• Support PF! Buy your school textbooks, materials and every day products Here!

Derivative in an abstract polynomial ring

  • Thread starter PingPong
  • Start date
62
0
"Derivative" in an abstract polynomial ring

1. Homework Statement
Let R be any ring and define D:R[X]-->R[X] by setting [itex]D[\sum a_nX^n]=\sum na_nX^{n-1}[/itex].

a) Check that, if [itex]f(X)=\sum a_nX^n[/itex] and [itex]g(X)=\sum b_nX^n[/itex], then D[f+g]=D[f]+D[g]
b) Check that [itex]D[a_ib_jX^{i+j}]=D[a_iX^i]b_jX^j+a_iX^iD[b_jX^j][/itex]
c) Now using the additivity in a), show that D[f(X)g(X)]=D[f(X)]g(X)+f(X)D[g(X)] for all polynomials f and g in R.

2. Homework Equations
None.

3. The Attempt at a Solution
I see the clear parallel to polynomial derivatives, but it's not defined in terms of epsilons and deltas - it's just a "black box" function that really has no meaning. But in any case, I've been able to get parts (a) and (b) without any problems. But I'm having trouble with (c).

I've chosen to write the product of [itex]f(X)=\sum_i a_iX^i[/itex] and [itex]g(X)=\sum_j b_jX^j[/itex] as just [itex]\sum_{i,j}a_ib_jX^{i+j}[/itex] so that it's similar to the result of part (b). Then
[tex]D[f(X)g(X)]=D[\sum_{i,j}a_ib_jX^{i+j}]=\sum_{i,j} D[a_ib_jX^{i+j}]=\sum_{i,j} \left(D[a_iX^i]b_jX^j+a_iX^iD[b_jX^j]\right)[/tex]
[tex]=\sum_{i,j}D[a_iX^i]b_jX^j+\sum_{i,j}a_iX^iD[b_jX^j]=D[f(X)]g(X)+f(X)D[g(X)].[/tex]

So does this work? I tried it another way by checking what's in the nth position on each side of the equation and showing them to be equal, but got bogged down in keeping track of indexes and stuff. My algebra professor said there was an easier way - is this it?

Thanks for your help in advance.
 

Answers and Replies

Dick
Science Advisor
Homework Helper
26,258
618
About all you really did there is restate, without really quite proving, part b). But if you actually did do part a) and b), then you are a pretty good shape. a) shows that the Leibniz rule (a common name for what you are trying to prove) is linear over sums. b) shows that it's true for products of monomials. That makes c) easy. f(X)*g(X) is really just a sum of products of monomials if you expand it out.
 

Related Threads for: Derivative in an abstract polynomial ring

Replies
2
Views
3K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
3
Views
3K
Replies
3
Views
651
  • Last Post
Replies
2
Views
978
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
1
Views
855
Top