[TI-M] Re: complex matrixs etc.
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[TI-M] Re: complex matrixs etc.
>
> I've got another question about "simple fraction" in algebra,
> i have a function like this
> 1/((x+1)x^2)
>
> and i want to write it as "simple fractions" (in polish we call it
> that), so it will come:
>
> 1/(x+1) - 1/x + 1/x^2
>
> Does anybody know function that can do it for me?
>
>
> Zupa
>
>
This is how to do it by hand:
(This is complicated, there's an example at the bottom.)
Say we have two polynomials, P(x) and Q(x), and we want to simplify
P(x)/Q(x).
First, deg(P) must be smaller than deg(Q). If it isn't already , you can
make it
smaller by division. That'll give you a R(x) and P'(x) such that R(x) +
P'(x)/Q(x)
equals P(x)/Q(x).
Then we will factor Q(x) into irreducibel polynomials. I won't give the
proof here,
put that will be factors of degree at most 2. So Q(x) = Q1(x)^N1 * Q2(x)^N2
*
... * Qm(x)^Nm (the 1, 2 and m are supposed to be subscripted, as they are
indices). For now we'll assume N1 = N2 = Nm = 1. (I'll explain the general
case
a little later.)
Now we make the fraction P1(x)/Q1(x) + P2(x)/Q2(x) + ... + Pm(x)/Qm(x), with
deg(Pi) = deg(Qi) - 1. As deg(Qi) is at most 2, P(x) is either the constant
a, or
the first degree polynomial a x + b.
Now we can calculate the coefficients of Pi. Multiply Pi by Q1, Q2, ...,
Qi-1, Qi+1,
..., Qm-1, Qm. This will give a polynomial in x, where the coefficients
depend on
a (if Pi(x) = a) or a and b (if Pi(x) = a x + b). Do this for all Pi's, and
add the
results. Now you can isolate the constant terms, the first degree terms,
etc. Now
all that's left is equating those expressions to the coefficients in the
original P'(x),
and solving the system of linear equations thus obtained.
Now the case where the Ni's are not all equal to 1. In that case, you'd use
the
dividers Qi(x)^1, Qi(x)^2, ..., Qi(x)^m.
(Okay, that was confusing, I'll try explaining it with an example. This
should be
viewed in a fixed width font.)
This is the fraction:
2x^4 + 3x^3 - 5x^2 + 2x - 3
----------------------------
x^3 - 3x^2 + x - 3
That gives these polynomials:
P(x) = 2x^4 + 3x^3 - 5x^2 + 2x - 3
Q(x) = x^3 - 3x^2 + x - 3
Since deg(P) is not smaller than deg(Q), we need to divide first.
x^3 - 3x^2 + x - 3 / 2x^4 + 3x^3 - 5x^2 + 2x - 3 \ 2x + 9
2x^4 - 6x^3 + 2x^2 - 6x
9x^3 - 7x^2 + 8x - 3
9x^3 -27x^2 + 9x -27
20x^2 - x +24
20x^2 - x + 24
So, 2x + 9 + -------------------- is the same as the original fraction.
x^3 - 3x^2 + x - 3
Reducing x^3 - 3x^2 + x - 3 gives ( x - 3 ) * ( x^2 + 1 ):
a b x + c
------- + ---------
x - 3 x^2 + 1
a * ( x^2 + 1 ) = a x^2 + a
(b x + c ) * ( x - 3 ) = b x^2 + c x - 3b x - 3c = b x^2 + ( c - 3b ) x - 3c
Adding these two, we obtain ( a + b ) x^2 + ( c - 3b ) x + ( a - 3c )
Remember, the original divider was 20 x^2 - x + 24, so the following
equalities
should hold: a + b = 20, c - 3b = -1, a - 3c = 24, or, in matrix
representation:
1 1 0 | 20
0 -3 1 | -1
1 0 -3 | 24
which solves to
1 0 0 | 201/10
0 1 0 | -1/10
0 0 1 | -13/10
2x^4 + 3x^3 - 5x^2 + 2x - 3
So, ---------------------------- is equal to
x^3 - 3x^2 + x - 3
201/10 1/10 x + 13/10
2x + 9 + -------- - ----------------
x - 3 x^2 + 1
I hope this helps,
Rob van Wijk
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