User:Egm6341.s10.team3.sa/Mtg11

Mtg 11: Thu, 21 Jan 10 [[media: Egm6341.s10.mtg11.djvu | Page 11-1]] 


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Find n at $$ \displaystyle \color{blue} \left|{f}_{n}^{T}(\frac{7\pi}{8}) - f(\frac{7\pi}{8})\right| \leq \left|{f}_{4}^{L}(\frac{7\pi}{8}) - f(\frac{7\pi}{8})\right| \leq \left| \frac{{q}_{5}(t)}{5!} \right| $$  for  $$ \displaystyle \color{blue} f(x) = sin(x) $$ and $$ \displaystyle \color{blue} x_{0} = 0, x_{1} = \frac{\pi}{4}, x_{2} = \frac{\pi}{2}, x_{3} = \frac{3\pi}{4}, x_{4} = 1 $$. 

Here T stands for Taylor series approximation and L stands for Lagrange interpolation. 


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[[media: Egm6341.s10.mtg11.djvu | Page 11-2]]   LAGRANGE INTERPOLATION ERROR We discussed about Lagrange interpolation error in the previous lecture. Recall the expression for G that we studied in [[media: Egm6341.s10.mtg10.djvu | Page 10-3]] Now we give a geometric interpretation for G(x)



Geometric interpretation of $$\displaystyle G(x)$$: As we know, the Lagrange interpolation function has to pass through all points $$\displaystyle x_0 \;x_1\; ... \; x_n$$ Here we are interested in the error $$\displaystyle E(x)\; = \;f-f_n(x) $$ which is the difference between the actual function and the Lagrange interpolation function at the point x.

Also in the figure is the Lagrange basis fuction This should be equal to 1 at $$\displaystyle x_i $$ and zero at all other points. And $$\displaystyle q_{n+1} $$ would pass through all points $$\displaystyle x_i\; through\; x_n $$

[[media: Egm6341.s10.mtg11.djvu | Page 11-3]]  

Let's now scale the error at t and x with the values of $$\displaystyle q_{n+1} $$ at t and x respectively. i.e, Scale, $$\displaystyle \frac{E(x)}{q_{n+1}(x)} \; and\; \frac{E(t)}{q_{n+1}(t)} $$

Now look at the difference between the scaled error at t and x. i.e, Define, $$\displaystyle d(x) \; := \; \frac{E(x)}{q_{n+1}(x)} \; - \; \frac{E(t)}{q_{n+1}(t)} $$ Now, multiply both sides by q_n+1(x) - And this product is called G(x). $$\displaystyle \underbrace{d(x) q_{n+1}(x)}_{G(x)} \; = \; E(x) \; - \; \frac{E(t)}{q_{n+1}(t)} q_{n+1}(x) $$

Let us talk about the particular case of derivative mean value theorem, i.e, the Rolle's theorem. It states that, For a continuous function f, If $$\displaystyle f(a) = f(b) = 0 $$ then there is atleast 1 point between a and b such that $$\displaystyle f'(\xi) = 0 $$ where, a and b are the endpoints and $$\displaystyle \xi \in (a,b).$$

In the following figure, f(a)=f(b)=0 and we have 6 points between them where,$$\displaystyle f'(\xi) = 0 $$ Image : Mtg11-Fig2.svg

Here, $$\displaystyle G^{n+1}(.)$$ is continuous since $$\displaystyle E^{n+1}(.)$$ and $$\displaystyle q_{n+1} ^ {n+1} (.)$$ are continuous.

It is important to note the following: $$\displaystyle \underbrace{G(x_i)}_{0} \; = \; \underbrace{E(x_i)}_{0} \; - \; \frac {E(t)}{q_{n+1}(t)} \underbrace{q_{n+1}(x_i)}_{0} $$