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You are not logged in. #1 2012-10-20 02:30:45
Factorising a^n + kb^nIs there a method of deducing whether or not a polynomial of the form is factorisable?For instance, we can see that and but something like is not factorisable. Is there a way to tell if we can factorise something of this form? Is there an easy way to do this, or would your best bet be just to write a general factorisation and solve for your general co-efficients? (for instance, writing one factor as (x^2 + ax + b) and finding a and b or something. #2 2012-10-20 05:07:32
Re: Factorising a^n + kb^nYou can try cyclotomic polynomials. I remember bobbym said they can be used for deriving such identities, so... The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “A secret's worth depends on the people from whom it must be kept.” ― Carlos Ruiz Zafón #3 2012-10-20 05:24:08
Re: Factorising a^n + kb^nYes, I saw that too, but I have no idea how to use those here. I can't find anything that discusses multivariable cyclotomic polynomials. #4 2012-10-20 07:55:01
Re: Factorising a^n + kb^nWell, Last edited by anonimnystefy (2012-10-20 08:04:11) The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “A secret's worth depends on the people from whom it must be kept.” ― Carlos Ruiz Zafón #6 2012-10-20 08:03:57
Re: Factorising a^n + kb^nFixed it. You can treat a/b like only one variable, and then calculate ((a/b)+k)^n and then multiply by b^n. The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “A secret's worth depends on the people from whom it must be kept.” ― Carlos Ruiz Zafón #7 2012-10-20 08:12:43
Re: Factorising a^n + kb^nAre you sure that is correct? I tested it with Sophie-Germain's identity (k = 4, n = 4) and I am not getting a^4 + 4b^4... unless I went wrong somewhere. #8 2012-10-20 10:49:24
Re: Factorising a^n + kb^nHi all; In mathematics, you don't understand things. You just get used to them. Probability is the most important concept in modern science, especially as nobody has the slightest notion what it means. 90% of mathematicians do not understand 90% of currently published mathematics. #9 2012-10-20 21:22:49
Re: Factorising a^n + kb^nShouldn't it be #10 2012-10-20 22:48:08
Re: Factorising a^n + kb^nThat is correct -- but I'm more interested in integer factors. In other words, the above is not desirable as a does not always divide b (for a = 2, b = 3 for instance). For example, suppose that I wanted to deduce whether or not something was prime -- finding a factorisation with fractions in it might not help. #11 2012-10-21 07:27:44
Re: Factorising a^n + kb^nHi; In mathematics, you don't understand things. You just get used to them. Probability is the most important concept in modern science, especially as nobody has the slightest notion what it means. 90% of mathematicians do not understand 90% of currently published mathematics. #12 2012-10-21 07:38:21
Re: Factorising a^n + kb^nWhy, though? I can't see why, for example, if a² - b² factorises to (a-b)(a+b), that one condition is that b divides a. Yet that is also of the form an + kbn. #13 2012-10-21 07:40:22
Re: Factorising a^n + kb^nHi; In mathematics, you don't understand things. You just get used to them. Probability is the most important concept in modern science, especially as nobody has the slightest notion what it means. 90% of mathematicians do not understand 90% of currently published mathematics. #15 2012-10-21 07:45:40
Re: Factorising a^n + kb^nDo you still have that computer program you used to compute factorisations? #16 2012-10-21 07:47:00
Re: Factorising a^n + kb^nHi; In mathematics, you don't understand things. You just get used to them. Probability is the most important concept in modern science, especially as nobody has the slightest notion what it means. 90% of mathematicians do not understand 90% of currently published mathematics. #17 2012-10-21 07:51:18
Re: Factorising a^n + kb^nSo you were able to pull that a^6 + 8b^6 factorisation off the top of your head? #18 2012-10-21 07:54:55
Re: Factorising a^n + kb^nHi; To start, did you read post #2? In mathematics, you don't understand things. You just get used to them. Probability is the most important concept in modern science, especially as nobody has the slightest notion what it means. 90% of mathematicians do not understand 90% of currently published mathematics. #19 2012-10-21 08:00:50
Re: Factorising a^n + kb^nYes but you never told me how you used cyclotomic polynomials to do that... #20 2012-10-21 08:04:55
Re: Factorising a^n + kb^nMore than a century ago, I came across a book with a big chart of Aurifeuillian Factorizations. I was amazed. In mathematics, you don't understand things. You just get used to them. Probability is the most important concept in modern science, especially as nobody has the slightest notion what it means. 90% of mathematicians do not understand 90% of currently published mathematics. #21 2012-10-21 08:17:39
Re: Factorising a^n + kb^nHow does that relate to this? The wiki article is saying it is a factorisation of the form 24n+2 + 1. #22 2012-10-21 08:20:15
Re: Factorising a^n + kb^nThe point is I had tables of them like a table of integrals or sums. In mathematics, you don't understand things. You just get used to them. Probability is the most important concept in modern science, especially as nobody has the slightest notion what it means. 90% of mathematicians do not understand 90% of currently published mathematics. #24 2012-10-21 08:24:32
Re: Factorising a^n + kb^nSo, I am guessing this book might be a bit difficult for me to find... #25 2012-10-21 08:26:51
Re: Factorising a^n + kb^nYes, I am ancient. You know that old quote:
In mathematics, you don't understand things. You just get used to them. Probability is the most important concept in modern science, especially as nobody has the slightest notion what it means. 90% of mathematicians do not understand 90% of currently published mathematics. |