# Randomness can be a useful tool for solving problems.

Algebra

### Re: Randomness can be a useful tool for solving problems.

Guest wrote:Updated Remark: The discrete graph of $$\triangle n(\lambda)$$ for $$1 \le \lambda < \frac{\sqrt{I }}{2}$$ 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 $$\triangle n(\lambda)$$ for almost all $$1 \le \lambda < \frac{\sqrt{I }}{2}$$ with some increased oscillations (rapid rise and fall of $$\triangle n(\lambda)$$). Those increased oscillations do, however, dissipate over the full interval of $$\lambda$$. Moreover, we observe those increased oscillations begin before $$\triangle n(\lambda =65) = 0$$, and they continue for awhile afterwards.
Guest

### Re: Randomness can be a useful tool for solving problems.

Remark: $$\triangle n(\lambda)$$ helps our programming.
Guest

### Re: Randomness can be a useful tool for solving problems.

Guest wrote:
Guest wrote:Updated Remark: The discrete graph of $$\triangle n(\lambda)$$ for $$1 \le \lambda < \frac{\sqrt{I }}{2}$$ 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 $$\triangle n(\lambda)$$ for almost all $$1 \le \lambda < \frac{\sqrt{I }}{2}$$ with some increased oscillations (rapid rise and fall of $$\triangle n(\lambda)$$). Those increased oscillations do, however, dissipate over the full interval of $$\lambda$$. Moreover, we observe those increased oscillations begin before $$\triangle n(\lambda =65) = 0$$, and they continue for awhile afterwards.
Attachments
Discrete Graph of Delta N for I = 127 x 257.pdf
Guest

### Re: Randomness can be a useful tool for solving problems.

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

### Re: Randomness can be a useful tool for solving problems.

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

### Re: Randomness can be a useful tool for solving problems.

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

### Re: Randomness can be a useful tool for solving problems.

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

For I = 577 * 631, we compute $$\triangle n(\lambda = 27) = 0$$.

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

Remarks: There are three regions, $$1 \le \lambda < \frac{\sqrt{I }}{6}$$,

$$\frac{\sqrt{I }}{6} \le \lambda < \frac{\sqrt{I }}{3}$$,

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

Hopefully, our other tools can limit our search for $$\triangle n(\lambda) = 0$$ accurately to just one region!
Attachments
Discrete Graph of Delta N for I = 577 x 631.pdf
Guest

### Re: Randomness can be a useful tool for solving problems.

'Floor Function',

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

'Floor and ceiling functions'

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

### Re: Randomness can be a useful tool for solving problems.

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

### Re: Randomness can be a useful tool for solving problems.

An Update:

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

Remarks: First part is $$1 \le \lambda \le \frac{\sqrt{I }}{6}$$;

Second part is $$\frac{\sqrt{I }}{6} \le \lambda \le \frac{\sqrt{I }}{3}$$;

Third part is $$\frac{\sqrt{I }}{3} \le \lambda < \frac{\sqrt{I }}{2}$$.
Guest

### Re: Randomness can be a useful tool for solving problems.

Dave wrote:An Update:

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

Remarks: First part is $$1 \le \lambda \le \frac{\sqrt{I }}{6}$$;

Second part is $$\frac{\sqrt{I }}{6} \le \lambda \le \frac{\sqrt{I }}{3}$$;

Third part is $$\frac{\sqrt{I }}{3} \le \lambda < \frac{\sqrt{I }}{2}$$.

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

Enjoy!
Attachments
Continued Fractions Factoring method.pdf
Guest

### Re: Randomness can be a useful tool for solving problems.

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

### Re: Randomness can be a useful tool for solving problems.

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

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

### Re: Randomness can be a useful tool for solving problems.

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

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

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

### Re: Randomness can be a useful tool for solving problems.

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

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

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

### Re: Randomness can be a useful tool for solving problems.

Remark: $$Floor(\sqrt{I + \lambda^{2}})$$ is a big headache!
Guest

### Re: Randomness can be a useful tool for solving problems.

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

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

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

Guest

### Re: Randomness can be a useful tool for solving problems.

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

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

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

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

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

$$\triangle n(I, \lambda) = \sqrt{I + \lambda^{2}} - Floor(\sqrt{I + \lambda^{2}})$$.

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

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

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

Question: What is the solution of $$\triangle n(I, \lambda) = 0$$ ?
Guest

### Re: Randomness can be a useful tool for solving problems.

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

$$\triangle n(I, \lambda) = \sqrt{I + \lambda^{2}} - Floor(\sqrt{I + \lambda^{2}})$$.

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

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

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

Question: What is the solution of $$\triangle n(I, \lambda) = 0$$ ?

'Continuity and series expansions',

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

Dave.
Guest

### Re: Randomness can be a useful tool for solving problems.

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

$$\triangle n(I, \lambda) = \sqrt{I + \lambda^{2}} - Floor(\sqrt{I + \lambda^{2}})$$.

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

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

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

Question: What is the solution of $$\triangle n(I, \lambda) = 0$$ ?

'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: $$\triangle n(I=91, \lambda ) = 0$$ imples $$\lambda = 3$$.
Attachments
Graph of Solution for ΔN(91, λ) = 0.pdf