Difference between revisions of "User:Negativebplusorminus"
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<cmath>\sqrt{a+bi}=\sqrt{\frac{a+\sqrt{a^2+b^2}}{2}}+i\sqrt{\frac{-a+\sqrt{a^2+b^2}}{2}}</cmath> | <cmath>\sqrt{a+bi}=\sqrt{\frac{a+\sqrt{a^2+b^2}}{2}}+i\sqrt{\frac{-a+\sqrt{a^2+b^2}}{2}}</cmath> | ||
I derived that equation myself, and I am quite proud of it. I have a similar one for the fourth roots of <math>a+bi</math> which can be derived from inputting that equation into itself. I have also found various roots of unity in their radical forms during my spare time. | I derived that equation myself, and I am quite proud of it. I have a similar one for the fourth roots of <math>a+bi</math> which can be derived from inputting that equation into itself. I have also found various roots of unity in their radical forms during my spare time. | ||
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==Inspirographs== | ==Inspirographs== | ||
Another amazing creation of mine. More can be found [http://www.artofproblemsolving.com/Forum/blog.php?u=93546& here] (but you might have to look through a few pages of other stuff, too). | Another amazing creation of mine. More can be found [http://www.artofproblemsolving.com/Forum/blog.php?u=93546& here] (but you might have to look through a few pages of other stuff, too). | ||
Revision as of 19:00, 22 September 2015
A AoPS member, National MathCounts qualifier, and USAJMO qualifier.
Contents
Contest Results
MathCounts
In 2011, as a 7th grader, I didn't qualified for the State Countdown Round. In 2012, as an 8th grader, I lost the National MathCounts.
In the National competition, and scores as the worst.
AMCs
2012: 17 on AMC 10A, 27.5 on AMC 10B, 1 on AIME, 37.5 index for USAJMO. The cutoff was a 999.5, so I did not qualify for the USAJMO. However, I got 0 on the USAJMO. That thing is hard.
Equations for the Roots of the Complex
![\[\sqrt{a+bi}=\sqrt{\frac{a+\sqrt{a^2+b^2}}{2}}+i\sqrt{\frac{-a+\sqrt{a^2+b^2}}{2}}\]](http://latex.artofproblemsolving.com/1/1/5/115b79b994447d87745338438f7a8dc8bf2c25ad.png)
I derived that equation myself, and I am quite proud of it. I have a similar one for the fourth roots of 
which can be derived from inputting that equation into itself. I have also found various roots of unity in their radical forms during my spare time.
Inspirographs
Another amazing creation of mine. More can be found here (but you might have to look through a few pages of other stuff, too). To view the entire collection, please visit negativebplusorminus.blogspot.com in the near future (the site will be updated soon). Below are a few samples. <asy2> import graph3; import grid3; import palette; size(400,300,IgnoreAspect); defaultrender.merge=true; real f(pair z) {return sin(z.y)*(z.x^2+1)^(0.1*log(z.y^2+1));} surface s=surface(f,(-30,-30),(30,30),70,Spline); s.colors(palette(s.map(zpart),Rainbow())); draw(s,render(compression=Low,merge=true)); grid3(XYZgrid);</asy2> <asy2> import graph3; import grid3; import palette;currentprojection=orthographic(1,5,0.2); size(400,300,IgnoreAspect); defaultrender.merge=true; real f(pair z) {return sin(z.x^2+z.y^2);} surface s=surface(f,(-2.95,-2.95),(2.95,2.95),70,Spline); s.colors(palette(s.map(zpart),Rainbow())); draw(s,render(compression=Low,merge=true)); grid3(XYZgrid);</asy2>
