What determines how we group in method of undetermined coefficients?

[Integrate]

Θ"-Θ=tsint








For polynomials it is fairly straight forward what we set equal too.

And for just sin and cos we match them just as easily as we do when we have a polynomial

But why do we combine them when we have both a polynomial and sin or cos?

why not all cosines and sines together or all powers together.

Why do we need to do each combination?

 

[blamocur]

Not sure I correctly understand your question, but I'll try to answer it. You have 4 different functions ([imath]\cos t, \sin t, t\cos t, t\sin t[/imath]), and their linear combination is uniquely defined by the coefficients. I.e., if two linear combinations are equal then each of the 4 coefficients must be the same on both sides of the equation:
[math]a_1 \cos t + a_2 \sin t + a_3 t \cos t + a_4 t \sin t =b_1 \cos t + b_2 \sin t + b_3 t \cos t + b_4 t \sin t[/math] if and only if
[math]a_1 = b_1 \wedge a_2 = b_2 \wedge a_3 = b_3 \wedge a_4 = b_4[/math]

 

[Integrate]

You understood my question correctly I think.

So what you are saying is that we group them in each combination between t, cost, and sint is that they are all different functions.

And there for coefficients will different impact on their value when given an input.

Makes sense.

Also just for reference.

I still don't understand what linearity means in differential equations. Just know how to identify it using its general form.

I also don't know what ^ means in your answer.


Teaching myself math and its going as well as you could imagine.

 

[fresh_42]

The crucial point here - as far as I understood the posts - is that the functions [imath] \{\cos(t),\sin(t),t\cos(t),t\sin(t)\} [/imath] are linearly independent (which has to be proven btw.). It means that [imath] 0=a_1\cos(t)+a_2\sin(t)+a_3t\cos(t)+a_4t\sin(t) [/imath] can only hold if [imath] a_1=a_2=a_3=a_4=0. [/imath] This allows us to compare the coefficients of equations like the ones in post #1.

It is not hard to prove it since we can simply insert suited values for [imath] t. [/imath]

There are many other linear independent functions. A large set is the so-called Legendre polynomials.

[imath] \wedge [/imath] is a logical AND and [imath] \vee [/imath] a logical OR.

 

[blamocur]

I still don't understand what linearity means in differential equations. Just know how to identify it using its general form.

I wasn't talking about linearity but about linear combination. A linear combination of, in our case, functions [imath]f_1,f_2,f_3,f_4[/imath] is a new function of the form [imath]a_1f_1 + a_2 f_2 + a_3 f_3 + a_4 f_4[/imath] where [imath]a_i[/imath] are all constants.

 

[Integrate]

What makes it a linear combination?

 

[Dr.Peterson]

What makes it a linear combination?

Here is the definition of linear combination:

In mathematics, a linear combination is an expression constructed from a set of terms by multiplying each term by a constant and adding the results (e.g. a linear combination of x and y would be any expression of the form ax + by, where a and b are constants). The concept of linear combinations is central to linear algebra and related fields of mathematics.Most of this article deals with linear combinations in the context of a vector space over a field, with some generalizations given at the end of the article.

Or look here, for a quicker version:

Definition of

Linear Combination​

Where we multiply each term by a constant then add them up.
Example: ax + by is a linear combination of x and y
Example: Acos(x) + Bsin(x) is a linear combination of cos(x) and sin(x)

Or are you asking why linear combinations are relevant in this context?

 

[Steven G]

What makes it a linear combination?

If you have a set of functions which you multiply each function by a scalar and then add these together you have a linear combination of theses functions.
ex: let f, g, h and j all be functions and 3, 4, 5 and -6 be scalars. Then a linear combination of the f, g, h and j is 3f + 4g + 5h - 6j.

 

[Dr.Peterson]

But why do we combine them when we have both a polynomial and sin or cos?

why not all cosines and sines together or all powers together.

Why do we need to do each combination?

So what you are saying is that we group them in each combination between t, cost, and sint is that they are all different functions.

And there for coefficients will different impact on their value when given an input.

I'm not sure whether your actual question has been answered fully.

The quote you showed appears to be just a book's solution to a problem you didn't show; you also haven't shown what was taught, or what sort of alternative you have in mind. what sort of "grouping" are you asking about?

If the grouping you mean refers to the parenthesized sums in the example, those don't matter at all; just distribute. If you mean what's multiplied (e.g. x by cosine, and not sine by cosine), that is based on thinking about how derivatives work.

The basic idea is that, in order to be able to get some particular function (say, [imath]x\sin(x)[/imath]) by differentiating, we look for what functions yield that sort of term in their derivatives. You can get a power from a higher power, and a sine or cosine from either a sine or a cosine; and you can get a product of functions from a product of functions (by the product rule, which will lead to a sum of terms with different combinations). Experience leads to certain standard "guesses".

Does the text you're learning from have both a list of standard guesses, and some sort of explanation for why they make sense? Here is one that does (much of the explanation is in the solutions):

https://tutorial.math.lamar.edu/classes/de/undeterminedcoefficients.aspx

 

[Integrate]

Here is the definition of linear combination:


Or look here, for a quicker version:


Or are you asking why linear combinations are relevant in this context?

This very much helps me understand what a linear combination is for sure. Something that I did not understand fully.


Though I am wondering why they apply in this situation.

 

[Integrate]

I'm not sure whether your actual question has been answered fully.

The quote you showed appears to be just a book's solution to a problem you didn't show; you also haven't shown what was taught, or what sort of alternative you have in mind. what sort of "grouping" are you asking about?

If the grouping you mean refers to the parenthesized sums in the example, those don't matter at all; just distribute. If you mean what's multiplied (e.g. x by cosine, and not sine by cosine), that is based on thinking about how derivatives work.

The basic idea is that, in order to be able to get some particular function (say, [imath]x\sin(x)[/imath]) by differentiating, we look for what functions yield that sort of term in their derivatives. You can get a power from a higher power, and a sine or cosine from either a sine or a cosine; and you can get a product of functions from a product of functions (by the product rule, which will lead to a sum of terms with different combinations). Experience leads to certain standard "guesses".

Does the text you're learning from have both a list of standard guesses, and some sort of explanation for why they make sense? Here is one that does (much of the explanation is in the solutions):

https://tutorial.math.lamar.edu/classes/de/undeterminedcoefficients.aspx

I was wondering why they grouped the way they did.

Which after working it through their way I see it is much easier to group with the corresponding trig function.


I was curious how they knew to do it this way.

I see now its pretty obvious that it just makes differentiation easier.