Putnam problem of the day

[Anonymous No.4568831]

Prove that there is a constant <span class="math">C[/spoiler] such that, if <span class="math">p(x)[/spoiler] is a polynomial of degree 1999, then
<div class="math">|p(0)|\leq C \int_{-1}^1 |p(x)|\,dx.</div>

[Anonymous No.4568852]

One constant C for all degree 1999 polynomials or for all degree 1999 polynomials one constant?

[Anonymous No.4568862]

>>4568852
There is a C such that for every polynomial of degree 1999. Otherwise it's trivial.

[Anonymous No.4568878]

This question is equivalent to saying:

prove that <span class="math"> \int_{-1}^1 |p(x)| dx [/spoiler] has an nonzero infimum if p(x) has a constant term.

But the space of polynomials is an infinite dimensional vector space therefore the basis <span class="math"> (x,x^2,x^3,...,x^1999) [/spoiler] has a nonzero perpendicular onto the constant function -3.

Done.

[Anonymous No.4568883]

>>4568878

lol, a few errors with this post:

>-3 should be -C
>999 should be in the exponent

[Anonymous No.4568894]

>>4568878
You mean a vector perpendicular to the constant function, as in <span class="math">\int_{-1}^1 p(x) \cdot constant dx = 0[/spoiler]? Yeah, it's trivial to find such a polynomial, but how does that prove anything?

[Anonymous No.4568906]

We want to show that there is a <span class="math"> m \in \mathbb{R} [/spoiler] such as for every polynomial p of degree 1999, <span class="math"> \int_{-1}^{1} |p(x)| dx \geq m [/spoiler]
for that, I suggest to show that the set of all the possible integrals of abs(p(x)) from -1 to 1 is a closed set.
Indeed, the integral is never equal to 0, but no one says that I cannot find one as small as I want. So if I show that the set of all those integrals is closed, it has a minimal value, which is not 0.
I'm gonna think about that and publish some other elements.

[Anonymous No.4568938]

>>4568906
OK.
Every convergent sequence of polynomials of degree 1999 converges towards a polynomial of degree 1999. Therefore the set of the polynomials of degree 1999 is closed.
Therefore, as the mapping
<span class="math"> p \mapsto \int_{-1}^{1} |p(x)| dx [/spoiler] is continuous, the codomain is closed.
As it does not contain 0, there exists an m such that the integral is greater than m for any polyniomial of degree 1999.
The existence of C follows trivially from that.

[Anonymous No.4568948]

>>4568938
>Therefore, as the mapping is continuous, the codomain is closed.
No

[Anonymous No.4568955]

>>4568948
yes.
Or please explain why?

[Anonymous No.4568960]

>>4568955
Obvious counterexample is f : R -> R of f(x) = 1/(x^2 + 1).

[Anonymous No.4568961]

>>4568955
Nevermind, brainfart

[Anonymous No.4568963]

>>4568955
In general continuous functions don't map closed sets to closed sets.
Consider f(x)=x^2/(1+x^2). It is continuous and maps the closed set R to [0,1[ but [0,1[ is not closed.

[Anonymous No.4569002]

hmm...how about that
p: [-1,1] -> R is continuous then by mean value theorem,
the integral equals (1-(-1))|p(a)|=2|p(a)|, for some a in (-1,1)
OK.
Every convergent sequence of polynomials of degree 1999 converges towards a polynomial of degree 1999. Therefore the set of the polynomials of degree 1999 is closed.

[Anonymous No.4569066]

>>4568906
I'm pretty sure if you're aim to find such an m is impossible.

if you have a polynomial and then just divide through by 10, the integral will become a 10th of the size. Obviously you won't get a lower bound for all polynomials. (Unless you only consider polynomials for a given p(0) such as p(0) = 1)

[Anonymous No.4569071]

The annoying part seems to be the absolute value, so maybe it would help to factor the polynomial?
<div class="math">\int_{-1}^1 |p(x)| dx = \int_{-1}^1 q(x) \cdot \prod_{i=1}^n |x-r_i| dx </div>where <span class="math">n \leq 1999 - order(q)[/spoiler], <span class="math">-1 < r_i < 1[/spoiler], and q(x) > 0 in (-1,1).

[Anonymous No.4569159]

Solved, gonna post the answer in 5 minutes, fixing my laptop

[Anonymous No.4569182]

>>4569066
yes you're right, I was actually looking for those.

[Anonymous No.4569255]

>>4569159
what happened with the solution...........

[Anonymous No.4569582]

I've done it. A few hints:

Either you can argue as above (note that a map does not have to be a closed map to be continuous) trying to prove the existence of m. It suffices to prove the result for polynomials p such that p(0) = 1. Without loss of generality we can also assume that the max over [-1,1] of p(x) is attained at 0. The set of p satisfying this condition forms a compact set because we are working on a finite dimensional vector space. The image of a compact set is compact under a continuous map, and the map defined by the integral is continuous. So the integral is bounded below by m, and since 0 is not attained, it is bounded below by m>0. Take <span class="math">\frac{1}{m} [/spoiler].

Alternative proof: You can consider the distances to each of the roots. There are finitely many roots, so inside [-1,1] there is an interval where we have no roots. Additionally, we can find a subinterval of this interval where p is some distance away from the roots (independent of p). Now if the roots are all small, then p(0) is also small, and an easy argument can be done. If the roots are very large you can argue that <span class="math"> \Pi_{r=1}^{1999} |X_i - a_i| \geq p(0)C [/spoiler] by expanding the former product and arguing that <span class="math"> \Pi_{r=1, r \neq j}^{1999} |a_i| \leq \Pi_{r=1}^{1999} |a_i| [/spoiler] divided by the minimum of the |a_i|.

[Anonymous No.4569625]

>>4569582
>Without loss of generality we can also assume that the max over [-1,1] of p(x) is attained at 0.
Why?

[Anonymous No.4569662]

>>4569625
Thinking about it, I'm not sure that simply saying WLOG suffices. Instead, use the argument to discuss the unit ball of polynomials under the supremum norm of degree 1999. This bounds the integral below by <span class="math"> m = m ||p||_\infty \geq m p(0) [/spoiler]

[Anonymous No.4569950]

I think I got it. With induction. It holds for a polynomial of any degree. The constant depends on the degree of the polynomial, which is okay because we only have to prove it for degree 1999.

I'm only going to prove the cases where p(0) = 1. I proved the linear case, so I'll assume that the statement holds for a polynomial of degree n-1 with p(0) = 1. Let p(x) = a_n*x^n + ... 1 = (r_0*x - 1)*(b_{n-1}*x^{n-1} + ... 1). Now consider the integral from -1 to 1 of p(x). Since |r_0*0 - 1| = 1, there must be a non-empty interval (-B, B) inside of [-1,1] centered at 0 where |r_0*x - 1| > 1/2. We know the integral of p(x) from -1 to 1 is greater than the integral of p(x) from -B to B. We know that since |r_0*x - 1| > 1/2 on this interval, that the integral is greater than 1/2 times the integral from -B to B of (b_{n-1}*x^{n-1} ... + 1). Now, use u substitution with u = x/B.This gives us integral from -1 to 1 of 1/B*(c_{n-1}*u^{n-1} + ... 1). Now we know by the induction hypothesis that this integral is greater than (1/B)*1/C_{n-1}. Thus, tracing through the "greater than or equal tos" we have the integral from -1 to 1 of |p(x)| >= 1/B*C_{n-1} or 1 <= B*C_{n-1}|p(x)|.

[Anonymous No.4569953]

>>4569950

Also, for future reference, how do you start writing in latex?

[Anonymous No.4569973]

>>4569953
Use TeX/jsMath with the <span class="math">[ math ][/spoiler] (inline) and <span class="math">[ eqn ][/spoiler] (block) tags. Double-click equations to view the source.

<span class="math">[ math ][/spoiler]E = mc^2<span class="math">[ /math ][/spoiler]
becomes
<span class="math">E = mc^2[/spoiler]

[Anonymous No.4569977]

>>4569953
Use TeX/jsMath with the <span class="math">[ math ][/spoiler] (inline) and <span class="math">[ eqn ][/spoiler] (block) tags. Double-click equations to view the source.

<span class="math">[ math ][/spoiler]E = mc^2<span class="math">[ /math ][/spoiler]
becomes
<span class="math">E = mc^2[/spoiler]

[Anonymous No.4570141]

>>4569950
I like this approach, but B seems to depend on p, which puts a hole in the proof. There's probably a way to patch it; I'll have to think about it.

[Anonymous No.4570437]

>>4568831
Take C = +infinity
QED

[Anonymous No.4570469]

how much did you guys take in college. this question is so simply stated but the answer seems like sorcery. one of the reasons i sometimes wish i had more than a capability to undergo mathematical rigor - the ability to think elegantly would be nice.

=

[Anonymous No.4570516]

Equivalence of norms on finite dimensional vector spaces kills this. Which of course makes it a lousy contest type problem, because it comes equipped with an instant victory condition for anyone who has taken a real analysis course.

It's still a nice exercise to work out exactly what the constant is for any fixed degree. I'll have to think about that some.

[Anonymous No.4570545]

> finite dimensional vector space
> Real Analysis

Isn't that more of an Algebra thing? We didn't make much reference to vector spaces in Real Analysis.

[Anonymous No.4570573]

>>4570545

Vector spaces are among the basic players in higher maths, and they show up everywhere. Normed vector spaces are pretty central in analysis. Depending on how material is arranged, you might not see them in a first course on real analysis, but only later in "functional analysis".

[Anonymous No.4571246]

>>4570573
The key term was >finite dimensional <span class="math">[/spoiler], I think

[Anonymous No.4571253]

Why can't nobody solve this? Is /sci/ so stupid?

[Anonymous No.4571300]

>>4571253
I haven't finished Algebra yet. So yes

[Anonymous No.4571363]

[Anonymous No.4571400]

>>4571363
I hate it when people take that expression literally.

The "radiance" or "light" of "a thousand suns" is a recurring element in Hindu mythology:

From the Bhagavad Gita (http://members.aol.com/heraklit1/gita.htm):
If the light of a thousand suns
should suddenly burst forth in the sky
it would be like the light
of that exalted one.

(text is from the translation of Franklin Edgerton, The Bhagavad Gita, Harper Torchbooks, New York, 1944)


From a site of Hindu sloka (religious proverbs) (http://www.geocities.com/Athens/Acropolis/5294/sloka.html):
Chapter 11, Verse 12 (of the Bhagavad Gita -- b)

Even If the radiance of a thousand Suns, bursts forth all at once in the heavens, it would still hardly approach the splendor of the mighty Lord.

Learn a little myths and history science fags.

[Anonymous No.4571420]

>>4571400
this should become the reply informational

[Anonymous No.4571444]

>>4571420
The expression light of 1000 suns became widespread to american culture after J. Robert Oppenheimer used it to describe the tests of the first atomic bombs.

Since he used it, we butchered that quote and made 1000 nerds angry because they take it in the literal sense.

[Anonymous No.4571533]

>>4571444
Shakespeare would like a word with you.

[Anonymous No.4571692]

>>4571363

wait. isn't it impossible to be faster than the speed of light? where did the calculations in this image go wrong? is it because it assumes that an increase in energy increases speed, instead of an increase in intensity.

[Anonymous No.4571721]

<span class="math">2^{2}[/spoiler]

[Anonymous No.4571746]

>>4571692
isn't it more like <span class="math">Wtot = W0 + Wkin[/spoiler] and
<span class="math">mrel = m0/sqrt(1-(v²/c²))[/spoiler] thus
<span class="math"> v = c* sqrt(1-m0²/mrel²) [/spoiler]

[Anonymous No.4571765]

>>4571746
forgot
<span class="math"> v = c* sqrt( 1- m0²/( m0²+Wkin/c²)²) [/spoiler]

[Anonymous No.4571767]

>>4571765
and the <span class="math"> ²[/spoiler] on the second <span class="math">m0²[/spoiler] should not be there