It is the 2nd row of letters in the QWERTY keyboard
NothingA b c d e f g h does not have a meaning. They are the first 8 letters of the Engish alphabet.
These ARE negative prefixes A dis- B il- C im- D in- E ir- F non- G un- This is the only positive prefix in the above list. H re-
Well, your two levels away from going to f, which is EVEN HARDER. Im on level M right now and trust me, its REALLY HARD. But you also asked what level d is. Its a level (be more specific next time.)
dedede
R i g h t - a n g l e d
a,a,g,g,h,h,g,f,f,d,d,s,s,a
NothingA b c d e f g h does not have a meaning. They are the first 8 letters of the Engish alphabet.
Recall that a linear transformation T:U-->V is one such that 1) T(x+y)=T(x)+T(y) for any x,y in U 2) T(cx)=cT(x) for x in U and c in R All you need to do is show that differentiation has these two properties, where the domain is C^(infinity). We shall consider smooth functions from R to R for simplicity, but the argument is analogous for functions from R^n to R^m. Let D by the differential operator. D[(f+g)(x)] = [d/dx](f+g)(x) = lim(h-->0)[(f+g)(x+h)-(f+g)(x)]/h = lim(h-->0)[f(x+h)+g(x+g)-f(x)-g(x)]/h (since (f+g)(x) is taken to mean f(x)+g(x)) =lim(h-->0)[f(x+h)-f(x)]/h + lim(h-->0)[g(x+h) - g(x)]/h since the sum of limits is the limit of the sums =[d/dx]f(x) + [d/dx]g(x) = D[f(x)] + D[g(x)]. As for ths second criterion, D[(cf)(x)]=lim(h-->0)[(cf)(x+h)-(cf)(x)]/h =lim(h-->0)[c[f(x+h)]-c[f(x)]]/h since (cf)(x) is taken to mean c[f(x)] =c[lim(h-->0)[f(x+h)-f(x)]/h] = c[d/dx]f(x) = cD[f(x)]. since constants can be factored out of limits. Therefore the two criteria hold, and if you wished to prove this for the general case, you would simply apply the same procedure to the Jacobian matrices corresponding to Df.
A
Darth Vader's Theme Song is Auctually called the Imperial March When two notes are together it means that you can play either note. ENJOY :) "H" = HIGH (C D E F G A B -HIGH STARTS HERE- C D E F G A B) H H H HH H H H H H H H H H H G G G Eb Bb Eb Bb D D D Eb Bb Gb Eb Bb G G G G G Gb F E Eb E Ab Db C B Bb A Bb H H H H H H H H H H H Eb Gb Eb Bb A Eb Bb D G G G G Gb F E Eb E Ab Db C B Bb A Bb Eb Gb Eb Bb G Eb Bb
I think H
,fa,d,e,d,e,f,g,h,a,d,e,f,d,c,c,d,f,e,s,a
A=1 B=1 C=1 D=1 E=1 f=0 G=0 and H=2
c c c d f c f g B natural f g h
h
k
# b # c # d # f # g # h So H it is!