At least one real zero?

[Ender]

I like this problem

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.

[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.

[Ender]

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

An interesting thing about math is that an important theorem or principle may be obvious, but the way it changes your thinking can be profound. Take the pigeonhole principle, for example. It's ridiculously intuitive and obvious, but it can help you solve some really tough theorems or problems.

[rabbit]

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

An interesting thing about math is that an important theorem or principle may be obvious, but the way it changes your thinking can be profound. Take the pigeonhole principle, for example. It's ridiculously intuitive and obvious, but it can help you solve some really tough theorems or problems.

IIRC Sidoh disproved the pigeon hole principle.

[d&q]

Some theorems are pretty cool and can be appreciated, but a theorem like Rolle's theorem is really intuitive and..bleh.

[Ender]

Some theorems are pretty cool and can be appreciated, but a theorem like Rolle's theorem is really intuitive and..bleh.

The whole point was that intuitive and obvious theorems and principles often deserve a lot of appreciation. MVT is very important