So we all know that you can factorise the difference of two squares easily, and also for two cubes, but is it possible to factorise the difference of any two numbers raised to the same power in the same way?
What about two numbers raised to different powers?
And what about the sum of two numbers raised to the same (or different) power(s)? Are there any powers that work every time?
Please excuse my use of paint.
read the binomial theorem.
>>8133668
[eqn]a^n - b^n = (a-b) (\sum_{i=1}^{n-1}a^i b^{n-1-i})[/eqn]
>>8133688
Correct me if I'm wrong but isn't binomial theorem is about (a+b)^n ?
I'm looking for stuff on (a^n)+(b^n)
obvious but...
a^2n - b^2n = (a^n - b^n)(a^n + b^n)
>>8133704
Sweet this is what I was looking for. Can anyone give me a name for this so I can do more research into it?
>>8133668
It's [eqn]a^n - b^n = (a-b) (a- \zeta b) \cdots (a- \zeta^{n-1} b)[/eqn] where [math] \zeta = e^{2 \pi i/n}[/math]
What subset of math is all this?
Where do you even learn it?
>>8133774
~Where you do begin in order to learn it?
>>8133714
It's called "algebra".
>>8133774
Algebra
Complex numbers
>>8133765
is there a prove online.
I can't seem to make induction work:
Assume
[math] a^n = b^n + (a-b)\prod_{k=1}^{n-1} (a-b\cdot{\mathrm e}^{2\pi i\frac{k}{n}}) [/math]
[math] a^{n+1} = a\, a^n = a\,b^n + a\,(a-b)\prod_{k=1}^{n-1} (a-b\cdot{\mathrm e}^{2\pi i\frac{k}{n}}) [/math]
[math] = (a-b) \left(\dfrac{a\,b^n}{a-b} + \prod_{k=1}^{n-1} (a-b\cdot{\mathrm e}^{2\pi i\frac{k}{n}}) \right) [/math]
...?
>>8134011
>I can't seem to make induction work:
Induction isn't a good path to take because zeta has a completely different value for each n.
Sum/difference of two nth powers over the field of the algebraic numbers
https://en.wikipedia.org/wiki/Factorization#Sum.2Fdifference_of_two_cubes
>>8134031
But I don't use the n-dependent symbol zeta. I use exp and your revelation that it's not a good path can be seen only after the fact.
>>8134035
Doesn't "actually answer my question" at all, mate.
The task is to prove the theorem. Yes, it's factored over C, duh.
(don't wanted to come off as an ass here, sorry, but c'mon)
>>8134079
>The task is to prove the theorem.
That proves the theorem. That polynomial factors as [math](x-b)(x-\zeta b)\cdots(x- \zeta^{n-1} b)[/math]. Now just evaluate the polynomial at a. Since the polynomials are the same, you get the same thing.
>>8134089
Okay, I see your argumentation now, thx
>>8133704
Am I being retarded right now or is this formula unable to generate the correct answers for a^2-b^2 and a^3-b^3?
>>8134103
[math] a^{n}-b^{n} = (a-b) \sum_{k=0}^{n-1} a^k\,b^{n-1-k} [/math]
Post more interesting factorizations, please.
>>8134944
ax + ay = a(x+y)
>>8134944
xj+xy+yk+jk=(x+k)(y+j)
>>8134103
You're right. The sum should start at i=0
This is called the binomial theorem. It has been known for over 2000 years, probably. It has many generalizations.
https://en.wikipedia.org/wiki/Binomial_theorem
>>8135784
Fuck you Simon. That's hardly a factoring 'trick'.
>>8133668
The general case is
[math]\displaystyle b^n-a^n = \prod_{k|n} a^{\varphi(k)} \Phi_k(\tfrac{b}{a})[/math]
where [math]\varphi(n)[/math] is the Euler totient function, [math]\Phi_n(x)[/math] is a cyclotomic polynomial and the product is over the divisors of n.
>>8136012
Excuse me you shitlord. Allow me to introduce you to the 'Fucking High School Retard Theorem' that says that for all complex n, a and b:
[math] b^n \pm a^n = (b \pm a)^n [/math]
I'm sure we have all seen this theorem in use before but for some reason most people ignore its brilliance.
>>8136024
you are excused
>>8135837
Yes indeed.
[math] \displaystyle b^n-a^n = \prod_{k|n} a^{\varphi(k)} \Phi_k(\tfrac{b}{a}) = (b - a)^n [/math]
>>8133668
[math]a^{2^{n+1}} - b^{2^{n+1}} = (a-b)\Pi_{i=0}^n (a^{2^i} + b^{2^i})[/math]
f a g
>>8140366
And for
[math]
a^n + b^n
[/math]
Is the same, it is just that
the plus-sign is just alternating, memorizing that formula saved me many times in calc 1.
>>8133688
I think you mean the fundamental theorem of algebra.
