User:Egm6341.s10.team3.ynahn81/HW2-13

= (13) Proof of $$\displaystyle G^{(2)}(0)=0 $$ in Simpson's Rule = Ref: Lecture notes [[media:Egm6341.s10.mtg15.pdf|p.15-2]]

Problem Statement
For the Simpson's rule, prove that $$\displaystyle G^{(2)}(0)=0 $$.

Solution
From [[media:Egm6341.s10.mtg14.pdf|p.14-3]] of lecture note, $$\displaystyle G(t) $$ can be expressed by:


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

G(t) = e(t)-t^{5}e(1) $$ $$  
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1)
 * }
 * }

where
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e(t)=\alpha(t)-\alpha_{2}(t)=\int_{-t}^{t}F(s)ds - \frac{t}{3}[F(-t)+4F(0)+F(t)] $$ $$  
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 2)
 * }
 * }

From [[media:Egm6341.s10.mtg15.pdf|p.15-1]] of lecture note, we already get the first derivatives of $$\displaystyle \alpha(x) $$ and $$\displaystyle \alpha_{2}(x) $$:


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\alpha^{(1)}(t)=F(-t)+F(t) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 3)
 * }
 * }


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\alpha_{2}^{(1)}(t)=\frac{1}{3}[F(-t)+4F(0)+F(t)]+\frac{t}{3}[F^{(1)}(-t)+F^{(1)}(t)] $$ $$  
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 4)
 * }
 * }

From the Eq.(3) and Eq.(4), we can obtain the second derivatives of $$\displaystyle \alpha(x) $$ and $$\displaystyle \alpha_{2}(x) $$:


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\alpha^{(2)}(t)=F^{(1)}(-t)+F^{(1)}(t) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 5)
 * }
 * }


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\alpha_{2}^{(2)}(t)=\frac{2}{3}[F^{(1)}(-t)+F^{(1)}(t)]+\frac{t}{3}[F^{(2)}(-t)+F^{(2)}(t)] $$ $$  
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 6)
 * }
 * }

From the Eq.(1), the second derivative of $$\displaystyle G(t) $$ is


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G^{(2)}(t) = e^{(2)}(t)-20t^{3}e(1) = \alpha^{(2)}(t)-\alpha_{2}^{(2)}(t)-20t^{3}e(1) $$ $$  
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 7)
 * }
 * }

Plug Eq.(5) and Eq.(6) into Eq.(7).


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G^{(2)}(t) = \frac{1}{3}[F^{(1)}(-t)+F^{(1)}(t)]-\frac{t}{3}[F^{(2)}(-t)+F^{(2)}(t)]-20t^{3}e(1) $$ $$ <br\> <br\>
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 8)
 * }
 * }

Now, let's explicitly express $$\displaystyle F^{(1)}(t) $$ and $$\displaystyle F^{(1)}(t) $$ as follows.


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F^{(1)}(t) = \frac{dF(t)}{dt} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 9)
 * }
 * }

For the expression of the derivation of $$\displaystyle F^{(1)}(-t) $$, let $$\displaystyle -t=\tau $$ and use chain rule.


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F^{(1)}(-t) = F^{(1)}(\tau)=\frac{dF(\tau)}{dt}=\frac{d\tau}{dt}\frac{dF(\tau)}{d\tau}=(-1)\frac{dF(\tau)}{d\tau} $$ $$ <br\> <br\>
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 10)
 * }
 * }

Now, Plug Eq.(9) and Eq.(10) into Eq.(8). Then,


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G^{(2)}(t) = \frac{1}{3}[(-1)\frac{dF(\tau)}{d\tau}+\frac{dF(t)}{dt}]-\frac{t}{3}[F^{(2)}(-t)+F^{(2)}(t)]-20t^{3}e(1) $$ $$ <br\> <br\>
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 11)
 * }
 * }

To obtain $$\displaystyle G^{(2)}(0) $$, note that $$\displaystyle t=0 \rightarrow \tau=0 $$ and the second and the third terms of RHS in Eq.(11) are going to be zero when $$\displaystyle t=0 $$. Then,


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G^{(2)}(0) = \frac{1}{3}[(-1)\left(\frac{dF(\tau)}{d\tau}\right)_{\tau=0}+\left(\frac{dF(t)}{dt}\right)_{t=0}] - 0 - 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 12)
 * }
 * }


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\Rightarrow G^{(2)}(0) = \frac{1}{3}[(-1)F^{(1)}(0)+F^{(1)}(0)] $$ $$ <br\> <br\>
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 13)
 * }
 * }

Therefore,


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$$\displaystyle G^{(2)}(0) = 0 $$ $$
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * <p style="text-align:right;">$$\displaystyle (Eq. 14)
 * }
 * }