User:Subbu.ase/HW2 pblm 15

Problem Statement
$$ -\frac{1}{90}F^{(4)}(\zeta _{4})=-\frac{(b-a)^{4}}{1440}f^{4}(\xi) $$

Ref: Lecture Notes [[media:Egm6341.s10.mtg15.pdf|p.15-3]]

Solution
Use chain Rule

$$ \displaystyle F(t)=f(x) $$

$$ \displaystyle x(t)=x_{1}+ht $$

$$ \displaystyle F^{(1)}(t)=\frac{dF(t)}{dt} = \frac{df(x)}{dx}\frac{dx}{dt} = f^{(1)}(x) \times h = hf^{(1)}(x) $$

$$ \because \frac{d\left(x(t)\right)}{dt} = \frac{d(x_{1}+ht)}{dt} = h $$

$$ \displaystyle F^{(2)}(t) = \frac{d\left(F(t)^{(1)}\right)}{dt} = \frac{d\left(hf(x)^{(1)}\right)}{dx}\frac{dx}{dt} = hf^{(2)}(x) \times h = h^{2}f^{(1)}(x) $$

$$ \displaystyle F^{(3)}(t) = \frac{d\left(F(t)^{(2)}\right)}{dt} = \frac{d\left(h^{2}f(x)^{(2)}\right)}{dx}\frac{dx}{dt} = h^{2}f^{(3)}(x) \times h = h^{3}f^{(3)}(x) $$

$$ \displaystyle F^{(4)}(t) = \frac{d\left(F(t)^{(3)}\right)}{dt} = \frac{d\left(h^{3}f(x)^{(3)}\right)}{dx}\frac{dx}{dt} = h^{3}f^{(4)}(x) \times h = h^{4}f^{(4)}(x) $$

Now, obtain the relationship between $$ \displaystyle \xi $$ and $$ \displaystyle \zeta_{4} $$

$$ \displaystyle x(\zeta_{4}) = x_{1} + h\zeta_{4} = \xi $$

And recall that $$ \displaystyle h=\frac{b-a}{2} $$ Then,

$$ \displaystyle -\frac{1}{90}F^{(4)}(\zeta _{4})=-\frac{1}{90}\left(h^{4}f^{(4)}(\xi)\right)=-\frac{1}{90}\left(\frac{(b-a)^{4}}{2^{4}}f^{(4)}(\xi)\right) $$

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

$$\displaystyle -\frac{1}{90}F^{(4)}(\zeta _{4})=-\frac{(b-a)^{4}}{1440}f^{4}(\xi) $$ where
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle \xi = x_{1}+h \zeta_{4} $$
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }