## Monday, 4 April 2011

### Equation involving multiple functions

Basic concept:
-Even function f(x) satisfies f(x) = f(-x)
-Odd function g(x) satisfies g(x) = -g(-x)
-degree of a polynomial (or a polynomial functions) depends on the degree of highest term in the function.

For simplicities, f(g(x)) is written as fg(x). (Different from f(x)g(x) = f(x) * g(x)), f(f(x)) is written as f_2(x).

Recell that if deg (fg(x)) = (deg f)(deg g), so mutiple linear functions is still linear.

Corollary 1: if f(x) is a linear function, then exist unique k for f_n(k) = k, where n is a natrual number.
The prove involves the techniques about geometric sequences:
Let f(x) = ax+b, f_n(x) = a^nx + a^(n-1)b + a^(n-2)b + ... + b
= a^n x + b(1+a+a^2+...+a^(n-1)) = x
b(1-a^n)/(1-a) = x(1-a^n)
x = b/(1-a), which is independent of n, so the root is unique.

Now consider two linear functions. f(x) = ax+b, g(x) = cx+d.
Does gfgfgf(x) = x and gfgfgfgf(x) = x gives the same root?
Method 1: Expanding gfgf...gf(x) (<- n "gf") = a^n c^n x + a^(n-1) c^n b + a^(n-1) c^(n-1) d +...+cb+d
= a^n c^n x + (cb+d)(1+ac+a^2c^2+...+a^(n-1)c^(n-1))
= a^n c^n x + (cb+d)(1-a^n c^n)/(1-ac) = x
x = (cb+d)/(1-ac) which is also independent of n, so it's also unique.

You may found that  the proofs are quite similar, so are there a quicker method?

Method 2:  Since they're both linear, by deg (fg(x)) = (deg f)(deg g), gf(x) is also linear.
gfgf...gf(x) = (gf)_n (x) = h_n(x) = x, where h(x) identically equal to gf(x), and is linear.
By corollary 1, this is true.

Now I'm going to introduce a graphical idea to solve such equation.

Theorem : Given f(x) and g(x). In order to solve f(g(x)) = 0 on the graph, draw the curve x = f(y) and y = g(x). Project the y-intercept of f to a horizontal line. The x value of intecepts between projected horizontal line and y = g(x) is the (real) root of the equation.

We can do the above method by algraical way:
solve f(x) = 0 and get the set of roots {a_i}, then solve the set of equations g(x) = a_i, all roots of it are answers.

Now check out corollary by the above method:
Root of ax+b=x <=> (a-1)x+b=0
The graph of y = (a-1)x+b is simply a stright line with y-intercept b, x-intercept (root) b/(1-a), and dy/dx = 1-a.
Now for a(ax+b)+b = b, we transform it into a((a-1/a)x+b)+b = 0
So g (x) = (a-1/a)x+b, f(x) = ax+b.
The graph y = g(x) is simple, with x-intercept b and dy/dx = a. How about x = f(y)?
Rotate the graph about 90 degrees and draw the graph, it has x-intercept b, and dx/dy = a, then dy/dx = 1/a.
Apply the above result and we can show that they have the same root.

Example : f, g are quadratic functions. Given that f(g(x)) = 0 gives 4 real roots, 1,3,4,p respectively. Find maximum possible p. (PC11, F5 final, #15)

Consider the above picture (ignore the numerical values), we've applied the above method, and we find that the roots are symmetrical about the symmetrical axis of f(x). So maxima occurs when 3 is symmetrical to 4, then 1 is symmetrical to 6 (which is the answer).

For higher degrees we can also apply the technique of shiting coordinate to make equations simplier. One of the example is adding odd and even functions up. You will find that the sum is neither odd nor even, but we are able to change it into odd functions.

Example: y = x^3 + x^2
The resulting polynomial has local maxima and minima 4/27 and 0 respectively, so we know that the points of infextion occurs at (-1/3, 2/27). And the graph is rotational symmetrical about that point, so it would be odd function if the function takes (-1/3, 2/27) as origin.
Take the origin with new coordinate x' and y', parallel with x and y axis respectively.
The new equation becomes (y'+2/27) = (x'-1/3)^3 + (x'-1/3)^2
Expanding gives y = x^3 - x/3, which is the sum of odd functions, so we are done.

Example: Can we turn y = x^2(1+cos x) + x into odd or even functions?
No. Since it's positve when x is big enough, we can't turn it into odd functions. Also compare the x-distance of the three local minima which gives different distance, so it's impossible to turn it into even functions too.

Exercise:
1) Consider the taylor expansion, determine whether the trigonometric functions are odd/even/neither odd nor even.
2) Define f_-1(x) as the inverse function of f, f_-2(x) = f_-1(f_-1(x)), etc. then for negative integer, does corollary 1 still valid?
3) Proof the basic assumption of the "shifting coordinate of cubic functions": The points of infletion occurs at the mid-point of local maxima and minima.
4) Given f(x) = x^3 + 6x^2 + 11x + 6. By choosing a suitable new origin, express the given equation in terms of equation concerning the new coordinates.
5) Given deg f = 2, deg g = 3, the equation f(g(x)) = 0 gives 5 real roots, 1,3,8,9,p respectively. Find all posssible value of p.
You can take the technique in pure maths about "rotating coordinate" as reference, and use a suitable shift of coordinates and change the follow equations into odd or even functions.
6) y^3 +x^3 = 3xy (Higher Maths Q4, 1981)
7) y = sin^2 x + x
8) Can we turn y = e^x - x^7 into even functions? You may consider the taylor expansions of e^x.