04-23-2014, 09:57 PM
Abandon dcents!
04-23-2014, 09:57 PM
Abandon dcents!
04-24-2014, 07:50 PM
04-24-2014, 09:12 PM
Abandon hope ;-;*
I found a way. As an exit slip of ORE, here is my way. First, let function f map a factorial to its number, like f(120) = 5. If I use normal math terms it'd be f: x! → x. Now, the binomial coeffecient of n and k is n!/k!(n-k)!. This can be represented differently, but let's use this definition because it has (n-k)!. In order to remove that ugly n! numerator we multiply the coeffecient by 1/n!. To remove the k! at the bottom, we multiply everything by k!. Now we have k!/n! * binom(n,k) = 1/(n-k)!, so let's do 1/binom(n,k) to get (n-k)!. Now we have n!/k! * 1/binom(n,k) = (n-k)!. To simplify this, we can remember reciprocals from pre-algebra (did that two months ago, fuck prealgebra) and we know a * 1/b = a/b, so we'd get (n!/k!)/binom(n,k) = (n-k)!. Now, remember our function f? To map (n-k)! without using subtraction, we do f((n!/k!)/binom(n,k)) = n-k. Now to transform n-k to n+k? Simple; take the conjugate of this. Writing this in equation form makes:
Well I'm lazy, go onto http://www.codecogs.com/latex/eqneditor.php and type in f\overline{\left (\frac{n!/k!}{\binom{n}{k}}\right )}=n+k in the thing. The line over everything is the conjugate symbol. Take that aft! Fuck... nobody will understand my post I just realized.
04-27-2014, 03:32 PM
fuck.
04-27-2014, 03:59 PM
04-27-2014, 08:55 PM
Huh?.
04-27-2014, 10:57 PM
Obviously if you were to evaluate it you would have to use the fundamental theorem of calculus and subtract anyway (and you subtract in some quantity in all the numerical methods too), but the general presentation doesn't imply any kind of additive relationship. (Since it's a rectangle, it almost feels more like multiplication).
04-28-2014, 01:28 PM
I'll just let you two figure it out, I'm outta here.
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