User:Egm6341.s11.team2.zou/HW1

=Problem 1.1=

Find
Plot $$ f\left(x\right) $$ and find:


 * $$ \lim_{x \to 0}f(x) $$

Given

 * $$ f(x) = \frac{x} $$ for $$ x \in [0,1] $$

Solution
Using l'Hôpital's rule,


 * $$ \lim_{x \to 0}\frac{x} = \lim_{x \to 0}\frac{1} = 1$$


 * Hw1aplot.jpg]

=Problem 1.2=

Find
$$\displaystyle {p_n}(x) $$ and $$\displaystyle {R_{n + 1}}(x) $$ of $$ \displaystyle f(x) = {e^x} $$ and $$\displaystyle f(x) = \frac{x} $$ at $$\displaystyle {x_0}=0 $$

Can someone please check why my equations appear like this?.. one of them looks good but the rest look so small... I have this problem in Solution section too..... Thank you... Amir

Fixed.

-Adam

Given
$$ {p_n}(x) = f({x_0}) + \frac{f^{(1)}}({x_0}) + \frac{f^{(2)}}({x_0}) + ... + \frac{f^{(n)}}({x_0}) $$

$$ {R_{n + 1}}(x) = \frac{f^{(n + 1)}}(\xi ) $$, $$ \xi \in [{x_0},x] $$

Solution
Note that for $$ \displaystyle f(x) = {e^x} $$

$$ \displaystyle {f^{(n)}}({x_0}) = {f^{(n)}}(0) = {e^0} = 1 $$

So,

$$ {p_n}(x) = f(0) + \frac{f^{(1)}}(0) + \frac{f^{(2)}}(0) + ... + \frac{f^{(n)}}(0) $$


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$$  \displaystyle {p_n}(x) = 1 + \sum\limits_{i = 1}^\infty {\frac} $$     (2-1)
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and

$$ {R_{n + 1}}(x) = \frac{f^{(n + 1)}}(\xi ) = \frac{e^\xi } $$


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$$  \displaystyle {R_{n + 1}}(x) = \frac{e^\xi } $$     (2-2)
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$$  \displaystyle {e^x} = {p_n}(x) + {R_{n + 1}}(x) = 1 + \sum\limits_{i = 1}^\infty {\frac}  + \frac{e^\xi } $$     (2-3)
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From Eq(2-3)

$$ {e^x} - 1 = \sum\limits_{i = 1}^\infty {\frac}  + \frac{e^\xi }$$

so,

$$ f(x) = \frac{x} = \frac{x} = \sum\limits_{i = 1}^\infty {\frac}  + \frac{e^\xi } $$


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$$  \displaystyle f(x) = {p_n}(x) + {R_{n + 1}}(x) = \frac{x} = \sum\limits_{i = 1}^\infty {\frac}  + \frac{e^\xi } $$     (2-4)
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So for $$ f(x) = \frac{x} $$


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$$  \displaystyle {p_n}(x)= \sum\limits_{i = 1}^\infty {\frac} $$
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$$  \displaystyle {R_{n + 1}}(x)=\frac{e^\xi } $$

where $$ \xi \in [{x_0},x] $$ (2-5)
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=Problem 1.3=

Required
Integral Mean Value Theorem:

Let $$ w(x)$$ be a nonnegative integrable function defined on $$ [a,b] $$

If $$ f(x) $$ is a continuous function defined on $$ [a,b] $$, then

for some $$ \xi $$ such that

Proof
Suppose $$ \exist $$ an $$ m $$ and $$ n $$ such that

This implies that $$ f(x) $$ is bounded below by $$ m $$ and bounded above by $$ M $$,

For a given weight function

and by simply multiplying the inequality in (4.5) by $$ w(x) $$ one obtains,

A corollary to the Fundamental Theorem of Calculus ensures that, upon integration of $$ \textbf{(4.7)} $$, the following relation is obtained

$$ \Rightarrow $$

The definite integral,

is just some numerical value therefore enabling one to divide (4.9) through by (4.10) to get

Furthermore, the Intermediate Value Theorem ensures for a continuous function $$ f(x) $$ on the interval $$ x \in [a,b] $$ that if there exists a value $$ Z $$ on the interval $$ [m,M] $$ then there is at least one such point $$ \xi $$ for which,

Recognizing that $$ Z $$ in (4.11) is

it then follows from (4.11) and (4.13) that

!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!Still needs work!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

=Problem 1.4=

From the lecture slide Mtg 5-3

Given
$$ f(x) = 3x + 5{x^3},x \in [0,1]; g(x) = \sinh x,x \in [0,1] $$

Find
Plot $$\displaystyle {f(x)}$$, $$\displaystyle {g(x)}$$ and find $$\displaystyle {\left\| f \right\|_\infty }$$, $$\displaystyle {\left\| g \right\|_\infty }$$ and $$\displaystyle {\left\| {f - g} \right\|_\infty }$$

Solution
Because


 * $$ {f^'}(x) = 3 + 15{x^2} > 0,\;\;x \in [0,1]

$$


 * $${g^'}(x) = {(\sinh x)^'} = {(\frac{2})^'} = \frac{2} > 0,\;\;x \in [0,1]

$$


 * $${\left( {f(x) - g(x)} \right)^'} = 3 + 15{x^2} - \cosh x > 0,\;\;x \in [0,1]

$$

Which means that $$\displaystyle {f(x)}$$ keeps increasing at [0,1], and $$\displaystyle {g(x)}$$ and $$\displaystyle {f(x) - g(x)}$$ as well

So
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$$ {\left\| f \right\|_\infty } = \max \left| {f(x)} \right| = {\left. {\left| {f(x)} \right|} \right|_{x = 1}} = 8 $$     (4-1)
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$$ {\left\| g \right\|_\infty } = \max \left| {g(x)} \right| = {\left. {\left| {g(x)} \right|} \right|_{x = 1}} = \frac{1}{2}(e - \frac{1}{e}) \approx 1.71520 $$     (4-2)
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$$ {\left\| {f - g} \right\|_\infty } = \max \left| {f(x) - g(x)} \right| = {\left. {\left| {f(x) - g(x)} \right|} \right|_{x = 1}} = 8 - \frac{1}{2}(e - \frac{1}{e}) \approx 6.82479 $$     (4-3)
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Using the following Fortran code

program main real,allocatable::x,f,g integer::i,n real::e=2.71828 open (unit=10, file="input", status="old") read(10,*)n allocate(x(n+1),f(n+1),g(n+1)) do i=1,n+1 x(i)=-1.0+2.0/n*(i-1) f(i)=3.0*x(i)+5*x(i)**3 g(i)=0.5*(e**x(i)-e**(-1.0*x(i))) end do  open (unit=11, file="hw1.4.plt", status="replace") do i=1,n+1 write(11,"(3F12.4)") x(i),f(i),g(i) end do   end program main

We can easily get the image of $$\displaystyle {f(x)}$$ and $$\displaystyle {g(x)}$$ as following





Conclusion remark
We can see from above that,


 * $$\displaystyle {\left\| {f - g} \right\|_\infty } = 6.8248$$


 * $$\displaystyle {\left\| f \right\|_\infty } - {\left\| g \right\|_\infty } = 8 - 1.71520 = 6.8248$$

thus in this case we have,


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$$ \displaystyle {\left\| {f - g} \right\|_\infty } = {\left\| f \right\|_\infty } - {\left\| g \right\|_\infty } $$     (4-4)
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In this case we can't prove the statement made in Lecture slide 5-3, it is believed to caused by the same monotonicity of $$\displaystyle f(x)$$, $$\displaystyle g(x)$$ and $$\displaystyle {f(x) - g(x)}$$.

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