Exploring Coefficient Pairing in Polynomials: A Scientific Approach

In summary, when pairing coefficients of polynomials with corresponding unknowns on the other side of an equation, the justification for this action is the definition of equality of polynomials. This can be seen in a similar way to adding vectors in the plane, where the only way to get a zero vector is if both vectors are zero. And for expressing that a polynomial has no constant term, we can use the notation P(0) = 0, meaning that the polynomial evaluated at 0 is equal to 0.
  • #1
aikismos
145
34
Just to double check, but if one wanted to, like in partial fraction decomposition, associate literal coefficients of polynomials with corresponding unknowns on the other side of the equation, the justification for this action is the definition of equality of polynomials?

EDIT: I know this isn't true. Let's see, ## a - b = 2 ## is as far as it can be reduced.
## 3x^2 + bx = ax^2 + 5x \rightarrow a = 3, b = 5 ##

Another related question is how do I express symbolically that in:

## P(x) = ax^2 + bx ##

P(x) has no constant term?

I'm kinda groping around for a rigorous way to express and justify the pairing of coefficients if you were to write them as two equivalent n-tuples.
 
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  • #2
I would put everything on the same side and group similar powers of ##x## like this: ##(3-a)x^2+(b-5)x = 0 ##. Since this is true for all ## x ##, and since ## x ## and ## x^2 ## are independent functions (their Wronskian does not vanish), their coefficients must separately each be zero. This is a lot like having, for arrow vectors in the plane, that ## (3-a)\mathbf{i}+(b-5)\mathbf{j} = 0 ##. The only way to add a vector in the ## \mathbf{i} ## direction to one in the ## \mathbf{j}## direction and get zero is if, in fact, both vectors are the zero vector.

For the second question, how about writing it as ## P(0) = 0 ##?
 
  • #3
aikismos said:
Just to double check, but if one wanted to, like in partial fraction decomposition, associate literal coefficients of polynomials with corresponding unknowns on the other side of the equation, the justification for this action is the definition of equality of polynomials?

EDIT: I know this isn't true. Let's see, ## a - b = 2 ## is as far as it can be reduced.
## 3x^2 + bx = ax^2 + 5x \rightarrow a = 3, b = 5 ##

Another related question is how do I express symbolically that in:

## P(x) = ax^2 + bx ##

P(x) has no constant term?

I'm kinda groping around for a rigorous way to express and justify the pairing of coefficients if you were to write them as two equivalent n-tuples.
In partial fractions decomposition, you're setting up an equation that is true for all values of the independent variable (except those that make any denominator zero). IOW, two expressions that are identically equal.
So, for ## 3x^2 + bx = ax^2 + 5x##, for all x, it must be the case that a = 3 and b = 5.
 
  • #4
Thanks guys!
 

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