User:Egm6341.s10.team3.sa/Mtg5

Mtg 5, 12 Jan 10

[[media: Egm6341.s10.mtg5.djvu |Page 5-1]]


 * {| style="width:100%" border="0" align="left"

HW:
 * style="width:100%; padding:10px; border:2px solid #8888aa" |
 * style="width:100%; padding:10px; border:2px solid #8888aa" |

Prove that: (i) $$ \displaystyle \color{blue} \int_{a}^{b}w(x)\cdot f(x)dx = f(\xi )\int_{a}^{b}w(x)dx $$  for $$ \displaystyle \color{blue} w(x)\geq 0 $$

(ii) Another version of IMVT : Also prove IMVT for $$ \displaystyle \color{blue} w(x)<0$$ for all $$ \displaystyle \color{blue} x\in [a,b] $$


 * style = |
 * }
 * }


 * {| style="width:100%" border="0" align="left"

HW:
 * style="width:100%; padding:10px; border:2px solid #8888aa" |
 * style="width:100%; padding:10px; border:2px solid #8888aa" |

We know from IMVT in [[media:Egm6341.s10.mtg2.pdf|Lecture p.2-3]] that,


 * {| style="width:100%" border="0" align="left"

with $$ \displaystyle \color{blue} w(x)\geqslant 0 \quad \forall \ x \in [a,b] $$
 * $$\displaystyle \color{blue} \int_{a}^{b} w(x)f(x)\ dt = f(\xi)\int_{a}^{b} w(x)\ dt $$
 * }
 * }

Use IMVT to show that (5) in [[media:Egm6341.s10.mtg2.pdf|Lecture p.2-2]] is equal to (1) in [[media:Egm6341.s10.mtg2.pdf|Lecture p.2-3]] i.e,


 * {| style="width:100%" border="0" align="left"

= \frac{1}{n!} \frac{(x-x_0)^{n+1}}{(n+1)!}f^{(n+1)}(\xi) $$ for $$ \displaystyle \color{blue} \quad \xi \in [x_0,x]$$
 * $$\displaystyle \color{blue} R_{n+1}(x) = \frac{1}{n!} \int_{x_0}^{x} (x-t)^n f^{(n+1)}(t)\ dt
 * }


 * style = |
 * }
 * }

[[media: Egm6341.s10.mtg5.djvu | Page 5-2]] PROOF OF TAYLOR SERIES This same technique will be used in error analyses such as that of the Trapezoidal rule

Consider the function f(x). It can be written as, $$ \displaystyle f(x) = f(x_0) + \int_{x_0}^{x} f^{(1)}(t) \,dt $$ $$\displaystyle (Eq. 1) $$ Note: $$ \displaystyle f^{(1)}(t) = \frac {df}{dt}$$ Clearly, $$ \displaystyle \int_{x_0}^{x} f^{(1)}(t) \,dt = [f(t)]_{x_0}^{x} $$ $$\displaystyle (Eq. 2) $$

Now consider the integral in Eq.(1) Using integration by parts, $$ \displaystyle \int_{x_0}^{x} f^{(1)}(t) \,dt = \int_{x_0}^{x} \underbrace{1}_{u^\prime} \underbrace{f^{(1)}(t)}_{v}\, dt  = [ uv ]_{x_0}^x - \int_{t=x_0}^{x} uv^\prime\, dt $$ $$ \displaystyle = [{t}{ f^{(1)}}(t)]_{t=x_0}^{t=x} + \underbrace{\int_{x_0}^{x} {t}{f^{(2)}}(t)\, dt}_{\alpha} $$ $$ \displaystyle = x f^{(1)}(x) - {x_0} f^{(1)}(x_0) - \alpha $$

In the Taylor series we have, the term $$ \displaystyle x-{x_0}f^{(1)}(x_0) $$ Since we have the term, $$ \displaystyle {x_0}f^{(1)}(x_0) $$, let us add and subtract $$ \displaystyle {x}f^{(1)}(x_0) $$ to the above equation.

Thus, we have, $$ \displaystyle = x f^{(1)}(x) - {x_0} f^{(1)}(x_0) - \alpha + {x}f^{(1)}(x_0) - {x}f^{(1)}(x_0) $$ Now combining the terms and rearranging, $$ \displaystyle = \underbrace {[x f^{(1)}(x) - {x}f^{(1)}(x_0)]}_{\beta} + (x-x_0) f^{(1)}(x_0) - \alpha $$

On observing carefully, we see that, $$ \displaystyle \beta $$ is nothing but ,$$ \displaystyle \int_{x_0}^{x}x f^{(2)}(t) \,dt $$ .Thus,

$$ \displaystyle  = \int_{x_0}^{x}x f^{(2)}(t) \,dt + (x-x_0) f^{(1)}(x_0) - \alpha $$ $$\displaystyle (Eq. 3) $$

[[media: Egm6341.s10.mtg5.djvu | Page 5-3]] Substituting Eqn(3) in Eqn(1) on Pg 5-2, we have the expression for the Taylor series : $$\displaystyle f(x) = f(x_0) + \frac{(x-x_0)}{1!} f^{(1)}(x_0) + \int_{x_0}^{x} (x-t)f^{(2)}(t)\, dt $$


 * {| style="width:100%" border="0" align="left"

HW:
 * style="width:100%; padding:10px; border:2px solid #8888aa" |
 * style="width:100%; padding:10px; border:2px solid #8888aa" |

$$\displaystyle \color{blue} f(x) = f(x_0) + \frac{(x-x_0)}{1!}\ f^{(1)}(x_0) + \int_{x_0}^{x} (x-t) f^{(2)}(t)\, dt $$ Repeat integration by parts to reveal the following items with remainder $$\displaystyle \color{blue} \frac{(x-x_0)^2}{2!}\ f^{(2)}(x_0) + \frac{(x-x_0)^3}{3!}\ f^{(3)}(x_0) $$ Note: Assume (4)and (5) in [[media:Egm6341.s10.mtg2.pdf|Lecture p.2-2]] are true and do integration by parts one more time


 * style = |
 * }
 * }