Partial fractions with generating functions

Click For Summary
To express q(z)/p(z) in partial fractions, the equation 1 = A(1 + 3z) + B(1 + z) is established. This leads to the system of equations A + B = 1 and 3A + B = 0. The confusion arises regarding the interpretation of coefficients in polynomial equality. The resolution clarifies that two polynomials are equal if their corresponding coefficients match, confirming that 1 + 0*z = 0. The discussion emphasizes the importance of understanding polynomial coefficient comparison in solving for constants A and B.
BigBoss22
Messages
2
Reaction score
0
Suppose that q(z) = 1, and p(z) = (1 + z)(1 + 3z).

We wish to express q(z)/p(z) in the form

where A and B are constants. To find them, we multiply through by p(z) =
(1 + z)(1 + 3z) and obtain
1 = A(1 + 3z) + B(1 + z)
= (A + B) + (3A + B)z

Im fine up to this point, But according to my notes it is obvious that A + B = 1 and (3A + B)z = 0.

I can not see why this is the case?

Any help would be greatly appreciated.
 
Physics news on Phys.org
Never mind figured it out:

1 + 0*z = ...
 
Two polynomials are equal iff their coefficients are equal.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

Similar threads

  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
4
Views
2K
Replies
6
Views
2K
  • · Replies 17 ·
Replies
17
Views
3K
Replies
3
Views
2K
Replies
9
Views
2K