University of Florida/Egm6321/f12.Rep5hid

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R5.1 Proof that exponentiation of Transverse of a Matrix equals the Transverse of the Exponentiation Expansion[edit]

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

Given[edit]

and

 

 

 

 

(Eq.(2)p.20-2b)

Show That[1][edit]

 

 

 

 

(Eq.5.1.1)

Solution[edit]

We will first expand the LHS, then the RHS of (Eq. 5.1.1 ) using (Eq.(2)p.20-2b ) and compare the two expressions.

Expanding the LHS,


But we know that


 

 

 

 

(Eq.5.1.2)


Now expanding the RHS,



Which on calculating, reduces to



or


 

 

 

 

(Eq.5.1.3)


Comparing (Eq. 5.1.2 ) and (Eq. 5.1.3 )

We conclude the LHS = RHS, Hence Proved.

R5.2. Exponentiation of a Complex Diagonal Matrix [2][edit]

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

Given[edit]

A Diagonal Matrix

, where, .

 

 

 

 

(Eq.(2)p.20-2b)

Problem[edit]

Show that

, where, .

 

 

 

 

(Eq.(3)p.20-2b)

Solution[edit]

We know, from Lecture Notes [3],

 

 

 

 

(Eq.(2)p.15-3)

Let us consider a Simple yet Generic 4x4 Complex Diagonal Matrix

 

 

 

 

(Eq.5.2.2)

where .

Applying (Eq.(2)p.15-3 ) to (Eq.5.2.2 ) and expanding,

 

 

 

 

(Eq.5.2.3)

Simplifying Term 2 and other higher power terms (upto Term k) in the following way,

 

 

 

 

(Eq.5.2.4)

Similarly,

 

 

 

 

(Eq.5.2.5)

Using (Eq.5.2.4 ) and (Eq.5.2.5 ) in (Eq.5.2.3 ) and carrying out simple matrix addition, we get,

 

 

 

 

(Eq.5.2.6)

But every diagonal term of the matrix is of the form,

 

 

 

 

(Eq.5.2.7)

Therefore, (Eq.5.2.6 ) can be rewritten as,

 

 

 

 

(Eq.5.2.8)

are nothing but the diagonal elements of the original matrix in (Eq.5.2.2 ). Hence,

 

 

 

 

(Eq.5.2.9)

Similarly it can be easily found for an complex diagonal matrix that

 

 

 

 

(Eq.5.2.10)

Hence Proved.

R5.3 Show form of Exponentiation of Matrix in terms of Eigenvalues of that matrix[edit]

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

Given[edit]

A matrix can be decomposed as

 

 

 

 

(Eq.(2) p.20-4)

where

is the diagonal matrix of eigenvalues of matrix

 

 

 

 

(Eq.(5) p.20-3)

and

is the matrix established by n linearly independent eigenvectors of matrix , that is,

 

 

 

 

(Eq.(2) p.20-3)

Problem[edit]

Show that

Solution[edit]

The power series expansion of exponentiation of matrix in terms of that matrix has been given as

 

 

 

 

(Eq.(2) p.15-3)

Since matrix can be decomposed as,

 

 

 

 

(Eq.(2) p.20-4)

Expanding the power of matrix yields

 

 

 

 

(Eq.5.3.1)

Where the factors which are neighbors of factors can be all cancelled in pairs, that is,

 

 

 

 

(Eq.5.3.2)

 

 

 

 

(Eq.5.3.3)

 

 

 

 

(Eq.5.3.4)

Thus, the equation (Eq. 5.3.1 ) can be expressed as

 

 

 

 

(Eq.5.3.5)

 

 

 

 

(Eq.5.3.6)

According to the equation (Eq.(2) p.15-3 ), now we have,

 

 

 

 

(Eq.5.3.7)

Referring to the conclusion obtained in R5.2, which is

 

 

 

 

(Eq.(3) p.20-2b)

Replacing the matrix with , the elements with ,where and then substituting into (Eq. 5.3.7 ) yields

 

 

 

 

(Eq.5.3.8)

R5.4 Show Decomposed Form of Matrix and its Exponentiation[edit]

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

Given[edit]

Exponentiation of a matrix can be decomposed as

 

 

 

 

(Eq.(3) p.20-4)

The matrix is defined in lecture note

 

 

 

 

(Eq.(1) p.20-2)

Problem[edit]

Show

 

 

 

 

(Eq.(1) p.20-5)

and

 

 

 

 

(Eq.(2) p.20-5)

Solution[edit]

To show the equation (Eq.(1) p.20-5 ), we should first find eigenvalues of matrix using the matrix equation as follow, indroduce to represent Identity matrix.

Since the two eigenvalues of matrix are both obtained, now solve for the corresponding two eigenvectors.

 

 

 

 

(Eq.5.4.1)

Thus, for the first value of , we have

 

 

 

 

(Eq.5.4.2)

Substituting into the equations above and solving yields, for the eigenvalue , that

 

 

 

 

(Eq.5.4.3)

Similarly, we have the equation which can be used for solving eigenvector corresponding to ,

 

 

 

 

(Eq.5.4.4)

Substituting into the equations above and solving yields, for the eigenvalue , that

 

 

 

 

(Eq.5.4.5)

Now we have obtained two eigenvectors and of matrix , where

,

 

 

 

 

(Eq.5.4.6)

Thus we have

 

 

 

 

(Eq.5.4.7)

Then, calculating the inverse matrix of matrix yields

 

 

 

 

(Eq.5.4.8)

Therefore we reach the conclusion that,

 

 

 

 

(Eq.(1) p.20-5)


According to the conclusion we have reached in R5.3, we have,

Doing the multiplication of matrices at the right side of equation above yields

 

 

 

 

(Eq.5.4.9)

Consider Euler’s Formula,[4]

 

 

 

 

(Eq.5.4.10)

Replacing with yields

 

 

 

 

(Eq.5.4.11)

Solve (Eq.5.4.10 ) together with (Eq.5.4.11 ), we have

 

 

 

 

(Eq.5.4.12)

Substituting (Eq.5.4.12 ) into (Eq.5.4.9 ) yields

 

 

 

 

(Eq.(2) p.20-5)

Obviously,

R*5.5 Generating a class of exact L2-ODE-VC [5][edit]

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

Given[edit]

A L2-ODE-VC [6]:

 

 

 

 

(Eq. 5.5.1)

The first intregal can also be expressed as:

 

 

 

 

(Eq. 5.5.2)

Problem [7][edit]

Show that(Eq. 5.5.1 ) and (Eq. 5.5.2 ) lead to a general class of exact L2-ODE-VC of the form:

 

 

 

 

(Eq. 5.5.3)

Solution[edit]

Nomenclature[edit]

Derivation of Eq. 5.5.3[edit]

The first exactness condition for L2-ODE-VC: [8]

 

 

 

 

(Eq. 5.5.4)

From (Eq. 5.5.1 ) and (Eq. 5.5.4 ), we can infer that

 

 

 

 

(Eq. 5.5.5)

Integrating (Eq. 5.5.5 ), w.r.t p, we obtain:

 

 

 

 

(Eq. 5.5.6)

Partial derivatives of w.r.t to x and y can be written as:

 

 

 

 

(Eq. 5.5.7)

 

 

 

 

(Eq. 5.5.8)

Substituting the partial derivatives of w.r.t x,y and p [(Eq. 5.5.7 ), (Eq. 5.5.8 ), (Eq. 5.5.6 )] into (Eq. 5.5.4 ), we obtain:

 

 

 

 

(Eq. 5.5.8)

Comparing (Eq. 5.5.8 ) with (Eq. 5.5.1 ), we can write:

 

 

 

 

(Eq. 5.5.9)

Thus

Integrating w.r.t x,

 

 

 

 

(Eq. 5.5.10)

Substituting the obtained in (Eq. 5.5.10 ) back into the expression for obtained in (Eq. 5.5.6 ), we obtain:

 

 

 

 

(Eq. 5.5.11)

The partial derivative of (Eq. 5.5.11 ) w.r.t y,

 

 

 

 

(Eq. 5.5.12)

But from (Eq. 5.5.1 ) and (Eq. 5.5.2 ), we see that .

So,

Since, is only a function of , so, we can now say that and .

Thus is a constant.

Hence we obtain the following expression for :

 

 

 

 

(Eq. 5.5.13)

which represents a general class of Exact L2-ODE-VC.

R*5.6 Solving a L2-ODE-VC[9][edit]

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

Given[edit]

 

 

 

 

(Eq. 5.6.1)

Problem[edit]

1. Show that (Eq. 5.6.1 ) is exact.

2. Find

3. Solve for

Solution[edit]

Nomenclature[edit]

Exactness Conditions[10][edit]

The exactness conditions for N2-ODE (Non Linear Second Order Differential Equation) are:

First Exactness condition

For an equation to be exact, they must be of the form


 

 

 

 

(Eq. 5.6.2)


 

 

 

 

(Eq. 5.6.3)


Second Exactness Condition

 

 

 

 

(Eq. 5.6.4)

 

 

 

 

(Eq. 5.6.5)

Work[edit]

We have


Where we can identify


and


Thus the equation satisfies the first exactness condition.


For the second exactness condition, we first calculate the various partial derivatives of f and g.


Substituting the values in (Eq. 5.6.4 ) we get



Therefore the first equation satisfies.

Substituting the values in (Eq. 5.6.5 ) we get



Therefore the second equation satisfies as well.

Thus the second exactness condition is satisfied and the given differential equation is exact.

Now, we have

Integrating w.r.t. p, we get

where h(x,y) is a function of integration as we integrated only partially w.r.t. p.


 

 

 

 

(Eq. 5.6.6)


Partially differentiating (Eq. 5.6.6 ) w.r.t x



Partially differentiating (Eq. 5.6.6 ) w.r.t y



From equation (Eq. 5.6.3 ), we have





We have established that


Comparing the two equations, we get,



On integrating,


Thus,



Thus we have




This N1-ODE can be solved using the Integrating Factor Method that we very well know.




R*5.7 Show equivalence to symmetry of second partial derivatives of first integral[11][edit]

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

Given[edit]

 

 

 

 

(Eq.(1) p.22-3)

where

 

 

 

 

(Eq.(3) p.21-7)

Problem[edit]

Show equivalence to symmetry of mixed second partial derivatives of first integral, that is

where

Solution[edit]

 

 

 

 

(Eq.(3) p.21-7)

 

 

 

 

(Eq.(2) p.22-4)

From (Eq.(2)p.22-4 ), we have,

 

 

 

 

(Eq. 5.7.1)

 

 

 

 

(Eq.(3) p.22-4)

Substituting (Eq.(3)p.21-7 ),(Eq. 5.7.1 ) and (Eq.(3)p.22-4 ) into (Eq.(1)p.22-3 ) yields

 

 

 

 

(Eq. 5.7.2)

Because

Thus

 

 

 

 

(Eq. 5.7.3)

Substituting (Eq. 5.7.3 ) into (Eq. 5.7.2 ) yields

 

 

 

 

(Eq. 5.7.4)

Because

 

 

 

 

(Eq. 5.7.5)

Substitute (Eq. 5.7.5 ) into (Eq. 5.7.4 ), we have

 

 

 

 

(Eq. 5.7.6)

Since and can be the second and first derivative of any solution function of any second order ODE in terms of which the equation is hold. That is, the factor , which consists of two derivatives of solution function and the derivative operater so that depends partly on the solution functin of ODE, can be arbitrary and thus linearly indepent of the derivative operater , which is a factor of the third term on left hand side of (Eq. 5.7.6 ).

Similarly, comparing the first and the third terms on left hand side of (Eq. 5.7.6 ) yields that the factor 1 (which can be treated as a unit nature number basis of function space) of the first term and the derivative operater (which is another basis of derivative function space) of the third term are linearly independent of each other.

For the left side of (Eq. 5.7.6 ) being zero under any circumstances, we should have,

 

 

 

 

(Eq. 5.7.7)

while

 

 

 

 

(Eq. 5.7.8)

 

 

 

 

(Eq. 5.7.9)

From (Eq. 5.7.7 ),since the factor is arbitrary, we obtain,

 

 

 

 

(Eq. 5.7.10)

Thus,

 

 

 

 

(Eq. 5.7.11)

From(Eq. 5.7.9 ), consider to be also a function of variables x,y and p, which can be represented as , thus,

 

 

 

 

(Eq. 5.7.12)

Since the partial derivative opraters are linearly independent, we have,

 

 

 

 

(Eq. 5.7.13)

 

 

 

 

(Eq. 5.7.14)

 

 

 

 

(Eq. 5.7.15)

Obviously the only condition by which the three equations above are all satisfied is that the function is a numerical constant.

Thus, we have

 

 

 

 

(Eq. 5.7.16)

where is a constant. To find the value of constant , try the process as follow.

 

 

 

 

(Eq. 5.7.17)

Find integral on both sides of (Eq. 5.7.17 ) in terms of x,

 

 

 

 

(Eq. 5.7.18)

where the term is an arbitrarily selected function of independent variables y and p. Then find integral on both sides of (Eq. 5.7.18 ) in terms of p,

 

 

 

 

(Eq. 5.7.19)

where the term is an arbitrarily selected function of variables x and y.

The first partial derivatives of both sides of (Eq. 5.7.19 ) in terms of x could be

 

 

 

 

(Eq. 5.7.20)

 

 

 

 

(Eq. 5.7.21)

Then find partial derivative of both sides of (Eq. 5.7.21 ) in terms of p,

 

 

 

 

(Eq. 5.7.22)

 

 

 

 

(Eq. 5.7.23)

 

 

 

 

(Eq. 5.7.24)

Because the right hand side of (Eq. 5.7.24 ) is a function of two variables y and p, while the left hand side is a function of p' only, the equation (Eq. 5.7.24 ) could not hold if the constant has a non-zero value. Thus, the only condition by which the equation (Eq. 5.7.24 ) will be satisfied is that while ,that is, .

Substituting into (Eq. 5.7.16 ) yields,

 

 

 

 

(Eq. 5.7.25)

Thus we have

 

 

 

 

(Eq. 5.7.26)

We are now left with

Thus

 

 

 

 

(Eq. 5.7.27)

R*5.8. Working with the coefficients in 1st exactness condition[edit]

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


Given[edit]

 

 

 

 

(Eq.(1) p.22-2)

Problem [12][edit]

Using The Coefficients in the 1st exactness condition prove that (Eq.(1)p.22-3 ) can be written in the form

Solution[edit]

Nomenclature[edit]



For an equation to be exact, they must be of the form

 

 

 

 

(Eq. 5.8.2)

using chain and product rule

 

 

 

 

(Eq. 5.8.3)

 

 

 

 

(Eq. 5.8.4)

plugging Eq(2),(3),&(40 into Eq(1)

after cancellation of the opposite term

Now, we can club the terms



and



Since 1 and q, i.e the second derivative of y, are in general non linear, for the equation to hold true, their coefficients must both be equal to zero.

Thus we say that



and



Which is the required proof.


R5.9: Use of MacLaurin Series[edit]

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

Problem[edit]

Use Taylor Series at x=0 (MacLaurin Series) to derive[13]


Solution[edit]

The Taylor's series [14] expansion of a function f(x) about a real or complex number c is given by the formula

 

 

 

 

(Eq. 5.9.1)

When the neighborhood for the expansion is zero, i.e c = 0, the resulting series is called the Maclaurin Series.

Part a[edit]

We have the function


Table for Maclaurin Series
And so on.. ..

Rewriting the Maclaurin series expansion,

 

 

 

 

(Eq. 5.9.2)


Substituting the values from the tables in (Eq. 5.9.2 ) we get


 

 

 

 

(Eq. 5.9.3)

 

 

 

 

(Eq. 5.9.4)

Where[15]



We can represent



(Eq. 5.9.4 ) can be written as , hence proved.

Part b[edit]

We have the function

We will use a slightly different approach here when compared to part a of the solution. We will expand and multiply the resulting expanded function with


Table for Maclaurin Series
And so on.. ..

Rewriting the Maclaurin series expansion,

 

 

 

 

(Eq. 5.9.5)


Substituting the values from the tables in (Eq. 5.9.5 ) we get


 

 

 

 

(Eq. 5.9.6)


Multiplying (Eq. 5.9.6 ) with



This expression does not match the expression that we have been asked to prove. This, we believe is because there has been a misprint and the expression to be found out must be


Expanding using Maclaurin's series

Table for Maclaurin Series
And so on.. ..


Rewriting the Maclaurin series expansion,

 

 

 

 

(Eq. 5.9.7)


Substituting the values from the tables in (Eq. 5.9.7 ) we get

 

 

 

 

(Eq. 5.9.8)


Multiplying (Eq. 5.9.8 ) with



 

 

 

 

(Eq. 5.9.9)

Which is the expression in the RHS.

R5.10 Gauss Hypergeometric Series[16][edit]

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

Problem[edit]

1. Use MATLAB to plot near x=0 to show the local maximum (or maxima) in this region.


2. Show that

 

 

 

 

((1) pg. 64-9b)

Solution[edit]

The MATLAB code, shown below, will plot the hypergeometric function over the interval: .

x = [0:0.01:0.8]';
plot(x,hypergeom([5,-10],1,x))

The plot of the hypergeometric function near x=0 reveals a local maximum of 0.1481 at x = 0.23.

Hypergeometric Function.jpg


The hypergeometric function can be expressed as using the Pochhammer Symbol


where  

 

 

 

 

(Eq. 5.10.1)

Here , and .


The hypergeometric series represented by terminates after the 11th term because the constant b = -10. This is because starting with the 12th term in the series the factor appears in the numerator.


For the 12th term in the series k = 11, so


The hypergeometric series represented by the function can be written in expanded form:


 

 

 

 

(Eq. 5.10.2)


If the expansion of agrees with (Eq. 5.10.2 ) then it is a valid representation of the hypergeometric function.



 

 

 

 

(Eq. 5.10.3)


    

Combining all like terms yields the following:



The expansion of (Eq. 5.10.3 ) agrees with the expanded form of the hypergeometric function (Eq. 5.10.2 ), which confirms that ((1) pg. 64-9b ) is true.

R 5.11 Calculation of Time Taken by a projectile to hit the Ground[edit]

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

Given[edit]

 

 

 

 

(Eq.(1)p.63-9)


Where is a Hypergeometric Function.

Problem[17][edit]

Consider the integral in (3) Pg.63-8 and (Eq.(1)Pg.63-9 )



 

 

 

 

(Eq.5.11.1)


Let n=3, a=2 and b=10

For each value of time (t), solve for altitude z(t), plot z(t) vs t, and find the time when projectile returns to ground.


Solution[edit]

The given integral is a reduced form of the integral (3) Pg 63.8 which relates the mass of a projectile, the forces acting upon it when moving in air ( the air resistance, which is a function of its height in air, and its own weight) and the time taken for the projectile to reach the ground. Thus it represents a real world problem whose solution must actually exist.

We have been given the values of n, a and b. Substituting the values in (Eq.5.11.1 ), we get:




The solution of the above Hypergeometric function contains complex terms according to Wolfram Alpha [18] which does not seem to make sense as the function represents a real world problem with real numbers.

When expanded, this is a series that goes to infinity as there is no negative term in the hypergeometric function which will make one of the terms go to zero. This computation is beyond our ability. Hence the problem could not be solved.

R5.12: Hypergeometric Function[edit]

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

Problem[edit]

1. Is (1)p.64-10 exact?

2. Is (1)p.64-10 in the power form of (3) p.21-1?

3. Verify that F(a,b;c;x) is indeed a solution of (1) p.64-10.

Given[edit]

 

 

 

 

((1)p.64-10)

 

 

 

 

((2)p.16-4)

 

 

 

 

((3)p.16-4)

 

 

 

 

((4)p.16-4)

 

 

 

 

((2)p.7-3)

 

 

 

 

((1)p.16-5)

 

 

 

 

((2)p.16-5)

 

 

 

 

((3)p.21-1)

Solution[edit]

1. In order for (1) p.64-10 to be exact, it must first be in the form of (2)p.16-4, with g and f defined in (3)-(4) p.16-4, as seen below.

 

 

 

 

(Eq. 12.1)

Therefore, the first exactness condition is satisfied.

In order to satisfy the second exactness condition, the following derivatives must be found.

 

 

 

 

(Eq. 12.2)

 

 

 

 

(Eq. 12.3)

 

 

 

 

(Eq. 12.4)

 

 

 

 

(Eq. 12.5)

 

 

 

 

(Eq. 12.6)

 

 

 

 

(Eq. 12.7)

 

 

 

 

(Eq. 12.8)


 

 

 

 

(Eq. 12.9)

 

 

 

 

(Eq. 12.10)

 

 

 

 

(Eq. 12.11)

 

 

 

 

(Eq. 12.12)

 

 

 

 

(Eq. 12.13)

 

 

 

 

(Eq. 12.14)

 

 

 

 

(Eq. 12.15)

 

 

 

 

(Eq. 12.16)


By substituting into the 1st relation, (1) p.16-5:

 

 

 

 

(Eq. 12.17)

 

 

 

 

(Eq. 12.18)

This is not true for all values of a and b, so the 1st relation is not valid.


By substituting into the 2nd relation, (2) p.16-5: