Question

in attempt to regain my latex ...

Answer 1

\(\normalsize\color{black}{{\bbox[5pt, black ,border:2px solid black ]{ \color{black}{\frac{ \frac{ \frac{ \color{cyan}{\rm \Huge General{\tiny~~~~~~}Setup.} \\[2.3em] }{ \color{white}{\color{white}{\Huge \nabla f=\lambda \nabla g} \\[2.3em] } }\\ }{ \color{white}{\Huge f_x=\lambda g_x} } }{ \frac{ \frac{\color{white}{\Huge f_y=\lambda g_y} }{ \frac{\color{white}{\Huge g(x,y)=d}\\[2.3em]}{ \color{white}{\Huge {\color{white}{\Huge {\rm where}{\tiny~~~~~}g(x,y)=d}}} }\\ }\\ }{ \color{white}{\Huge {\rm is~the~constraint.}} } }} }}}\)

Answer 2

Not bad :P

Answer 3

Yeah, I'm learning to use tricks on this site every once in a while ... tnx

Answer 4

\(\Large \color{black}{\displaystyle \underbrace{x^2+x+1+\frac{1}{x}+\frac{1}{x^2}}_{\LARGE \underbrace{x^2+x+1+\frac{1}{x}+\frac{1}{x^2}}_{\huge \underbrace{x^2+x+1+\frac{1}{x}+\frac{1}{x^2}}_{\underbrace{x^2+x+1+\frac{1}{x}+\frac{1}{x^2}}_{\underbrace{x^2+x+1+\frac{1}{x}+\frac{1}{x^2}}_{\huge\text{______________} } } } } } }\)

Answer 5

\(\normalsize\color{blue}{{\bbox[5pt,yellow,border:2px solid blue ]{ \underbrace{2x^2-5x-11=e^2}_{\rm \large quadratic{\normalsize~~}equation} }}}\)

Answer 6

\(\large \color{cyan }{{\bbox[5pt, cyan ,border:2px solid black ]{
\frac{\frac{\color{black}{\large \displaystyle D[f(x)g(x)]=f'(x)g(x)+f(x)g'(x)}}{\color{black}{\large \displaystyle \int D[f(x)g(x)]\mathrm{d}x=\int f'(x)g(x)\mathrm{d}x+\int f(x)g'(x)\mathrm{d}x+C}}}{\frac{\color{black}{\large \displaystyle f(x)g(x)=\int f'(x)g(x)\mathrm{d}x+\int f(x)g'(x)\mathrm{d}x+C} }{\color{black}{\large \displaystyle \int f'(x)g(x)\mathrm{d}x=f(x)g(x)-\int f(x)g'(x)\mathrm{d}x+C} }} }}}\)

Answer 7

\(\Huge\color{black}{{\bbox[5pt, black ,border:2px solid black ]{ \color{white}{■}\color{yellow}{■}\color{orange}{■}\color{red}{■}\color{lime}{■}\color{blue}{■}\color{magenta}{■}}}}\)

Answer 8

\(\color{black}{\displaystyle {\rm Every}{\tiny~~~}n{\rm th}{\tiny~~~}{\rm Fibonacci{{\tiny~~~}}}f_n{\tiny~~~}{\rm number{{\tiny~~~}}satisfies{{\tiny~~~}}}f_n=\frac{\varphi ^n-\psi^n}{\varphi -\psi },\\
{\rm where{\tiny~~~}\varphi{\tiny~~~}and{\tiny~~~}\psi{\tiny~~~}are{\tiny~~~}the{\tiny~~~}two{\tiny~~~}roots{\tiny~~~}of{\tiny~~~}}z^2-z-1.}\)\(\tiny\\[2.0em]\)
\(\tiny\color{black}{{\bbox[5pt, blue ,border:2px ]{ \quad }}}\)\(\tiny\color{black}{{\bbox[5pt, orange ,border:2px ]{ \quad }}}\)\(\tiny\color{black}{{\bbox[5pt, blue ,border:2px ]{ \quad }}}\)\(\tiny\color{black}{{\bbox[5pt, orange ,border:2px ]{ \quad }}}\)\(\tiny\color{black}{{\bbox[5pt, blue ,border:2px ]{ \quad }}}\)\(\tiny\color{black}{{\bbox[5pt, orange ,border:2px ]{ \quad }}}\)\(\tiny\color{black}{{\bbox[5pt, blue ,border:2px ]{ \quad }}}\)\(\tiny\color{black}{{\bbox[5pt, orange ,border:2px ]{ \quad }}}\)\(\tiny\color{black}{{\bbox[5pt, blue ,border:2px ]{ \quad }}}\)\(\tiny\color{black}{{\bbox[5pt, orange ,border:2px ]{ \quad }}}\)\(\tiny\color{black}{{\bbox[5pt, blue ,border:2px ]{ \quad }}}\)\(\tiny\color{black}{{\bbox[5pt, orange ,border:2px ]{ \quad }}}\)\(\tiny\color{black}{{\bbox[5pt, blue ,border:2px ]{ \quad }}}\)\(\tiny\color{black}{{\bbox[5pt, orange ,border:2px ]{ \quad }}}\)\(\tiny\color{black}{{\bbox[5pt, blue ,border:2px ]{ \quad }}}\)\(\tiny\color{black}{{\bbox[5pt, orange ,border:2px ]{ \quad }}}\)\(\tiny\color{black}{{\bbox[5pt, blue ,border:2px ]{ \quad }}}\)\(\tiny\\[1.8em]\)
\(\color{black}{\displaystyle \rm Proof{\tiny~~~~}(by{\tiny~~~}{\rm Induction).}}\)\(\tiny\\[1.8em]\)
\(\color{black}{\displaystyle {\rm Let{{\tiny~~~}}}z(n)=\frac{\varphi ^n-\psi^n}{\varphi -\psi }{\rm.{{\tiny~~~}}Then,{{\tiny~~~}}the{{\tiny~~~}}statement{{\tiny~~~}}translates{{\tiny~~~}}to{{\tiny~~~}}} \\ \forall n\in\mathbb{N}\left[z(n)=f_n \right]. \tiny\\[1.8em]}\)
\(\color{black}{\displaystyle z(1)=\frac{\varphi ^1-\psi^1}{\varphi -\psi }=1,{\rm{\tiny~~~}and{\tiny~~~}}f_1=1,{\rm{\tiny~~~}thus{\tiny~~~}}z(1)=f_1.}\)\(\tiny \\[2.9em] \)
\(\color{black}{\displaystyle z(2)=\frac{\varphi ^2-\psi^2}{\varphi -\psi }.{\tiny~~~}{\rm Since{{\tiny~~~}}\varphi{{\tiny~~~}}and{{\tiny~~~}}\psi{{\tiny~~~}}satisfy{{\tiny~~~}}}z^2-z-1=0,\\{\rm therefore,}{\tiny~~~}\varphi=\varphi^2-1{\tiny~~~}{\rm and }{\tiny~~~}\psi=\psi^2-1.{\tiny~~~}{\rm Substitutions{\tiny~~~}give,} \\ \displaystyle z(2)=\frac{\varphi ^2-\psi^2}{(\varphi ^2-1) -(\psi ^2-1) }=\frac{\varphi ^2-\psi^2}{\varphi ^2 -\psi ^2 }=1.{\tiny~~~}{\rm Also, }{\tiny~~~}f_2=1, \\ {\rm therefore,{\tiny~~~}}z(n)=f_n{\tiny~~~}{\rm for}{\tiny~~~}n=1{\tiny~~~}{\rm and}{\tiny~~~}n=2. }\)
\(\tiny \\[1.3em] \)
\(\color{black}{\displaystyle {\rm Assume}{\tiny~~~}z(n)=f_n{\tiny~~~}{\rm for{\tiny~~~}all{\tiny~~~}integers}{\tiny~~~}k\le n .{\tiny~~~}{\rm Then,}{\tiny~~~}z(k)=\frac{\varphi ^k-\psi^k}{\varphi -\psi }. }\)
\(\color{black}{\displaystyle {\rm By{\tiny~~~}definition{\tiny~~~}of{\tiny~~~}Fibonacci{\tiny~~~}sequence,}{\tiny~~~}f_{k+1}=f_{k}+f_{k-1},{\tiny~~~}{\rm and}}\)\(\tiny \\[0.5em] \)
\(\color{black}{\displaystyle {\rm by{\tiny~~~}this{\tiny~~~}assumption}{\tiny~~~}f_{k}=\frac{\varphi^{k}-\psi^{k}}{\varphi -\psi }{\tiny~~~}{\rm and}{\tiny~~~}f_{k-1}=\frac{\varphi^{k-1}-\psi^{k-1}}{\varphi -\psi },}\)
\(\color{black}{\displaystyle {\rm thus,}{\tiny~~~}f_{k+1}=\frac{\varphi^{k}-\psi^{k}}{\varphi -\psi }+\frac{\varphi^{k-1}-\psi^{k-1}}{\varphi -\psi }=\frac{\varphi^{k-1}(\varphi+1) -\psi^{k-1}(\psi+1)}{\varphi -\psi }.}\)

\(\color{black}{\displaystyle {{\rm Then,{\tiny~~~}since{\tiny~~~}\varphi {\tiny~~~}and{\tiny~~~}\psi {\tiny~~~}satisfy }{{\tiny~~~}}z^2-z-1=0,{{\tiny~~~}}{\rm therefore}} \\
\varphi+1=\varphi^2{{\tiny~~~}}{\rm and}{{\tiny~~~}}\psi+1=\psi ^2 ,{\tiny~~~}{\rm and{\tiny~~~}thus,} \\ \displaystyle f_{k+1}=\frac{\varphi^{k-1}(\varphi^2) -\psi^{k-1}(\psi^2)}{\varphi -\psi }=\frac{\varphi^{k+1} -\psi^{k+1}}{\varphi -\psi }=z(k+1). }\)

\(\color{black}{\displaystyle {\rm In{\tiny~~~}conclusion,{\tiny~~~}since}{\tiny~~~} \left[z(1)=f_1\right]\wedge\left[z(2)=f_2\right] {\tiny~~~}{\rm and } }\)
\(\color{black}{\displaystyle \left[z(k-1)=f_{k-1}\right]\wedge\left[z(k)=f_k\right]{\tiny~}\Longrightarrow {\tiny~} \left[z(k+1)=f_{k+1}\right], }\)
\(\color{black}{\displaystyle {\rm therefore}{\tiny~~~} z(n)=f_n {\tiny~~~}{\rm for {\tiny~~~}all{\tiny~~~}}n\in\mathbb{N}. \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad ■}\)

Answer 9

i can never do that..

Answer 10

@SolomonZelman You are good!