my average rank was 1559

HERE I WILL PUT FILES THAT ARE DESERVING OF AWARDS IN STEPMANIA FILES YOU SHOULD CARE IF YOUR FILE IS PUT IN HERE THIS LIST BECAUSE I MADE MAGICAL 8BIT TOUR BYE


Pewter Award
- Magical 8-bit Tour
- Milky Blue

Talc Award
- POP

Pyrite Award


Glass Award
- let's nyancing

Silica Award


Bad Award
- zaghurim files
- ringtone files i mean piano demo files
- files with notes beyond 64ths except if they're used in rolls

Talkative Award
-robertsona

Watashi Wa Bakane Desu~
- gamepro's files


IF YOU WANT ME TO PLAY YOUR FILE TO BE NOMINATED FOR A STEPMANIA AWARD POST HERE

how do you beat lenneth shes so cute i dont want to hurt her T__T

(yet another alternate and admittedly less elegant solution for b2 that I found yesterday while avoiding studying for exams, but it's still pretty awesome because it uses taylor's formula hehe! :D)

we're given a continuous function f:[0,1]->R such that \int_0^1 f(x) dx = 0. so define g by g(x)= \int_0^a f(t) dt. then clearly g(0)=g(1)=0 and g'(x) is just f(x).

now we use taylor's formula to arrive at the solution. normally we would use that to say something like "f(x)=f(0)+f'(0)*(x-0)+.5*f"(c)*(x-0)^2 " for some c on the interval, but here we need to consider the formula about x with respect to 0 and 1. thus, g(0)=g(x)+g'(x)*(0-x)+.5*g"(a)*(0-x)^2 and g(1)=g(x)+g'(x)*.5*g"(b)*(1-x)^2, where a and b are of course in [0,1].

subtracting, you get:

g'(x) - .5*g"(a) + .5*[g"(b)-2x*g"(b)+g"(b)] = 0.

But wait! since g'(x)=f(x), g"(x) = f'(x), which is continuous on a closed interval and thus bounded by some constant K. now we are getting somewhere......

0 $/leq$ f(x) - xK + K/2, or
-f(x) $/leq$ K(1/2-x). aha! now we integrate both sides from 0 to a to get:

/int_0^a -f(x) $/leq$ (K/2)*(a-a^2). since a is in [0,1] we are done.

so yeah, kinda lengthy compared to the other ones we came up with, but extra methods never hurt anybody :P

seriously i've never ever heard of that