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.

cosh(x+1)+x^2

=(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

Exercise:

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.

My God ! A Prodigy. So many faculties with a rare combination in profile. A genius in maths. Hope continue more frequently. Thank you for the spirit of sharing. With Ever Best Wishes.

ReplyDeleteThanks for reading & commenting here :d

ReplyDeleteI just want to share some interesting academic stuffs to everyone~