Putnam Problem of the Day

[http://math.harvard.edu/putnam(...) !!84u21X0KjW+ No.10510739]

[math]
\text{Let }f\text{ be a real function on the real line with continuous third derivative. Prove}
\\
\text{that there exists a point }a\text{ such that}
\\
\qquad \qquad \qquad \qquad \qquad f(a)\cdot f'(a) \cdot f''(a) \cdot f'''(a)\geq 0 .
[/math]

[http://math.harvard.edu/putnam(...) !!84u21X0KjW+ No.10510740]

Previous Thread >>10510005

[Anonymous No.10511147]

>>10510739
The plan is to use proof by contradiction.
Assume no a exists.
f * f' * f'' * f''' < 0 for all inputs.
For any input, 3 of the terms have one sign and the other term has the opposite sign.
If any of terms change sign as the input varies, it must pass through zero (by continuity) which would contradict the assumption.
Suppose the terms never change signs or equal 0.

f(x) satisfies this condition iff -f(x), f(-x), and -f(-x) also satisfy it.
Choose g(x) to be one of {f(x), -f(x), f(-x), -f(-x)}
such that g > 0 and g' > 0.

Can g'' be less than 0?
No, since convexity gives g(-a) <= g(0) - a*g'(0) which could be made negative for large enough a.

Using the same argument on g' and g'' tells us that g''' must also be greater than 0 which is a contradiction since at least one of them are negative by assumption.

I hope this isn't too confusing.
Using {f(x), -f(x), f(-x), -f(-x)} was just to make things convenient.
You could have enumerated all of the cases for the signs of f and f'.

[Anonymous No.10511169]

>>10510739
f''''(x) = x ^4
f'''(x) = 4x^3
f''(x) = 12x^2
f'(x) = 24x
f(x) = 24

f(1) * f'(1) * f''(1) * f'''(1) = 24 * 24 * 12 * 4 * 1 = 27648 > 0

There exists a point, that point is 1. QED.

[Anonymous No.10511181]

>>10511169
Wut.
You got your 's wrong.
You went up to 4 for some reason.
You showed it was true for a specific function and not all of them.

[Anonymous No.10511191]

>>10511181
not the guy that did that solution, but doesn't his still work? It said prove that a point exists, and they proved that at least one point exists fitting this description, even if they did an extra derivative.

[Anonymous No.10511194]

>>10511191
>showed it was true for a specific function and not all of them.

[Anonymous No.10511210]

>>10511147
Nice.

[Anonymous No.10511211]

>>10511194
Where does it specify that it should be for all of them? And yes, I would 100% accept solving a putnam problem on a grammatical basis.

[Anonymous No.10511223]

>>10511169
f(x) = -e^(-x)
f'(x) = e^(-x)
f''(x) = -e^(-x)
f'''(x) = e^(-x)
f''''(x) = -e^(-x)

f(x)*f'(x)*f''(x)*f'''(x)*f''''(x) = -e^(-5x) < 0

[Anonymous No.10511226]

>>10510739
Is this shit even true? Let [math]f'''[/math] be a strictly negative function whose integral [math]\int_{\infty}^x f'''[/math] is bounded below by some constant C. Let [math]f''[/math] be this integral, plus C, and define [math]f'[/math] and [math]f[/math] as integrals in the obvious way. Then [math]f, f', f''[/math] are strictly positive but [math]f'''[/math] is strictly negative... what am I missing?

[Anonymous No.10511228]

Doesn't this >>10511223 disprove this >>10511194?

It is not true for all of them.

[Anonymous No.10511233]

>>10511226
Since f''' is strictly negative, f' is strictly concave. But a strictly concave function can't be strictly positive.
(draw any tangent line that isn't horizontal and the whole function has to lie under it, but this tangent line will intersect the x-axis)

[Anonymous No.10511242]

>>10511211
Let f be a real function (with conditions)
Means f is ANY real function (with those conditions)

>>10511228
No. Don't go up to f''''
Only go up to the third derivative.

[Anonymous No.10511251]

>>10511226
[math]\int_{-\infty}^x f'''[/math] vanishes for [math]x\to - \infty[/math]. If we then define [math]f''[/math] like you did, it has to go to [math]C[/math] in this limit. This implies that [math]f' = \int_{-\infty}^x f''[/math] does not exist as the integrand does not approach zero.

[Anonymous No.10511263]

>>10511242
Yeah. Any function. Not every function.

[Anonymous No.10511281]

>>10511263
I can't tell if you are trolling or are just ignorant.

[Anonymous No.10511327]

aX^n where x\ge 0
and the function is differentiable to that height, or the absolute value of n is/ge 3

[Anonymous No.10511518]

>>10511281
Are you illiterate? Read the problem out loud or something, damn.

[Anonymous No.10511629]

>>10511518

[Anonymous No.10511730]

>>10511629

[Anonymous No.10511738]

This seems to be too simple to be a correct proof, but I can't see any mistake in it. Continuity is used when I assume increasing -> positive derivative.

[Anonymous No.10511779]

>>10511738
You don't argue why f' couldn't be decreasing.

[Anonymous No.10511783]

>>10511779
Isn't that handled by f(a) =/= 0 for all a?

[Anonymous No.10511792]

>>10511783
It is, but if you don't show or even mention it that makes the proof incomplete. Also f itself could just as well be decreasing and then f' would have to be negative.

[Anonymous No.10511809]

>>10511792
I agree that I should add in why f' can't be decreasing, although it's a minor thing.

Like I mention at the end the case for a decreasing f is basically the same. You get f decreasing and never 0 -> f' <= 0, if f' = 0 we are done, otherwise f' strictly decreasing - > f'' strictly decreasing -> f''' strictly decreasing. It's not different enough or complicated enough to actually write down, simply stating that it's handled in the same way is fine, I think.

[Anonymous No.10511816]

>>10511809
Do you also know why f' decreasing implies f will be 0 at some point, because even if you say it's a minor thing, it is what the entire proof rests on.

[Anonymous No.10511854]

>>10511809
e^(-x) is positive, strictly decreasing, and never 0.

[Anonymous No.10511878]

>>10511816
>>10511854
Yeah, I just realized that myself. I could probably write a more bruteforce proof (but that's actually correct), but I don't want fifty-eleven different cases. I'll see if I come up with something a bit more elegant.

[Anonymous No.10511892]

>>10511878
Brute force isn't needed, I hope you can figure it out.

[Anonymous No.10511903]

>>10511211
it should be for all functions on the real line with continuous third derivative.

[Anonymous No.10512040]

>>10511169
>d/dx 24 = 24x
>d/dx 24x = 12 x^2
How the fuck did you fuck up so bad

[Anonymous No.10512041]

>>10511878
>57 cases
You mean four.

[Anonymous No.10512045]

>>10512040
stfu i was making a point, not accurate calculations. i couldn't be bothered putting it into a derivative calculator

[Anonymous No.10512072]

>>10510739
This is the function composition operator, right? Not just a multiplication symbol?

[Anonymous No.10512075]

>>10512045
What the fuck are you on about and what are you doing on this board retard
Your point at no point coincided with any form of relevance

[Anonymous No.10512440]

>>10512072
Wrong, [math]f(a)[/math] is a real number, not a function.

[Anonymous No.10512836]

bump

[Anonymous No.10512864]

>>10511518
The only information you have about f is that it's a real function with continuous 3rd derivative. It could be any function satisfying those conditions and you don't know which one. Given that you don't know which particular function it is, to then prove something about it is equivalent to proving something about all functions satisfying those conditions

[Anonymous No.10512879]

>>10512864
He knows that, he's trying to make a joke on the semantic ambiguity that wouldn't be there if using formal quantifiers.

[Anonymous No.10512918]

>>10511792
>>10511809
>>10511816
>>10511854
>>10511878
>>10511892
OK, came back to it now, hopefully this updated one is actually correct. Again I only show the case for strictly positive f, since all the logic holds for strictly negative f by just setting g = -f.

[Anonymous No.10512931]

>>10512440

f(a) is a function
not a number
wtf

[Anonymous No.10512939]

>>10512931
[math]f(a)[/math] is the the function [math]f[/math] evaluated at point [math]a[/math], a real number.

[Anonymous No.10513862]

>>10510739
Who wrote this?
>[math]f(a)[/math]
>eff-aayyyyy
Contrast with
[math]f(\alpha)[/math]
eff-alpha
Alpha is also easier to write by hand than a. I hate academics who ignore the feng-shui of the symbols they use.

[Anonymous No.10513872]

>>10511169
what is this garbage

[Anonymous No.10514107]

if f(a)f'(a)f''(a)f'''(a) < 0 for all a, then wlog lim f(x) ,x->inf is 0.

so f is always negative and approaches zero in the limit. this means that each of the derivatives approach zero in the limit too

so f is negative and increasing, f' positive and decreasing, f'' must be negative and increasing, but then f''' must be positive, which contradicts the assumption.

[Anonymous No.10514123]

>>10510739
Can you omit the cartoon in your image so I don't look like a faggot looking at that in public?

[Anonymous No.10514131]

suppose f(a)f'(a)f''(a)f'''(a) < 0. f must be strictly monotonic and have no roots. wlog we can assume it's negative, increasing, and approaches some constant <= 0. this means that f',f'',f''' go to zero in the limit.

so f is negative and increasing, f' positive and decreasing, f'' must be negative and increasing, but then f''' must be positive, which contradicts the assumption.

[Anonymous No.10514172]

>>10512918

pretty sure you can make some assumptions on f wlog. i followed similar logic here:

>>10514131

the fact that f can't have a root and must be strict monotonic clearly follow from the assumption. just take f to be negative and increasing. then you don't have to awkwardly break it down by cases.

[Anonymous No.10514577]

Suppose [math] f(x)\cdot f^{(1)}(x)\cdots f^{(n)}(x) < 0 [/math] for all [math] x\in\mathbb{R}[/math], then an odd number of terms must be negative. If [math] n \equiv 0 \text{ mod } 4[/math], then there will be an [math] 0\leq m\leq n-2[/math] such that [math]f^{(m)}[/math] and [math] f^{m+2}[/math] have opposing signs. Since neither can be zero we see that [math] f^{(m)}[/math] must be either a strictly concave strictly positive function or a strictly convex strictly negative function, but such a function does not exist. For other [math] n [/math] one of [math] e^{-x}[/math] and [math] -e^{-x}[/math] is an example.