User:Egm6341.s10.team3.sa/Mtg3

 Mtg 3: Thu, 07 Jan 10 [[media: Egm6341.s10.mtg3.djvu | Page 3-1]]  NORM  Norm is a measure of the magnitude of a vector or a function.

We are already familiar with the 2-norm denoted by $$\displaystyle ||. ||_2 $$ The norm that is more relevant to our discussion is the infinity norm denoted by $$\displaystyle ||. ||_\infty $$

Let us denote the vector v as $$ \displaystyle \mathbf v $$, and let $$ \mathbf v \in \mathbb {R} ^n$$ Generally a tensor is represented as a matrix. A vector is a column matrix.

Based on the coordinate system used, there can be different representations for a given vector. So let us define a orthonormal basis ,namely $$\displaystyle \mathbf {e_i}$$ i.e., {$$\displaystyle \mathbf e_i $$} = { $$\displaystyle e_1,e_2,.....,e_n $$ } This is a family of vectors whose magnitude is one and these are orthogonal to one another $$\displaystyle \mathbf e_i \in \mathbb R $$ And, we write, $$\displaystyle \mathbf v = v_i \cdot e_i $$.

[[media: Egm6341.s10.mtg3.djvu | Page 3-2]] The 2-norm of a vector is defined as, $$\displaystyle || \mathbf v||_2 = {\Bigg[ \sum_{i=1}^{n} {v_i}^2 \Bigg]}^{1/2} $$ $$\displaystyle = {(v_i \cdot v_i)}^{1/2} $$ 

The infinity norm of a vector v is the absolute maximum of among all the components of $$\displaystyle v $$. $$\displaystyle || \mathbf v||_\infty $$ = max $$\displaystyle |v_i|$$ Based on this definition, if $$\displaystyle || \mathbf v||_ \infty \rightarrow 0, \;\; v_i \rightarrow 0 \;\;\forall\;i=1,2...,n.$$ i.e., If the largest component of a vector tends to 0, then all the components of the vector are zero.

Now, let us generalize the norm to functions The dot product of two functions f and g is given by, $$ \displaystyle (f,g) =  = \int_{a}^{b} f(x)g(x) \,dx $$

As seen above, the 2-morm of a vector is obtained by multiply the vector with itself and then taking the square root. Applying the same logic, to functions, the 2-norm of the function f is:

$$\displaystyle ||f||_2 = {(f \cdot f)}^{1/2} $$ Similarly, $$\displaystyle ||f||_\infty $$ = max $$\displaystyle |f(x)|$$

[[media: Egm6341.s10.mtg3.djvu | Page 3-3]] Now let us apply the concept of infinity norm to the actual function f and the approximated function $$\displaystyle f_n $$ By definition, $$\displaystyle ||f - f_n|\_\infty $$ = $$\displaystyle max |f(x)-f_n(x)| $$ Now if $$\displaystyle max |f(x)-f_n(x)| \rightarrow 0$$, then our approximation of the function $$ f_n $$ will be very close to the function f, which is precisely what we need.<br\><br\>

Note: $$\displaystyle max |f(x)-f_n(x)| \neq \underbrace{max|f(x)|}_{f(x^*)}$$ $$\displaystyle - \underbrace{max|f_n(x)|}_{f(x^{**})} $$


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HW: Plot f(x)= sin(x) and g(x) = - cos(x) $$\displaystyle\color{blue} {\in} [0,{\pi}] $$ and also find $$\displaystyle\color{blue} ||f(x)||_{\infty}, ||g(x)||_{\infty} and ||f(x)-g(x)||_{\infty}$$
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<p style="text-align:left;">[[media: Egm6341.s10.mtg3.djvu | Page 3-4]]

INTEGRAL MEAN VALUE THEOREM(IMVT) - Continued from Pg 2-4 From Eqn(3) on Pg 2-4: <br\><br\> Dividing the entire equation by b-a,we get, $$ \displaystyle m \;\leq\; \underbrace {\frac{1}{b-a} \int_{a}^{b}f(x)\,dx}_{\zeta}\;\leq\; M $$<br\><br\>

Then, by Intermediate Mean value theorem,there exists $$ \displaystyle \xi \in [a,b]$$ such that $$ \displaystyle f(\xi) = \zeta$$<br\><br\>

$$ \displaystyle \Rightarrow f(\xi) = \frac{1}{b-a} \int_{a}^{b}f(x)\,dx $$<br\> This is the proof for the special case of IMVT where the weighting function,w(x)=1.