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Mar
11
comment square root / factor problem $(A/B)^{13} - (B/A)^{13}$
@crash Sorry, I hit "enter" at the wrong time. Take another look at my message, which shows you how to get to 94642 by algebraic manipulation without using a calculator, as you asked for a couple of days ago.
Mar
11
comment square root / factor problem $(A/B)^{13} - (B/A)^{13}$
@crash $\frac{A}{B} = 1 + \sqrt 2 $ and $\frac{B}{A} = - 1 + \sqrt 2 $ so ${\left( {\frac{A}{B}} \right)^{13}} - {\left( {\frac{B}{A}} \right)^{13}} = {\left( {1 + \sqrt 2 } \right)^{13}} - {\left( {1 - \sqrt 2 } \right)^{13}}$. Applying the binomial theorem and Pascal's triangle, the $\sqrt 2 $ diappears and we get ${\left( {\frac{A}{B}} \right)^{13}} - {\left( {\frac{B}{A}} \right)^{13}} = 2\sum\limits_{k = 0}^6 {{2^k}\left( {\begin{array}{*{20}{c}} {13}\\ {2k} \end{array}} \right)} = 2 + 312 + 5720 + 27456 + 41184 + 18304 + 1664= 94642$.
Feb
25
comment Most ambiguous and inconsistent phrases and notations in maths
When I grew up in Britain, the decimal point was centralised, so the lower dot was used for multiplication. This is definitely different to US usage.
Sep
24
awarded  Autobiographer
Sep
10
awarded  Commentator
Sep
10
awarded  Supporter
Sep
10
comment Notation for repeated composition of functions
I can't use K as that's already in use for continued fractions, because K = Kettenbruch, the German for Continued Fraction.
Sep
10
asked Notation for repeated composition of functions
May
18
comment Splitting polynomials
@Jyrki: Yes, and lhf seems to be suggesting that too.
May
18
comment Splitting polynomials
@lhf: My original question was motivated by Paul Garrett's paper, which is why I'm looking for a different sort of solution. I admit that what I'm looking for may be unobtainable, but I figure there's no harm in looking.
May
18
comment Splitting polynomials
I think I could prove it if I put my mind to it. Howeveer, I'm not too bothered, as this is not the solution I'm looking for. (Maybe what I'm looking for is unobtainable). Your second comment here mirrors what's in Paul Garrett's paper. See my comments to the next solution.
May
18
awarded  Self-Learner
May
18
awarded  Teacher
May
18
answered Splitting polynomials
May
18
comment Splitting polynomials
It's not the sort of solution I was expecting. I was hoping to see something along the lines of extracting square roots by long division (yes, I know this is not a square root, but there are some things in common with square roots), and this solution doesn't do that.
May
17
comment Splitting polynomials
Another example, for $p=11$ is$$161051 + 161051x + 73205x^2 + 14641x^3 -1331x^4 -1331x^5 -121x^6 + 121x^7 + 55x^8 + 11x^9 + x^{10}$$ together with its companion where the signs of the coefficients of the odd powers of x are reversed.
May
17
comment Splitting polynomials
@Jyrki: Thanks for all your help, it is most appreciated.
May
17
comment Splitting polynomials
Oh my gosh; I seem to have found some sort of solution. It involves Lucas's formula for Cyclotomic Polynomials (see Riesel, Table 24).
May
17
comment Splitting polynomials
Sorry, @Gerry, I should have mentioned: yes it is all related to Aurifeuillian factorization and cyclotomic polynomials.
May
17
comment Splitting polynomials
Thanks; this is very helpful. I assume that the product should go from $k=1$ and not $j=1$. I am definitely looking for a closed formula for the coefficients.