Partial fraction decomposition help
- Thread starter Mechaman
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Harry_the_cat
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To get from the first equation to the second equation, both sides have been multiplied by (z-2)(z-1).
As lev888 has said, then expand the RHS and collect the z terms and the constant terms.
Then equate the coefficient of z and the constant term on the right with the coeff of z and the constant term on the left.
This will give you 2 simultaneous equations in A and B which you can solve in the usual way.
Give it a go.
As lev888 has said, then expand the RHS and collect the z terms and the constant terms.
Then equate the coefficient of z and the constant term on the right with the coeff of z and the constant term on the left.
This will give you 2 simultaneous equations in A and B which you can solve in the usual way.
Give it a go.
HallsofIvy
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There are several different ways to do this. One is to do the indicated multiplications on the right:
z- 3= A(z- 1)+ B(z- 2)= Az- A+ Bz- 2B= (A+ B)z- A- 2B. Since this is to be true for all z, the corresponding coefficients must be the same. We have A+ B= 1 and -A- 2B= -3. Add the two equations and the "A" cancels leaving -B= -2 so B= 2. Then A+ 2= 1 so A= -1.
Another is to choose specific values for z. We can do this because, again, this equation is to be true for all z. Taking, say, z= 0 we get 0- 3= A(0-1)+ B(0-2) or -3= -A- 2B. Taking z= -1 we get -1-3= A(-1-1)+ B(-1-2) or -4= -2A- 3B. Again, the solution to those 2 equations is A= -1, B= 2.
The simplest way is to choose z to make the parenthetical terms, z- 1 and z- 2, 0. That is, we choose z=1 we have 1- 3= A(1- 1)+ B(1- 2) or -2= -B so B= 1 Immediately. If we choose z= 2 we have 2- 3= A(2- 1)+ B(2- 2) or -1= A.
z- 3= A(z- 1)+ B(z- 2)= Az- A+ Bz- 2B= (A+ B)z- A- 2B. Since this is to be true for all z, the corresponding coefficients must be the same. We have A+ B= 1 and -A- 2B= -3. Add the two equations and the "A" cancels leaving -B= -2 so B= 2. Then A+ 2= 1 so A= -1.
Another is to choose specific values for z. We can do this because, again, this equation is to be true for all z. Taking, say, z= 0 we get 0- 3= A(0-1)+ B(0-2) or -3= -A- 2B. Taking z= -1 we get -1-3= A(-1-1)+ B(-1-2) or -4= -2A- 3B. Again, the solution to those 2 equations is A= -1, B= 2.
The simplest way is to choose z to make the parenthetical terms, z- 1 and z- 2, 0. That is, we choose z=1 we have 1- 3= A(1- 1)+ B(1- 2) or -2= -B so B= 1 Immediately. If we choose z= 2 we have 2- 3= A(2- 1)+ B(2- 2) or -1= A.