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Paper.tex

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\title{Complex Tangent-Magnitude Proportionality\\מתכונתיות משיק-עוצמה מרוכבת}
\author{Elisha Hollander}
\date{January 2021}
\newtheorem*{theorem}{Theorem}
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\begin{document}
    \maketitle
    \selectlanguage{english}
    \begin{center}
        \textbf{Theoretical Doctoral Research Proposal}
    \end{center}
    \addcontentsline{toc}{section}{Main page}
    \newpage
    \tableofcontents
    \addcontentsline{toc}{section}{Contents}
    \newpage
    \addcontentsline{toc}{section}{Introduction}
    \begin{center}
        \Large{Introduction}
    \end{center}
    \begin{theorem}[Main]
        Let $z$ be an one dimensional function of $\Re \left(z\right)$ over $\mathbb Z$, then $\mid z\mid\not\propto\frac{d\left(z-\bar z\right)}{d\left(z+ \bar z\right)}$
    \end{theorem}
    \begin{center}
        To read it, you need to be familiar with: Calculus, Bessel-functions, Basic arithmetic, Induction, Real analysis, Discrete mathematics, Summation theory, Ricatti-equations, Infinite series, Complex analysis, Partial derivatives and Logarithms.
    \end{center}
    \newpage
    \begin{center}
        \Large{$z\bar z\propto\frac{d\left(z-\bar z\right)}{d\left(z+\bar z\right)}$ is possible}
    \end{center}
    \addcontentsline{toc}{section}{Similar case}
    \begin{proof}[Proof]
        First of all, we apply a complex to two dimensional real function transformation ($z\mapsto f$).\\In this case we can see that $f$ will have the property for a ratio of one if and only if:$$f\left(\sqrt{2x}\right)=\sqrt{\frac x2}\frac{J_{5/4}\left(x\right)+J_{3/4}\left(x\right)-J_{-5/4}\left(x\right)-J_{-3/4}\left(x\right)}{J_{1/4}\left(x\right)+J_{-1/4}\left(x\right)}-\frac 1{\sqrt{8x}}$$the complete workout is complex but it is easy to verify the outcome.\\To finish the proof we will transform $f$ into $z$ to get:$$z=\sqrt{2x}+\sqrt{\frac {-x}2}\frac{J_{5/4}\left(x\right)+J_{3/4}\left(x\right)-J_{-5/4}\left(x\right)-J_{-3/4}\left(x\right)}{J_{1/4}\left(x\right)+J_{-1/4}\left(x\right)}+\frac 1{\sqrt{-8x}}$$
    \end{proof}
    \newpage
    \clearpage
    \begin{center}
        \Large{$Ratio=1$}
    \end{center}
    \addcontentsline{toc}{section}{Ratio of one}
    \begin{small}
    \begin{proof}[Proof]
        First of all, we apply a complex to two dimensional real function transformation ($z\mapsto f$).\\Let us define a new function$:\\g\left(x,y\right)\coloneqq$\begin{cases}$\lim_{a\rightarrow y\lfloor\frac xy\rfloor^-}g\left(a\right)\left(1+x-y\lfloor\frac xy\rfloor\right)$&$x\cancel{=}0\\f\left(0\right)$ & \text{else}\end{cases}\\Now, it is easy to see that $g$ is continuous and:\begin{equation}g\left(y\lfloor \frac xy \rfloor,y\right)=f\left(0\right)\left(y+1\right)^{\lfloor \frac xy \rfloor}\end{equation}\begin{equation}g'\left(x\right)=g\left(y\lfloor \frac xy \rfloor\right)\le g\left(x\right)\end{equation}and by definition: \begin{equation}g\left(0\right)=f\left(0\right)\end{equation}So by using equations 2 and 3 we get:\\$\forall_{y\in\mathbb{R}}f\left(x\right)\ge g\left(x,y\right)$\\and using equation 1 we get:\\$\forall_{y\in\mathbb{R}}g\left(x,y\right)\ge \left(1+y\right)^{\lfloor \frac xy \rfloor}f\left(0\right)$\\so overall we get:\\$\forall_{y\in\mathbb{R}}f\left(x\right)\ge g\left(x,y\right)\ge \left(1+y\right)^{\lfloor \frac xy \rfloor}f\left(0\right)$\\so by using $y=-1$ we get:\\$f\left(x\right)\ge g\left(x,-1\right)\ge \lim_{y\rightarrow -1}\left(1+y\right)^{\lfloor -x\rfloor}f\left(0\right)=\lim_{y\rightarrow 0}\frac{f\left(0\right)}{y^{\lfloor x\rfloor}}$\\So if we want $f$ to be continues, and it has to be, because it's actually $z$, then $f\left(0\right)=0$, by using this we can see that $f^{\left(n\right)}$ is zero for any $n$ which is not divisible by 4, then we define a new series:\\$b_n:=\frac{f^{\left(n\right)}\left(0\right)}{n!}$\\so we can inductively derive $f$ to get:\\$b_n=\sum_{k=1}^{n-1}\frac {\left(n-2\right)!}{\left(k-1\right)!\left(n-k-1\right)!}b_kb_{n-k}$\\So because of the properties of Maclaurin series:\\$f\left(x\right)=\sum_{n=1}^\infty\frac {b_{4n}x^{4n}}{\left(4n\right)!}$\\we will now define $a_n$ as $b_{4n}$ we can see that $a_1$ is 2 by deriving $f$ four times\\now we can define a new function $y$:\\\begin{equation}y\coloneqq\sum_{n=1}^\infty\frac {a_nx^n}{\left(4n-1\right)!}\end{equation}\\and using a new sum from 2 we can get:\\$\sum_{n=2}^\infty\frac{a_nx^n}{\left(4n-1\right)!}\left(4n-1\right)=\sum_{n=2}^\infty\sum_{k=1}^{n-1}\frac{a_ka_{n-k}x^kx^{n-k}}{\left(4k-1\right)!\left(4\left(n-k\right)-1\right)!}$\\and we know:\\$4xy'-y-\frac{a_1x}{2}=y^2$\\and since $a_1=2$:\\$4xy'=x+y+y^2$\\using the solving steps of Riccati equations:\\$\left(\ln u\right)'\coloneqq-\frac y{4x}\Rightarrow u''+\frac3{4x} u'+\frac1{16x}u=0$\\then we can get the general solution:\\$u=C_1x^{\frac 18}J_{-\frac 14}\left(\frac {\sqrt x}2\right) +C_2x^{\frac18}J_{\frac14}\left(\frac {\sqrt x}2\right)$\\and we know that $y/x$ is defined at zero so:\\$u=C_1x^{\frac 18}J_{-\frac 14}\left(\frac {\sqrt x}2\right)$\\so by using the definition of $u$ we get:\\$y=-4x\frac {\left(x^{\frac18}J_{-\frac14}\left(\frac{\sqrt x}2\right)\right)'}{x^{\frac 18}J_{-\frac 14}\left(\frac {\sqrt x}2\right)}$\\Since $J_v'\left(x\right)=\frac12 \left(J_{v-1}\left(x\right)-J_{v+1}\left(x\right)\right)$ we know:\\$y=\frac{-\frac 12 x^{\frac 18}J_{-\frac14}\left(\frac {\sqrt x}2\right)-\frac 12 x^{\frac 58}J_{-\frac 54}\left(\frac {\sqrt x}2\right)+\frac 12 x^{\frac 58}J_{\frac 34}\left(\frac {\sqrt x}2\right)}{x^{\frac 18}J_{-\frac 14}\left(\frac {\sqrt x}2\right)}=\frac{\sqrt{x}\left(J_{\frac34}\left(\frac {\sqrt{x}}2\right)-J_{-\frac54}\left(\frac{\sqrt{x}}2\right)\right)-J_{-\frac14}\left(\frac{\sqrt{x}}2\right)}{2J_{-\frac14}\left(\frac{\sqrt{x}}2\right)}$\\now we can use $x^4$ in place of $x$ to get:\begin{equation}y\left(x^4\right)=\frac{x^2\left(J_{\frac34}\left(\frac {x^2}2\right)-J_{-\frac54}\left(\frac{{x^2}}2\right)\right)-J_{-\frac14}\left(\frac{x^2}2\right)}{2J_{-\frac14}\left(\frac{x^2}2\right)}\end{equation}\\using equation number 4 we can see that $f\left(x\right)=\int\frac{y\left(x^4\right)}xdx$, we need the fourth power for an obvious reason and the integration and division by $x$ to divide each term in $y$ by $4n$ so using the fifth equation:\\$f\left(x\right)=\int\frac{x^2\left(J_{\frac34}\left(\frac {x^2}2\right)-J_{-\frac54}\left(\frac{{x^2}}2\right)\right)-J_{-\frac14}\left(\frac{x^2}2\right)}{2xJ_{-\frac14}\left(\frac{x^2}2\right)}dx=\int\left(\frac{x\left(J_{\frac 34}\left(\frac {x^2}2\right)-J_{-\frac 54}\left(\frac {x^2}2\right)\right)}{2J_{-\frac 14}\left(\frac {x^2}2\right)}-\frac 1{2x}\right)dx$\\we can add $\frac{\ln x}2$ to both sides and get:\\$f\left(x\right)+\frac{\ln x}2=\int\frac{x\left(J_{\frac 34}\left(\frac {x^2}2\right)-J_{-\frac 54}\left(\frac {x^2}2\right)\right)}{2J_{-\frac 14}\left(\frac {x^2}2\right)}dx$\\we know that $J_v'\left(x\right)=\frac12\left(J_{v-1}\left(x\right)-J_{v+1}\left(x\right)\right)$ so we can see:\\\begin{small}$f\left(x\right)+\frac{\ln x}2=-\int\frac x{J_{-\frac14}\left(\frac{x^2}2\right)}\frac{dJ_{-\frac 14}\left(\frac {x^2}2\right)}{d\frac{x^2}2}dx=-\int \frac{dJ_{-\frac14}\left(\frac{x^2}2\right)}{J_{-\frac14}\left(\frac {x^2}2\right)}=c-\ln\left(J_{-\frac14}\left(\frac{x^2}2\right)\right)\Rightarrow f\left(x\right)=c-\ln\left(J_{-\frac 14}\left(\frac {x^2}2\right)\right)-\ln\left(\sqrt x\right)$\end{small}\\so $f\left(0\right)$ is undefined, but we know $f\left(0\right)=0$
    \end{proof}
    \end{small}
    \newpage
    \begin{center}
        \Large{$Ratio\neq0,1$}
    \end{center}
    \begin{proof}[Proof]
        For every ratio $r$, we apply a complex to two dimensional real function transformation ($z\mapsto f_r$), and then, we can find $g_r$ to prove $f_r\left(0\right)=0:\\g_r\left(x,y\right)\coloneqq$\begin{cases}$\lim_{a\rightarrow y\lfloor\frac xy\rfloor^-}g_r\left(a\right)\left(r+x-y\lfloor\frac xy\rfloor\right)$&$x\cancel{=}0\\f_r\left(0\right)$ & \text{else}\end{cases}\\This will result in $f_r$ being derived from $f$ and its derivatives, and can be easily proved to be impossible as well.
    \end{proof}
    \addcontentsline{toc}{section}{Other ratios}
    \newpage
    \begin{center}
        Consequence
    \end{center}
    \addcontentsline{toc}{section}{Consequence}
    \begin{theorem}[Consequence]
        If we look at the five elements describing a 2d analytic function plot, the length $(L)$, the slope $(M)$, the distance from the origin $(D)$ and the coordinates $(x,\,y)$, we discover that $M\neq D\land L\neq y\land D\neq y$ is the only definite quality.
        To prove it, I will show, for every other equality, a way to find a case where it applies and finally prove the statement itself.
    \end{theorem}
    \begin{proof}[M=L]
        $$f'(x)=\int_0^x\sqrt{1+f'(x')^2}dx'\Rightarrow f''(x)=\sqrt{1+f'(x)^2}\Rightarrow \ln\left(\sqrt{1+f'(x)^2}+f'(x)\right)=\int\frac{\partial f'(x)}{\sqrt{1+f'(x)^2}}=\int\partial x=x$$$$\Rightarrow \sqrt{1+f'(x)^2}+f'(x)=e^x\Rightarrow 1=e^{2x}-2f'(x)e^x\Rightarrow f'(x)=\frac{e^{2x}-1}{2e^x}\Rightarrow f(x)=\frac12\left(e^x+e^{-x}\right)+C$$
    \end{proof}
    \begin{proof}[L=D]
        $$\sqrt{x^2+f(x)^2}=\int_0^x\sqrt{1+f'(x')^2}dx'\Rightarrow\frac{2x+2f(x)f'(x)}{2\sqrt{x^2+f(x)^2}}=\sqrt{1+f'(x)^2}\Rightarrow$$$$ 4x^2+4f(x)^2+4x^2f'(x)^2+4f(x)^2f'(x)^2=4x^2+8xf(x)f'(x)+4f(x)^2f'(x)^2$$$$\Rightarrow f(x)^2-2xf(x)f'(x)+x^2f'(x)^2=0\Rightarrow \left(f(x)-xf'(x)\right)^2=0\Rightarrow\frac{f'(x)}{f(x)}=x\Rightarrow\ln f(x)=\int\frac{\partial f(x)}{f(x)}=\int x\partial x=\frac{x^2}2$$$$\Rightarrow f(x)=\sqrt {e^{x^2}}$$
    \end{proof}
    \begin{proof}[M=x]
        $$f'(x)=x\Rightarrow f(x)=\frac{x^2}2+C$$
    \end{proof}
    \begin{proof}[M=y]
        $$f'(x)=f(x)\Rightarrow f(x)=Ce^x$$
    \end{proof}
    \begin{proof}[L=x]
        $$x=\int_0^x\sqrt{1+f'(x')^2}dx'\Rightarrow 1=\sqrt{1+f'(x)^2}\Rightarrow f(x)=C$$
    \end{proof}
    \begin{proof}[D=x]
        $$x=\sqrt{x^2+f(x)^2}\Rightarrow f(x)=0$$
    \end{proof}
    \begin{proof}[$M\neq D\land L\neq y\land D\neq y$]
        $$f(x)\neq\int_0^x\sqrt{1+f'(x')^2}dx'\Leftrightarrow f'(x)\neq\sqrt{1+f'(x)^2}\Leftrightarrow1\neq0$$
        $$f(x)=\sqrt{x^2+f(x)^2}\Leftrightarrow x=0$$
        And for the last part we need to apply the transformation ($z\mapsto f$) and get the $M\neq D$ part from the main theorem.
    \end{proof}
\end{document}