^\xi, \quad \xi \in [0, x] $$ $$
 * <p style="text-align:right"> $$ \displaystyle (Eq. 3.10)

Combining equation 3.5 and 3.10

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

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

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

$$ f(x) = \sum_{i=0}^{n} \frac{x^{i}}{(i+1)!} + \frac{R_{n+1}[e^x;x]}{n+2} = \sum_{i=0}^{n} \frac{x^{i}}{(i+1)!} + \frac{1}{n+2} \frac{x^{n+1}}{(n+1)!} e^\xi, \quad \xi \in [0, x] $$ $$
 * <p style="text-align:right"> $$ \displaystyle (Eq. 3.12)

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

= HW 2.4 =

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

$$ I := \int_{-1}^{+1} \frac {e^x - 1}{x} dx $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \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

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

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

So the integration could be derived as

<span id="(1)">
 * {| 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 $$ yields,
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 4.2)
 * }

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

$$ I_n = \sum_{i=0}^{k} \frac{2}{(2i+1)(2i+1)!} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \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 :

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

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

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

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

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

The weight functions are given by :

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

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

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

$$ P_{n}(x) = \sum_{i=0}^{[n/2]} (-1)^i\frac{(2n-2i)!x^{n-2i}}{2^ni!(n-i)!(n-2i)!} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \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 $$ P_n(x) \quad $$

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

$$ P_n(x) = \sum_{i=0}^{n/2} (-1)^i \frac{(2n-2i)! x^{(n-2i)}}{2^ni!(n-i)!(n-2i)!} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \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 $$
 * <p style="text-align:right;"> $$ \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) $$
 * <p style="text-align:right;"> $$ \displaystyle

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

0

-0.9062

-0.5385

0.9062

0.5385 $$
 * <p style="text-align:right;"> $$ \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 $$
 * <p style="text-align:right;"> $$ \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))**** $$
 * <p style="text-align:right;"> $$ \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 $$
 * <p style="text-align:right;"> $$ \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

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

$$ f_2(x) = p_2(x) = c_2x^2 + c_1x + c_0 \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \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:

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

$$ p_2(x_0) = f(x_0) = c_2x_0^2 + c_1x_0 + c_0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

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

$$ p_2(x_1) = f(x_1) = c_2x_1^2 + c_1x_1 + c_0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

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

$$ p_2(x_2) = f(x_2) = c_2x_2^2 + c_1x_2 + c_0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Cleaning up and grouping

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

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

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

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

Isolating $$ c_1 \quad $$

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

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

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

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

Equating:

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

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

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

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

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

$$ c_2 = \frac {(x_0 - x_2)[f(x_0) - f(x_1)] - (x_0 - x_1)[f(x_0) - f(x_2)]}{(x_0 - x_1)(x_2^2 - x_0^2) - (x_0 - x_2)(x_1^2 - x_0^2)} $$ $$
 * <p style="text-align:right;"> $$ \displaystyle (Eq. 10.4)

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

$$ c_1 = \frac {[f(x_0) - f(x_2)] + [\frac {(x_0 - x_2)[f(x_0) - f(x_1)] - (x_0 - x_1)[f(x_0) - f(x_2)]}{(x_0 - x_1)(x_2^2 - x_0^2) - (x_0 - x_2)(x_1^2 - x_0^2)}(x_2^2 - x_0^2)]}{x_0 - x_2} $$ $$
 * <p style="text-align:right;"> $$ \displaystyle (Eq. 10.5)

and $$ c_0 \quad $$

$$ c_0  = f(x_0) - (\frac {(x_0 - x_2)[f(x_0) - f(x_1)] - (x_0 - x_1)[f(x_0) - f(x_2)]}{(x_0 - x_1)(x_2^2 - x_0^2) - (x_0 - x_2)(x_1^2 - x_0^2)} x_0^2 + $$ $$
 * <p style="text-align:right;"> $$ \displaystyle

$$ \qquad + \frac {[f(x_0) - f(x_2)] + [\frac {(x_0 - x_2)[f(x_0) - f(x_1)] - (x_0 - x_1)[f(x_0) - f(x_2)]}{(x_0 - x_1)(x_2^2 - x_0^2) - (x_0 - x_2)(x_1^2 - x_0^2)}(x_2^2 - x_0^2)]}{x_0 - x_2}x_0)$$ $$
 * <p style="text-align:right"> $$ \displaystyle (Eq. 10.6)

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

= HW 2.11 =

Given
Given from lecture notes:

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

$$ p_2(x)=\sum_{i=0}^{n=2} l_{i,2}(x)f(x_i) $$ $$ <span id="(1)">
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 11.1)
 * }
 * {| style="width:100%" border="0"

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

Find
Simple simpson's rule.

Solution
From definition:

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

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

With

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

$$ x_0=a;x_1=\frac {a+b}{2};x_2=b. $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

from (Eq. 11.2), we can obtain:

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

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

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

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

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

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

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

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

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

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

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

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

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

With

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

$$ h=\frac {b-a}{2} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

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

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

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

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

$$ \Rightarrow I_2(f) = \frac {h}{3} (f(x_0) + 4f(x_1) + f(x_2)) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

Which is right the simple simpson's rule.

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

= HW 2.12 =

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

$$ f(x)=\frac {e^x-1}{x} on [-1,1], x_0=-1,x_n=1 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \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:

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

$$ f_n(x)=p_n(x)=\sum_{i=0}^{n} l_{i,n}(x)f(x_i) $$ $$ <span id="(1)">
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 12.2)
 * }
 * {| style="width:100%" border="0"

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

For n=1,

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

$$ \quad x_0=-1,x_1=1 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

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

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

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

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

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

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

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

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

For n=2,

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

$$ \quad x_0=-1,x_1=0,x_2=1 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

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

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

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

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

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

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

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

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

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

$$ f(x_1) = \frac {e^{x_1}-1}{x_1} = 1 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

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

$$ f(x_2) = \frac {e^{x_2}-1}{x_2} = e-1 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

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

$$ \Rightarrow f_2(x) = l_{0,2}(x)f(x_0) + l_{1,2}(x)f(x_1) +  l_{2,2}(x)f(x_2)$$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \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,

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

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

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

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

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

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

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

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

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

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

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

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

<span id="(1)">
 * {| 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 $$ \quad j=0,1,2,...,n $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \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:

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

$$ I_n(f) = \int_{a}^{b} f_n(x)dx = \sum_{i=0}^{n} w_{i,n}f(x_i) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \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:

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

$$ \quad x_i = -x_{n-i} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

From (Eq. 12.3), we could derive

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

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

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

$$ = \prod_{j=0,n-j \neq i}^{n} \frac{-x-x_{n-j}}{x_{i}-x_{n-j}} = \prod_{k=0,k \neq i}^{n} \frac{-x-x_{k}}{x_{i}-x_{k}} = l_{i,n}(-x) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \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:

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

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

(b) Simple Simptson's rule:

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

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

Find
(a) Derive the composite Trapezoidal rule.

(b) Derive the Simpson's composite rule.

Solution
(a) Trapezoidal rule

Simple rule:

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

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

Composite rule:

<span id="(1)">
 * {| 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%" |

$$=\sum^n_{j=0}\dfrac{x_j-x_{j-1}}{2}[f(x_{j-1})+f(x_j)]$$ $$
 * <p style="text-align:right"> $$ \displaystyle
 * }

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

<span id="(1)">
 * {| 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%" |

$$=h[\frac{1}{2}f(x_1)+f(x_2)+\ \cdot\cdot\cdot\ +f(x_{n-1})+\frac{1}{2}f(x_n)]$$ $$
 * <p style="text-align:right"> $$ \displaystyle
 * }

(b) Simpson's rule

Simple rule:

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

$$I_2=\dfrac{h}{3}[f(x_0)+4f(x_1)+f(x_2)]$$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \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

<span id="(1)">
 * {| 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%" |

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

$$=\dfrac{h}{3}[f(x_0)+4f(x_1)+2f(x_2)+4f(x_3)+2f(x_4)+ \cdot\cdot\cdot\ +2f(x_{n-2})+4f(x_n)+f(x_n)]$$ $$
 * <p style="text-align:right"> $$ \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 <span id="(1)">
 * {| style="width:100%" border="0"

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

$$ g(b)=bC_1+C_2=1 \quad $$ $$
 * <p style="text-align:right"> $$ \displaystyle
 * }

Solving for $$C_1 \quad $$ and $$C_2 \quad $$, we have <span id="(1)">
 * {| style="width:100%" border="0"

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

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

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

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

$$\bar f(y)=\frac{b-a}{2}f(\frac{b-a}{2}y-\frac{1}{2})$$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \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)