I need to make two definitions before I get to my question, which is at the bottom.
Domain: is the X-axis (variable)
Range: is the Y-axis
$$f(Domain) = Range$$ I think semiprimes can be factored by using the slopes of curves to calculate $P$ directly. Given $N$ we want to find $P$ and $Q$, where $P<Q$ and both are odd.
Given: $$f(N) = \sqrt{N}$$ The slope is: $$f'(N) = \frac{1}{2\sqrt{N}}$$ Both of these functions are in the $N-Domain$ (N varies) and each produce real numbers.
Now suppose I create a system where I hold $Q$ constant and I only vary $P$. I let $P$ start at $1$ and I let $P$ approach $Q$. As $P$ approaches $Q$ I calculate $N = PQ$ for each $P$. Now we have a system where we are in the $P-Domain$ because $P$ is the only variable.
My attempt is to plot the distances $f_1$, $f_2$, and $f_3$ below as $P$ approaches $Q$ and to use their slopes to calculate $P$: $$f_1(P) = \sqrt{N}-P$$ $$f_1(P) = \sqrt{P}\sqrt{Q}-P$$ $$f'_1(P) = \frac{\sqrt{Q}}{2\sqrt{P}} - 1$$ and: $$f_2(P) = Q - \sqrt{N}$$ $$f_2(P) = Q - \sqrt{P}\sqrt{Q}$$ $$f'_2(P) = - \frac{\sqrt{Q}}{2\sqrt{P}}$$ and: $$ f_3(N) = \sqrt{N}$$ $$ f'_{3o}(N) = m_{3o} = \frac{1}{2\sqrt{N}}$$ $$ f'_{3i}(N) = m_{3i} = \frac{Q}{2\sqrt{N}}$$
Where...$ m_{3o}$ is the slope in the $N-Domain$, and $m_{3i}$ is the slope in the $P-Domain$. It's a linear transformation where $m_{3i} = Qm_{3o}$.
$\textbf{See the figure below}$. I have created a plot with 3 functions superimposed on top of each other. $f_1$ and $f_2$ already reside in the $P-Domain$... but I have also pulled into the plot, the function $f_3(N) = \sqrt{N}$ from the $N-Domian$.
This is okay to pull $\sqrt{N}$ int0 the $P-Domain$ as long as we remember two rules.
- The magnitude $f_3(N)$ is unchanged when pulling into $P-Domain$.
- The slope $f'_{3o}(N)$ must be multiplied by $Q$ to fit properly, now $f'_{3i}(N)$ = $m_{3i}$.
In the plot, there is a vertical line inside a red box that shows all 3 slopes share the same $(P)$ value and this is the $(P)$ we are looking for.
Doing this allows us to work with the slopes in the $P-Domain$. But here's my problem. If I set $m_2 = m_{3i}$ and try to solve for $P$ using algebraic simplification, I simply get: $$\frac{1}{2\sqrt{N}} = \frac{1}{2\sqrt{N}}$$ Here's an example with data, but reducing to an equation with variable $P$ results in the above: $$-f'_2(P) = \frac{\sqrt{37}}{2\sqrt{13}} = f'_{3i}(N) = \frac{37}{2\sqrt{481}}$$ $\textbf{Question}$ Is there a way to use these slopes to isolate $P$ in such away we can solve for $P$ directly? Is there a way to use related rates between the 3 sets of curves to calculate $P$?
$\textbf{Things Investigated}$
Looking at the ratio of slopes, the ratio $\frac{m_1}{m_{3i}}$ was examined to see if there was something that will help in isolating $P$. I was unsuccessful.
$$\frac{m_1}{m_{3i}} = \frac{\sqrt{N} - 2P}{\sqrt{N}}$$
This is equivalent to:
$$1 - \frac{2P}{\sqrt{P}\sqrt{Q}} = 1 - \frac{2\sqrt{P}}{\sqrt{Q}}$$
Differentiating with respect to $P$ gives:
$$-\frac{1}{\sqrt{P}\sqrt{Q}} = -\frac{1}{\sqrt{N}}$$
Looking at the Area under the curve $f_1(P) = \sqrt{N}-P$ is:
$$\frac{Q^2}{6}$$
Need to see if there is a relationship between slope to area. Is there a rate of change between their rates of change? A double derivative?
