Let's Solve A Nice Polynomial Equation | Math Olympiads

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PLAYLISTS 🎵 :

[teamcowgill]

A=0 is a valid solution as well. It just means you have a trivial solution of P(x) = 0. And this satisfies the first equation given as well.

[donniemorrow]

6a^2*x^3 should have been 18a^2*x^3, leading to a final result of 4x^3/9 instead of 4x^3/3

[scottleung]

I found e^x to be a solution, even though it's not a polynomial.

[ianfowler]

Found one. Let f(x) = -2sin(x)
1) f(2x) = -2sin(2x) = -4sin(x)cos(x)
2) f '(x) = -2cos(x) and f ' '(x) = 2sin(x) ====> f '(x)*f ' (x) = -4sin(x)cos(x)
Thanks Scott.

[Raj.MathsKaDar]

For a General Function:
P(x) belongs to the set {e^x, (4/3)x³, 0, -2*sin(x), ...} But Denoted Function Are Fairly Easy To Guess

[actions-speak]

Taking the derivative of a constant doesn't reduce its degree, so n = 2n - 3 doesn't hold when the degree of p is less than 2.
If p(x) = ax + b then p(2x) = 2ax + b = p'(x) p''(x) = a * 0 = 0 and a = b = 0.

[martinhosek]

time 8:00 8a=18a*a, a=4/9: 8:10 mistake, 6*3=6.
I don't mind this, everyone makes mistakes,
I follow him and blackpenn, the Chinese guy, and it's enough for me.

[ianfowler]

Doesn't your solution a = 0 give you a solution P(x) = 0 ?

[dandeleanu]

p(x) = 2^.5/2 x^2 + c and p(x) = -2^.5/2 x^2 + c are also solutions

[georgesbv]

lol, you are ready to read the disclaimer on pharmaceutical commercials ;-P

[ianfowler]

Something interesting: e^x - e^(-x), -2sin(2x) and a y = 4x^3/9 are all symmetrical about their inflection points and they all have an inflection point at (0, 0) Hmmm. So symmetrical about the origin which is an IP.

Addendum: I have just proved that the conditions described above are sufficient but not necessary!!
1) f(x) = -f(-x) ====> f(2x) = -f(-2x)
2) f '(x) = f '(-x)
3) f ' '(x) = - f ' ' (-x)
We know: f(2x) = f '(x)f ' '(x)
Substituting: -f(-2x) = f '(-x)* [-f '(-x)]
====> -f(-2x) = -f '(-x)f ' '(-x)
====> f(-2x) = f '(-x)f ''(-x)
Let t = -x
f(2t) = f '(t)f ' '(t) which is the original diff eq. Bloody marvelous.