User:Egm6341.s10.team3.sa/Mtg1

Mtg 1: Tue, 05 Jan 10  [[media: Egm6341.s10.mtg1.djvu | Page 1-1]]

NUMERICAL INTEGRATION   Define a function f such that, $$\displaystyle f: \underbrace {[a,b]}_{domain} \Rightarrow \underbrace \mathbb {R}_{Range}$$ where, $$ \displaystyle \mathbb R $$ is the set of real numbers.  

Now, define I, such that, $$\displaystyle I:= \;\; \int_{a}^{b}f(x) \,dx$$ $$\displaystyle (Eq. 1) $$  

Approximate f(.) by f_n(.) i.e, $$\displaystyle f \approx f_n$$ $$\displaystyle (Eq. 2) $$ Generally f_n(.) is a polynomial.  

Now, $$\displaystyle I \approx I_n := \; \int_{a}^{b}f_n(x) \,dx $$ $$\displaystyle (Eq. 3) $$  

We need convergence of $$\displaystyle f to f_n $$ so that the approximation is close to the actual function f $$\displaystyle || f - f_n ||_{\infty} \Rightarrow 0 \; as \; n \Rightarrow \infty $$ $$\displaystyle (Eq. 4) $$ <br\> <br\>

Note: $$\displaystyle || g ||_{\infty} = max \,|g(x)| $$ is defined as the infinity norm of the function g(x).

<p style="text-align:left;">[[media: Egm6341.s10.mtg1.djvu | Page 1-2]] Error of the numerical integration of f is defined by, $$\displaystyle E_n := \;\; I-I_n = \int_{a}^{b}f(x) \,dx - \int_{a}^{b}f_n(x) \,dx$$ <p style="text-align:right;">$$\displaystyle (Eq. 1) $$ <br\><br\> $$\displaystyle = \int_{a}^{b}(f-f_n(x)) \,dx$$ <br\><br\>

Depending on the function, the error $$E_n$$ can be positive, negative or both(which is mostly the case). Hence we need an upper bound for the error.

$$\displaystyle |E_n| \leq \int_{a}^{b}|(f-f_n(x))| \,dx$$ $$\displaystyle \leq (b-a)|| f - f_n ||_{\infty} $$ <p style="text-align:right;">$$\displaystyle (Eq. 2) $$ <br\><br\>

To understand the above equation, let's look into the following figures. <br\> Fig 1(a) : g(x) vs x Fig 1(b) : |g(x)| vs x

Note: $$\displaystyle \int_{a}^{b} g(x)\,dx  \leq  \int_{a}^{b}|g(x)| \,dx $$ <p style="text-align:right;">$$\displaystyle (Eq. 3) $$ <p style="text-align:left;">[[media: Egm6341.s10.mtg1.djvu | Page 1-3]]

Since, $$\displaystyle || g ||_{\infty} = max \,|g(x)| $$ <br\> it is clear that $$\displaystyle |g(x)| \leq || g ||_{\infty} $$.<br\>

And therefore, $$\displaystyle \int_{a}^{b}|g(x)| \,dx \leq \int_{a}^{b} || g ||_{\infty} \,dx $$ <br\> $$\displaystyle = || g ||_{\infty} \int_{a}^{b} \,dx $$ $$\displaystyle = (b-a) || g ||_{\infty}$$

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Applying the same logic to the function $$f_n$$, we will end up getting the Eqn (2) on Pg 1-2.