User:Egm6321.f12.team5.report 5

= Report 5, team 5 =

Given
(From lecture notes sec 20)

Find
Show that

Solution
On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions.

If $$ \mathbf{A} $$ is diagonalizable, then

where $$\mathbf{D}$$ is a diagonal matrix of the eigenvalues of $$\mathbf{A}$$, and $$\mathbf{P}$$ is made up of the corresponding eigenvectors. Therefore,

and it follows from ($$) that

From R5.3, we know that

and ($$) can be written as

Using ($$)~($$) and using the fact that $$(\mathbf{A}\mathbf{B}\mathbf{C})^T = \mathbf{C}^T\mathbf{B}^T\mathbf{A}^T$$, it can be shown that

Since we have shown that ($$) and ($$) are equal, we have shown that ($$) is true.

Comments
Initial solution by --Egm6321.f12.team5.kim (talk) 03:12, 28 October 2012 (UTC)

Given
(From lecture notes sec 20) Consider a diagonal matrix $$\mathbb C^{n \times n} $$ set of $$n \times n $$ matrices with complex coefficients

Find
Show that

Solution
On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions. If we have a diagonal matrix According to the property of a diagonal matrix, we have the following relationship (From lecture notes sec 20) We know that So Simplify the equation, we get (From lecture notes sec 20) So So we get

Comments
Initial solution by Yu.

Given
Given the following equations: Here $$\Phi=[\phi_1, \ldots, \phi_n]$$, are n linearly independent eigenvectors.

Find
Show that:

Solution
According to (5.3.2),

Consider $$(\mathbf \Phi \mathbf \Lambda \mathbf \Phi^{-1})^k$$, it's expansion is:

And as we know $$\mathbf \Phi \mathbf \Phi^{-1}=I$$, (5.3.6) can rewritten as:

Base on the result above, (5.3.5) equals to:

Because $$\mathbf \Lambda = \text{Diag}[\lambda_1, \ldots,\lambda_n]$$, we can find:

Comments
I think $$\mathbf \Phi \mathbf \Phi^{-1}=I$$ in (5.3.6) Yu

Given
(From lecture notes sec 20) The matrix $$\mathbf{B}t $$ with $$\mathbf{B} $$ defined in (1) p.20-2 can be decomposed as follows:

Find
Show ($$) and ($$)

Solution
On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions.

Take the Eigenvalue problem of a matrix $$ \mathbf{B} \in  \mathbb{R}^{nxn} $$

There are 2 linearly independent eigenvectors

Which can be rewritten as and be used to find the eigenvalues of $$ \mathbf{B} $$

Solving for the Eigenvalues of $$ \mathbf{B} $$

Putting these two eigenvalues into a matrix gives

Putting the eigenvalues back into the original equations, we can solve for $$ \Phi$$ when $$ \lambda = -i, i $$, we get

Solving for the $$\phi$$ components

Inverting the matrix $$ \boldsymbol{\Phi}$$ ,

Therefore, matrix $$\mathbf{B} $$ can be decomposed as follows

where

Where shown in problems 5.1, 5.2, 5.3 mentioned earlier, we can also show that

Comments
Not sure where to go after finding $$ \Phi $$ to get the inverse of phi and the 1/2i part! ~ Anh 10/28/2012 3:08pm

Added Tae's Solution to find phi^-1 ~ Anh 10/28/2012 5:59pm

Given
A general class of L2 - ODE - VC represented by the following equation,

where,

and the following function which creates a class of exact L2 - ODE - VC

(From lecture notes sec 21)

Find
Show that ($$) and ($$) leads to ($$)

Solution
On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions.

From ($$) and ($$) we can write,

Since it is given that $$ P(x) = \phi_p $$, ($$) becomes

The above equation represents an L2 - ODE - VC which satisfies the first exactness condition. Now, We know that,

Therefore,

Differentiating the above expression for $$ \phi $$ with respect to $$ x $$ to get the L2 - ODE - VC (G)

Since $$ p = y' $$ and $$ p' = y'' $$, we can write the above equation as,

Comparing the above equation with ($$) and ($$), We get,

Now substituting the above equation into Equation ($$)

Setting $$ \int^x R(s)ds = T(x) $$, The above equation can be modified as,

Differentiating the above equation with respect to $$ y $$.

But from Equations ($$) and ($$), We know that,

Equating the two equations above,

which gives,

or

Substituting in ($$),

Hence it is proved that ($$) and ($$) leads to ($$)

Comments
Initial Solution by Vishnu

Given
(From lecture notes sec 21) L2-ODE-VC

Find
1. Show (5.6.1) is exact 2. Find $$\phi$$ 3. Solve for $$y(x)$$

Solution
On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions.

Part one
If we substitute $$p=y'$$

Then we get

which is in a particular form, so the operator safisfies the first exactness condition.

The operator satisfies the second exactness condition, if

In this case,

$$f_{xx}=\frac{\partial ^2f(x,y,p)}{\partial x^2}=-\cos x$$

$$pf_{xy}=p\frac{\partial ^2f(x,y,p)}{\partial x \partial y}=0$$

$$p^2f_{yy}=p^2\frac{\partial ^2f(x,y,p)}{\partial y^2}=0$$

$$g_{xy}=\frac{\partial ^2g(x,y,p)}{\partial x \partial y}=2x-\cos x$$

$$pg_{py}=p\frac{\partial ^2g(x,y,p)}{\partial p \partial y}=0$$

$$-g_{y}=-\frac{\partial g(x,y,p)}{\partial y}=-2x$$

$$-\cos x=-\cos x$$, so (5.6.5) is satisfied

$$f_{xp}=\frac{\partial ^2f(x,y,p)}{\partial x \partial p}=0$$

$$pf_{py}=p\frac{\partial ^2f(x,y,p)}{\partial p \partial y}=0$$

$$2f_{y}=2\frac{\partial f(x,y,p)}{\partial y}=0$$

$$g_{pp}=\frac{\partial ^2g(x,y,p)}{\partial p^2}=0$$

$$0=0$$, so (5.6.6) is satisfied

So the function G satisties both first exactness condition and the second exactness condition. (5.6.1) is exact.

Part two
So $$h_x+h_yp=px^2+2xy$$

By comparing left side and right side, we got $$h_x=2xy$$ and $$h_y=x^2$$

So $$k'_1(y)=0$$. Thus $$k_1(y)=k_1$$

Part three
Substitute $$\frac{x^2}{\cos x}=a_0(x)$$ and $$\frac{k}{\cos x}=b(x)$$

We get

(From lecture notes Sec11, Pg 4 and 5.)

Comments
Initial Solution by Yu

Given
(From lecture notes sec 22)

Solution
On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions.

From ($$),we know:

Comments
Initial Solution by Enze

Given
(From lecture notes sec 22) Equivalent form of 2nd Exactness Condition for N2-ODEs:

Detailed explanation of the Coefficient equalities:

Find
Show the following equality is true:

Solution
On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions. Transform the($$) with ($$), ($$) and ($$), we can get:

Which can be further differential as following:

Further simplify($$), we can get:

Hence Proved

Comments
Initial Solution by Enze

Given
(From lecture notes sec 64) The Talyor series expansion at $$ x = 0 $$ (aka Maclaurin Series)

Where $$ f^{n} $$ represents the nth derivative of $$ f(x) $$,

Find
Derive

and

Solution
On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions.

(a). Considering Equation $$ We have $$ f(x) = (1-x)^{-a} $$. We need to find derivatives with respect to $$ x $$ and plug it in $$, to obtain the result.

Differentiating with respect to $$ x $$ at $$ x = 0 $$,

Similarly,

Now, Substituting the derivatives at $$ x = 0 $$ onto Equation $$

Therefore

Hence Proved (b). Considering Equation $$ We have $$ f(x) = \frac{1}{x}\arctan(1+x) $$. We have to assume that $$ |1-x| < 1 $$ for this function to be real and convergent. Writing the function as a power series as follows,

So,

Now we know the individual taylor series expansions of $$ \frac{1}{x} $$ and $$ \arctan(x) $$

Comparing, $$, and $$,We get

Therefore, $$, can be written as,

Hence Proved

Comments
Initial solution by Vishnu

Given
(From lecture notes sec 64)

Find
Use matlab to plot $$F(5,-10;1;x)$$ near $$x=0$$ to display the local maximum (or maxima) in this region. Show that

Solution
On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions.

Using Matlab's 'hypergeom' function, $$F(5,-10;1;x)$$ is plotted from for $$0 \leq x \leq 1.2.$$ $$F(5,-10;1;x)$$ steadily increases for $$x \leq 0$$ and $$x \geq 1.2$$, so the local maxima are located within the range [0, 1.2].

Plotting the polynomial in ($$) on the same graph shows that two are identical for $$ x = [0, 1.2]$$.

Similar plot for a larger range of $$x$$ from -10 to 10, shows that the ($$) and ($$) are equal in the larger interval as well.

Matlab code used to generate the plots is shown below.

The problem is solved

Given
Consider the following integrals: Let n=3, a=2, b=10.

Find
For each value fo time t, solve for altitude z(t), then plot z(t) versus t, Find the time when the projectile returns to the ground.

Solution
Since n=3, a=2, b=10, combine (5.11.1) and (5.11.2) into:

Because

(5.11.3) can be rewritten as:

Let z=0 when t=0, we can easily find the value of k equals to zero.

So we can solve for altitude z(t) in Matlab:



Comments
initial solution by --Egm6321.f12.team5.kim (talk) 05:04, 28 October 2012 (UTC)

Seems something wrong

The problem is that find the time when the projectile returns to the ground, so besides t=0, z(t)=0, we should find another t that satisfies z(t)=0. Am I right?--Enze

Given
(From lecture notes sec 64)

Find
Is (5.12.1) exact

Is (5.12.1) in the power form in(3) p.21-1

Verify that F(a,b;c;x) in (1) p.64-4 is indeed a solution of (5.12.1)

Solution
On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions.

Part one
(From lecture notes sec 16) So it satisfies the first exactness condition.We substite y' to p.The operator satisfies the second exactness condition, if

In this case,

$$f_{xx}=\frac{\partial ^2f(x,y,p)}{\partial x^2}=-2$$

$$pf_{xy}=p\frac{\partial ^2f(x,y,p)}{\partial x \partial y}=0$$

$$p^2f_{yy}=p^2\frac{\partial ^2f(x,y,p)}{\partial y^2}=0$$

$$g_{xp}=\frac{\partial ^2g(x,y,p)}{\partial x \partial y}=-(a+b+1)$$

$$pg_{py}=p\frac{\partial ^2g(x,y,p)}{\partial p \partial y}=0$$

$$-g_{y}=-\frac{\partial g(x,y,p)}{\partial y}=ab$$

$$ -2\neq ab-(a+b+1)$$

(5.12.2) is not satisfied, so (5.12.1) is not exact.

Part two
(From lecture notes sec 21) is not satisfies the form For each coefficient of terms is not just consists of a constant multiplied by arbitrary power of x.

Part three
In this case

So we get

By adding (5.12.11), (5.12.13), (5.12.14), (5.12.16), (5.12.17) together, we get So is a solution of

Comments
Initial solution by Yu.

Given
(From lecture notes sec 27) Consider the following governing L2-ODEs-VC for some classical special function: Legendre:

Hermite:

Find
1. Verify the exactness of the designated L2-ODEs-VC. For the 2nd exactness condition, use 2 methods:
 * 1a. (1)-(2) p.16-5 (sec16)


 * 1b. (1) p.22-3 (sec22)

2. If ($$) is not exact, check whether it is in power form, and see whether if it can be made exact using IFM with $$h(x,y) = x^m y^n$$; see p.21-2 (sec21). 3. The first few Hermite polynomials are

Verify that From ($$)-($$) are homogeneous solutions of the Hermite differential equation ($$).

Solution
Sections 1 and 2: 1.13.3.1 Verify Exactness of the designated L2-ODEs-VC 1.13.3.2 Making the Hermite function exact through power form On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions.

Section 3: 1.13.3.3 Verify homogeneous solutions of Hermite Equation In this part of the assignment, we referred to an external website. Please see this section and the 'Comments' section for further details.

Verify Exactness of Legendre Function
First Exactness The first exactness condition for an N2-ODE states that it must have the particular form:

Evaluating the Legendre Function given in ($$), we can replace $$ y' $$ with $$ p $$ and rearrange the equation to put it into the particular form.

Therefore the first exactness condition for the Legendre function has been met. Second Exactness: Method 1 The second exactness condition for an N2-ODE states that it must satisfy the following conditions:

Taking the partial derivatives of ($$), we get the following sets of terms:

Evaluating the Second Exactness Condition,

Although the second relation of the Second Exactness Condition is met, the first relation is not met. Therefore, this equation is not exact. Second Exactness: Second Method 2 To evaluate the Second Exactness condition under the second method, we first identify the following:

Plugging in ($$), ($$) and ($$) into the second exactness equation ($$) we can evaluate:

Therefore, the Legendre Function meets the second exactness condition when n=0 or n=-1.

Verify Exactness of Hermite Function
First Exactness Again, the first exactness condition for an N2-ODE states that it must have the particular form:

Evaluating the Hermite Function given in ($$), we can replace $$ y' $$ with $$ p $$ and rearrange the equation to put it into the particular form.

Therefore the first exactness condition for the Hermite function has been met. Second Exactness: Method 1 Again, the second exactness condition for an N2-ODE states that it must satisfy the following conditions:

Taking the partial derivatives of ($$), we get the following sets of terms:

Evaluating the Second Exactness Condition,

Although the second relation of the Second Exactness Condition is met, the first relation is not met. Therefore, this equation is not exact. Second Exactness: Second Method 2 To evaluate the Second Exactness condition under the second method, we first identify the following:

Plugging in ($$), ($$) and ($$) into the second exactness equation ($$) we can evaluate:

Therefore, the Hermite Function meets the second exactness condition when n=-1.

Making the Hermite function exact through power form
For an L2-ODE-VC to be in power form, it must be in the particular form:

Evaluating ($$), we can identify the following:

Since the equation is in the special power form, we can find an integrating factor of the form $$ h(x,y) = x^a y^b $$ Substituting $$ y' = p $$ and multiplying the equation by $$ h(x,y) $$ the equation can be made exact

Putting the equation in the first exactness particular form, we can derive the following terms:

Plugging in the terms from ($$) and ($$) into the second exactness conditions, we can derive two equations to find values of a and b

Since we know that b=0, the equation can be reduced to

Solving, we get

Therefore the integrating factor for the Hermite equation is $$ x^a y^b = x^1y^0 = x$$ and the equation to satisfy the second exactness condition will be

To evaluate the Second Exactness condition under the second method for this new equation, we first identify the following:

Plugging in ($$), ($$) and ($$) into the second exactness equation ($$) we can evaluate:

Therefore, the Legendre Function meets the second exactness condition when n=-2.

Verify homogeneous solutions of Hermite Equation
Referring to Series Solutions: Hermite's Equation from sosmath.com, Hermite's Equation has a polynomial solution of

Since we have $$ n $$ in our equation, we will use the term $$ c $$ instead to avoid any mistakes. Additionally, to verify the homogeneous solutions, we must find the first and second derivatives of the polynomial solution.

Now substitute ($$), ($$), and ($$) into our original function ($$)

To combine the terms into one sum, we shift the summation up by two units

and evaluate the 0th term for the second sum

since the 0th term for the second sum is zero, the sum when c=0 and c=1 are the same. Therefore, we can change the equation to start at c=0 rather than c=1 for the second sum

Now that all of the sums are start in the same index, we can combine the sums

From inspection, the recurrence becomes

Then further simplifying the recurrence, we get the following equation that can be used to verify the homogeneous solutions

From the given initial conditions, we can evaluate for $$ c=0 $$ when $$ H_0(x) = 1 $$ where

Since $$ a_2 = 0 $$ and $$ a_3 = 0 $$ then $$ a_4, a_5, ...,a_{\infty} = 0 $$ verifying that the first Hermite Polynomial $$ H_0(x) = 1 $$ From the given initial conditions, we can evaluate for $$ c=1 $$ when $$ H_1(x) = 2x $$ where

Since $$ a_2 = 0 $$ and $$ a_3 = 0 $$ then $$ a_4, a_5, ...,a_{\infty} = 0 $$ verifying that the second Hermite Polynomial $$ H_1(x) = 2x $$ From the given initial conditions, we can evaluate for $$ c=2 $$ when $$ H_2(x) = 4x^2-2 $$ where

Since $$ a_3 = 0 $$ and $$ a_4 = 0 $$ then $$ a_5, a_6, ...,a_{\infty} = 0 $$ verifying that the third Hermite Polynomial $$ H_2(x) = 4x^2-2 $$

Comments
Parts 1 and 2 were solved on my own. However, Part 3: "Verifying the homogeneous solutions of the Hermite differential equation" was solved while referring to Series Solutions: Hermite's Equation from sosmath.com. From that website, I specifically referred to the polynomial solution equation and the process in which to make the indexes all start from 0. In terms of how my solution differs from the website was that I added a little more detail and explanation and equations. ~ Anh 10/28/2012 11:49AM

Given
(From lecture notes sec 30)

Find
The expressions for $$ X(x)$$ In terms of $$ cosKx $$,$$ sinKx $$, $$ coshKx $$, $$ sinhKx $$

Solution
On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions.

From ($$) and ($$),we can obtain final solution X(x) as:

Since we already know:

Then ($$) can be transformed as :

With further simplified, ($$) can be changed as:

Which is a combination of $$ cosKx $$,$$ sinKx $$, $$ coshKx $$, $$ sinhKx $$ Hence solved

Comments
Initial Solution by Enze

What does "cosh" mean? Is it "cos"?Yu

I have added that to the solution.Enze

= Contributors = Report Leader: Yu Hou Report Co-Leaders : Yi Zhongwen