Monday 18 April 2011

Variables and parametric equations

Recall that last time we switch the origin of reference corrdinate to make y = x^3 + x^2 into an even function, which is (y'+2/27) = (x'-1/3)^3 + (x'-1/3)^2 <=> y' = x'^3 - x'/3.

Now for a given random polynomial [with integer coefficient], is it possible to transform it into even/odd function?
The answer is: depends on the degree of the polyomial. It is possible for deg g =< 3, and more restriction [hence not "for all"] when the degree is larger than 3.

The proof is somehow like that:
1) The leading term decide that whether it's odd or even.
Proof: deg f = n, f = O(x^n).
(Do you remember the big O notation? f(x) = O(g(x))  implies that there exist reals a, k such that f(x)=< a(g(x)) for all x>= k.)

2)Under random polynomial, Substituting x -> (x+a) only eliminates one of the terms.
Proof: Consider the general form of real polynomial, if x is substituted by (x+a), then the new coefficient can be expressed in some binomial terms of the original ones. If one of the coefficient is sets to be zero, the a is unique. There does not exist a such that two of the cofficient disappears unless the original coefficient is specified.

For example, x^4 + 11x^3 + 45x^2 + 83x + 60 can be transformed into x^4 + ax^3 + bx by substituting x by x-3, while that's impossible for x^4 + 12x^3+45x^2+83x+60.

3) Consider the group relation of odd/even function:
odd*odd = even       (e.g. x (odd) * x^3 (odd) = x^4 (even)
proof: Let f(x) = -f(-x), g(x) = -g(-x)
f(x)g(x) = (-f(-x))(-g(-x)) = f(-x)g(-x)
even*even = even     (e.g. x^2 * x^4 = x^6)
odd*even = odd       (e.g. x^2 * x = x^3)
proof: Let f(x) = -f(-x), g(x) = g(-x)
f(-x)g(-x) = -f(x)g(x)
odd + odd = odd     (x^3 + x is still odd)
even + even = even

The difficulty is that : ODD + EVEN = ????

It's neither odd nor even.

4) When deg f =1,2,3 what you'll have to do is shift the origin to the symmetrical point. For linear functions, it's the y-intercept, for quadratic functions, it's the vertex, and for the cubic functions, it's the point of inflexion.

Ignore the constant term. there are 2 even and 2 odd terms in the polynomial, so it's impossible to eliminate both two terms.
For example: x^4 + 0.5x^2 + x
Though we see that the coefficient of x^3 is zero, but when we substitute a suitable k that the coefficient of x is zero, the coefficient of x^3 will be non-zero.

Though the coefficient elimination method is not useful here, it still get lots of application.

As many of you knows, eliminating the second leading term is almost a must in solving polynomials:
For x^n + a_(n-1) x^(n-1) +..., sub. x by (x-a_(n-1) / n) would eliminate that term by binomial theorem.
when it's eliminated, by fundamental theorem of algebra, we can assume that the polynomial is a product of a series of quadratic or linear factors. When one of the coefficient is zero that grealy reduces the difficulty in calculation.

e.g Simplify sin x + cos x.
sin x + 2cos x
= (sin x + cos x)
= (sin x + sin (x+pi/2))
= 2sin pi/4 sin x+pi/4
= rt2 sin (x+pi/4)

e.g. Show that cosh (x+1) +x^2 can't be rewritten into even function.
=(e^(x+1) + e^(-x-1))/2 + x^2
=(exp x + exp -x)/2e + (e-1/e)(exp x) + x^2
The first and third term is obviously even, but the middle term is asymmetrical. Since exp x > O(x^2), there's no way for writing the above equation.

e.g. Show that for positive real a,b,c, (a+b)^2 + (a+b+4c)^2 >= 100abc/(a+b+c). (HK team selection test)
What to do with variable elimination?
Divide both sides by c^2:
(a/c + b/c)^2 + (a/c + b/c + 4)^2 >= 100(a/c)(b/c)/(a/c + b/c + c/c)
(x+y)^2 + (x+y+4)^2 >= 100xy/(x+y+1)
(One interesting question here: can you see deg(LHS) = 2 > 1 = deg (RHS)? What's happening? Anything wrong?)
Use a stronger form:
(x+y)^2 (x+y+4)^2 >= 25(x+y)^2/(x+y+1)
z = x+y
(z+1)(z^2+(z+4)^2) -25z^2 >= 0 is trivially true.
Consider its [positive] zero is 4, so the equality stands at z = a/c+b/c = 4
a+b = 4c
By symmetrical reason a = b
So equality stands at a = b = 2c.

Parametic equation:
The euqation is written in forms of functions of t.
i.e., x = f(t), y=g(t)
When we have to transform it into x-y equation form, we write t = f^-1 (x) (If inverse function exist)
Then y = g(t) = g(f^-1(x)).
Example : x = t + 1, y = t + 3
Then t = x-1, y = t+3 = (x-1)+3 = x+2

Example : x = 2t, y = 1- t^2
y = 1-t^2 = 1 - (x/2) ^2
4y = 4-x^2

Example : x = sin t, y = cos t
Observe that sin ^2 t + cos ^2 t = 1
x^2 + y^2 = 1 (circle)
alternative method:
x = arcsin t
y = cos (arcsin x) = rt(1-x^2)
x^2 + y^2 = 1

Here's a question from Tokyo University Entrance Exam
Let P(1/2, 1/4) be a fixed point. Q(a,a^2) and R(b,b^2) is variable points which PQ=PR. Find locus of centroid of triangle PQR.
Equating PQ=PR, we get a^4+a^2/2-a = b^4 + b^2/2 - b.
Group the term:
2(a-b)(a+b)(a^2+b^2) + (a-b)(a+b) - 2(a-b)=0
since a =/= b,
(a+b)(2a^2+b^2+1) = 2
Let a^2 + b^2 = t, t>0
then a+b = 2/(2t+1)
x-value of centroid: (a+b+1/2)/3 = 1/6 + 2/(6t+3)
y-value of centroid: (a^2+b^2+1/4)/3 = 1/12 + t/3
Rewrite the parametric equations then we get x = 1/6 + 4/(36y+3).
But this is a parametric equations from t = a^2 + b^2 > 0, so t > 0
Moreover, the limit of the centroid is (1/2,1/4), so it starts from 1/2
Therefore we can conclude that the locus is x = 1/6 + 4/(36y+3) where 1/2 > x > 1/6

1) By subsituting a suitable variable, eliminate the functions/coefficient:
a) x^3 + 2x + 2, eliminate the constant term
b) x^4 + 3x^3 + 5x^2 +2x, eliminate the x^3 term
c) sin x + 2 cos x, simplify it into one function
Extended information: Hilbert's 13th challenge which states that "Solve all 7-th degree equations using continuous functions of two parameters.". It has been given a positive answer.
2) Is it possible to rewrite the positive part of Re(ln x + ln(ln x) + ln(ln(lnx))) into even functions?
3) Rewrite x = t^3 + 2t -1 , y = t^3 +2t^2 -1 into x-y form. Hence answer this question : "The x^3 term in both x and y can be eliminated each other. But why the resulting x-y form is still cubic?
4) Rewrite x = sin t, y = sin 3t in x-y form, sketch the resulting graph. Hence sketch the graph for x = cos t, y = sin 3t.
5) Show that the curve x = sin t, y = sin nt is enclosed if and only iff n is even non-zero integer. Hence, evaluate the length of curve (1 loop) in terms of n.
Extended information:  x = sin at, y = cos bt is a series of graph which is useful in physics too. Find the appliacble field and sketch the relavent graph.


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  2. Thanks for reading & commenting here :d
    I just want to share some interesting academic stuffs to everyone~