Why do we count 0 twice when factoring a polynomial for differential equations?

[Noesis]

Well, strange in my opinion, hah.

This is actually for my differential equations class...but I figure it's stuff that I should've already learned.

Now for something that we are doing, the number of roots are important.

So upon factoring something like (m^4 - 2m^2), it goes to m^2(m^2 - 2)

So the roots are supposedly, 0, 0, -2 and 2.

Now WHY ON EARTH do we include zero twice? My teacher said that it is crucial to do so.

This doesn't really make sense...it's the same number...isn't this redundant?

It makes a huge difference in the DE application...so there must be something to this.

If anybody could please explain. Thank you.

 

[HallsofIvy]

Well, strange in my opinion, hah.

This is actually for my differential equations class...but I figure it's stuff that I should've already learned.

Now for something that we are doing, the number of roots are important.

So upon factoring something like (m^4 - 2m^2), it goes to m^2(m^2 - 2)

So the roots are supposedly, 0, 0, -2 and 2.

No, the roots are 0, 0, -\sqrt{2}, and \sqrt{2}.

Now WHY ON EARTH do we include zero twice? My teacher said that it is crucial to do so.

This doesn't really make sense...it's the same number...isn't this redundant?

It makes a huge difference in the DE application...so there must be something to this.

If anybody could please explain. Thank you.

It is "crucial" for some purposes. I would say it is crucial for you to do so because your teacher insists on it! There is a very nice theorem that says that any n^th degree polynomial equation has exactly n solutions if they are counted correctly: we have to include complex number solutions and count "multiple" roots. That can also be stated in terms of factors: every n^th degree polynomial can be factored into n linear factors. But again, you have to count multiple roots. In the case of m⁴- 2m² it equals (m)(m)(m-\sqrt{2})(m+\sqrt{2})- fourth degree so four factors. Another example would be x⁴+ x²= (x)(x)(x-i)(x+i).

 

[arildno]

It is, indeed crucial to count multiplicity of roots when doing diffeq.s
This is because you need to generate additional independent solutions for each "multiplicity" beyond the first.

 

[Noesis]

Okay, thanks a lot HallsofIvy and Arildno.

And yes, indeed, it is the square root of 2, was a little quick on my typing hah.

It makes sense, I suppose the theory in DE is related in some form or another to the theory you were saying about an nth degree polynomial have n amount of roots, so that must be the reason.

Thanks again guys for clearing it up.

 

[Noesis]

It is, indeed crucial to count multiplicity of roots when doing diffeq.s
This is because you need to generate additional independent solutions for each "multiplicity" beyond the first.

You know, as I'm looking over this again in my book...I'm still having trouble understanding exactly why.

How does taking zero to be a root of multiplicity two have any bearing on the answer?

I'm trying to think of it in terms of the differential equation and the function that solves it; or the function and its relation to the derivative.

It's all just a bit hazy.

By the way, the particular application here is to solve linear homogeneous equations with constant coefficients.

So we use it to find the power that e is raised to in order to find the solutions.

Thanks for shedding light in mathematical darkness guys.

 

[Hurkyl]

the particular application here is to solve linear homogeneous equations with constant coefficients.

The full solution to a 4-th degree homogeneous ODE requires you to find four linearly independent fundamental solutions. Each root of your characteristic polynomial gives you one of your fundamental solutions. If you forget to count 0 twice, you'll only be able to find three of your fundamental solutions!