Solving polynomial using matrix methods

They should sum to 1.So, for λ = 20:-20a4 = 0,=> a4 = 0-12a3 = 0=> a3 = 0-6a2 + 12a4 = 0=> a2 = 0-2a1 + 6a3 = 0=> a1 = 020a0 + 2a2 = 0=> a0 = (-1/10)a2 = (3/35)a4Condition: a0 + a1 + a2 + a3 + a4 = 1a0 = 0a1 = 0a2
  • #1
unscientific
1,734
13

Homework Statement



Let y(x) = a0 + a1x + a2x2 + a3x3 + a4x4

Represent the differential equation

(d/dx)[ (1-x2)(dy/dx) ] + λy = 0

in matrix form. Find the values of λ for which there is a solution to the matrix equation, and find the solutions for the smallest and largest values of λ. Normalize these solutions with the condition y(1) = 1.


Homework Equations





The Attempt at a Solution



I got as far as finding the values of λ = 0, 2, 6, 12, 20. Am i supposed to plug in the values of λ into the differential equation and solve it by finding the characteristic equation by assuming y = Aeβx? But the question gives y as a polynomial: y(x) = a0 + a1x + a2x2 + a3x3 + a4x4

6nz8nl.png
 
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  • #2
unscientific said:

Homework Statement



Let y(x) = a0 + a1x + a2x2 + a3x3 + a4x4

Represent the differential equation

(d/dx)[ (1-x2)(dy/dx) ] + λy = 0

in matrix form. Find the values of λ for which there is a solution to the matrix equation, and find the solutions for the smallest and largest values of λ. Normalize these solutions with the condition y(1) = 1.


Homework Equations





The Attempt at a Solution



I got as far as finding the values of λ = 0, 2, 6, 12, 20. Am i supposed to plug in the values of λ into the differential equation and solve it by finding the characteristic equation by assuming y = Aeβx? But the question gives y as a polynomial: y(x) = a0 + a1x + a2x2 + a3x3 + a4x4

6nz8nl.png

Check your matrix. You are missing a λ in the first row. That will give you another value for λ. And you are supposed to substitute the values of λ into your matrix and find corresponding a's.
 
  • #3
Dick said:
Check your matrix. You are missing a λ in the first row. That will give you another value for λ. And you are supposed to substitute the values of λ into your matrix and find corresponding a's.

Oops i wrote my matrix wrongly from my working on paper. The smallest value of λ = 0.

I don't understand what you mean by "finding the corresponding a's".

Say i sub in λ = 0. All i get is: BA = 0, where B is the matrix with the subbed value of λ and A is the column with all the a0, a1, ...

For example, let's consider [5,5].

(-20)(a4) = 0.

This doesn't mean anything other than a4 = 0. Then it would mean a0 = a1 = a2 = a3 = a4 = 0
 
  • #4
unscientific said:
Oops i wrote my matrix wrongly from my working on paper. The smallest value of λ = 0.

I don't understand what you mean by "finding the corresponding a's".

Say i sub in λ = 0. All i get is: BA = 0, where B is the matrix with the subbed value of λ and A is the column with all the a0, a1, ...

For example, let's consider [5,5].

(-20)(a4) = 0.

This doesn't mean anything other than a4 = 0. Then it would mean a0 = a1 = a2 = a3 = a4 = 0

You are jumping to conclusions. Just because a4=0 doesn't mean ALL of they a's have to be zero. They don't. Which one doesn't?
 
  • #5
Dick said:
You are jumping to conclusions. Just because a4=0 doesn't mean ALL of they a's have to be zero. They don't. Which one doesn't?

Taking λ = 0,

Let's start with the bottom row.

-20a4 = 0,

=> a4 = 0

-12a3 = 0

=> a3 = 0

-6a2 + 12a4 = 0

=> a2 = 0

-2a1 + 6a3 = 0

=> a1 = 0

and finally 2a2 = 0
 
  • #6
unscientific said:
Taking λ = 0,

Let's start with the bottom row.

-20a4 = 0,

=> a4 = 0

-12a3 = 0

=> a3 = 0

-6a2 + 12a4 = 0

=> a2 = 0

-2a1 + 6a3 = 0

=> a1 = 0

and finally 2a2 = 0

You forgot a0.
 
  • #7
Dick said:
You forgot a0.

I put in λ = 0 into the matrix at [1,1]. This gives a0 = 0 as well..
 
  • #8
No, it doesn't.
 
  • #9
vela said:
No, it doesn't.

0 * a0 + 2a1 = 0

does this mean a0 can be any real number?
 
  • #10
To satisfy that equation, yes. But it needs to be non-zero because you're looking for a non-trivial solution to the system of equations.
 
  • #11
vela said:
a

Then when i sub in λ = 20,

I have a4 = any real number,

a3 = 0,

a2 = -(6/7)a4

a1 = 0

a0 = 0

is this right?
 
  • #12
You shouldn't get ##a_0=0##. The rest are right.
 
  • #13
vela said:
You shouldn't get ##a_0=0##. The rest are right.

For λ = 0

-20a4 = 0,

=> a4 = 0

-12a3 = 0

=> a3 = 0

-6a2 + 12a4 = 0

=> a2 = 0

-2a1 + 6a3 = 0

a1*0 = 0

a1 [itex]\subset[/itex] ℝ


is it because from the condition:

y(1) = 1

a0 + a1 + a2 + a3 + a4 = 1

So, a0 = 1


For λ = 20

=> a4 [itex]\subset[/itex] ℝ,

=> a3 = 0

14a2 + 12a4 = 0
=> a2 = (-6/7) a4

18a1 + 6a3 = 0
=> a1 = 0

20a0 + 2a2 = 0
=> a0 = (-1/10)a2 = (3/35)a4

Using a0 + a1 + a2 + a3 + a4 = 1



a0 = 3/8
a1 = 0
a2 = -48/245
a3 = 0
a4 = 35/8
 
Last edited:
  • #14
unscientific said:
For λ = 0

-20a4 = 0,

=> a4 = 0

-12a3 = 0

=> a3 = 0

-6a2 + 12a4 = 0

=> a2 = 0

-2a1 + 6a3 = 0

a1*0 = 0

a1 [itex]\subset[/itex] ℝis it because from the condition:

y(1) = 1

a0 + a1 + a2 + a3 + a4 = 1

So, a0 = 1For λ = 20

=> a4 [itex]\subset[/itex] ℝ,

=> a3 = 0

14a2 + 12a4 = 0
=> a2 = (-6/7) a4

18a1 + 6a3 = 0
=> a1 = 0

20a0 + 2a2 = 0
=> a0 = (-1/10)a2 = (3/35)a4

Using a0 + a1 + a2 + a3 + a4 = 1
a0 = 3/8
a1 = 0
a2 = -48/245
a3 = 0
a4 = 35/8

All correct, except your final a's for λ=20 don't sum to 1.
 

FAQ: Solving polynomial using matrix methods

What is the basic concept behind solving polynomials using matrix methods?

The basic concept behind solving polynomials using matrix methods is to convert the polynomial equation into a matrix form, where the unknown variables are represented by the elements of the matrix. By using matrix operations, such as multiplication and inversion, the solution to the polynomial equation can be obtained.

What are the advantages of using matrix methods to solve polynomials?

One advantage is that matrix methods can handle equations with multiple unknown variables, which can be difficult to solve using traditional methods. Additionally, matrix methods provide a systematic and efficient approach to solving polynomials, making it easier to find the solution.

What are the common steps involved in solving polynomials using matrix methods?

The first step is to convert the polynomial equation into a matrix form. Then, the matrix is manipulated using operations such as multiplication, inversion, and elimination to isolate the unknown variables. Finally, the values of the unknown variables are determined and substituted back into the original equation to check the solution.

Are there any limitations to using matrix methods for solving polynomials?

Yes, matrix methods may not be suitable for solving polynomials with complex roots. Also, the size of the matrix increases as the degree of the polynomial increases, making it more computationally intensive for higher degree polynomials.

Can matrix methods be applied to all types of polynomial equations?

Matrix methods can be applied to linear, quadratic, and higher degree polynomials. However, the process and complexity may vary depending on the degree of the polynomial. For higher degree polynomials, numerical methods may be more efficient for finding the solution.

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