So anyway, I understand and can visualize how composition of shift and rotation would always have a fixed point, but how can it remain fixed, when you take a homothety of < 1 after that, unless of course you specifically choose it's center to be that fixed point?
I've seen some proof on stackexchange, but it included linear algebra which is obviously unacceptable, since it's a FUCKING ANALYSIS book - intro chapter of the first volume in fact - and you're clearly supposed to be able to solve it without linear algebra.
How do you prove Newton's binomial formula by induction without the use of combinatorics or pascal's triangle or any other such shit? You're cleary supposed to be able to do so.
> Check that ANY composition of shift, rotation and homothety of a plane has a fixed point if k of homothety is less than 1
Ryan Gonzalez
Assume [math](x+y)^n = \sum_{0 \le k \le n} {n \choose k} x^k y^{n-k}[/math] Write [math](x+y)^{n+1} = (x+y)(x+y)^n[/math] and use the formula [math]{n \choose k} + {n \choose {k+1}} = {{n+1} \choose {k+1}}[/math]
Isaac Butler
Rewrite pls
Daniel Peterson
you rewrite
Thomas Young
Your tex doesn't work, you fucked it up.
Gabriel Scott
u w0t m8 ?
Brody Bennett
> without the use of combinatorics or pascal's triangle or any other such shit
Christopher Wright
Um if you want to prove an identity involving binomial coefficients, it stands to reason that at some point you are going to have to use properties of the binomial coefficients.
Alexander Ward
There's SPECIFICALLY no combinatorics notation in the book so that retards know you can solve it with no combinatorics.
Lincoln Murphy
Ugh.. binomial notation is a *notation*. That thing I wrote could be rewritten as [math]\frac{n(n-1)\dots n-k+1)}{k!} + \frac{n(n-1)\dots (n-k)}{(k+1)!} = \frac{(n+1)\dots (n-k+1)}{(k+1)!}[/math]. Just do exactly what I said and replace binomials by that disgusting expression with factorials and you'll get your "combinatorics-free" proof.
Juan Long
Nigger, the formula you use is from COMBINATORICS. And in the exercise there's specifically factorial notation, so you know you don't have to use combinatorics here.
Tyler Collins
Dude I don't know how I can explain it any further. There is no use of combinatorics in that formula above, it's a literal manipulation of fractions. Just use assume the formula g) is true, multiply it by a+b and then use the formula I wrote above. I dunno if this is some kind of elaborate bait or something but I'm done explaining
John Hill
You learn this formula in the combinatorics course HENCE it's combinatoric. You see now ffs?
Anthony Johnson
Factorials are combinatorics since n! is defined as the number of permutation of a set with n elements.
Christian Torres
BUMP
Oliver Allen
Bump
Logan Phillips
bump
Aaron Nguyen
Are you seriously retarded? Then don't use that notation and exand everything into products if numbers and do it manually so you don't get triggered.
Easton Morgan
Retard, it's not about notation - this FORMULA is a part of combinatorics course which is obviously something you can avoid using to prove the theorem.
Aiden Campbell
>This formula is part of a cominatorics course Nigger, that identify is something universally used. Grab the definition of n chose k and do the arithmetic involved. If you don lime that theb again just right it as n!/(n-k)!k! And try it directly. And factorial is just notation for a string of multiplications. You could have used the gamma function for all I care. The only other proof I can remember that isn't by induction uses REAL combibatorial principles not definitions and identities.
Jack Wright
Fucking, sorry for the spelling. Me and my phone are retarded.
