Randomness can be a useful tool for solving problems.

[Guest]

Updated Remark: The discrete graph of [tex]\triangle n(\lambda)[/tex] for [tex]1 \le \lambda < \frac{\sqrt{I }}{2}[/tex] is quite interesting and valuable!

For a given case, I = 127 * 257, we observe that are oscillations between zero and one, exclusively, in the values of [tex]\triangle n(\lambda)[/tex] for almost all [tex]1 \le \lambda < \frac{\sqrt{I }}{2}[/tex] with some increased oscillations (rapid rise and fall of [tex]\triangle n(\lambda)[/tex]). Those increased oscillations do, however, dissipate over the full interval of [tex]\lambda[/tex]. Moreover, we observe those increased oscillations begin before [tex]\triangle n(\lambda =65) = 0[/tex], and they continue for awhile afterwards.

[Guest]

Remark: [tex]\triangle n(\lambda)[/tex] helps our programming.

[Guest]

Updated Remark: The discrete graph of [tex]\triangle n(\lambda)[/tex] for [tex]1 \le \lambda < \frac{\sqrt{I }}{2}[/tex] is quite interesting and valuable!

For a given case, I = 127 * 257, we observe that are oscillations between zero and one, exclusively, in the values of [tex]\triangle n(\lambda)[/tex] for almost all [tex]1 \le \lambda < \frac{\sqrt{I }}{2}[/tex] with some increased oscillations (rapid rise and fall of [tex]\triangle n(\lambda)[/tex]). Those increased oscillations do, however, dissipate over the full interval of [tex]\lambda[/tex]. Moreover, we observe those increased oscillations begin before [tex]\triangle n(\lambda =65) = 0[/tex], and they continue for awhile afterwards.

[Guest]

Remarks: The effective use of [tex]\triangle n(\lambda)[/tex] may require some sampling theory/Fourier analysis when I = p * q is large. And our search for [tex]\triangle n(\lambda) = 0[/tex] must be very selective and fast.

[Guest]

Remark: We have other tools to help us solve our integer factorization problem. Please see the past posts.

[Guest]

Remark: By combining the right tools, we shall achieve our goals.

[Guest]

Updated Remark: The discrete graph of [tex]\triangle n(\lambda)[/tex] for [tex]1 \le \lambda < \frac{\sqrt{I }}{2}[/tex] is quite interesting and valuable!

For I = 577 * 631, we compute [tex]\triangle n(\lambda = 27) = 0[/tex].

Please see the attached file for the graph of [tex]\triangle n(\lambda)[/tex] for [tex]1 \le \lambda < \frac{\sqrt{I }}{2}[/tex].

Remarks: There are three regions, [tex]1 \le \lambda < \frac{\sqrt{I }}{6}[/tex],

[tex]\frac{\sqrt{I }}{6} \le \lambda < \frac{\sqrt{I }}{3}[/tex],

and [tex]\frac{\sqrt{I }}{3} \le \lambda < \frac{\sqrt{I }}{2}[/tex], that we must consider when searching for [tex]\triangle n(\lambda) = 0[/tex] without the assistance other tools.

Hopefully, our other tools can limit our search for [tex]\triangle n(\lambda) = 0[/tex] accurately to just one region!

[Guest]

Relevant Reference Links:

'Floor Function',

https://mathworld.wolfram.com/FloorFunction.html;

'Floor and ceiling functions'

https://en.wikipedia.org/wiki/Floor_and_ceiling_functions

[Guest]

Question: Can the continued fraction factorization of [tex]\sqrt{I + \lambda^{2}}[/tex] helps us decide what region (first part, second part, or third part) over [tex]1 \le \lambda < \frac{\sqrt{I}}{2}[/tex] we should investigate?

[Guest]

An Update:

Question: Can the continued fraction factorization of [tex]\sqrt{I}[/tex] helps us decide what region (first part, second part, or third part) over [tex]1 \le \lambda < \frac{\sqrt{I}}{2}[/tex] we should investigate?


Remarks: First part is [tex]1 \le \lambda \le \frac{\sqrt{I }}{6}[/tex];

Second part is [tex]\frac{\sqrt{I }}{6} \le \lambda \le \frac{\sqrt{I }}{3}[/tex];

Third part is [tex]\frac{\sqrt{I }}{3} \le \lambda < \frac{\sqrt{I }}{2}[/tex].

[Guest]

An Update:

Question: Can the continued fraction factorization of [tex]\sqrt{I}[/tex] helps us decide what region (first part, second part, or third part) over [tex]1 \le \lambda < \frac{\sqrt{I}}{2}[/tex] we should investigate?


Remarks: First part is [tex]1 \le \lambda \le \frac{\sqrt{I }}{6}[/tex];

Second part is [tex]\frac{\sqrt{I }}{6} \le \lambda \le \frac{\sqrt{I }}{3}[/tex];

Third part is [tex]\frac{\sqrt{I }}{3} \le \lambda < \frac{\sqrt{I }}{2}[/tex].

We have reattached a file that describes the continued fraction factorization method/algorithm in great detail.

Enjoy!

[Guest]

Question: Can the power series expansion of [tex]\sqrt{I + \lambda^{2}}[/tex] about [tex]\lambda_{0 }[/tex] ( i.e. [tex]\lambda_{0 } = \left \lfloor{\frac{\sqrt{I}}{4}} \right \rfloor[/tex] ) helps us find [tex]\triangle n(\lambda) = 0[/tex] in the interval, [tex]1 \le \lambda < \frac{\sqrt{I}}{2}[/tex] ?

[Guest]

Question: Can the power series expansion of [tex]\sqrt{I + \lambda^{2}}[/tex] about [tex]\lambda_{0 }[/tex] ( i.e. [tex]\lambda_{0 } = \left \lfloor{\frac{\sqrt{I}}{4}} \right \rfloor[/tex] ) helps us find [tex]\triangle n(\lambda) = 0[/tex] in the interval, [tex]1 \le \lambda < \frac{\sqrt{I}}{2}[/tex] ?

Remark: We are seeking any formulation that yields a global minimum over [tex]1 \le \lambda < \frac{\sqrt{I}}{2}[/tex].

[Guest]

Question: Can the power series expansion of [tex]\sqrt{I + \lambda^{2}}[/tex] about [tex]\lambda_{0 }[/tex] ( i.e. [tex]\lambda_{0 } = \left \lfloor{\frac{\sqrt{I}}{4}} \right \rfloor[/tex] ) helps us find [tex]\triangle n(\lambda) = 0[/tex] in the interval, [tex]1 \le \lambda < \frac{\sqrt{I}}{2}[/tex] ?

Remark: We are seeking any formulation that yields a global minimum over [tex]1 \le \lambda < \frac{\sqrt{I}}{2}[/tex].

Question: What is the power series expansion of [tex]\sqrt{I + \lambda^{2}}[/tex] about [tex]\lambda_{0 } = \left \lfloor{\frac{\sqrt{I}}{4}} \right \rfloor[/tex] ?

[Guest]

Updates:

Question: Can the power series expansion of [tex]\sqrt{I + \lambda^{2}} - Floor(\sqrt{I + \lambda^{2}})[/tex] about [tex]\lambda_{0 }[/tex] ( i.e. [tex]\lambda_{0 } = \left \lfloor{\frac{\sqrt{I}}{4}} \right \rfloor[/tex] ) helps us find [tex]\triangle n(\lambda) = 0[/tex] in the interval, [tex]1 \le \lambda < \frac{\sqrt{I}}{2}[/tex] ?

Remark: We are seeking any formulation that yields a global minimum over [tex]1 \le \lambda < \frac{\sqrt{I}}{2}[/tex].

Question: What is the power series expansion of [tex]\sqrt{I + \lambda^{2}} - Floor(\sqrt{I + \lambda^{2}})[/tex] about [tex]\lambda_{0 } = \left \lfloor{\frac{\sqrt{I}}{4}} \right \rfloor[/tex] ?

[Guest]

Remark: [tex]Floor(\sqrt{I + \lambda^{2}})[/tex] is a big headache!

[Guest]

Updates:

Question: Can the power series expansion of [tex]\sqrt{I + \lambda^{2}} - Floor(\sqrt{I + \lambda^{2}})[/tex] about [tex]\lambda_{0 }[/tex] ( i.e. [tex]\lambda_{0 } = \left \lfloor{\frac{\sqrt{I}}{4}} \right \rfloor[/tex] ) helps us find [tex]\triangle n(\lambda) = 0[/tex] in the interval, [tex]1 \le \lambda < \frac{\sqrt{I}}{2}[/tex] ?

Remark: We are seeking any formulation that yields a global minimum over [tex]1 \le \lambda < \frac{\sqrt{I}}{2}[/tex].

Question: What is the power series expansion of [tex]\sqrt{I + \lambda^{2}} - Floor(\sqrt{I + \lambda^{2}})[/tex] about [tex]\lambda_{0 } = \left \lfloor{\frac{\sqrt{I}}{4}} \right \rfloor[/tex] ?

Remarks: Our questions (above) are redundant, and they have dead-end answers!

[Guest]

\begin{cases} x + y = 4 \\ x - y = 0 \end{cases}

Updates:

Question: Can the power series expansion of [tex]\sqrt{I + \lambda^{2}} - Floor(\sqrt{I + \lambda^{2}})[/tex] about [tex]\lambda_{0 }[/tex] ( i.e. [tex]\lambda_{0 } = \left \lfloor{\frac{\sqrt{I}}{4}} \right \rfloor[/tex] ) helps us find [tex]\triangle n(\lambda) = 0[/tex] in the interval, [tex]1 \le \lambda < \frac{\sqrt{I}}{2}[/tex] ?

Remark: We are seeking any formulation that yields a global minimum over [tex]1 \le \lambda < \frac{\sqrt{I}}{2}[/tex].

Question: What is the power series expansion of [tex]\sqrt{I + \lambda^{2}} - Floor(\sqrt{I + \lambda^{2}})[/tex] about [tex]\lambda_{0 } = \left \lfloor{\frac{\sqrt{I}}{4}} \right \rfloor[/tex] ?

Remarks: Our questions (above) are redundant, and they have dead-end answers!

Tentative Remark: Yes! The power series expansion is not appropriate here. But the Fourier series expansion is appropriate here.

[tex]\triangle n(I, \lambda) = \sqrt{I + \lambda^{2}} - Floor(\sqrt{I + \lambda^{2}})[/tex].

Moreover, [tex]\triangle n(I, \lambda) = \frac{1}{2} - \frac{1}{\pi} * \sum_{k=1}^{\infty }\frac{sin(2 \pi k \sqrt{I + \lambda^{2}})}{k}[/tex]

where [tex]1 \le \lambda < \frac{\sqrt{I}}{2}[/tex] for a given I such that I = p * q.

Tentative Remark: If our explicit definition of [tex]\triangle n(I, \lambda)[/tex] holds, then we have almost solved our integer factorization problem!

Question: What is the solution of [tex]\triangle n(I, \lambda) = 0[/tex] ?

[Guest]

Tentative Remark: Yes! The power series expansion is not appropriate here. But the Fourier series expansion is appropriate here.

[tex]\triangle n(I, \lambda) = \sqrt{I + \lambda^{2}} - Floor(\sqrt{I + \lambda^{2}})[/tex].

Moreover, [tex]\triangle n(I, \lambda) = \frac{1}{2} - \frac{1}{\pi} * \sum_{k=1}^{\infty }\frac{sin(2 \pi k \sqrt{I + \lambda^{2}})}{k}[/tex]

where [tex]1 \le \lambda < \frac{\sqrt{I}}{2}[/tex] for a given I such that I = p * q.

Tentative Remark: If our explicit definition (Fourier series expansion) of [tex]\triangle n(I, \lambda)[/tex] holds, then we have almost solved our integer factorization problem!

Question: What is the solution of [tex]\triangle n(I, \lambda) = 0[/tex] ?

Relevant Reference Link:

'Continuity and series expansions',

https://en.wikipedia.org/wiki/Floor_and_ceiling_functions#Continuity_and_series_expansions.

Dave.

[Guest]

Tentative Remark: Yes! The power series expansion is not appropriate here. But the Fourier series expansion is appropriate here.

[tex]\triangle n(I, \lambda) = \sqrt{I + \lambda^{2}} - Floor(\sqrt{I + \lambda^{2}})[/tex].

Moreover, [tex]\triangle n(I, \lambda) = \frac{1}{2} - \frac{1}{\pi} * \sum_{k=1}^{\infty }\frac{sin(2 \pi k \sqrt{I + \lambda^{2}})}{k}[/tex]

where [tex]1 \le \lambda < \frac{\sqrt{I}}{2}[/tex] for a given I such that I = p * q.

Tentative Remark: If our explicit definition (Fourier series expansion) of [tex]\triangle n(I, \lambda)[/tex] holds, then we have almost solved our integer factorization problem!

Question: What is the solution of [tex]\triangle n(I, \lambda) = 0[/tex] ?

Relevant Reference Link:

'Continuity and series expansions',

https://en.wikipedia.org/wiki/Floor_and_ceiling_functions#Continuity_and_series_expansions.

Dave.

Remark: The answer to our former question is quite technical.

Please refer to the attachment below for an example. Example: [tex]\triangle n(I=91, \lambda ) = 0[/tex] imples [tex]\lambda = 3[/tex].