User:Egm6341.s10.team4.anandankala/HW1-11

=Problem 11=

Show That

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 * $$P_2(x_j)= \displaystyle\sum_{i=0}^{2} \underbrace{l_i(x_j)}_{\delta_{ij}}f(x_i)= f(x_j)$$
 * }.
 * }.

Solution

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 * $$\underset{i=0}{l_0}= \displaystyle\prod_{\underset{j\neq i}{j=0}}^{2}\frac{x-x_j}{x_i-x_j}= \frac{(x-x_1)(x-x_2)}{(x_0-x_1)(x_0-x_2)}$$
 * }.
 * }.


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 * $$\underset{i=1}{l_1}= \displaystyle\prod_{\underset{j\neq i}{j=0}}^{2}\frac{x-x_j}{x_i-x_j}= \frac{(x-x_0)(x-x_2)}{(x_1-x_0)(x_1-x_2)}$$
 * }.
 * }.


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 * $$\underset{i=2}{l_2}= \displaystyle\prod_{\underset{j\neq i}{j=0}}^{2}\frac{x-x_j}{x_i-x_j}= \frac{(x-x_0)(x-x_1)}{(x_2-x_0)(x_2-x_1)}$$
 * }.
 * }.

@x=x0
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 * $$l_0= \frac{(x_0-x_1)(x_0-x_2)}{(x_0-x_1)(x_0-x_2)}= 1$$
 * }.
 * }.


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 * $$l_1= \frac{ \cancelto{0}{(x_0-x_0)}(x_0-x_2)}{(x_1-x_0)(x_1-x_2)}= 0$$
 * }.
 * }.


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 * $$l_2= \frac{ \cancelto{0}{(x_0-x_0)}(x_0-x_1)}{(x_2-x_0)(x_2-x_1)}= 0$$
 * }.
 * }.

Similarly @x=x1


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 * $$l_0= \frac{ \cancelto{0}{(x_1-x_1)}(x_1-x_2)}{(x_0-x_1)(x_0-x_2)}= 0$$
 * }.
 * }.


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 * $$l_1= \frac{(x_1-x_0)(x_1-x_2)}{(x_1-x_0)(x_1-x_2)}= 1$$
 * }.
 * }.


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 * $$l_2= \frac{(x_1-x_0) \cancelto{0}{(x_1-x_1)}}{(x_2-x_0)(x_2-x_1)}= 0$$
 * }.
 * }.

Similarly @x=x2


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 * $$l_0= \frac{(x_2-x_1) \cancelto{0}{(x_2-x_2)}}{(x_0-x_1)(x_0-x_2)}= 0$$
 * }.
 * }.


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 * $$l_1= \frac{(x_2-x_0) \cancelto{0}{(x_2-x_2)}}{(x_1-x_0)(x_1-x_2)}= 0$$
 * }.
 * }.


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 * $$l_2= \frac{(x_2-x_0)(x_2-x_1)}{(x_2-x_0)(x_2-x_1)}= 1$$
 * }.
 * }.


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 * $$ p_2 (j=0)_{x=x_0}= l_0.f(x_0)+\cancelto{0}{l_1}.f(x_1)+\cancelto{0}{l_2}.f(x_2)= 1.f(x_0)= f(x_j)$$
 * }.
 * }.


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 * $$ p_2 (j=1)_{x=x_1}= \cancelto{0}{l_0}.f(x_0)+l_1.f(x_1)+\cancelto{0}{l_2}.f(x_2)= 1.f(x_1)= f(x_j)$$
 * }.
 * }.


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 * $$ p_2 (j=2)_{x=x_2}= \cancelto{0}{l_0}.f(x_0)+\cancelto{0}{l_1}.f(x_1)+l_2.f(x_2)= 1.f(x_2)= f(x_j)$$
 * }.
 * }.

Hence proved.