User:Egm6341.s11.team2/hwk1

=Problem 1.1: The Limit and Plot of a Given Function f(x)=

From the lecture slide Mtg 3-1

Problem Statement
Given:

i) Determine $$ \lim_{x \to 0}f(x) $$ and

ii) Plot $$ f\left(x\right) $$

Solution
i) Finding the limit of $$\displaystyle f(x) $$

Since it is possible to determine the limit of both the numerator and denominator independently as $$\displaystyle x $$ tends to $$\displaystyle 0 $$ as shown below and It is therefore clear that we have indeterminate form of type $$\displaystyle \frac{0}{0} $$. Limits of functions having this indeterminate form are easily found using L'Hôpital's Rule. Thus using L'Hôpital's Rule we find that,

ii) Plot of f(x)

Shown below is the plot of $$\displaystyle f(x) $$ on the interval $$\displaystyle [0,1] $$ which graphically verifies our solution in part i) of the limit of $$\displaystyle f(x) $$ as $$\displaystyle x $$ tends to $$\displaystyle 0 $$.


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=Problem 1.2: Determination of Taylor Polynomial and Remainder=

From the lecture slide Mtg 3-4

Given: Taylor's Theorem
Taylor's Theorem

Let $$\displaystyle f(x) $$ be a function with derivatives $$\displaystyle f^{(1)},f^{(2)},...,f^{(n)} $$ that are continuous on $$\displaystyle I:=[a,b] $$ and where $$\displaystyle f^{(n+1)} $$ exist on $$\displaystyle (a,b) $$.

If $$\displaystyle x_{0} \in I $$, then for any point $$\displaystyle x $$ in $$\displaystyle I $$ there exists a point $$\displaystyle \xi $$ between $$\displaystyle x $$ and $$\displaystyle x_{0} $$ such that

Where $$\displaystyle p_{n}(x) $$ is called the Taylor Polynomial for $$\displaystyle f $$ at $$\displaystyle x_{0} $$

and $$\displaystyle R_{n+1}(x) $$ is the remainder of $$\displaystyle f $$ at $$\displaystyle x_{0} $$

Find
The Taylor Polynomial $$\displaystyle {p_n}(x) $$ and remainder $$\displaystyle {R_{n + 1}}(x) $$ at $$\displaystyle {x_0}=0 $$ for

i) $$ \displaystyle g(x) = {e^x} $$

ii) $$ \displaystyle f(x) = \frac{x} $$

Solution
i) For $$ \displaystyle g(x) = {e^x} $$ all derivative are identically equal to the original function. Evaluating these derivatives for $$\displaystyle n=1,2,...,n,n+1 $$ at $$\displaystyle x_{0} = 0 $$ results in the following

Then by the definition of the Taylor Polynomial $$\displaystyle p_{n}(x) $$ provided in (2.2) we can write

This can be more concisely written in summation notation as

Using the definition of the remainder $$\displaystyle R_{n+1}(x) $$ shown in (2.3) and the $$\displaystyle n+1 $$-th derivative found in (2.4) we can write


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$$ \displaystyle \therefore {e^x} = {p_n}(x) + {R_{n + 1}}(x) = 1 + \sum\limits_{i = 1}^\infty {\frac}  + \frac{e^\xi } $$     (2.10)
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ii) Next we will consider $$\displaystyle f(x) $$.

Using the Taylor Polynomial and Remainder for $$\displaystyle e^{x} $$ shown in (2.10), we will subtract $$\displaystyle 1 $$ from each side to get

Dividing (2.11) through by $$\displaystyle x $$ results in the desired function f(x) for which we wish to obtain the Taylor Polynomial $$\displaystyle p_{n}(x) $$ and remainder $$\displaystyle R_{n+1}(x) $$ results in

So in summary, the function $$\displaystyle f(x) $$ has the Taylor Polynomial $$\displaystyle p_{n}(x) $$ and remainder $$\displaystyle R_{n+1}(x) $$
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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 $$ \xi \in [{x_0},x] $$ (2.14)
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=Problem 1.3: Prove the IMVT=

''' This problem was solved without referring to S10 homework. '''

From the lecture slide Mtg 4-3

Prove: The Integral Mean Value Theorem (IMVT)
Prove the Integral Mean Value Theorem (IMVT) for

a) $$\displaystyle w\left(x\right) \geq 0, \quad\forall x \in \left[a,b\right]$$ and

b) $$\displaystyle w\left(x\right) \neq 0, \quad\forall x \in \left[a,b\right]$$

using same technique used in lecture 4 to prove IVMT for $$\displaystyle w\left(x\right) = 1 $$.

Required
Integral Mean Value Theorem:

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

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

for some $$\displaystyle \xi $$ such that

a) $$\displaystyle w\left(x\right) \geq 0, \quad\forall x \in \left[a,b\right]$$
Suppose $$\displaystyle \exist $$ an $$\displaystyle m $$ and $$\displaystyle n $$ such that

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

For a given weight function

and by simply multiplying the inequality in (3.5) by $$\displaystyle w(x) $$ one obtains,

A corollary to the Fundamental Theorem of Calculus ensures that, upon integration of (3.7) the following relation is obtained

$$\displaystyle \Rightarrow $$

The definite integral,

is just some numerical value therefore enabling one to divide (3.9) through by (3.10) to get

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

Recognizing that $$\displaystyle Z $$ in (3.11) is

it then follows from (3.11) and (3.13) that

b) $$\displaystyle w\left(x\right) \neq 0, \quad\forall x \in \left[a,b\right]$$
This can really be broken down into two cases $$\displaystyle w(x) > 0 $$ and $$\displaystyle w(x) < 0 $$

Case 1: $$\displaystyle w(x)>0 $$

Observe that $$\displaystyle w(x) > 0 $$ is really just a more restricted case than $$\displaystyle w(x) \geq 0 $$ since it does not include $$\displaystyle w(x) = 0 $$. Because of this it is not necessary to reprove what has already been proven in the first part. Therefore by virtue of the case where $$\displaystyle w(x) \geq 0 $$ the case for $$\displaystyle w(x) > 0 $$ can be reasonably inferred from part a.

Case 2: $$\displaystyle w(x)<0 $$

Consider $$\displaystyle w\left(x\right) f\left(x\right) $$ continuos on $$\displaystyle\left[a,b\right] .$$

Let $$\displaystyle m $$ and $$\displaystyle M $$ be what was previously defined in (3.4). Multiplying (3.5) by $$\displaystyle |w\left(x\right)| $$

Integrating (3.15) over $$\displaystyle \left[a,b\right] $$

$$\displaystyle m $$ and $$\displaystyle M $$ are both constants and can be pulled out of the integrals.

Let

Divide (3.18) by $$\displaystyle \alpha $$

By the Intermediate Value Theorem (IVT), $$\displaystyle \exists \xi  \in \left[a,b\right] $$ such that $$\displaystyle f\left(\xi\right)=Z. $$

Since $$ \displaystyle w\left(x\right) < 0 ,$$

Therefore,

= Problem 1.4 Infinity Norm of Functions =

From the lecture slide Mtg 5-3

Problem Statement
Given:

and

i) Find $$\displaystyle {\left\| f \right\|_\infty }$$, $$\displaystyle {\left\| g \right\|_\infty }$$ and $$\displaystyle {\left\| {f - g} \right\|_\infty }$$.

ii) Plot $$\displaystyle {f(x)}$$, $$\displaystyle {g(x)}$$ and $$\displaystyle f(x)-{g(x)}$$

iii) Then establish a relationship between the infinity norms of the two functions $$\displaystyle f(x) $$ and $$\displaystyle g(x) $$.

Solution
i)

Taking the first derivative of $$\displaystyle f(x) $$ and $$\displaystyle g(x) $$ indicates that both function are increasing on the interval $$\displaystyle [0,1] $$ and

Similar analysis shows that the difference of the functions $$\displaystyle f(x) $$ and $$\displaystyle g(x) $$ results in an increasing behavior as seen here

Which means that $$\displaystyle {f(x)}$$ and $$\displaystyle {g(x)}$$ and therefore $$\displaystyle {f(x) - g(x)}$$ all achieve their maximal value at $$\displaystyle x=1 $$. Hence
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$$ {\left\| f \right\|_\infty } = \max \left| {f(x)} \right| = {\left. {\left| {f(x)} \right|} \right|_{x = 1}} = 8 $$     (4.6)
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$$ {\left\| g \right\|_\infty } = \max \left| {g(x)} \right| = {\left. {\left| {g(x)} \right|} \right|_{x = 1}} = \frac{1}{2}(e - \frac{1}{e}) \approx 1.17520 $$     (4.7)
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$$ {\left\| {f - g} \right\|_\infty } = \max \left| {f(x) - g(x)} \right| = {\left. {\left| {f(x) - g(x)} \right|} \right|_{x = 1}} = 8 - \frac{1}{2}(e - \frac{1}{e}) \approx 6.82479 $$     (4.8)
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ii) Using the following Fortran code

program main real,allocatable::x,f,g,fg 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),fg(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))) fg(i)=f(i)-g(i) end do  open (unit=11, file="hw1.4.plt", status="replace") do i=1,n+1 write(11,"(4F12.4)") x(i),f(i),g(i) end do   end program main

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





iii) Concluding remark on the relationship between the infinity norms of the given functions

We can see from above that,


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


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$$ \displaystyle {\left\| f \right\|_\infty } - {\left\| g \right\|_\infty } = 8 - 1.17520 = 6.8248 $$


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thus in this case we have,


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$$ \displaystyle {\left\| {f - g} \right\|_\infty } = {\left\| f \right\|_\infty } - {\left\| g \right\|_\infty } $$     (4.9)
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In this case we can't prove the assertion made in Lecture slide 5-3. The relationship shown in (4.9) is a special case and it arises from the monotonically increasing behavior of $$\displaystyle f(x)$$ and $$\displaystyle g(x)$$. Because both functions are monotonically increasing on $$\displaystyle [0,1] $$ then the difference between the two will also be monotonically increasing and thus $$\displaystyle {f(x) - g(x)}$$ has an infinity norm at $$\displaystyle x = 1 $$ on this domain.

= References =

=Contributing Members=