At least one real zero?

[Ender]

I like this problem 8)

Given that a + b/2 + c/3 + d/4 + e/5 = 0,

does the polynomial p(x) = a + bx + cx^2 + dx^3 + ex^4 have at least one real zero?

Feel free to post your progress / ask for hint.

[Nate]

What are a,b,c,d,e?

[Joe]

0 + 0/2 + 0/3 + 0/4 + 0/5 = 0

boom

[Sidoh]

0 + 0/2 + 0/3 + 0/4 + 0/5 = 0

boom

He's asking for a general solution, not a solution of one in infinitely many cases. :P

[Ender]

0 + 0/2 + 0/3 + 0/4 + 0/5 = 0

boom

The question is whether p(x) has at least one real zero.

You gave a solution to a + b/2 + c/3 + d/4 + e/5 = 0, i.e. (0, 0, 0, 0, 0), but this is entirely unrelated. You need to show that there is a solution to p(x) = 0.

And you also need not come up with an explicit solution. You can just as well show that there is one, without explicitly saying what this solution is.

[Nate]

Im not going to bother but since the constraint seems to indicate P(1)=0, then I would say yea, there probably a zero of p(x)...too lazy to do the math to prove it though.

P(0)=0 as well but 0<x<1 for P(x) isn't zero...so its a parabola and the inflection point of that parabola is a zero of p(x) or something like that.

[Ender]

Incorrect. p(1) is not necessarily 0, nor is p(0). In particular, p(1) = a + b + c + d + e, and p(0) = a. The constraint doesn't say that p(1) = 0.

[Nate]

P(x) as in the anti derivative of p(x).  Your given constraint is P(1)=0.

[Ender]

P(x) as in the anti derivative of p(x).  Your given constraint is P(1)=0.

Oh -- I didn't notice your notation.

You've made good progress, but it's not complete yet. P(0) = 0 and P(1) = 0. Now what?

[Nate]

for 0 < x < 1, P(x) <> 0, therefore, p(x) needs to signs between 0 and 1 crossing the x-axis...I am sorry my math doesn't translate to words very well.

[Ender]

Yep, that's basically it =) Though there's a particular calculus theorem that says what you just said succinctly.

[Ender]

Forgot to say -- Good job, Nate!

[Ender]

So yeah, now that Nate got it, I'll just say the calculus theorem. It's Rolle's Theorem.

[Nate]

Calculus theorems are pointless, they can all be easily re derived on the spot if you don't see a solution intuitively.

[Newby]

Calculus theorems are pointless, they can all be easily re derived on the spot if you don't see a solution intuitively.

Unless you're a math nerd like Ender or Rule, in which case they're amazing. :P