User:Egm6341.s10.team3.sa/Mtg21

Mtg 21: Tue, 16 Feb 10

 Typeset of transparencies, not lecture transcript. Subramanian Annamalai 13:54, 12 August 2010 (UTC)

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Continued from [[media: Egm6341.s10.mtg20.djvu | Page 20-3]]

$$\displaystyle E_n^1 = \sum_{k=0}^{n-1} \Bigg[\int_{x_k}^{x_{k+1}} f(x)\,dx - \frac{h}{2} \bigg[{f(x_k)+f(x_{k+1})} \bigg] \Bigg] $$ $$\displaystyle (Eq. 1) $$

$$\displaystyle x_k = a+ kh \;\;\; h=(b-a)/n $$ $$\displaystyle (Eq. 2) $$

Transform the integral $$\displaystyle \big[x_k,x_{k+1} \big]\; to\; [-1,1].$$

$$\displaystyle x(t) := t\frac{h}{2} + \frac{x_k + x_{k+1}}{2} \;\; ,t \in [-1,1].$$ $$\displaystyle (Eq. 3) $$

$$\displaystyle x(-1) = x_k \;\;\; x(0) = \frac{x_k+x_{k+1}}{2}\;\;\; x(1)=x_{k+1} $$ $$\displaystyle (Eq. 4) $$

$$\displaystyle \therefore E_n^1 = \frac{h}{2} \sum_{k=0}^{n-1} \Bigg[\int_{-1}^{1} g_k(t)\,dt - \bigg[g_k(-1)+g_k(1) \bigg] \Bigg] $$ $$\displaystyle (Eq. 5) $$

[[media: Egm6341.s10.mtg21.djvu | Page 21-2]] 

$$\displaystyle g_k(t) = f(x(t)) \;\; x \in [x_k,x_{k+1}] $$ $$\displaystyle (Eq. 1) $$

Step 1: $$ \displaystyle \int_{-1}^{1}(-t)g^{(1)}(t)\,dt = \int_{-1}^{1}(-t)g(t)\,dt - [g(-1) + g(1)] =: E $$ $$\displaystyle (Eq. 2) $$


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HW:
 * style="width:100%; padding:10px; border:2px solid #8888aa" |
 * style="width:100%; padding:10px; border:2px solid #8888aa" |

In the proof of the Higher order error of the Trapezoidal rule, show that,
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\int_{-1}^{1}(-t)g^{(1)}(t)\,dt = \int_{-1}^{1}(-t)g(t)\,dt - [g(-1) + g(1)] $$
 * $$ \displaystyle \color{blue}
 * $$ \displaystyle \color{blue}
 * }


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Step 2a: $$ \displaystyle \underbrace{g_k^{(i)}}_{\frac{d^{(i)}g(t)}{dt^{(i)}}} = \bigg(\frac{h}{2}\bigg) ^{(i)} \cdot f^{(i)}(x(t)) \,\,\,\,\,\,\,\, \in [x_k, x_{k+1}] $$ $$\displaystyle (Eq. 3) $$


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HW:
 * style="width:100%; padding:10px; border:2px solid #8888aa" |
 * style="width:100%; padding:10px; border:2px solid #8888aa" |

In the proof of the Higher order error of the Trapezoidal rule, show that,
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g_k^{(i)} = \bigg(\frac{h}{2}\bigg) ^{(i)} \cdot f^{(i)}(x(t)) \,\,\,\,\,\,\,\, \in [x_k, x_{k+1}] $$
 * $$ \displaystyle \color{blue}
 * $$ \displaystyle \color{blue}
 * }


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Note: To obtain higher powers of h,i.e, higher order derivatives of $$\displaystyle g_k(t) $$, we carry out successive integration by parts. 

$$\displaystyle \therefore E:= \int_{-1}^{1}(-t)g^{(1)}(t)\, dt = [p_2(t)g^{(1)}(t)]_{-1}^{1} - \int_{-1}^{1} p_2(t) g^{(2)}(t) \,dt  $$ $$\displaystyle (Eq. 4) $$


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HW:
 * style="width:100%; padding:10px; border:2px solid #8888aa" |
 * style="width:100%; padding:10px; border:2px solid #8888aa" |

Prove that:

$$\displaystyle \color{blue} \therefore E:= \int_{-1}^{1}(-t)g^{(1)}(t)\, dt = [p_2(t)g^{(1)}(t)]_{-1}^{1} - \int_{-1}^{1} p_2(t) g^{(2)}(t) \,dt  $$

where

$$\displaystyle \color{blue} p_1(t)= -t $$ and $$\displaystyle \color{blue} p_2(t)= \int_{-1}^{1} p_1(t) \,dt $$


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[[media: Egm6341.s10.mtg21.djvu | Page 21-3]]  Note:

$$\displaystyle p_2(t) = \int p_1(t) \,dt = -\frac{t^2}{2} + \alpha $$ $$\displaystyle = c_1 \frac{t^2}{2}+c_3 $$ $$\displaystyle (Eq. 1) $$ and <p style="text-align:right;">$$\displaystyle (Eq. 2) $$ $$\displaystyle c_2 = 0 $$

Here, $$\displaystyle \alpha $$ is the integration constant used to cancel the odd powers of h.

Step 2b: Integration by parts $$\displaystyle E= \bigg[ p_2(t)g^{(1)}(t) \bigg]_{-1}^{1} - \underbrace{\int_{-1}^{1}p_3(t)g^{(2)}(t) \,dt}_{A} + \int_{1}^{1}p_3(t)g^{(3)}(t) \,dt $$ <p style="text-align:right;">$$\displaystyle (Eq. 3) $$

Note: $$\displaystyle p_3(t) = \int p_2(t) \,dt = -\frac{t^3}{6} + \alpha t +\beta $$ Here $$\displaystyle \beta $$ is the new integration constant. <br\>

Now we have to find $$\displaystyle \alpha \;and\; \beta $$ to cancel the terms with $$\displaystyle g^{(2)} $$ (even order derivatives of g i.e, the odd powers of h).