User:Nothing2Do.fr

I'm a french poet (i want to be), a mathematic enthousiast, a physician (I'd liked to understand), and so, and so ... My website is here : http://www.nothing2do.fr I give you a formula to resolve everything in real world (but you aren't "imposed" to hear). Howto resolve equation for newbies (or computer) The first degrees equation has to be resolved like this : $$ \left(a_{1}x+b_{1}\right) $$ (with untheorical solution : $$ x_{1}=-\frac{b_{1}}{a_{1}} $$, a1 and b1, we can follow back the thread) The second degree equation has to be resolved like this : $$ \left(\left(\frac{k_{2}}{k_{0}}*\frac{1}{a_{1}}\right)x+\left(\frac{k_{1}}{k_{2}}-b_{1}\right)\right) $$ (with solution added : $$x_{2}=-\frac{\frac{k_{1}}{k_{2}}+b_{1}}{\frac{k_{2}}{k_{0}}*\frac{1}{a_{1}}}=-\frac{b_{2}}{a_{2}}) $$ The third degrees has to be resolved like this : $$\left(\left(\frac{1}{\left(\frac{k_{2}}{k_{3}}-b_{1}-b_{2}\right)b_{1}b_{2}}*\frac{1}{a_{1}a_{2}}\right)x+\left(\frac{k_{2}}{k_{3}}-b_{1}-b_{2}\right)\right)(x_{3}=-\frac{\frac{k_{2}}{k_{3}}-b_{1}-b_{2}}{\frac{1}{\left(\frac{k_{2}}{k_{3}}-b_{1}-b_{2}\right)b_{1}b_{2}}*\frac{1}{a_{1}a_{2}}}=-\frac{b_{3}}{a_{3}}) $$ p is the degree of considered equation The equation of degree p has to be resolved like this : $$\left(k_{p}x+k_{0}\right) $$ with $$a_{p}=\frac{1}{k_{p}}$$ and $$b_{P}=-k_{0}$$ (WARNING, this can't be resolved now). Considering you've read the last sentence, I've to give you this one : $$ x_{p}=-\frac{b_{p}}{a_{p}}$$ so, every equation is like this : $$\sum_{n}^{p}\left(k_{n}x^{n}\right)=\prod_{n}^{p}\left(a_{n}x+b_{n}\right)=\left(k_{p}x+k_{0}\right)\sum_{n=1}^{p-1}\left(\frac{k_{n}}{k_{p}}x^{n-1}\right)$$ To have only real solution, we need to have (for "classical" method, this will define "delta") : $$\sum_{n}^{p}\left(b_{n}\right)=\prod_{n}^{p}\left(a_{n}\right)\Rightarrow\Delta=0$$ elsewhere : All solution will be real (but won't work ...) except the last (this depending on a1 and b1 ). the result is that : every équation of p degrees have (p or p-1) real solutions but differences with "classical" method will appear with the third degrees. to laugh : if $$a_{1}a_{2}=b_{1}+b_{2}$$ then $$\triangle=0$$ !!! If you want to be considered as a genious, it's the same things for every degrees “Everything that we know, is that we know nothing” Nothing2Do