Wikipedia:Reference desk/Archives/Mathematics/2012 September 5#Project Euler question

There is a Project Euler problem (whose number I am not quoting so the answer is not given away directly here), where one needs integer solutions to 2b^2 - 2b - n^2 + n = 0. b=3 and n=4 is one solution to this equation. The equations b' = 3b + 2n - 2 and n' = 4b +3n - 3 generate b' and n' which are also solutions to the equation. Two questions please: (1) How do I prove that the b' and n' equations are valid? (2) How does one do Diophantine magic and generate those equations in the first place? -- SGBailey (talk) 08:46, 5 September 2012 (UTC)

: The equation is equivalent to {{nowrap|1=2b(b − 1) = n(n − 1)}}. Now it becomes obvious {{nowrap|1=b = n = 1}} is another solution... --CiaPan (talk) 09:22, 5 September 2012 (UTC)

: I know the equations work, I've got a dozen solutions and could have a dozen more if I wanted them (they get quite big...). But I'm interested in the proof and creation of the "next set of solutions" equations. Yes 1,1 is a solution. So is 0,1. But so is 3,4. -- SGBailey (talk) 09:36, 5 September 2012 (UTC)

:I don't know where they come from, but it's simple enough to show that they work: just plug your formulas for b' and n' into the original equation and simplify. You'll get 2b^2 - 2b - n^2 + n = 0. Under the assumption that you started with a solution, this shows that b' and n' are a solution.--121.73.35.181 (talk) 10:25, 5 September 2012 (UTC)

:: (1) solved. You are right thanks - when I tried that previously, I went wrong; but it worked ok second time. So (2) where did the generating equations come from??? -- SGBailey (talk) 11:26, 5 September 2012 (UTC)

:::Convergents (p,q) to the continued fraction expansion of sqrt(2) are (1,1), (3,2), (7,5), (17,12), (41,29) etc (see Pell number). Alternate convergents (1,1), (7,5), (41,29) etc. satisfy

::::$p^2\; =\; 2q^2\; -\; 1$

:::Successive convergents in this sequence are related by the recurrence relations

::::$p\text{'}\; =\; 4q\; +\; 3p\; \backslash quad\; q\text{'}\; =\; 3q\; +\; 2p$

:::Use the transform

::::$n\; =\; \backslash frac\{p+1\}\{2\}\; \backslash quad\; b\; =\; \backslash frac\{q+1\}\{2\}$

::: and you get the (n,b) sequence (1,1), (4,3), (21, 15) etc. which satisfies

::::$(2n-1)^2\; =\; 2(2b-1)^2\; -\; 1\; \backslash Rightarrow\; 4n^2\; -\; 4n\; +\; 1\; =\; 8b^2\; -\; 8b\; +\; 1\; \backslash Rightarrow\; n^2\; -\; n\; =\; 2b^2\; -\; 2b$

:::with the recurrence relations

::::$n\text{'}\; =\; 4b\; +\; 3n\; -\; 3\; \backslash quad\; b\text{'}\; =\; 3b\; +\; 2n\; -\; 2$

:::Gandalf61 (talk) 12:08, 5 September 2012 (UTC)

:::: Thank you (I think). I didn't understand most of that, but I'll study it for a few days and see if it makes any sense then. -- SGBailey (talk) 13:11, 5 September 2012 (UTC)

Hi. I'm working in a book, and looking at a claim that goes as follows: For $|e^z\; -\; 1|\backslash leq\; \backslash frac\{1\}\{2\}$, we have $\backslash frac\{1\}\{2\}|z|\; \backslash leq\; |e^z\; -\; 1|\; \backslash leq\; 2|z|$. I've checked this by doing a change of variables $\backslash zeta=e^z-1$, parameterizing the circle $|\backslash zeta|=\backslash frac\{1\}\{2\}$, and checking that $\backslash frac\{1\}\{2\}|\backslash log(\backslash zeta\; +\; 1)|\; \backslash leq\; |\backslash zeta|=\backslash frac\{1\}\{2\}\; \backslash leq\; 2|\backslash log(\backslash zeta\; +\; 1)|$ all around the circle (for the principal branch). I checked just by graphing the left and right as functions of a real variable, and sure enough, they stayed away from $\backslash frac\{1\}\{2\}$. This is terribly awkward, though. Does anyone know an easier way to see this? Thanks in advance. -GTBacchus^(talk) 19:54, 5 September 2012 (UTC)

:Unless I’m missing something, the first part of the inequality is false.

:I will type * for times, and ^ for ”to the power of”.

:Because, suppose that z = 2pi*i. Then e^z = e^(2pi*i) = cos(2pi) + i*sin(2pi) = 1 + 0 = 1. So then |e^z - 1| = 0, which is less than 1/2. So z = 2pi*i satisfies the assumption.

:But then the first part of the inequality, (1/2)|z| <= |e^z -1|, becomes (1/2)(2pi) <= 0, or pi <= 0, which is just not true. Cardamon (talk) 07:00, 8 September 2012 (UTC)

::Yeah, I noticed that too. The inequality applies for z close to 0, not close to other multiples of 2pi*i. It can be fixed by adding the condition that |z|<1 or something.

::I seem to have it figured out. The trick is to do the substitution $\backslash zeta=e^z-1$, and assume that $|\backslash zeta|\backslash leq\backslash frac\{1\}\{2\}$; this obviates the problem you mention. Then you really want to show that $\backslash frac\{1\}\{2\}|\backslash zeta|\backslash leq|\backslash log(\backslash zeta\; +\; 1)|\backslash leq\; 2|\backslash zeta|$. This can be accomplished by playing with the series expansion of $\backslash log(\backslash zeta\; +\; 1)$ and the triangle inequality. -GTBacchus^(talk) 01:59, 11 September 2012 (UTC)