Author Topic: At least one real zero?  (Read 6858 times)

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Offline Ender

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At least one real zero?
« on: April 10, 2008, 09:10:01 PM »
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.

Offline Nate

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Re: At least one real zero?
« Reply #1 on: April 10, 2008, 09:46:50 PM »
What are a,b,c,d,e?

Offline Joe

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Re: At least one real zero?
« Reply #2 on: April 10, 2008, 10:18:29 PM »
0 + 0/2 + 0/3 + 0/4 + 0/5 = 0

boom
I'd personally do as Joe suggests

You might be right about that, Joe.


Offline Sidoh

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Re: At least one real zero?
« Reply #3 on: April 10, 2008, 10:20:12 PM »
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

Offline Ender

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Re: At least one real zero?
« Reply #4 on: April 10, 2008, 10:32:22 PM »
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.

Offline Nate

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Re: At least one real zero?
« Reply #5 on: April 10, 2008, 11:28:28 PM »
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.
« Last Edit: April 10, 2008, 11:34:07 PM by Nate »

Offline Ender

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Re: At least one real zero?
« Reply #6 on: April 11, 2008, 12:00:35 AM »
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.

Offline Nate

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Re: At least one real zero?
« Reply #7 on: April 11, 2008, 01:26:27 AM »
P(x) as in the anti derivative of p(x).  Your given constraint is P(1)=0.

Offline Ender

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Re: At least one real zero?
« Reply #8 on: April 11, 2008, 01:50:36 AM »
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?

Offline Nate

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Re: At least one real zero?
« Reply #9 on: April 11, 2008, 02:19:03 AM »
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.

Offline Ender

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Re: At least one real zero?
« Reply #10 on: April 11, 2008, 02:38:05 AM »
Yep, that's basically it =) Though there's a particular calculus theorem that says what you just said succinctly.

Offline Ender

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Re: At least one real zero?
« Reply #11 on: April 11, 2008, 02:44:30 AM »
Forgot to say -- Good job, Nate!

Offline Ender

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Re: At least one real zero?
« Reply #12 on: April 11, 2008, 01:48:23 PM »
So yeah, now that Nate got it, I'll just say the calculus theorem. It's Rolle's Theorem.

Offline Nate

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Re: At least one real zero?
« Reply #13 on: April 11, 2008, 05:33:56 PM »
Calculus theorems are pointless, they can all be easily re derived on the spot if you don't see a solution intuitively.

Offline Newby

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Re: At least one real zero?
« Reply #14 on: April 11, 2008, 06:01:53 PM »
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
- Newby
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Quote
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[17:32:54] * xar sets mode: +o newby
[17:32:58] <xar> new rule
[17:33:02] <xar> me and newby rule all

I'd bet that you're currently bloated like a water ballon on a hot summer's day.

That analogy doesn't even make sense.  Why would a water balloon be especially bloated on a hot summer's day? For your sake, I hope there wasn't too much logic testing on your LSAT.