/sci/ - Science & Math

[Anonymous No.4263656]

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

Define <span class="math">C(\alpha) to be the coefficient of <span class="math">x^{1992}[/spoiler] in the power series about <span class="math">x=0[/spoiler] of <span class="math">(1+x)^\alpha[/spoiler]. Evaluate
<div class="math">\int_0^1 \left( C(-y-1) \sum_{k=1}^{1992} \frac{1}{y+k} \right) dy.</div></spoiler>

[Anonymous No.4263662]

[spoiler]1992</spoiler>

[Anonymous No.4263685]

[spoiler]I think alone Maxima can do this..</spoiler>

[Anonymous No.4263787]

[spoiler]Induction and integration by parts? Suggesting this but clearly not going to bother doing it. I certainly hope it's not the method we're supposed to use (if it even works).</spoiler>

[Anonymous No.4263809]

[spoiler]1992
proof is trivial</spoiler>

[Anonymous No.4263828]

[spoiler]This is the kind of question I would look at and then curl up in a ball in a corner of the room.

I wish I had the analysis skills to tackle this beast.</spoiler>

[Psistar !!bRma1Ao62vX No.4264256]

[spoiler]ok I got it, wait I'll type the answer, it's rather beautiful at the end.</spoiler>

[Anonymous No.4264258]

[spoiler]>>4263656
simplifies to 1.8354

too simple.</spoiler>

[Anonymous No.4264281]

[spoiler]u cant evaluate it cuz its not a continuis function

QED</spoiler>

[Anonymous No.4264301]

[spoiler]>>4264281
It is continuous because it is a polynomial.</spoiler>

[Psistar !!bRma1Ao62vX No.4264312]

let's first find <span class="math">C(-1-y):
<span class="math">(1+x)^(-1-y)=1+ \sum_{k=1}^\infty (\frac{(-1-y)...(-1-y-k+1)}{k!} x^k)[/spoiler]
the term in front of <span class="math">x^1992[/spoiler] is <span class="math">\frac{(-1-y)...(-y-1992)}{1992!} = \frac{(y+1)...(y+1992)}{1992!} [/spoiler]

So we're looking for the integral of:
\frac{(y+1)...(y+1992)}{1992!} \times \sum_{k=1}^1992 \frac{1}{y+k}

But I do recognize in the sum on the left something: it's the partial fraction expansion of <span class="math">\frac{P'}{P}[/spoiler], where <span class="math">P[/spoiler] is the polynom: <span class="math">\prod_{i=1}^1992 (X+i) [/spoiler]. Therefore, the sum is nothing else than <span class="math">\frac{P'(y)}{P(y)}[/spoiler], and the first part of the integral is nothing else than <span class="math">\frac{P(y)}{1992!}[/spoiler]

Thus, what we're looking for is the integral of <span class="math"> \frac{P'(y)}{1992}[/spoiler] between 0 and 1, which is <span class="math"> \frac{P(1)-P(0)}{1992!} = \frac{1993!-1992!}{1992!} = \frac{1992! * (1993-1)}{1992!} = 1992 [/spoiler]</spoiler>

[Anonymous No.4264364]

[spoiler]guy you seem smart, do you study maths?</spoiler>

[Anonymous No.4264372]

[spoiler]>>4264364
I'm majoring in women's studies. (female btw)</spoiler>

[Psistar !!bRma1Ao62vX No.4264374]

>>4264312
>we're looking for is the integral of <span class="math">\frac{P'(y)}{1992}
forgot that "!": <span class="math"> \frac{P'(y)}{1992!}[/spoiler]</spoiler>

[Anonymous No.4265165]

[spoiler]>>4264372</spoiler>

[Anonymous No.4265529]

[spoiler]>>4264312

how u so smart?</spoiler>

[Anonymous No.4265743]

[spoiler]>>4265529
He's not, this is just an easy one.

About two threads ago he brushed off one of the hardest Putnam probs ever as just a simple matter of taking a derivative.</spoiler>

[Psistar !!bRma1Ao62vX No.4265930]

[spoiler]>>4265529
I major in meat science.

Seriously though, it's just a matter of not being affraid of trying, I found the correct path to follow in like 30sec after deciding to write things down.</spoiler>

[Anonymous No.4266001]

[spoiler]>>4265930
>meat science

i honestly hope you're not joking, that would be sweet</spoiler>

[Anonymous No.4266040]

[spoiler]>>4265743
I have a deja-vu feeling about that comment. Have you posted it like two days in a row? Cause that would be amusing. The kind of "amusing" that would make me want to observe you more to see if there are other clues that you want to kill your dad and fuck your mom, etc. Then again, anonymity wouldn't let me...</spoiler>

[Anonymous No.4266076]

[spoiler]>>4266040
I didn't actually say which tripfag it was yesterday.</spoiler>

[Psistar !!bRma1Ao62vX No.4266239]

[spoiler]>>4266040
nope, wasn't me.</spoiler>

[Anonymous No.4266533]

[spoiler]<div class="math"> \def \d#1{{{#1}\atop{#1}}}\def \e#1{\d{\d{\d{\d{\d{#1}}}}}}\def \f#1{\e{\e{#1}}}\smash{\f{{-------------------------------------------------------------------------
----------------------------------------------------------------------------------------------------
-----}}} </div></spoiler>

[Anonymous No.4266537]

[spoiler]<div class="math"> \def \d#1{{{#1}\atop{#1}}}\def \e#1{\d{\d{\d{\d{\d{#1}}}}}}\def \f#1{\e{\e{#1}}}\smash{\f{{-------------------------------------------------------------------------
----------------------------------------------------------------------------------------------------
--------------------------------------------------------------------------------------------------}}
} </div></spoiler>

[Anonymous No.4266538]

[spoiler]<div class="math"> \def \d#1{{{#1}\atop{#1}}}\def \e#1{\d{\d{\d{\d{\d{#1}}}}}}\def \f#1{\e{\e{#1}}}\smash{\f{{-------------------------------------------------------------------------
----------------------------------------------------------------------------------------------------
----------------------------------------------------------------------------------------------------
----------------------------}}} </div></spoiler>

[Anonymous No.4266544]

[spoiler]<div class="math"> \def \d#1{{{#1}\atop{#1}}}\def \e#1{\d{\d{\d{\d{\d{#1}}}}}}\def \f#1{\e{\e{#1}}}\smash{\f{{.........................................................................
....................................................................................................
....................................................................................................
........}}} </div></spoiler>

[Anonymous No.4266556]

You aint stopping me fucking mods, enjoy the lag
<div class="math"> \def \d#1{{{#1}\atop{#1}}}\def \e#1{\d{\d{\d{\d{\d{#1}}}}}}\def \f#1{\e{\e{#1}}}\smash{\f{{.........................................................................
....................................................................................................
....................................................................................................
....................................................................................................
.......}}} </div></spoiler>