Guest wrote:Suppose Case 2 is true, but [tex]\lambda < \frac{1 + \sqrt{ \frac{(\sqrt{2} - 1) * I}{2}}}{2}[/tex].
Case 2.1: [tex]\beta_{21}(I) * (\frac{1 + \frac{1 + \sqrt{ \frac{(\sqrt{2} - 1) * I}{2}}}{2}}{2}) \le \lambda < \frac{1 + \sqrt{ \frac{(\sqrt{2} - 1) * I}{2}}}{2}[/tex] when [tex]0 < \beta_{21}(I) \le 1[/tex].
Now only one of the two following cases is true!
Case 2.1A: [tex]\lambda >\frac{1 + \frac{1 + \sqrt{ \frac{(\sqrt{2} - 1) * I}{2}}}{2}}{2}[/tex] when [tex]\beta_{21}(I) = 1[/tex].
Case 2.1B: [tex]\lambda = \beta_{21}(I) * (\frac{1 + \frac{1 + \sqrt{ \frac{(\sqrt{2} - 1) * I}{2}}}{2}}{2})[/tex] when [tex]0 < \beta_{21}(I) \le 1[/tex].
Remarks: The development process is ongoing, and it is quickly becoming convoluted too. We stress caution and accuracy.
Guest wrote:Remarks: We must make sure there's no gap in our programming. All cases must be fully developed and verified. We must test and do more testing. Math is hard work!
Guest wrote:Guest wrote:Remarks: We must make sure there's no gap in our programming. All cases must be fully developed and verified. We must test and do more testing. Math is hard work!
For each case of [tex]\lambda[/tex], is [tex]I + \lambda^{2}[/tex] becoming a perfect square?
And we expect the two factors of I to be prime.
Guest wrote:Relevant Reference Link:
'General number field sieve',
https://en.wikipedia.org/wiki/General_number_field_sieve.
Dave wrote:FYI: 'Continued fraction',
https://en.wikipedia.org/wiki/Continued_fraction#Infinite_continued_fractions_and_convergents.
The generalized continued fraction of [tex]\sqrt{I + \lambda^{2}}[/tex] offers great promise. So between the power series expansion of [tex]\sqrt{I + \lambda^{2}}[/tex] and its generalized continued fraction representation, we hope to solve [tex]\triangle n(\lambda) = 0[/tex]...
Good Luck!
A Tentative Update: [tex]\beta(I) * (\frac{1 + \sqrt{ \frac{(\sqrt{2} - 1) * I}{2}}}{2}) \le \lambda < \sqrt{ \frac{(\sqrt{2} - 1) * I}{2}}[/tex] where [tex]0 < \beta(I) \le 1[/tex].
Remark: We want to construct a convergent sequence that solves [tex]\lambda[/tex].
Depending on I, only one of the following two cases is true!
Case 1: [tex]\lambda > \frac{1 + \sqrt{ \frac{(\sqrt{2} - 1) * I}{2}}}{2}[/tex] when [tex]\beta(I) = 1[/tex].
Case 2: [tex]\lambda = \beta(I) * (\frac{1 + \sqrt{ \frac{(\sqrt{2} - 1) * I}{2}}}{2})[/tex] when [tex]0 < \beta(I) \le 1[/tex].
Dave wrote:Let's ponder the following sum of two functions, [tex]n(\lambda)[/tex] and [tex]∆n(\lambda)[/tex]:
[tex]n(\lambda) + ∆n(\lambda) = \sqrt{I + \lambda^{2}}[/tex] for some positive integers, [tex]n(\lambda)[/tex] and [tex]\lambda[/tex], where [tex]0 \le ∆n(\lambda) < 1[/tex].
Remark: [tex]1 \le \lambda < \frac{\sqrt{I}}{2}[/tex] and I = p * q where I is known.
Key Ideas: Power Series Expansion, Convergence, Rate of Convergence, etc.
Relevant Reference Link:
'Power series',
https://en.wikipedia.org/wiki/Power_series.]
Can we explicitly identify [tex]n(\lambda)[/tex] in the expression,[tex]\sqrt{I + \lambda^{2}}[/tex] ?
And can we explicitly identify [tex]∆n(\lambda)[/tex] in the expression, [tex]\sqrt{I + \lambda^{2}}[/tex] ?
Guest wrote:Dave wrote:Let's ponder the following sum of two functions, [tex]n(\lambda)[/tex] and [tex]∆n(\lambda)[/tex]:
[tex]n(\lambda) + ∆n(\lambda) = \sqrt{I + \lambda^{2}}[/tex] for some positive integers, [tex]n(\lambda)[/tex] and [tex]\lambda[/tex], where [tex]0 \le ∆n(\lambda) < 1[/tex].
Remark: [tex]1 \le \lambda < \frac{\sqrt{I}}{2}[/tex] and I = p * q where I is known.
Key Ideas: Power Series Expansion, Convergence, Rate of Convergence, etc.
Relevant Reference Link:
'Power series',
https://en.wikipedia.org/wiki/Power_series.]
Can we explicitly identify [tex]n(\lambda)[/tex] in the expression,[tex]\sqrt{I + \lambda^{2}}[/tex] ?
And can we explicitly identify [tex]∆n(\lambda)[/tex] in the expression, [tex]\sqrt{I + \lambda^{2}}[/tex] ?
[tex]n(\lambda) = \left \lfloor{\sqrt{I + \lambda^{2}}} \right \rfloor[/tex];
[tex]\triangle n(\lambda) = \sqrt{I + \lambda^{2}} - \left \lfloor{\sqrt{I + \lambda^{2}}} \right \rfloor[/tex].
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