To prove that the determinant of the Vandermonde matrix V is det ( V ( x 0 , x 1 , ... , x n ) ) = ∏ 0 ≤ j < k ≤ n − 1 ( x k − x j )

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter D, Problem 1P

Program Plan Intro

To prove that the determinant of the Vandermonde matrix V is det(V(x0,x1,...,xn))=∏0≤j<k≤n−1(xk−xj)

Expert Solution & Answer

Explanation of Solution

Given information:

Vandermonde matrix V is given by V(x0,x1,...xn−1)=(1x0x02x0n−11x1x12x1n−1............1xn−1xn−12xn−1n−1)

Explanation:

Start with induction on n .

For n = 1, the Vandermonde matrix is of size 1×1 with only one entry i.e. 1.It is easy to note that this matrix’s determinant is 1. Also product is one as no distinct i and j can be picked in the product on the right hand side.

Now, assume that the given condition is true for n×n matrix. Using this let’s show that it is true for (n+1)×(n+1) matrices.

Start from second last column on right side that is n−1 and go towards left adding (−x0) times that column to the one to its right.

The determinant will not be changed from this operation. All entries in the top row, except the leftmost one, have become zero as x0j−1−x0j−2⋅x0=0 . This means that expansion by minors along the top row can be done in order to compute the determinant. The top left column will be the only non-zero term in this operation. Thus determinant remains the same as the determinant of that one minor. Also, the factor of every item in i^t^h of that minor is (xi−x0) . This implies that taking all those factors out will give expression ∏i=1n−1(xi−x0) i.e. all the factors in the formula that include x0 .

After taking factors out,a n×n Vandermonde matrix on the variables x1,...,xn−1 is left.This will only contain all the other factors that are needed to get the right hand side.

Thus, this completes the induction.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

Please include comments and docs comments on the program. The two other classes are Attraction and Entertainment.

Object-Oriented Programming In this separate files. ent, you'll need to build and run a small Zoo in Lennoxville. All classes must be created in Animal (5) First, start by building a class that describes an Animal at a Zoo. It should have one private instance variable for the name of the animal, and one for its hunger status (fed or hungry). Add methods for setting and getting the hunger satus variable, along with a getter for the name. Consider how these should be named for code clarity. For instance, using a method called hungry () to make the animal hungry could be used as a setter for the hunger field. The same logic could be applied to when it's being fed: public void feed () { this.fed = true; Furthermore, the getter for the fed variable could be named is Fed as it is more descriptive about what it answers when compared to get Fed. Keep this technique in mind for future class designs. Zoo (10) Now we have the animals designed and ready for building a little Zoo! Build a class…

1.[30 pts] Answer the following questions: a. [10 pts] Write a Boolean equation in sum-of-products canonical form for the truth table shown below: A B C Y 0 0 0 1 0 0 1 0 0 1 0 0 0 1 1 0 1 0 0 1 1 0 1 0 1 1 0 1 1 1 1 0 a. [10 pts] Minimize the Boolean equation you obtained in (a). b. [10 pts] Implement, using Logisim, the simplified logic circuit. Include an image of the circuit in your report. 2. [20 pts] Student A B will enjoy his picnic on sunny days that have no ants. He will also enjoy his picnic any day he sees a hummingbird, as well as on days where there are ants and ladybugs. a. Write a Boolean equation for his enjoyment (E) in terms of sun (S), ants (A), hummingbirds (H), and ladybugs (L). b. Implement in Logisim, the logic circuit of E function. Use the Circuit Analysis tool in Logisim to view the expression, include an image of the expression generated by Logisim in your report. 3.[20 pts] Find the minimum equivalent circuit for the one shown below (show your work): DAB C…

Answer 2

Question

Chapter D, Problem 1P

Program Plan Intro

To prove that the determinant of the Vandermonde matrix V is det(V(x0,x1,...,xn))=∏0≤j<k≤n−1(xk−xj)

Expert Solution & Answer

Explanation of Solution

Given information:

Vandermonde matrix V is given by V(x0,x1,...xn−1)=(1x0x02x0n−11x1x12x1n−1............1xn−1xn−12xn−1n−1)

Explanation:

Start with induction on n .

For n = 1, the Vandermonde matrix is of size 1×1 with only one entry i.e. 1.It is easy to note that this matrix’s determinant is 1. Also product is one as no distinct i and j can be picked in the product on the right hand side.

Now, assume that the given condition is true for n×n matrix. Using this let’s show that it is true for (n+1)×(n+1) matrices.

Start from second last column on right side that is n−1 and go towards left adding (−x0) times that column to the one to its right.

The determinant will not be changed from this operation. All entries in the top row, except the leftmost one, have become zero as x0j−1−x0j−2⋅x0=0 . This means that expansion by minors along the top row can be done in order to compute the determinant. The top left column will be the only non-zero term in this operation. Thus determinant remains the same as the determinant of that one minor. Also, the factor of every item in i^t^h of that minor is (xi−x0) . This implies that taking all those factors out will give expression ∏i=1n−1(xi−x0) i.e. all the factors in the formula that include x0 .

After taking factors out,a n×n Vandermonde matrix on the variables x1,...,xn−1 is left.This will only contain all the other factors that are needed to get the right hand side.

Thus, this completes the induction.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Chapter D, Problem 1P

Program Plan Intro

To prove that the determinant of the Vandermonde matrix V is det(V(x0,x1,...,xn))=∏0≤j<k≤n−1(xk−xj)

Expert Solution & Answer

Explanation of Solution

Given information:

Vandermonde matrix V is given by V(x0,x1,...xn−1)=(1x0x02x0n−11x1x12x1n−1............1xn−1xn−12xn−1n−1)

Explanation:

Start with induction on n .

For n = 1, the Vandermonde matrix is of size 1×1 with only one entry i.e. 1.It is easy to note that this matrix’s determinant is 1. Also product is one as no distinct i and j can be picked in the product on the right hand side.

Now, assume that the given condition is true for n×n matrix. Using this let’s show that it is true for (n+1)×(n+1) matrices.

Start from second last column on right side that is n−1 and go towards left adding (−x0) times that column to the one to its right.

The determinant will not be changed from this operation. All entries in the top row, except the leftmost one, have become zero as x0j−1−x0j−2⋅x0=0 . This means that expansion by minors along the top row can be done in order to compute the determinant. The top left column will be the only non-zero term in this operation. Thus determinant remains the same as the determinant of that one minor. Also, the factor of every item in i^t^h of that minor is (xi−x0) . This implies that taking all those factors out will give expression ∏i=1n−1(xi−x0) i.e. all the factors in the formula that include x0 .

After taking factors out,a n×n Vandermonde matrix on the variables x1,...,xn−1 is left.This will only contain all the other factors that are needed to get the right hand side.

Thus, this completes the induction.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter D, Problem 1P

Program Plan Intro

To prove that the determinant of the Vandermonde matrix V is det(V(x0,x1,...,xn))=∏0≤j<k≤n−1(xk−xj)

Expert Solution & Answer

Explanation of Solution

Given information:

Vandermonde matrix V is given by V(x0,x1,...xn−1)=(1x0x02x0n−11x1x12x1n−1............1xn−1xn−12xn−1n−1)

Explanation:

Start with induction on n .

For n = 1, the Vandermonde matrix is of size 1×1 with only one entry i.e. 1.It is easy to note that this matrix’s determinant is 1. Also product is one as no distinct i and j can be picked in the product on the right hand side.

Now, assume that the given condition is true for n×n matrix. Using this let’s show that it is true for (n+1)×(n+1) matrices.

Start from second last column on right side that is n−1 and go towards left adding (−x0) times that column to the one to its right.

The determinant will not be changed from this operation. All entries in the top row, except the leftmost one, have become zero as x0j−1−x0j−2⋅x0=0 . This means that expansion by minors along the top row can be done in order to compute the determinant. The top left column will be the only non-zero term in this operation. Thus determinant remains the same as the determinant of that one minor. Also, the factor of every item in i^t^h of that minor is (xi−x0) . This implies that taking all those factors out will give expression ∏i=1n−1(xi−x0) i.e. all the factors in the formula that include x0 .

After taking factors out,a n×n Vandermonde matrix on the variables x1,...,xn−1 is left.This will only contain all the other factors that are needed to get the right hand side.

Thus, this completes the induction.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter D, Problem 1P

Program Plan Intro

To prove that the determinant of the Vandermonde matrix V is det(V(x0,x1,...,xn))=∏0≤j<k≤n−1(xk−xj)

Expert Solution & Answer

Explanation of Solution

Given information:

Vandermonde matrix V is given by V(x0,x1,...xn−1)=(1x0x02x0n−11x1x12x1n−1............1xn−1xn−12xn−1n−1)

Explanation:

Start with induction on n .

For n = 1, the Vandermonde matrix is of size 1×1 with only one entry i.e. 1.It is easy to note that this matrix’s determinant is 1. Also product is one as no distinct i and j can be picked in the product on the right hand side.

Now, assume that the given condition is true for n×n matrix. Using this let’s show that it is true for (n+1)×(n+1) matrices.

Start from second last column on right side that is n−1 and go towards left adding (−x0) times that column to the one to its right.

The determinant will not be changed from this operation. All entries in the top row, except the leftmost one, have become zero as x0j−1−x0j−2⋅x0=0 . This means that expansion by minors along the top row can be done in order to compute the determinant. The top left column will be the only non-zero term in this operation. Thus determinant remains the same as the determinant of that one minor. Also, the factor of every item in i^t^h of that minor is (xi−x0) . This implies that taking all those factors out will give expression ∏i=1n−1(xi−x0) i.e. all the factors in the formula that include x0 .

After taking factors out,a n×n Vandermonde matrix on the variables x1,...,xn−1 is left.This will only contain all the other factors that are needed to get the right hand side.

Thus, this completes the induction.

Answer 6

Chapter D, Problem 1P

Program Plan Intro

To prove that the determinant of the Vandermonde matrix V is det(V(x0,x1,...,xn))=∏0≤j<k≤n−1(xk−xj)

Expert Solution & Answer

Explanation of Solution

Given information:

Vandermonde matrix V is given by V(x0,x1,...xn−1)=(1x0x02x0n−11x1x12x1n−1............1xn−1xn−12xn−1n−1)

Explanation:

Start with induction on n .

For n = 1, the Vandermonde matrix is of size 1×1 with only one entry i.e. 1.It is easy to note that this matrix’s determinant is 1. Also product is one as no distinct i and j can be picked in the product on the right hand side.

Now, assume that the given condition is true for n×n matrix. Using this let’s show that it is true for (n+1)×(n+1) matrices.

Start from second last column on right side that is n−1 and go towards left adding (−x0) times that column to the one to its right.

The determinant will not be changed from this operation. All entries in the top row, except the leftmost one, have become zero as x0j−1−x0j−2⋅x0=0 . This means that expansion by minors along the top row can be done in order to compute the determinant. The top left column will be the only non-zero term in this operation. Thus determinant remains the same as the determinant of that one minor. Also, the factor of every item in i^t^h of that minor is (xi−x0) . This implies that taking all those factors out will give expression ∏i=1n−1(xi−x0) i.e. all the factors in the formula that include x0 .

After taking factors out,a n×n Vandermonde matrix on the variables x1,...,xn−1 is left.This will only contain all the other factors that are needed to get the right hand side.

Thus, this completes the induction.

Answer 7

Chapter D, Problem 1P

Program Plan Intro

To prove that the determinant of the Vandermonde matrix V is det(V(x0,x1,...,xn))=∏0≤j<k≤n−1(xk−xj)

Answer 8

Expert Solution & Answer

Explanation of Solution

Given information:

Vandermonde matrix V is given by V(x0,x1,...xn−1)=(1x0x02x0n−11x1x12x1n−1............1xn−1xn−12xn−1n−1)

Explanation:

Start with induction on n .

For n = 1, the Vandermonde matrix is of size 1×1 with only one entry i.e. 1.It is easy to note that this matrix’s determinant is 1. Also product is one as no distinct i and j can be picked in the product on the right hand side.

Now, assume that the given condition is true for n×n matrix. Using this let’s show that it is true for (n+1)×(n+1) matrices.

Start from second last column on right side that is n−1 and go towards left adding (−x0) times that column to the one to its right.

The determinant will not be changed from this operation. All entries in the top row, except the leftmost one, have become zero as x0j−1−x0j−2⋅x0=0 . This means that expansion by minors along the top row can be done in order to compute the determinant. The top left column will be the only non-zero term in this operation. Thus determinant remains the same as the determinant of that one minor. Also, the factor of every item in i^t^h of that minor is (xi−x0) . This implies that taking all those factors out will give expression ∏i=1n−1(xi−x0) i.e. all the factors in the formula that include x0 .

After taking factors out,a n×n Vandermonde matrix on the variables x1,...,xn−1 is left.This will only contain all the other factors that are needed to get the right hand side.

Thus, this completes the induction.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Ch. D.1 - Prob. 1E Ch. D.1 - Prob. 2E Ch. D.1 - Prob. 3E Ch. D.1 - Prob. 4E Ch. D.2 - Prob. 1E Ch. D.2 - Prob. 2E Ch. D.2 - Prob. 3E Ch. D.2 - Prob. 4E Ch. D.2 - Prob. 5E Ch. D.2 - Prob. 6E

To prove that the determinant of the Vandermonde matrix V is det ( V ( x 0 , x 1 , ... , x n ) ) = ∏ 0 ≤ j &lt; k ≤ n − 1 ( x k − x j )

To prove that the determinant of the Vandermonde matrix V is det(V(x0,x1,...,xn))=∏0≤j<k≤n−1(xk−xj)

Explanation of Solution

Want to see more full solutions like this?

Chapter D Solutions

To prove that the determinant of the Vandermonde matrix V is det ( V ( x 0 , x 1 , ... , x n ) ) = ∏ 0 ≤ j < k ≤ n − 1 ( x k − x j )