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/sci/ - Science & Math

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10499837 No.10499837 [Reply] [Original]

[math]
\text{Let }a_{m,n}\text{ denote the coefficient of $x^n$ in the expansion of}
\\
(1+x+x^2)^m\text{. Prove that for all [integers] }
\\
k\geq 0\text{,}
0\leq \sum_{i=0}^{\lfloor \frac{2k}{3}\rfloor} (-1)^i a_{k-i,i}\leq 1.
[/math]

>> No.10499839

Previous Thread >>10497705

>> No.10499873

bump

>> No.10499881

Computer Science students need not apply

>> No.10499925

>>10499837
Someone made that image.

>> No.10500035

>>10499925
a labor of love, no doubt

>> No.10500048

tfw math major and can't do it.
I assume its some sort of induction but im stuck.

>> No.10500405

thinly veiled homework thread

>> No.10500766
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10500766

this is getting out of hand

>> No.10500808

>>10499837
Let F(x,t) = 1/(1 - t(1 + x + x^2)) = sum (t^m)*(x^n)*a(m,n)
Think k = m+n and i = n.
First try to get the (-1)^i term.
F(-x,t) = sum (t^m)*(x^n)*a(m,n)*(-1)^n = 1/(1 - t(1 - x + x^2))

Let t=x so that the coefficient of x^k will encode the desired sum.
F(-x,x) = sum (x^(m+n))*a(m,n)*(-1)^n = 1/(1 - x(1 - x + x^2))
= 1/(1 - x + x^2 - x^3) = 1/[(1 + x^2)(1 - x)] = (1/2)[(1+x)/(1+x^2) + 1/(1-x)]

For a given k, the sum is (1/2)(1 + (-1)^floor(k/2)) which is 0 or 1.

>> No.10501603

>>10499837
I'm convinced anime-guy doesn't have a left arm.

>> No.10501617

>>10500035
a labour of cringe, no doubt

>> No.10501619

>>10501603
You know that this is not just any anime shirt, right? The madman actually wore an erotic visual novel shirt.

>> No.10501988

bump

>> No.10502661

>>10499837
I can't do it.