Solving a system of linear equations

In summary: But there are an infinite number of values of c and d that will do that.In summary, Fred is trying to determine two third degree polynomials p and q from given conditions. Using substitution, he obtains a system of linear equations and solves it to find the coefficients of p and q. However, due to the number of unknowns, there are an infinite number of solutions to the system.
  • #1
Mathman23
254
0
Hi

I have obtain two third degree polynomials p and q which are determint by the following conditions:

p(-1) = 1 , p'(-1) = 0

q(1) = 3 , q'(1) = 0

p(0) = q(0) , p'(0) = q'(0)

where p = a_1 * x^3 + b_1 *x^2 + c_1 *x + d_1
q = a_2 * x^3 + b_2 *x^2 + c_2 *x + d_2

I then end up with the the following linear equations by inserting the conditions into the equations above:

-a_1 + b_1 - c_1 + d_1 = 1

3* a_1 - 2 * b-1 + c_1 = 0

c_1 = d_1

3* a_2 + b_2 + c_2 + d_2 = 3

3 * a_2 + b_2 + c_2

d_1 = d_2

By the use of substitution I obtain the result, that

a_1 = 1/5 , b_1 = 34/5 , c_1 = d_1 = -15/5

My question is it correct to use substitution? If yes can I use approach to obtain a_2, b_2, c_2 and d_2 ?

Sincerley
Fred
 
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  • #2
I think your approach is, at least in principle, sound. However, it doesn't seem like you're getting the right equations.

You have:
[tex]p_1(x)=a_1x^3+b_1x^2+c_1x+d_1[/tex]
and
[tex]p_2(x)=a_2x^3+b_2x^2+c_2x+d_2[/tex]

Then you can convert your other conditions:
[tex]p_1(0)=p_2(0) \Rightarrow d_1=d_2[/tex]
like you have, but I don't see
[tex]p_1'(0)=p_2'(0) \Rightarrow c_1=c_2[/tex]
Which may be due to a transcription error. (I'm not going to speculate further on that.)

Moving on from there, you can certainly solve the system of equations using substitution, or some other method.
 
  • #3
Hi and thanks for Your answer,

My problem concerns cubic splines. I'm suppose to determine two third degree polynomials p and q from the conditions given in my earlier post.

My linear Algebra book doesn't explain this process very well, so I would very much appricate if someone could direct me to a website which explains this process and which hopefully could contain a couple of examples.

Thanks in advance,

Sincerely
Fred

NateTG said:
I think your approach is, at least in principle, sound. However, it doesn't seem like you're getting the right equations.

You have:
[tex]p_1(x)=a_1x^3+b_1x^2+c_1x+d_1[/tex]
and
[tex]p_2(x)=a_2x^3+b_2x^2+c_2x+d_2[/tex]

Then you can convert your other conditions:
[tex]p_1(0)=p_2(0) \Rightarrow d_1=d_2[/tex]
like you have, but I don't see
[tex]p_1'(0)=p_2'(0) \Rightarrow c_1=c_2[/tex]
Which may be due to a transcription error. (I'm not going to speculate further on that.)

Moving on from there, you can certainly solve the system of equations using substitution, or some other method.
 
  • #4
When you've got a polynomial like
[tex]p(x)=ax^3+bx^2+cx+d[/tex]
and something like
[tex]p(1)=5[/tex]
you can just substitute the value in for x. I.e.
[tex]p(1) \equiv a1^3+b1^2+c1+d[/tex]
You shouldn't have to deal with solving any kind of cubic to do this problem if you substitute in the x in all of the expressions.
 
  • #5
NateTG said:
I think your approach is, at least in principle, sound. However, it doesn't seem like you're getting the right equations.

You have:
[tex]p_1(x)=a_1x^3+b_1x^2+c_1x+d_1[/tex]
and
[tex]p_2(x)=a_2x^3+b_2x^2+c_2x+d_2[/tex]

Then you can convert your other conditions:
[tex]p_1(0)=p_2(0) \Rightarrow d_1=d_2[/tex]
like you have, but I don't see
[tex]p_1'(0)=p_2'(0) \Rightarrow c_1=c_2[/tex]
Which may be due to a transcription error. (I'm not going to speculate further on that.)
The original post (maybe it had been edited before I saw it)
said that p1[/sup]'(0)= 0, p2'(0)= 0
That would certainly give c1= 0, c2= 0!
 
  • #6
You mean I need to put it into a matrix in order to solve it?

/Fred

NateTG said:
When you've got a polynomial like
[tex]p(x)=ax^3+bx^2+cx+d[/tex]
and something like
[tex]p(1)=5[/tex]
you can just substitute the value in for x. I.e.
[tex]p(1) \equiv a1^3+b1^2+c1+d[/tex]
You shouldn't have to deal with solving any kind of cubic to do this problem if you substitute in the x in all of the expressions.
 
  • #7
Mathman23 said:
You mean I need to put it into a matrix in order to solve it?

/Fred

Not really, no. (Although that is an option.) There are many different ways to solve systems of linear equations.

If you have
[tex]p(1)=5[/tex]
and
[tex]p(x)=ax^3+bx^2+cx+d[/tex]
then you have
[tex]5=p(1)=a1^3+b1^2+c1+d=a+b+c+d[/tex]
so
[tex]5=a+b+c+d[/tex]

What I mean is that in order to solve your system, you don't need any cubics or quadratics since you can just substitute into get a linear system of equations in the cooeficients.
 
  • #8
Hi

If I assume that [itex]c_{1} = c_{2} = d_{1} = d_{2}=0 [/itex]

Then the system of equations is reduced to:

[itex]1 = -a_{1} + b_{1}[/itex]

[itex]0 = 3a_{1} - 2b_{1}[/itex]



[itex]3 = a_{2} + b_{2}[/itex]

[itex]0 = 3a_{2} + 2b_{2}[/itex]

I then solve the above as two systems of equations and then obtain the following values [itex]a_{1}= 2 , b_{1} = 3 , b_{2} = -6 , a_{2} = 9[/itex]

This in turn generates two 3 degree polynomials: [itex]p(x) = 2 x^{3} + 3x^2[/itex] and [itex]q(x) = -6 x^{3} + 9 x^{2}[/itex]

Finally I preform a test by inserting the values: p(-1) = 1, q(1) = 3, p(0) = q(0), p'(-1) = 0, q'(1) = 0, p'(0) = q'(0) into their respective polynomials.

I then discover that the polynomials give the desired result mentioned above.

My question is: Doesn't that mean that my approach is correct?

Sincerley
Fred
 
  • #9
Mathman23 said:
My question is: Doesn't that mean that my approach is correct?

You have found a solution to the system of equations. Since you have 6 equations, and 8 unknowns, there will be an infinite number of solutions, or no solutions at all.

p(-1) = 1,
p'(-1) = 0,
q(1) = 3,
q'(1) = 0,
p(0) = q(0),
p'(0) = q'(0)
[tex]p(x)=a_1x^3+b_1x^2+c_1x+d_1[/tex]
[tex]q(x)=a_2x^3+b_2x^2+c_2x+d_2[/tex]
So, chose any [tex]c,d[/tex] that you like. Then:
[tex]c_1=c_2=c[/tex]
[tex]d_1=d_2=d[/tex]
[tex]a_1=c-2d+2[/tex]
[tex]b_1=2c-3d+3[/tex]
[tex]a_2=2c+2d-6[/tex]
[tex]b_2=9-3c-3d[/tex]
Is going to solve your system of equations.
 

1. What is a system of linear equations?

A system of linear equations is a set of equations that contain multiple variables and must be solved simultaneously. These equations can be represented graphically as lines, and the solution of the system is the point where all the lines intersect.

2. How do you solve a system of linear equations?

To solve a system of linear equations, you can use a variety of methods such as substitution, elimination, or graphing. These methods involve manipulating the equations to isolate a variable and solve for its value. Once you have the value for one variable, you can substitute it into the other equations to find the values of the remaining variables.

3. Can a system of linear equations have more than one solution?

Yes, a system of linear equations can have one, infinite, or no solutions. If the equations represent parallel lines, there will be no intersection and no solution. If the equations represent the same line, there will be infinite solutions as every point on the line satisfies both equations. Otherwise, there will be a unique solution where all the equations intersect at one point.

4. What is the importance of solving systems of linear equations?

Solving systems of linear equations is crucial in many areas of science, technology, and engineering. It allows us to model and understand real-world situations that involve multiple variables and find the optimal solution. It is also the foundation for more complex mathematical concepts and techniques.

5. Can systems of linear equations be solved using technology?

Yes, there are many tools and software available that can solve systems of linear equations, such as graphing calculators, computer programs, and online solvers. These tools can help solve complex systems efficiently and accurately, saving time and effort in the problem-solving process.

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