University of Florida/Egm6321/f12.Rep5hid

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

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

Given

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and

 

 

 

 

 

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

Show That[1]

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(Eq.5.1.1)

Solution

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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]

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

Given

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A Diagonal Matrix

 , where,  .

 

 

 

 

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

Problem

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Show that

 , where,  .

 

 

 

 

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

Solution

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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

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

Given

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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

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Show that

 

Solution

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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

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

Given

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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

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Show

 

 

 

 

 

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

and

 

 

 

 

 

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

Solution

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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]

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

Given

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A L2-ODE-VC [6]:

 

 

 

 

 

(Eq. 5.5.1)

The first intregal   can also be expressed as:

 

 

 

 

 

(Eq. 5.5.2)

Problem [7]

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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

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Nomenclature

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Derivation of Eq. 5.5.3

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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]

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

Given

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(Eq. 5.6.1)

Problem

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

2. Find  

3. Solve for  

Solution

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Nomenclature

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Exactness Conditions[10]

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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

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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]

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

Given

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

where

 

 

 

 

 

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

Problem

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

 

where

 

Solution

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(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

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


Given

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

Problem [12]

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

 

Solution

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Nomenclature

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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

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

Problem

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

 


 

Solution

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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

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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

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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]

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

Problem

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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

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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.

 


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

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

Given

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


Where   is a Hypergeometric Function.

Problem[17]

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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

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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

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

Problem

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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

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((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

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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:

 

 

 

 

 

(Eq. 12.19)

 

 

 

 

 

(Eq. 12.20)

This is true, so the 2nd relation is valid.


One of the relations is not valid, therefore the second exactness condition is not satisfied.

Hence, (1) p.64-10 is not exact.


2. The following equalities must be true for (1) p.64-10 to be in power form of (3) p.21-1.


 

 

 

 

 

(Eq. 12.21)

 

 

 

 

 

(Eq. 12.22)

 

 

 

 

 

(Eq. 12.23)


Since there are no values of   that make these equalities true, then (1) p.64-10 is not in power form.


3. In order to verify that F(a,b;c;x) is a solution of (1) p.64-10, we select the example of  .

 

 

 

 

 

(Eq. 12.24)

Next, the first and second derivatives of y must be found.

 

 

 

 

 

(Eq. 12.25)

 

 

 

 

 

(Eq. 12.26)

Substituting into (1) p.64-10:

 

 

 

 

 

(Eq. 12.27)

 

 

 

 

 

(Eq. 12.28)

 

 

 

 

 

(Eq. 12.29)

 

 

 

 

 

(Eq. 12.30)

 

 

 

 

 

(Eq. 12.31)

This equation is valid, therefore, F(a,b;c;x) is a solution for (1) p.64-10.

R*5.13 Exactness of Legendre and Hermite equations [19]

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

Given

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Given

Legendre equation:

 

 

 

 

 

(Eq. 5.13.1)

Hermite equation:

 

 

 

 

 

(Eq. 5.13.2)


Problem

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1. Verify the exactness of the Legendre (Eq. 5.13.1) and Hermite (Eq. 5.13.2) equations.

2. If Hermite equation is not exact, check whether it is in power form, and see whether it can be made exact using IFM with  .

3. The first few Hermite polynomials are:

 
 
 

Verify that these are homogeneous solutions to the Hermite differential equation (Eq. 5.13.2).

Solution

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Nomenclature

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Exactness of Legendre equation

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To satisfy the first exactness condition, the Legendre equation (Eq. 5.13.1) should be of the form:

 

 

 

 

 

(Eq. 5.13.3)

 

Hence (Eq. 5.13.1) satisfies the first exactness condition.

The second exactness condition can be checked in two ways.

Method 1

The second exactness condition is satisfied if (Eq. 5.13.1) satisfies (Eq. 5.13.4) and (Eq. 5.13.5):

 

 

 

 

 

(Eq. 5.13.4)


 

 

 

 

 

(Eq. 5.13.5)

Computing derivatives,

 

Substituting these into (Eq. 5.13.4) and (Eq. 5.13.5),


(Eq. 5.13.4)  .

(Eq. 5.13.5)  .

Hence, the second exactness condition is satisfied when   or  .

Method 2

The second exacness condition is met if (Eq. 5.13.1) satisfies:

 

 

 

 

 

(Eq. 5.13.6)

where  

Computing the derivatives,

 

Substituting these in (Eq. 5.13.6) yields,

 
 

Again we see that the second exactness condition is satisfied when   or  .

Exactness of Hermite equation

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To satisfy the first exactness condition, the Hermite equation (Eq. 5.13.2) should be of the form (Eq. 5.13.3):


 

Hence (Eq. 5.13.2) satisfies the first exactness condition.

The second exactness condition can be checked in two ways.

Method 1

The second exactness condition is satisfied if (Eq. 5.13.2) satisfies (Eq. 5.13.4) and (Eq. 5.13.5):

Computing the derivatives,

 

Substituting these into (Eq. 5.13.4) and (Eq. 5.13.5),


(Eq. 5.13.4)  .

(Eq. 5.13.5)  .

Hence, the second exactness condition is satisfied only when   This is a necessary condition.

Method 2

The second exacness condition is met if (Eq. 5.13.2) satisfies (Eq. 5.13.6).

Computing the derivatives,


 

Substituting these in (Eq. 5.13.6) yields,

 
 

The second exactness condition is satisfied only when  .

Power form and making the Hermite equation exact using IFM

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We have seen that the Hermite equation (Eq. 5.13.2) is not exact when  .

The power form of L2-ODE-VC is

 

 

 

 

 

(Eq. 5.13.7)

Comparing (Eq. 5.13.2) with (Eq. 5.13.7), we can see that the Hermite equation is of the power form with:

 
 

Hence, we can consider an integrating factor which is in power form,  

Replacing the 'n' term in (Eq. 5.13.2) with   to avoid confusion, we need to find  , such that the following N2-ODE is exact:

 

 

 

 

 

(Eq. 5.13.8)

The Hermite equation (Eq. 5.13.2) can be written as:

 

 

 

 

 

(Eq. 5.13.9)

(Eq. 5.13.9) should satisfy (Eq. 5.13.4) and (Eq. 5.13.5) to meet the second exactness condition.

Computing the derivatives,

 

Substituting the derivatives in (Eq. 5.13.4) and (Eq. 5.13.5),

(Eq. 5.13.4)  

(Eq. 5.13.5)  

We can see that the second exactness condition can be satisfied only when  . When  ,

 .

Hence we can say

 
 

Therefore,   is a solution.

Hence, (Eq. 5.13.2) can be made exact using the integrating factor  

Verification of homogeneous solutions of the Hermite equation

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Case 1

 
 
 

Substituting in (Eq. 5.13.2),  


Case 2

 
 
 

Substituting in (Eq. 5.13.2),  

Case 3

 
 
 

Substituting in (Eq. 5.13.2),  

Hence the given first three Hermite polynomials are homogeneous solutions of the Hermite equation.

R*5.14 Expressions for X(x) [20]

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

Given

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Given

 

 

 

 

 

(Eq. 5.14.1)

 

 

 

 

 

(Eq. 5.14.2)

Where

 
 

Problem

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Find expressions for   in terms of  

Solution

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By definition,

  and
 

Hence, (Eq. 5.14.2) can be written as:

 
 

 

 

 

 

(Eq. 5.14.3)


where,

 
 
 
 

Hence, (Eq. 5.14.3) is a generic expression for X(x).

Contributing Members

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Problem Contribution
Problem number Solved by Typed by Proofread by
R5.1 Ramchandra, Mohammed Ramchandra Linghan
R5.2 Amrith, Fabian Amrith Ramchandra
R5.3 Linghan, Sarah Linghan Amrith
R5.4 Linghan, Rahul Linghan Fabian
R*5.5 Rahul, Mohammed, Fabian Rahul Amrith
R*5.6 Ramchandra, Amrith, Linghan Ramchandra Fabian
R*5.7 Linghan, Sarah, Mohammed Linghan, Ramchandra Ramchandra
R*5.8 Mohammed, Amrith Mohammed Rahul
R5.9 Ramchandra, Rahul Ramchandra Sarah
R5.10 Fabian, Amrith Fabian Ramchandra
R5.11 Fabian, Ramchandra, Amrith Ramchandra ----
R5.12 Sarah, Fabian, Mohammed Sarah Amrith
R*5.13 Rahul, Ramchandra, Sarah Rahul Mohammed
R*5.14 Rahul, Sarah, Linghan Rahul Mohammed

References

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  1. Lecture Notes Setion 20 Pg. 20-2b
  2. Lecture Notes Section 20 Pg 20-2b
  3. Lecture Notes Section 20 Pg 20-1
  4. Euler Formula
  5. Lecture Notes Section 21 Pg 21-5
  6. Lecture Notes Section 21 Pg 21-4
  7. Lecture Notes Section 21 Pg 21-5
  8. Lecture Notes Section 21 Pg 21-4
  9. Lecture Notes Section 21 Pg 21-6
  10. Lecture Notes Section 16
  11. Lecture Notes Section 22 Pg 22-4
  12. Lecture Notes Section 22 Pg 22-6
  13. Lecture Notes Section 64 Pg 64-7b
  14. Taylor Series Wiki
  15. Lecture Notes Section 64 Pg 64-4
  16. Lecture Notes Section 64 Pg 64-9b
  17. Lecture Notes Section 64 Pg. 64-9b
  18. Wolfram Alpha's Solution to Hypergeometric Function
  19. Lecture Notes Section 27 Pg 27-1, 27-2
  20. Lecture Notes Section 30 Pg 30-3