## Wednesday, 10 November 2010

### Theory of equations?

(I'm interested whether we can insert LaTeX here lol)

In pure maths algebra course, theory of equations (polynomials) has been a popular topic so far. Here're the theorems we may use:
1)Factor theorem: f(x) is divisible by (x-h), so (x-h) is a factor of f(x) iff f(h)=0.
2)Repeating roots theory:
If there're n repeated root a for f(x)=0, then f''''...'(-a)=0, where f is differentiated n-1 times.

The proof of the above theorem is easy and should be well-known among books, so I'll skip them.

Factor theorem is in fact the special case of remainder theorem:
f(a) is the remainder then f(x) is divided by (x-a).

Now combining them together:
1)Remainder theorem of quardratic divisors, or repeated linear factors?

The significance of quadratic factor is shown by "every polynomisl can be factorized into linear or quadratic factor with real coefficient".

Firstly we consider "when the quadratic factor can be factorized into (x-a)(x-b)".
f(x) = (x-a)Q1(x)+f(a) = (x-a)(x-b)Q3(x)+(x-a)R+f(a)
f(x) = (x-b)Q2(x)+f(b)=(x-a)(x-b)Q3(x)+(x-b)R+f(b)
when a and b is given they can be easily solved.
whenever the quadratic factor can be factorized into two real linear factor, the above method can be used.
What if the quadratic factor is irreducible in R[x]?
Ex. 1 - Show that the method is also applicable over complex number.

"When f(x) is divised by (x-h)^2, the remainder is f'(x)(x-h)+f(h)"
It's a question from pure maths paper. In fact we can prove this by using factor theorem twice.
i.e., f(x)-R(x) is divisible by (x-h), and (f-R)/(x-h) is also divisible by (x-h).
Whenever remainder is involved, we apply factor theorem is this way:
f(h)-R(h)=0.
f(x)-f'(x)(x-h)-f(h)=0 This is obviously true for x=h.
When it's divided by (x-h),
(f(x)-f(h))/(x-h) - f'(x) = 0
Note that we can't put x=h directly as it gives indeterminate form, therefore we limit x to h.
Then the first term, in forms of the first principle, becomes f'(x).
f'(x)-f'(x)=0, so (f(x)-f(h))/(x-h) - f'(x) is divisible by (x-h),
therefore (x-h)[(f(x)-f(h))/(x-h) - f'(x)] = f(x)-f'(x)(x-h)-f(h) is divisible by (x-h)^2.

Now think it reversely...
Ex. 2 Assume
f(x) = (x-h)Q(x)+f(h)
Q(x) = (x-h)R(x)+Q(h)
Then f(x) = R(x)(x-h)^2 + Q(h)(x-h) + f(h)
Ex. 2a Prove or disprove Q(h)=f'(h).
Ex. 3 Deduce remainder when f(x) is divided by (x-h)^3 or 4 or higher degree.
Ex. 2b Prove or disprove R(h)=Q'(h), and similar relationships in higher degree.
Ex. 4 Deduce the divisibility of f(x) divided by (x-h)^2, and hence the general divisibility of f(x) by a quadratic factor.