/sci/ - Science & Math

>>8500645
>is pic related legit?
wtf kind of question is that? no calculator?
i can prove this for you but i wont, im to smart.

>>8500646
no. is this how post numbers are assigned to each post?

TRUMP POSTERS GET THE FUCK OFF MY PLANET AWOOOOOOOOOOOOOOOOOOO

>>8500705

are people really calling dubs "numerical ouija" now?

>>8500645
>is pic related legit?
Yes, for any natural number base b and any i <= b we have 123...i * (b-2) + i = (b-1)(b-2)...(b-i).
Or more precisely, [math](\sum_{j=1}^i jb^{i-j})(b-2) + i=\sum_{j=1}^i(b-j)b^{i-j}[/math] which we abbreviate to P(i).
Proof by induction: [math]1 \times (b-2) + 1=b-1[/math] so P(1) is true. Suppose P(i) is true for fixed i, then the LHS of P(i+1) is:
[eqn](\sum_{j=1}^{i+1} jb^{i+1-j})(b-2) +i+1 = \left( b(\sum_{j=1}^ijb^{i-j}) +(i+1)b^0\right)(b-2) +i+1[/eqn]
[eqn]=b\left( (\sum_{j=1}^i(b-j)b^{i-j}) -i\right)+ (i+1)(b-2)+i+1[/eqn]
[eqn]=\left(\sum_{j=1}^i(b-j)b^{i+1-j} \right)+(b-[i+1])b^0 =\sum_{j=1}^{i+1}(b-j)b^{i+1-j}[/eqn]
which = RHS of P(i+1).
Incidentally, P(0) is also true, as is P(i) for negative i if we adopt the usual convention that an empty summation equals 0.

>>8500645
When you see this shit just ask yourself
"does it works outside of base 10? "
if it doesn't it's shit.

>>8500705
>>8501862

>>8502750
But it does, see >>8502104

>>8501862
what pol is actually doing

>>8501862
>>8503473

>>8503476
>>8503473
>>8503055
>>8501862
>>8500705

>/sci/ political memes
I'm going back to /g/ before this gets weirder

>>8503055
>analysis

>>8503672
>anal