User:Egm6341.s11.team3/sub3

= 3.1 Taylor Series vs. Lagrange approximation error =

Given
Given a function


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

$$ f(x) = sin(x) \quad $$ in interval $$ x \in [0, \pi] \quad $$, $$
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle (Eq. 3.1.1)
 * }

that is to be approximated by Lagrange polynomial, $$ f_4^L(x) \quad $$, at 5 nodes. Nodes were chosen to be equidistant $$ x_0 = 0, x_1 = \frac{\pi}{4}, x_2 = \frac{\pi}{2}, x_3 = \frac{3\pi}{4}, x_4 = \pi $$.

Moreover, the same function $$ f(x) \quad $$ was approximated with Taylor series, $$ f_n^T(x) \quad $$, around a point $$ \frac{3\pi}{8} \quad $$.

Find
At a chosen point $$ x = \frac{7\pi}{8} $$, find what degree 'n' of Taylor polynomial approximation would be needed to produce an error, lesser then Lagrange polynomial at 5 nodes.

or

Find 'n' such that $$ |e_n^T(\frac{7\pi}{8})| \le |e_4^L(\frac{7\pi}{8})| $$

Solution
We start with an error estimation for Lagrange interpolation at a point $$ x = \frac{7\pi}{8} $$.

From we see that


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

$$ f(x) - f_n^L(x) = e_n^L(x) =\frac{q_{n+1}(x)}{(n+1)!}f^{n+1}(\xi) $$ $$ where 
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle (Eq. 3.1.2)
 * }
 * {| style="width:100%" border="0"

$$ q_{n+1}(x) = \prod_{j=0}^{n}(x-x_j) = (x-x_0)(x-x_1)(x-x_2).......(x-x_n) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle (Eq. 3.1.3)
 * }

for $$ n = 4 \quad $$, nodes of $$ x_0 = 0, x_1 = \frac{\pi}{4}, x_2 = \frac{\pi}{2}, x_3 = \frac{3\pi}{4}, x_4 = \pi $$, and $$ f(x) = sin(x) \quad $$


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$$ e_4^L(x) =\frac{q_{5}(x)}{5!}f^{5}(\xi) = \frac{(x-0)(x-\frac{\pi}{4})(x-\frac{\pi}{2})(x-\frac{3\pi}{4})(x-\pi)}{5!}cos(\xi) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle\!
 * }

since $$ \xi \in [x_0, x_n] = [x_0, x_5] = [0, \pi] \Rightarrow cos(\xi) \in [0, 1] \quad $$.

Assuming error to be maximum


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

$$ max(e_4^L(x)) \Rightarrow cos(\xi) = 1 \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle (Eq. 3.1.4)
 * }

Therefore,


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$$ e_4^L(x) = \frac{(x-0)(x-\frac{\pi}{4})(x-\frac{\pi}{2})(x-\frac{3\pi}{4})(x-\pi)}{5!} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle (Eq. 3.1.5)
 * }

and at a point of an interest $$ x = \frac{7\pi}{8} $$


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

$$ e_4^L(x) = \frac{(\frac{7\pi}{8}-0)(\frac{7\pi}{8}-\frac{\pi}{4})(\frac{7\pi}{8}-\frac{\pi}{2})(\frac{7\pi}{8}-\frac{3\pi}{4})(\frac{7\pi}{8}-\pi)}{5!} = -0.0082 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle\!
 * }

Therefore,

$$ |e_4^L(x)| = 0.0082 \quad $$

Consequently, we start solving for error for Taylor Series. Error in Taylor Series is expressed as reminder $$ R \quad $$


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$$ R_{n+1}(x) := \frac{(x-\hat x)^{n+1}}{(n+1)!}f^{n+1}(\xi) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle (Eq. 3.1.6)
 * }

as it was mentioned before, since $$ \xi \in [x_0, x_n] = [x_0, x_5] = [0, \pi] \quad $$ and $$ |f^{(n)}(x)| \quad $$ is either $$ cos(x) \quad $$ or $$ sin(x) \Rightarrow |f^{(n+1)}(\xi)| \in [0, 1] \quad $$

Again, we assume a maximum error, that is $$ |f^{(n+1)}(\xi)| = 1 \quad $$

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

$$ e_n^T = \frac{(x-\hat x)^{n+1}}{(n+1)!} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 3.1.7)
 * }

Since we know the point around witch we are computing Taylor series from a given parameters $$ (\hat x = \frac{3\pi}{8} )$$ and the point that we want to find an approximation for $$( x = \frac{7\pi}{8} )$$, we have

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

$$ e_n^T = \frac{(\frac{7\pi}{8}-\frac{3\pi}{8})^{n+1}}{(n+1)!} = \frac{(\frac{\pi}{2})^{n+1}}{(n+1)!}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 3.1.8)
 * }

Then we equate the errors to find the order of Taylor Series polynomial that we need in order to produce results at least as good as 5 node Lagrange polynomial.

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

$$ |e_n^T(\frac{7\pi}{8})| \le |e_4^L(\frac{7\pi}{8})| $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle\!
 * }

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

$$ \Rightarrow \frac{(\frac{\pi}{2})^{n+1}}{(n+1)!} \le 0.0082 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 3.1.9)
 * }

Since this can't be solved by analytical methods, we used Matlab to find 'n' value for us

$$ n = 6, \quad |e_n^T(\frac{7\pi}{8})|=0.0047 $$

Therefore, you need Taylor Series of $$ 6^{th} \quad $$ order $$ (n = 6) \quad $$ to approximate a function at least as good as Lagrange Series at $$ 4^{th} \quad $$ order $$ (n = 4) \quad $$.

--Egm6341.s11.team3.JA 22:29, 9 February 2011 (UTC)

= 3.2 Error Analysis of Lagrange Interpolation at nodes =

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

$$ G(x) = q_{n+1}(x) d(x) \frac{}{}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 3.2.1)
 * }

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

$$ q_{n+1}(x) = \prod_{j=0}^{n}(x-x_j) = (x-x_0)(x-x_1)(x-x_2).......(x-x_n) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 3.2.2)
 * }

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

$$ d(x) = \frac{e(x)}{q_{n+1}(x)} - \frac{e(t)}{q_{n+1}(t)}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 3.2.3)
 * }

Find
<span id="(1)">
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$$ G(x_i) \frac{}{}$$ for $$ i=0,1,...,n \frac{}{}$$
 * style="width:95%" |
 * style="width:95%" |
 * }

Solution
Substituting the point of interest into Equation 3.2.2 yields:

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

$$ q_{n+1}(x_i) = \prod_{j=0}^{n}(x_i-x_j) = (x_i-x_0)(x_i-x_1)(x_i-x_2).......(x_i-x_n) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 3.2.4)
 * }

When variable i is substituted into Equation 3.2.4, the result becomes:

i=0: <span id="(1)">
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$$ q_{n+1}(x_0) = \prod_{j=0}^{n}(x_0-x_j) = (0)(x_0-x_1)(x_0-x_2).......(x_0-x_n) = 0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 3.2.4.1)
 * }

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

$$ q_{n+1}(x_1) = \prod_{j=0}^{n}(x_1-x_j) = (x_1 - x_0)(0)(x_1-x_2).......(x_1-x_n) = 0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 3.2.4.2)
 * }

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

$$ q_{n+1}(x_n) = \prod_{j=0}^{n}(x_n-x_j) = (x_n - x_0)(x_n - x_1)(x_n-x_2).......(0) = 0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 3.2.4.n)
 * }

Thus, for each value of variable i Equation 3.2.4 becomes: <span id="(1)">
 * {| style="width:100%" border="0"

$$ q_{n+1}(x_i) = \prod_{j=0}^{n}(x_i-x_j) = 0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 3.2.5)
 * }

Substituting the point of interest into Equation 3.2.1 yields:

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

$$ G(x_i) = q_{n+1}(x_i) d(x_i) \frac{}{}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 3.2.6)
 * }

Substituting Equation 3.2.5 into Equation 3.2.6 yields the final desired answer: <span id="(1)">
 * {| style="width:100%" border="0"

$$ G(x_i) = (0) d(x_i) = 0 \frac{}{}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 3.2.7)
 * }

$$\therefore G(x_i) = 0$$

Egm6341.s11.team3.russo 01:43, 15 February 2011 (UTC)

= 3.3.1 Plotting figures similar to proof of LIET =

Given
Given a function

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

$$ f(x) = \frac{exp(x)-1}{x} \quad $$ in interval $$ x \in [-1, 1] \quad $$, $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 3.3.1)
 * }

Find
Find by Lagrange polynomial,weights and $$ q(n+1) \quad $$ for, $$ L_2,4^L(x) \quad $$, at 5 nodes. Nodes were chosen to be equidistant $$ x_0 = -1, x_1 = \frac{-1}{2}, x_2 = \frac 0, x_3 = \frac{1}{2}, x_4 = 1 $$.

also find errors at points $$ x = 0.75 \quad $$ and $$ t = 5 \quad $$

Solution
Consider a Lagrange method on approximating exact function:


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

f_n^L\left( x \right)=\,\sum_{i=0}^{n}l_{i,n}\left ( x \right)f\left ( x_i \right)$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 3.3.2)
 * }
 * }

For given 5 nodes at $$ n = 4 \rightarrow x_{0} = -1, x_{1} = -0.5, x_{2} = 0, x_{3} = 0.5, x_{4} = 1 $$.


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

f_{4}^L=\sum_{i=0}^{4}l_{i,4}(x)f(x_{i})=l_{0,4}(x)f(x_{0})+l_{1,4}(x)f(x_{1})+l_{2,4}(x)f(x_{2})+l_{3,4}(x)f(x_{3})+l_{4,4}(x)f(x_{4}) $$ $$ where
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle\!
 * }
 * }
 * {| style="width:100%" border="0" align="left"

l_{i,n}=\prod_{j=0,j\neq 0}^{4}\frac{x-x_{j}}{x_{i}-x_{j}} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 3.3.3)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

l_{0,4}=\prod_{j=0,j\neq 0}^{4}\frac{x-x_{j}}{x_{0}-x_{j}} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

l_{1,4}=\prod_{j=0,j\neq 1}^{4}\frac{x-x_{j}}{x_{1}-x_{j}} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

l_{2,4}=\prod_{j=0,j\neq 2}^{4}\frac{x-x_{j}}{x_{2}-x_{j}} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

l_{3,4}=\prod_{j=0,j\neq 3}^{4}\frac{x-x_{j}}{x_{3}-x_{j}} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

l_{4,4}=\prod_{j=0,j\neq 4}^{4}\frac{x-x_{j}}{x_{4}-x_{j}} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

For example:


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

l_{0,4}=\prod_{j=0,j\neq 0}^{4}\frac{x-x_{j}}{x_{0}-x_{j}} = \frac{x-x_1}{x_{0}-x_{1}}\frac{x-x_{2}}{x_{0}-x_{2}}\frac{x-x_{3}}{x_{0}-x_{3}}\frac{x-x_{4}}{x_{0}-x_{4}} = \left(\frac{x-(-0.5)}{(-1)-(-0.5)} \right) \left(\frac{x-0}{(-1)-0} \right ) \left( \frac{x-0.5}{(-1)-0.5} \right) \left (\frac{x-1}{(-1)-1} \right ) $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Since matlab does not form the matrices/vectors for the symbolic and numeric values, we did a sample run for $$ l_{0,4}= \quad $$ to check the output for the expected numerical values above. Note, since we used loops to populate for values of $$ l \quad $$ we denoted $$ x_j = x_i \quad $$ case to be $$ 1 \quad $$ so that we could run the loops smoothly and we would not get the answer of $$ \frac{x-x_j}{\inf} $$.

These values did came out to be the same as the expected values from the equation above

Noting that $$ q_{n+1} \quad $$ is

$$\displaystyle q_{n+1}(x) := (x - x_0)(x - x_1)(x - x_2)(x - x_3)\cdots(x - x_n) $$ <p style="text-align:right;">$$\displaystyle (Eq. 3.3.4) $$

and that error is

$$\displaystyle e_T^L = f(x) - f^L(x) \quad $$ <p style="text-align:right;">$$\displaystyle (Eq. 3.3.5) $$

we generated matlab code to calculate and plot parameters such as $$ q_{n+1}, \ e_T^L(x), \  e_T^L(t), l_{i,n}, f^L \quad $$

as you may see, the answers came out to be



it is clear that q part hits all roots at specified nodes. Also, it may be seen that true and approximate values are really close to each other and overlap. From that image is hard to see if the approximation does cross the exact function at the nodes (note, we used the same function for n=8 where more oscillations are present and it clearly shows approximate function passing through the nodes).

We attached another figure showing the numerical values along with approximate values, just to show that the 2 graphs do follow 1 another really close



Errors at x = 0.75 and t = 5 came out to be:

$$ e_4^L(x = 0.75) = -1.6274*10^{-4} \quad $$

$$ e_4^L(t = 5) = 11.0239 \quad $$

It can be clearly seen that error inside the bound interval is really small. Therefore, functions are close to each other. On the other hand for t= 5 (outside the function bounds), error was relatively huge (note the function maximum value was only around 1.8). This suggests that Lagrange interpolation is a poor tool to extrapolate the function.

--Egm6341.s11.team3.JA 20:47, 16 February 2011 (UTC)

= 3.3.2 Runge-Phenomenon =

Given
Given a function

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

$$ f(x) = \frac{1}{1+4x^2}\quad $$ in interval $$ x \in [-5,5] \quad $$, $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 3.3.2)
 * }

Find
Find the error estimate for the function given above and plot, $$ L_3,8^L(x) \quad $$, for n=8 at equidistant nodes.

Solution
For $$ n = 8 \quad $$ nodes are:

$$x_{0} = -5, x_{1} = -3.75, x_{2} = -2.5, x_{3} = -1.25, x_{4} = 0, x_{5} = 1.25, x_{6} = 2.5, x_{7} = 3.75, x_{8} = 5 \quad $$.


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


 * $$\displaystyle
 * $$\displaystyle

f_{8}^L=\sum_{i=0}^{8}l_{i,8}(x)f(x_{i})=l_{0,8}(x)f(x_{0})+l_{1,8}(x)f(x_{1})+l_{2,8}(x)f(x_{2})+l_{3,8}(x)f(x_{3})+l_{4,8}(x)f(x_{4})+l_{5,8}(x)f(x_{5})+l_{6,8}(x)f(x_{6})+l_{7,8}(x)f(x_{7})+l_{8,8}(x)f(x_{8}) $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq.3.3.2.1)
 * }
 * }

where,


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

l_{3,8}=\prod_{j=0,j\neq 3}^{8}\frac{x-x_{j}}{x_{3}-x_{j}} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Using the same basic equations as in part 1 $$( Eq. 3.3.2 - 5 )  \quad $$ and relying on previous matlab code, we generated another matlab code



As you may see, high degree oscillations are apparent in this plot and oscillations do increase in frequency towards the endpoints of the function interval as well as increase in amplitude. This phenomenon (increase in oscillations towards the endpoints as number of members/nodes is increased) is known as Runge's phenomenon

Also note from the plot in 1st part and 2nd part(this part), that at whatever node we choose to plot $$ l_{i,n} \quad $$, for example $$ l_{3,8} \quad $$, as in this case, all the nodes except for $$ l_{3,8} \quad $$ node hit zero while the node $$ l_{3,8} \quad $$ hits value of $$ 1 \quad $$. The same was for $$ l_{2,4} \quad $$ node in 1st part of Problem 3.3.

$$ e_8^L(x) = -0.3786 \quad $$

$$ e_8^L(t) = 1.7840 \quad $$

--Egm6341.s11.team3.JA 20:48, 16 February 2011 (UTC)

= 3.4 Area of Biofolium =

Given
Cartesian coordinate

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

$$((x^2+y^2)^2)-4axy^2=0 \quad $$, $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right;">$$\displaystyle\!
 * }
 * }

when $$ a= 0.5 \quad $$

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

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

Polar Coordination

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

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

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

$$y=rsin(t) \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle\!
 * }

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

$$r=2cos(t){sin(t)}^{2} \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle\!
 * }

Find
1)Do literature search to find history and application of this classic curve

2)Find area in one leaf to $${10}^{-6}$$ accuracy

2.1) Trapozoial Rule

2.2) Simpson's Rule

2.3) Summation of Triangles

Solution
1) Refer the Web below 2) Area of Leaf

From Wolframalpha



Way to Calculation by Trapozoial Rule and Simpson's Rule

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

$${({x}^{2}+{y}^{2})}^{2}=2x{y}^{2} \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 3.4.1)
 * }

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

$$y=\sqrt{{-x}^{2}+\sqrt{{x}^{2}-2{x}^{3}}+x}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 3.4.2)
 * }

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

$$y=\sqrt{{-x}^{2}-\sqrt{{-x}^{2}(2x-1)}+x}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 3.4.3)
 * }

2.1)Trapozoial Rule

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

$${I}_{1}=\frac{b-a}{2}[f(a)+f(b)]$$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 3.4.4)
 * }

by (Eq. 3.4.2)

$$f(0)=0, f(0.5)=0.5\quad$$

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

$${I}_{1}=\frac{0.5-0}{2}[0+0.5]=0.125$$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 3.4.5)
 * }

by (Eq 3.4.3)

$$f(0)=0, f(0.5)=0.5\quad$$

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

$${I}_{1}=\frac{0.5-0}{2}[0+0.5]=0.125$$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 3.4.6)
 * }

Area of One Leaf by Trapozoial Rule $$0.125-0.125=0\quad$$

$$
 * <p style="text-align:right;"> $$ \displaystyle\!

2.2) Simpson's Rule

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

$${I}_{2}=\frac{h}{3}[{f}_{0}+4{f}_{1}+{f}_{2}]$$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 3.4.7)
 * }

$${x}_{0}= a = 0, {x}_{1}= \frac{{x}_{2}+{x}_{0}}{2}=0.25, {x}_{2}= b = 0.5\quad$$

$$h=\frac{b-a}{2}=0.25$$

by Eq(3.4.2)

$$f({x}_{0})= 0, f({x}_{1})= 0.603553, f({x}_{2})= 0.5 \quad$$

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

$${I}_{2}=\frac{0.25}{3}[0+4(0.603553)+0.5] = 0.242851 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 3.4.8)
 * }

by (Eq 3.4.3)

$$f({x}_{0})= 0, f({x}_{1})= 0.103553, f({x}_{2})= 0.5 \quad$$

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

$${I}_{2}=\frac{0.25}{3}[0+4(0.103553)+0.5] = 0.076184 \quad$$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 3.4.9)
 * }

Area of One Leaf by Simpson's Rule

$$ 0.242851-0.076184 = 0.166667 \quad$$ $$
 * <p style="text-align:right;"> $$ \displaystyle\!

2.3)Summation of Triangles



$${x}_{0}=0,{x}_{1}=0.125,{x}_{2}=0.25,{x}_{3}=0.375,{x}_{4}=0.5\quad$$

by (Eq 3.4.2)

$$f(0)=0, f(0.125)=0.466506, f(0.25)=0.603553, f(0.375)=0.649519, f(0.5)=0.500000 \quad$$

Length between two points

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

$$\sqrt{{({x}_{1}-{x}_{0})}^{2}+{({y}_{1}-{y}_{0})}^{2}}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 3.4.10)
 * }

$$\overline{ox_1}=0.482962$$

$$\overline{ox_2}=0.653281$$

$$\overline{ox_3}=0.749999$$

$$\overline{ox_4}=0.707106$$

$$\overline{{x}_{1}{x}_{2}}=0.185490$$

$$\overline{{x}_{2}{x}_{3}}=0.133183$$

$$\overline{{x}_{3}{x}_{4}}=0.194886$$

Heron's formula

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

$$ A=\sqrt{s(s-a)(s-b)(s-c)}$$
 * style="width:95%" |
 * style="width:95%" |

$$ s=\frac{a+b+c}{2}$$

$$ upperline area
 * <p style="text-align:right"> $$ \displaystyle (Eq. 3.4.11)
 * }

$$ \overline{o{x}_{1}{x}_{2}}= 0.020590$$

$$ \overline{o{x}_{2}{x}_{3}}= 0.031976 $$

$$ \overline{o{x}_{3}{x}_{4}}= 0.068629 $$

Summation of the Triangles of Upper Parts

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

$$Area = 0.020590 + 0.031976 + 0.068629=0.121195\quad$$
 * style="width:95%" |
 * style="width:95%" |

$$
 * <p style="text-align:right"> $$ \displaystyle (Eq. 3.4.12)
 * }

by (Eq 3.4.3)

$$f(0)=0, f(0.125)=0.033493, f(0.25)=0.0.03553, f(0.375)=0.216506, f(0.5)=0.500000\quad$$

Length between two points

by (Eq 3.4.10)

$$\overline{o{x}_{1}}=0.129409$$

$$\overline{o{x}_{2}}=0.270597$$

$$\overline{o{x}_{3}}=0.433012$$

$$\overline{o{x}_{4}}=0.707106$$

$$\overline{{x}_{1}{x}_{2}}=0.143294$$

$$\overline{{x}_{2}{x}_{3}}=0.168473$$

$$\overline{{x}_{3}{x}_{4}}=0.309828$$

by (Eq 3.4.11)

$$ \overline{o{x}_{1}{x}_{2}}= 0.002285$$

$$ \overline{o{x}_{2}{x}_{3}}= 0.007646 $$

$$ \overline{o{x}_{3}{x}_{4}}= 0.039623 $$

Summation of the Triangles of lower Parts

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

$$Area = 0.002285 + 0.007646 + 0.039623 = 0.049554\quad$$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 3.4.13)
 * }

Sum of (Eq 3.4.12) and (Eq 3.4.13)

Total Area by Sum of the Triangles

Area of Leaf = 0.121195 + 0.049554 = 0.170749 $$
 * <p style="text-align:right;"> $$ \displaystyle\!

Results

Area by Trapozoial Rule = 0

Area by Simpson's Rule = 0.166667

Area by Sum of Triangles = 0.170749

$$
 * <p style="text-align:right;"> $$ \displaystyle\!

– Hard to get exact area of leaf by Trapozoial Rule –

Egm6341.s11.team3.sungsikkim 03:39, 14 February 2011 (UTC) Egm6341.s11.team3.sungsikkim 07:07, 15 February 2011 (UTC)

= 3.5 Deriving the n+1th order derivative of En  =

Given
Given a function

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

$$
 * style="width:95%" |$$E_{n}=I(F)-I_{n}(F) \quad$$,
 * <p style="text-align:right"> $$ \displaystyle\!
 * <p style="text-align:right"> $$ \displaystyle\!
 * }

Find
Show that

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

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

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

$$ E_{n}^{n+1}\left ( x \right )=F^{n+1} \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 3.5.1)
 * }

We start with an general equation for error estimation

From we learn that

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

$$E_{n} = I-I_{n} \quad $$ $$ the first derivative of $$E_{n}\,$$ would yield us the following result for an interval (0,1) <span id="(1)">
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 3.5.2)
 * }
 * {| style="width:100%" border="0"

$$E_n^{1}=\int_{0}^{1}f^{1}\left ( x \right )dx-\int_{0}^{1}f_{n}^{1}\left ( x \right )dx\, \quad $$ $$ the second derivative of $$E_{n}\,$$ would yield us the following result for an interval (0,1) <span id="(1)">
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 3.5.3)
 * }
 * {| style="width:100%" border="0"

$$E_n^{2}=\int_{0}^{1}f^{2}\left ( x \right )dx-\int_{0}^{1}f_{n}^{2}\left ( x \right )dx\, \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 3.5.4)
 * }

the third derivative of $$E_{n}\,$$ would yield us the following result for an interval (0,1)

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

$$E_n^{3}=\int_{0}^{b}f^{3}\left ( x \right )dx-\int_{0}^{b}f_{n}^{3}\left ( x \right )dx\, \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 3.5.5)
 * }

the fourth derivative of $$E_{n}\,$$ would yield us the following result for an interval (0,1)

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

$$E_n^{4}=\int_{0}^{1}f^{4}\left ( x \right )dx-\int_{0}^{1}f_{n}^{4}\left ( x \right )dx\, \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 3.5.6)
 * }

the nth derivative of $$E_{n}\,$$ would yield us the following result for an interval (0,1)

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

$$E_n^{n}=\int_{0}^{1}f^{n}\left ( x \right )dx-\int_{0}^{1}f_{n}^{n}\left ( x \right )dx\, \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 3.5.7)
 * }

the n+1 th derivative of $$E_{n}\,$$ would yield us the following result for an interval (0,1)

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

$$E_n^{n+1}=\int_{0}^{1}f^{n+1}\left ( x \right )dx-\int_{0}^{1}f_{n}^{n+1}\left ( x \right )dx\, \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 3.5.8)
 * }

we have learned that

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


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

$$f_{n}\left ( x \right )=\sum _{i=0}^{n}l_{i,n}\left ( x \right ) f\left ( x_i \right)\,$$ $$
 * <p style="text-align:right"> $$ \displaystyle (Eq. 3.5.9)
 * }

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


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

where $$l_{i,n}\left ( x \right)= \prod_{j=0,i \neq j}^{n}\frac{x-x_{j}}{x_{i}-x_{j}}\,$$ $$
 * <p style="text-align:right"> $$ \displaystyle (Eq. 3.5.10)
 * }

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

$$ \begin{align} \\n=1, \quad & l_{i,1} (x)=\frac{x-x_{1}}{x_{0}-x_{1}}  l_{0}=c_1x+c_0 \\ \\n=2, \quad & l_{i,1} (x)=\frac{x-x_{1}}{x_{0}-x_{1}}\frac{x-x_{2}}{x_{0}-x_{2}}  l_{0}=c_2x^2 + c_1x + c_0
 * style="width:95%" |
 * style="width:95%" |

\\.\, \\.\, \\n=n, \quad & l_{i,n} ( x )=\frac{x-x_{1}}{x_{0}-x_{1}}\frac{x-x_{2}}{x_{0}-x_{2}}...\frac{x-x_{n-1}}{x_{0}-x_{n-1}}\frac{x-x_{n}}{x_{0}-x_{n}}  l_{0}= c_nx^n+c_{n-1}x^{n-1}++c_{n-2}x^{n-2}+......+c_{1}x + c_0 \end{align} $$ $$
 * <p style="text-align:right"> $$ \displaystyle\!
 * }

where $$ c_n, c_{n-1}, c_{n-2},.... c_2, c_1, c_0 \quad $$ are constants

since $$f\left( x_{i} \right)\,$$ is constant, hence forth  $$f_n\left( x \right)\,$$ is is of highest order of $$x^{n}\,$$

Consequently,

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

$$f_n^{n+1}\left ( x \right )= \frac{d}{dx^{n+1}}(c_nx^n+c_{n-1}x^{n-1}++c_{n-2}x^{n-2}+......+c_{1}x + c_0) = 0\,$$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 3.5.11)
 * }

Finally,

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

$$E_n^{n+1}=\int_{0}^{1}f^{n+1}\left ( x \right )dx-\int_{0}^{1}f_{n}^{n+1}\left ( x \right )dx\,$$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 3.5.12)
 * }

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

$$E_n^{n+1}=\int_{0}^{1}f^{n+1}\left ( x \right )dx-\int_{0}^{1} 0 dx\,$$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 3.5.13)
 * }

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

$$ E_n^{n+1}=\!F^{n+1}\left ( x \right )dx- 0 \,$$ $$ Therefore,
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 3.5.14)
 * }

$$ E_n^{n+1}=\!F^{n+1}\left ( x \right )dx- 0 \,$$

Egm6341.s11.team3.rakesh 20:32, 11 February 2011 (UTC)

= 3.6 Deriving the n+1th order derivative of Q n+1th order =

Find
Prove that $$q_{n+1}^{(n+1)}\left ( x \right )={(n+1)}!$$

Given
$$q_{n+1}\left ( x \right )=(x-x_{0})(x-x_{1})(x-x_{2})...(x-x_{n}) $$

Solution
We know from our lectures

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

$$q_{n+1}\left ( x \right )=(x-x_{0})(x-x_{1})(x-x_{2})...(x-x_{n})$$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 3.6.1)
 * }

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

$$n=0 \,$$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle\!
 * }

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

$$q_{1}\left ( x \right )=(x-x_{0})=x^1+ a_0 \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 3.6.2)
 * }

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

$$q_{1}^{(1)}\left ( x \right )=1$$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 3.6.3)
 * }

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

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

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

$$q_{2}\left ( x \right )=(x-x_{0})(x-x_{1})= x^2 + a_1x^1+ a_0 \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 3.6.4)
 * }

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

$$q_{2}^{(2)}\left ( x \right )=2\times 1=2!$$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 3.6.5)
 * }

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

$$n=2 \,$$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle\!
 * }

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

$$q_{3}\left ( x \right )=(x-x_{0})(x-x_{1})(x-x_{2})= x^3 + a_2x^2 + a_1x^1+ a_0 \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 3.6.6)
 * }

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

$$q_{3}^{(3)}\left ( x \right )=3\times 2\times 1=3!$$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 3.6.7)
 * }

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

$$n=3 \,$$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle\!
 * }

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

$$q_{4}\left ( x \right )=(x-x_{0})(x-x_{1})(x-x_{2})(x-x_{3})$$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 3.6.8)
 * }

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

$$q_4 = (x-x0)(x-x1)(x-x3)(x-x4) = x^4 + a_3x^3 + a_2x^2 + a_1x^1+ a_0 \quad $$ $$ <span id="(1)">
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 3.6.9)
 * }
 * {| style="width:100%" border="0"

$$q_{4}^{(4)}\left ( x \right )=4\times 3\times 2\times 1=4!$$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 3.6.10)
 * }

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

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

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

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

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

$$q_{n+1}\left ( x \right ):=(x-x_{0})(x-x_{1})(x-x_{2})(x-x_{3}...(x-x_{n}) = x^n + a_{n-1}x^{n-1} + ....+ a_2x^2 + a_1x^1+ a_0 \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 3.6.11)
 * }

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

$$q_{n+1}^{(n+1)}\left ( x \right )=n\times (n-1)\times (n-2)...\times 1=(n+1)!$$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 3.6.12)
 * }

where constants $$ a_0, a_1, a_2,....., a_n \quad $$ are combination of constants $$ x_0, x_1, x_2,....., x_n \quad $$

Therefore,

$$ P_{n+1}^{(n+1)}\left ( x \right )={(n+1)}!$$

Egm6341.s11.team3.rakesh 20:17, 11 February 2011 (UTC)

= 3.7 Integration Derivation for Error of Trap. rule =

Given
Give the Integration for Error of Trap. rule: <span id="(1)">
 * {| style="width:100%" border="0"

$$ \frac{M_2}{2!} \int_{a}^{b} \left | (x-a)(x-b) \right | dx $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 3.7.1)
 * }

Find
Derive the value of the integration in terms of $$ a \quad $$, $$ b \quad $$ and $$ M_2 \quad $$.

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

$$ \frac{M_2}{2!} \int_{a}^{b} \left | (x-a)(x-b) \right | dx $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle\!
 * }

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

$$ = \frac{M_2}{2!} \int_{a}^{b} (x-a)(b-x) dx $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle\!
 * }

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

$$ = \frac{M_2}{2} \int_{a}^{b} [-x^2+(a+b)x-ab] dx $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle\!
 * }

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

$$ = \frac{M_2}{2} [-\frac{1}{3}x^3+\frac{1}{2}(a+b)x^2-abx]\mid _{a}^{b}  $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle\!
 * }

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

$$ = \frac{M_2}{2} [-\frac{1}{3}(b^3-a^3)+\frac{1}{2}(a+b)(b^2-a^2)-ab(b-a)] $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle\!
 * }

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

$$ = \frac{M_2}{12} (b^3-3b^2a+3a^2b-a^3) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle\!
 * }

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

$$ = \frac{M_2}{12} (b-a)^3 $$ $$ Therefore,
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle\!
 * }

$$ \frac{M_2}{2!} \int_{a}^{b} \left | (x-a)(x-b) \right | dx = \frac {(b-a)^3}{12} M_2 $$

Egm6341.s11.team3.Xia 19:00, 11 February 2011 (UTC)

= 3.8 Integration Derivation for Error of Simple Simpson's rule =

Given
Give the Integration for Error of Simple Simpson's rule: <span id="(1)">
 * {| style="width:100%" border="0"

$$ \frac{M_3}{3!} \int_{a}^{b} \left | (x-a)(x-\frac{a+b}{2})(x-b) \right | dx = \frac {(b-a)^4}{192} M_3 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 3.8.1)
 * }

Find
Derive the value of the integration in terms of $$ a \quad $$, $$ b \quad $$ and $$ M_3 \quad $$.

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

$$ \frac{M_3}{3!} \int_{a}^{b} \left | (x-a)(x-\frac{a+b}{2})(x-b) \right | dx $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle\!
 * }

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

$$ = \frac{M_3}{3!} [\int_{a}^{\frac{a+b}{2}} (x-a)(x-\frac{a+b}{2})(x-b) dx + \int_{\frac{a+b}{2}}^{b} (x-a)(x-\frac{a+b}{2})(b-x) dx $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle\!
 * }

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

$$ = \frac{M_3}{3!} (\int_{a}^{\frac{a+b}{2}} - \int_{\frac{a+b}{2}}^{b}) [x^3-\frac{3}{2}(a+b)x^2+\frac{a^2+4ab+b^2}{2}x- \frac{a^2b+b^2a}{2}] dx $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle\!
 * }

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

$$ = \frac{M_3}{3!} [\frac{1}{4}x^4-\frac{1}{2}(a+b)x^3+\frac{a^2+4ab+b^2}{4}x^2- \frac{a^2b+b^2a}{2}x] (\mid_{a}^{\frac{a+b}{2}} + \mid_{b}^{\frac{a+b}{2}}) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle\!
 * }

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

$$ = \frac{2M_3}{6} [\frac{1}{64}(a+b)^4-\frac{1}{16}(a+b)(a+b)^3+\frac{a^2+4ab+b^2}{16}(a+b)^2- \frac{a^2b+b^2a}{4}(a+b)] $$ $$ - \frac{M_3}{6} [\frac{1}{4}a^4-\frac{1}{2}(a+b)a^3+\frac{a^2+4ab+b^2}{4}a^2- \frac{a^3b+b^2a^2}{2}] $$ $$ - \frac{M_3}{6} [\frac{1}{4}b^4-\frac{1}{2}(a+b)b^3+\frac{a^2+4ab+b^2}{4}b^2- \frac{a^2b^2+b^3a}{2}] $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle\!
 * }

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

$$ = \frac{M_3}{192} (a^4-4a^3b-10a^2b^2-4ab^3+b^4) - \frac{M_3}{6} (-\frac{a^2b^2}{2}) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle\!
 * }

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

$$ = \frac{M_3}{192} (a^4-4a^3b+6a^2b^2-4ab^3+b^4) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle\!
 * }

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

$$ = \frac{M_3}{192} (b-a)^4 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle\!
 * }

Therefore,

$$ \frac{M_3}{3!} \int_{a}^{b} \left | (x-a)(x-\frac{a+b}{2})(x-b) \right | dx = \frac {(b-a)^4}{192} M_3 $$

Egm6341.s11.team3.Xia 20:00, 11 February 2011 (UTC)

= References =

= Contributing Members =

--Egm6341.s11.team3.rakesh 15:18, 14 February 2011 (UTC)

--Egm6341.s11.team3.sungsikkim 17:12, 14 February 2011 (UTC)

--Egm6341.s11.team3.ren 14:22, 15 February 2011 (UTC)

--Egm6341.s11.team3.Xia 18:35, 15 February 2011 (UTC)

--Egm6341.s11.team3.elango 20:00, 15 January 2011 (UTC)

--Egm6341.s11.team3.JA 20:48, 16 February 2011 (UTC)

--Egm6341.s11.team3.russo 20:57, 16 February 2011 (UTC)