/sci/ - Science & Math

[Anonymous No.4318469]

Putnam/Olympiad problem of the day from
http://www.math.harvard.edu/putnam/

Show that
<div class="math">\int_{-100}^{-10} \left( \frac{x^2-x}{x^3-3x+1} \right)^2 dx +
</div><div class="math">\int_{\frac{1}{101}}^{\frac{1}{11}} \left( \frac{x^2-x}{x^3-3x+1} \right)^2 dx +
</div><div class="math">\int_{\frac{101}{100}}^{\frac{11}{10}} \left( \frac{x^2-x}{x^3-3x+1} \right)^2 dx
</div> is a rational number.

[Anonymous No.4318485]

Seriously? Just integrate that shit and see there's nothing irrational in there.
Looks more like basic calc I homework.

[Anonymous No.4318484]

It is 11131110/107634259 which is rational

[EK !EKFQOBUFnQ No.4318491]

no, prove it yourself you lazy cunt.

sage, reported, hidden, etc for homework thread

[Anonymous No.4318501]

>>4318491
>saging a sticky

SURE IS INTELLIGENCE IN HERE

[Anonymous No.4318523]

This is a horrible question, can anyone see something clever that I don't see? I'm guessing they don't want alpha...

[Anonymous No.4318526]

>>4318485
Either there's indeed nothing irrational in any of the integrals (but I doubt that), or there's a simplification to do after solving each separately (I still doubt it, but it would be boring so I wouldn't bother doing it), or there's a clever idea to have (but I don't like analysis so I'm certainly not going to have it or even bother searching).

[Anonymous No.4318617]

>>4318491
I refuse to believe that anyone can be this retarded, so you must be trolling

[Anonymous No.4318629]

>>4318485
I'm trusting you haven't actually tried to integrate that function. Can anyone see any way of approaching the integral? I'm guessing there must be a trick here to avoid that.

[Yo, I got this one No.4318642]

I've got some ideas. The numbers in the bounds must be significant. I suspect that under the right substitutions, the three integrals get the same bounds, and the three new integrands add to an expression that can be simplified and then easily integrated and evaluated.

I'll think more about this.

[Anonymous No.4318634]

I'm just going to make some wild claims that the sum of rationals is rational; the square of a rational is a rational; and the integral of a rational (at rational bounds) is a rational.

That last one is harder to prove, but it seels like putting rationals in for x has to yeild another. We've got no square rooting and no trigonometry...

[Anonymous No.4318647]

>>4318634
>the integral of a rational (at rational bounds) is a rational.

Have you tried integrating 1/x recently?

[Anonymous No.4318691]

>>4318642
Sounds good to me. I'm still too lazy to do it but I'm with you on that approach.

[Anonymous No.4318736]

Transforming the second integral with x -> 1 - 1/x and the third one with x -> 1/(1-x)
You get the integral of [(x²-x)/(x³-3x+1)²] * (1+1/x²+1/(1-x)²) over [-100,-10]
What do now?

[Anonymous No.4318745]

>>4318736
Whoops, forgot a square.

[Anonymous No.4318849]

>>4318642
Uhmm look at the inverse of both bound in the third integral. I think you may be on to something here.

Still I can't solve this because I decided I was too cool to learn how to properly integrate shit and failed Calc 1.

[Yo, I got this one No.4318898]

>>4318642
Let <div class="math">f(x) = \frac{x^2-x}{x^3-3x+1}.</div>

Without being sure if it would be helpful, I evaluated <span class="math">f[/spoiler] at the bounds of the three integrals. It turns out the values are the same at all the lower bounds and the same at all the upper bounds: <div class="math">f(-100)=f(1/101)=f(101/100),\quad\quad
f(-10)=f(1/11)=f(11/10).</div>

This can't be a coincidence.

I'll try to see if substituting <span class="math">u = f(x)[/spoiler] does any good beyond giving the three integrals the same bounds.

(I cheated and used a calculator [I always have MATLAB running] to do these calculations. But they are doable — if somewhat tedious — by hand and should only take a few minutes.)

[Anonymous No.4318956]

>>4318898
>(I cheated and used a calculator [I always have MATLAB running] to do these calculations. But they are doable — if somewhat tedious — by hand and should only take a few minutes.)
I don't think that counts as cheating.

Also:
>It turns out the values are the same at all the lower bounds and the same at all the upper bounds
Makes me think of many things, but none actually helps directly...

[Yo, I got this one No.4318988]

>>4318898
Also possibly helpful: letting <span class="math">a_1,a_2,a_3[/spoiler] denote the lower bounds of the 1st, 2nd and 3rd integrals, and <span class="math">b_1,b_2,b_3[/spoiler] denote the upper bounds, we see <div class="math">a_3 = 1- \frac{1}{a_1}\quad,\quad
b_3 = 1- \frac{1}{b_1}\quad,\quad
a_2 = 1- \frac{1}{a_3}\quad,\quad
b_2 = 1- \frac{1}{b_3}.</div> So substituting with the function <span class="math">x\mapsto x-\frac{1}{x}[/spoiler] might be useful.

I thought I saw a post on here that had that same idea, but it's gone.

>>4318956
>Makes me think of many things, but none actually helps directly...
Same here.

[Yo, I got this one No.4319101]

>>4318988
> So substituting with the function <span class="math">x\mapsto x-\frac{1}{x}[/spoiler] might be useful.

I meant the function <span class="math">x\mapsto 1-\frac{1}{x}[/spoiler]. Denote this function by <span class="math">\phi[/spoiler]. Then <div class="math">a_1=\phi(a_2)\quad,\quad
a_2=\phi(a_3)\quad,\quad
a_3=\phi(a_1),</div> and similarly for the <span class="math">b_j[/spoiler]'s.

[Anonymous No.4319278]

is there any pracitcal relevance in solving that?

[Anonymous No.4319399]

>>4319278
Its a nerd game. We're not solving cancer, just having fun with maths. Think of it as masturbation.

[Anonymous No.4319458]

>>4318469

0.1027404785

[Yo, I got this one No.4319503]

>>4319101
So here's what I have so far:

If we substitute <span class="math">u=\phi(x)[/spoiler] into the third integral and <span class="math">v=\phi^{-1}(x)[/spoiler] into the first, we find that the sum of the first, second and third integrals is <div class="math"> \int_{a_2}^{b_2} \frac{(v-1)^2}{(v^3-3v+1)^2} dx
+\int_{a_2}^{b_2} \frac{x^2(x-1)^2}{(x^3-3x+1)^2} dx
+\int_{a_2}^{b_2} \frac{u^2}{(u^3-3u+1)^2} dx
= \int_{a_2}^{b_2} \frac{(x-1)^2+x^2(x-1)^2 + x^2}{(x^3-3x+1)^2} dx. </div> Unfortunately, I don't see how to simplify this integral any further (to show that its value must be rational).

Another observation: since <div class="math">x^3-3x+1 = -(1-x)^3 + 3(1-x) -1</div>, the integrand is symmetric in <span class="math">x[/spoiler] and <span class="math">1-x[/spoiler]. Not sure what use that may be.

[Yo, I got this one No.4319511]

>>4319503
should have been <span class="math">dv[/spoiler] in the first integral and <span class="math">du[/spoiler] in the third, of course.

[Yo, I got this one No.4319555]

>>4319458

[Anonymous No.4319562]

>>4319458
>show that
>show

>rational number
>rational

[Anonymous No.4319568]

>>4319562
All terminating decimals are rational.

[Anonymous No.4319587]

a bump so there isnt a pony on the front page

[Anonymous No.4319725]

>>4319568
reported for pony enjoy your ban

[Anonymous No.4319741]

>>4319503
Not sure if this helps, but

<div class="math">(x-1)^2+x^2(x-1)^2+x^2 = x^2-2x+1+x^4-2x^3+x^2+x^2 = x^4-2x^3+3x^2-2x+1 = (x^2-x+1)^2</div>

so the integral is

<div class="math">\int_{a_2}^{b_2} \left(\frac{x^2-x+1}{x^3-3x+1}\right)^2 dx</div>

[Anonymous No.4319815]

Forgive me for maybe sounding stupid, but is just plainly solving these integrals (or whatever the correct verb is) not the easiest and by far the simplest way to show that X in case is a rational number?

[Anonymous No.4319863]

>>4319815

its possible, but that brute force method is longer (a lot) and harder than anything else suggested in the thread.

that's why people are trying to come up with a different proof.

[Anonymous No.4319865]

>>4319815
you can't just simply solve the later two, since they dont converge.

[Anonymous No.4319872]

>>4319865
reported for pony enjoy your ban

[Anonymous No.4319883]

>>4319872
contribute something usefull, tits, or gtfo.

[Anonymous No.4319899]

>>4319741
Me again.

So I was screwing around with the function <div class="math">f(x) = \frac{x^2-x}{x^3-3x+1}</div>
and I found that <div class="math">f'(x) = -\left(\frac{x^2-x+1}{x^3-3x+1}\right)^2</div>

So I guess the problem is solved. I still have no idea why that should be true though. I just stumbled upon it by accident.

[Anonymous No.4319921]

>>4319899
>I still have no idea why that should be true though. I just stumbled upon it by accident.
Clever design. Can't be evolution here. The problem must have been created by someone with intelligence.

Also, nice work. I'm half-trolling but I still appreciate the result.

[Anonymous No.4319987]

-0,2968

[Yo, I got this one No.4319995]

>>4319741
It might help, but I haven't yet figured out how.

By the way, here's how one can find that factorization: First, you try the four possible rational roots (±1, ±1/2), and once you see those don't work, you guess at a factorization of the form <div class="math">x^4-2x^3+3x^2-2x+1= (x^2 + Ax + \delta)\left(x^2 + Bx + \frac{1}{\delta}\right)</div>
(if one exists, it must have this form). After expanding and equating coefficients, you'll see that it would be natural to try taking <span class="math">\delta=1[/spoiler]. Solving for suitable <span class="math">A[/spoiler] and <span class="math">B[/spoiler] is then easy.

>>4319899
Um, you just happened to be screwing around with the function <span class="math">x\mapsto \frac{x^2-x}{x^3-3x+1}[/spoiler]? Gosh, what are the chances?

Here's a more plausible explanation: you entered the integral into Wolfram Alpha. And that, my friend, is cheating.

I'm still interesting in finding a method for solving the problem.

[Anonymous No.4320043]

>>4319995
>Gosh, what are the chances?
It's the function that's squared in the integrand of the original integrals.

[Yo, I got this one No.4320058]

>>4320043
Oh snap, you're right.

My apologies. I now believe it is plausible that one could stumble upon that solution. But it's still pretty unsatisfying. There has to be a better (more reasonable) method.

[Yo, I got this one No.4320277]

>>4320058
Here's a possible route to the solution: Since we are asked to show the integral is rational, we suspect <div class="math"> \frac{\eta(x)}{q(x)^2} \quad\quad
[\eta(x)= x^4-2x^3+3x^2-2x+1 = (x^2-x+1)^2, \quad\quad
q(x) = x^3-3x+1] </div>
is the antiderivative of a rational function <span class="math">r(x)[/spoiler]. Because we know the Quotient Rule, we suspect that <span class="math">r(x)[/spoiler] has the form <div class="math"> r(x) = \frac{p(x)}{q(x)},</div> so that <div class="math"> q(x)p'(x) - p(x)q'(x) = \eta(x).\quad\quad\quad(1)</div> This suggests that <span class="math">\deg p(x) = 2[/spoiler], so we try writing <span class="math">p(x)[/spoiler] as a general, unknown 2nd degree polynomial: <div class="math"> p(x)= Ax^2 +Bx+C.</div> We may collect coefficients in (1) to get <div class="math">
-(A+1)x^4 -2(B-1)x^3+ 3(A+C+1)x^2+2(A+1)x +(B+3C-1)=0.</div> We solve to find <span class="math">A=-1, B=1, C=0[/spoiler]; i.e., <span class="math">p(x) = -x^2+x[/spoiler].

We then check directly that <span class="math">r(x) = p(x)/q(x)[/spoiler] indeed satisfies <span class="math">r'(x)=\eta(x)/q(x)^2[/spoiler]. Consequently, the original definite integral is a rational number.

I'm satisfied with this method. It seems reasonably well motivated, which is to say it doesn't require pulling a rabbit out of a hat. Though I would be curious if there are any other possible approaches.

[Anonymous No.4320382]

>>4320277
Piss off with your solution manual

[Yo, I got this one No.4320455]

>>4320382
I assume you're implying (with "solution manual") that my solution is well written. Thanks. Why does that irritate you?

[Anonymous No.4320674]

I still don't get why they would state the problem in terms of a sum of three integrals.

As far as I can tell, the three integrals all have the same value anyway.

[Anonymous No.4320684]

>>4320277
Glad to see we finally got a workable solution but it's kinda disappointing that we didn't exploit something with the integral itself. Not very elegant but it works.

[Yo, I got this one No.4320720]

>>4320684
Agreed. I strongly suspect the problem-writing committee had a more elegant solution in mind.

If the anon who posted a screenshot of several solutions to yesterday's problem (from a book of solutions) is still around, would you mind posting the solutions to this one?

Sadly, I think he may have been banned for posting ponies. All his posts from yesterday's thread have disappeared.

[Anonymous No.4320889]

There is some long-ass super complicated formula for finding cubic roots. We could find the roots of the denominator, two of which are complex, then break it apart into fractions in the form of a/(x-r), where r is a complex root, which would give us some logs, which we could combine over certain intervals, which would hopefully give us an answer. It's a super long complicated method, which could work, but they probably want us to do something simpler.

[Anonymous No.4320896]

>>4320720
>Sadly, I think he may have been banned for posting ponies.
>implying bans do fucking anything

Too lazy to cap, but yeah, that's basically the way to do it.

[Anonymous No.4320905]

>>4320896
reported for ban evasion enjoy your permaban

[Anonymous No.4320906]

>>4320896
reported for cheating
reported for posting a pony
enjoy your ban

[Anonymous No.4320909]

>>4320905
>thinks a permaban is any more effective than a regular ban

[Anonymous No.4320911]

>>4320909
reported for ban evasion. enjoy your permaban

[Anonymous No.4320921]

>>4320918
reported for ban evasion. reported for offtopic. enjoy your permaban

[Anonymous No.4320918]

>>4320911

[Anonymous No.4320965]

stop giving bronies a bad reputation

there is only two on this board that know what the fuck they are doing the rest are shit posters

[Anonymous No.4320973]

>there is only two
>is only two
reported enjoy your ban

>stop giving bronies a bad reputation
reported enjoy your ban

[Anonymous No.4320974]

>>4320965
y u mad tho?

[Anonymous No.4320976]

>>4320974
reported for pony enjoy your ban

[Anonymous No.4320979]

>>4320974
reported for ban evasion. reported for trolling. enjoy your ban

[Yo, I got this one No.4320985]

>>4320889
I thought of this too, but I imagine it would be incredibly tedious.

>>4320896
So they didn't write a more clever solution... That's disappointing.

[Anonymous No.4321009]

>>4320985
Didn't seem too bad to me.

[Anonymous No.4321022]

>>4321009
reported for ban evasion. enjoy your permaban

[Anonymous No.4321021]

>>4321009
Stop doing that.

There is no reason to post that image. You could have responded without an image.

At least use a tripcode so I can filter you manchildren.

Enjoy your ban.

[Anonymous No.4321026]

>>4321021
>saging a sticky

[Anonymouse !70ikdWM2E6 No.4321032]

>>4321026
Sage

[Anonymous No.4321029]

>>4321026
reported for offtopic. reported for trolling. reported for ban evasion. enjoy your permaban

[Anonymous No.4321034]

>>4321026
Fuck off ye slimey cunt

[Yo, I got this one No.4321039]

>>4321009
There's something about having to guess at a form of the antiderivative that rubs me the wrong way. But I suppose it wasn't so unreasonable.

[Anonymous No.4321077]

>>4321039
That's how math is, d00d. There's hardly ever a step-by-step, pre-written algorithm to solve a problem.

[Anonymous No.4321079]

>>4321077
reported for ban evasion. enjoy your permaban

[Anonymous No.4321093]

>>4320277

You're solution is beautifully typed, elegant, intelligent, but also completely wrong. The numerator is <span class="math"> (x^2 - x)^2 [/spoiler], not <span class="math"> (x^2 - x + 1)^2 [/spoiler]. It could have been a good start though because once we subtract your solution, we're left with proving that <span class="math"> \int \frac{1}{x^3-3x+1}[/spoiler], is a rational number, which is much simpler. Personally I think that the bounds are clues that may lead to an elegant solution that would be hard to notice without them.

[Anonymous No.4321156]

>>4319399
Guy has a point. It's kind of like masturbation, except you have to work alot harder to get to the climax, (which you very likely may never get to), and you have to resort to using a bunch of anonymous dudes on the internet for help. I think I'm gonna give up and go try good old-fashioned masturbation that I'm used to for now. Peace out, bitches!

[Anonymous No.4321233]

>>4320889
Actually this cubic has three real roots and they're not too bad since the cubic doesn't have an order 2 term. They can be written like this:
<div class="math">x^3 -3x +1 = (x+2 \cos 20 ^{\circ} )(x - 2 \cos 40 ^{\circ} )(x - 2 \cos 80 ^{\circ} )</div>
Not that this actually helps get to a solution, since the roots are obviously irrational. I just thought it was interesting.

[Yo, I got this one No.4321299]

>>4321093
>You're solution is beautifully typed, elegant, intelligent, but also completely wrong. The numerator is <span class="math">(x^2 - x)^2[/spoiler], not <span class="math">(x^2 - x + 1)^2
[/spoiler].

I'm guessing you missed my prior posts. Here's the order:
>>4318642
>>4318898
>>4318988
>>4319101
>>4319503
>>4319511
someone else: >>4319741
someone else: >>4319899
>>4319995
someone else: >>4320043
>>4320058
and then, finally, the one you saw: >>4320277

The integrand I was referring to is the integrand of the rightmost integral in post>>4319503
not the integrand common to the three integrals in the original statement of problem.

> It could have been a good start though because once we subtract your solution, we're left with proving that <span class="math">\int \frac{1}{x^3-3x+1}
[/spoiler], is a rational number, which is much simpler.

Actually, I don't think it's a rational number when evaluated over any of the three separate pairs of bounds, <span class="math">[-100,-10],\, [1/101,1/11],\, [101/100,11/10][/spoiler]. Try evaluating it on Wolfram Alpha. I'd be willing to bet that final expression does not sum to a rational number.

[Anonymous No.4321317]

the worst thing about these threads is that probablitistically you most likely won't see it before it has been solved. So you get all fucking excited to solve it and someone already has.... ugh. hate these threads

[Anonymous No.4321323]

I have a most elegant proof, but alas, I cannot fit it in this comment box

[Anonymous No.4321328]

>>4321317

http://www.math.harvard.edu/putnam/

This website. Go to it to find the problems before they're put on /sci/. Tomorrow's one is already up, and it looks cooler than this one.

[Anonymous No.4322392]

>>4321233
This can't be the correct factorization.

2*cos(20^pi/180) * (-2*cos(40*pi/180)) * (-2*cos(80*pi/180)) ≠ 1

[Anonymous No.4322992]

>>4321323
The joke is beyond tired.

[Anonymous No.4323196]

>>4322992
I have an hilarious joke, but alas, I cannot fit it in this comment box

[Anonymous No.4323202]

>>4323196
I had an uproarious laughter, but alas, I cannot fit it into my vocal range.

[Anonymous No.4323302]

>>4323196
>>4323202
OK, those are better.

[Anonymous No.4324913]

>>4322392
I know, it seems ludicrous, right? Nevertheless, it is true. http://en.wikipedia.org/wiki/Morrie%27s_law

[Yo, I got this one. No.4325001]

>>4324913
Not ludicrous. There was a typo in that previous post:
>2*cos(20^pi/180) * (-2*cos(40*pi/180)) * (-2*cos(80*pi/180)) ≠ 1
>2*cos(20^pi/180)
>20^pi
> ^ instead of *

[Anonymous No.4326615]

>>4321079
Why does the pony guy keep getting banned?

His posts are on topic, at least until he's baited with "Reported for [some questionable reason]" replies.

And then his relevant posts end up getting removed. It seems silly.

[Anonymous No.4326641]

>>4326615
It's a meme!

lol ^^

[Anonymous No.4327143]

>>4326641
what are you referring to?