User:Egm6341.s10.team3.heejun/Meeting 6 transcript

 MTG 6: Tue. 12 Jan 10

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$$ \displaystyle Contruct\,\, Taylor\,\, Series\,\, of\,\, f(\cdot)\,\, around\,\, x_{0}= \frac {\pi}{4}\,\, for\,\, n=0,1,...,10. $$ $$ Plot\,\, these\,\, series\,\, (for\,\, each\,\, n). $$ $$ \displaystyle R(x):\,\, Find\,\, (estimate\,\, the\,\, max)\,\, of\,\, R(x)\,\, at\,\, x= \frac {\pi}{2}. $$

$$ \begin{align} \displaystyle R(x)&= \frac {1}{n!} \int_{x_{0}}^{x} (x-t)^{n} f^{(n+1)}(t)dt\\ &=\frac {1}{n!} f^{(n+1)}(\xi) \int_{x_{0}}^{x} (x-t)^{n} dt \\ &=\frac {(x-x_{0})^{(n+1)}}{(n+1)!} f^{(n+1)} (\xi),\,\,\,\, \xi \in [x_{0}, x] \\\end{align}\ $$

$$ \displaystyle \left| R(x) \right| \leq \frac {(x-x_{0})^{(n+1)}}{(n+1)!},\,\,\,\, max \left| f^{(n+1)}(t) \right|\,\,\,\, \underbrace{t \in [x_{0}, x]}_{\color{blue}{\leq 1}}\,\,\,\, $$
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Note(Mini lecture plan): Motivation for proof of Taylor series expansion (similar technic will be used)

* Higher order error analysis of trapezoidal rule (not in our text book, A.)

* Richardson extrapezoidal rule

* Clenshaw-Curtis  quadrature

* Chebyshev polynomial (orthogonal)

* Recent development using Chebyshev polynomial to solve  L2-ODE_VC (Linear 2nd order Ordin. Diff. Eq. Varying Coefficient) $$ \displaystyle \Rightarrow $$ Eventually, Recent develpoment means to basically combine the symbolic and numeric methods. (Symbolic + Numeric)

What is the meaning of Quadrature?

Note: Quadrature = numerical integration

In Greek: means to measure area

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If you want to measure the below area, how to do?



It's another name is Cubature


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 * $$ \displaystyle Area\,\, = \sum Quad $$
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 * $$ \displaystyle Volume\,\, = \sum Cubes $$
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Let's go back to the Numerical Integration using Taylor series Recall, Eq. (1)  p.2-2


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 * $$ \displaystyle f(x)=\frac{e^{x}-1}{x} = \sum_{j=1}^{\infty}\frac{x^{j-1}}{j!} $$
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 * $$ \displaystyle I_{n}= \int_{0}^{1} f_{n}(x) dx = \int_{0}^{1} \sum_{j=1}^{n} \frac{x^{j-1}}{j!}dx = \sum_{j=1}^{n} \frac{1}{j!j} $$
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 * $$ \displaystyle f(x) - f_{n}(x) = R_{n}(x) = \frac{(x-0)^n}{(n+1)!}exp[\xi]\,\,\,\, \xi \in [0,x] $$
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 * $$ \displaystyle \xi\,\, is\,\, a\,\, function\,\, of\,\, x\,\, \Rightarrow \xi_{x} = \xi(x) $$
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&=\int_{0}^{1}[f(x)-f_{n}(x)]dx \\ &=\int_{0}^{1} \underbrace{\frac{x^{n}}{(n+1)!}}_{\color{blue}{w(x)}} \underbrace{exp[\xi(x)]}_{\color{blue}{g(x)}} dx \\ & \displaystyle \color{blue}{By\,\,\,\, IMVT} \\ &= \underbrace{g(x)}_{\color{blue}{exp[\xi(\alpha)]}} \underbrace{\int_{0}^{1}w(x)dx}_{\color{blue}{\frac{1}{(n+1)!(n+1)}}}\,\,\,\, \alpha \in [0,1] \,\,\,\, \xi \in [0,1] \\\end{align}\ $$
 * $$ \begin{align} \displaystyle \underbrace{E_{n}}_{\color{blue}{Error\, of\, nth\, order}} &:= I - I_{n} \\
 * $$ \begin{align} \displaystyle \underbrace{E_{n}}_{\color{blue}{Error\, of\, nth\, order}} &:= I - I_{n} \\
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$$ \displaystyle \color{blue}{The\,\, grape\,\, of\,\, exp[\xi(\alpha)]} $$



$$ \displaystyle min\,g(\alpha) = 1\,\,\,\, \alpha \in [0,1] $$ $$ \displaystyle max\,g(\alpha) = e\,\,\,\, \alpha \in [0,1] $$

Thus, $$ \displaystyle E_{n} $$ is placed in between $$ \displaystyle \frac {min\,g(x)}{(n+1)!(n+1)}\,\, and\,\, \frac {max\, g(x)}{(n+1)!(n+1)} $$ $$ \displaystyle \frac{1}{(n+1)!(n+1)} \leq E_{n} \leq \frac{e}{(n+1)!(n+1)} $$

6-5

If you compute, then $$ \displaystyle E_{n} $$ is placed in $$  \displaystyle I_{6} = 1.3178...\,\,\,\, \color{blue}{A. p.250} $$ $$ \displaystyle 2.83\times10^{-5} \leq E_{6}=I-I_{6} \leq 7.70\times10^{-5} $$


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$$ \displaystyle Use\,\, 3\,\, methods\,\, to\,\, find\,\, I_{n}: $$ $$ \displaystyle 1)\,\, Taylor\,\, series\,\, expansion\,\, (f_{n}) $$ $$ \displaystyle 2)\,\, Composite\,\, Trapzoidal\,\, rule\,\, $$ $$ \displaystyle 3)\,\, Composite\,\, Simpson\,\, rule\,\, $$ $$ \displaystyle n=2,4,8,...\,\, until\,\, error\,\, of\,\, order\,\, 10^{-6}\,\,\,\, $$
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