User:A math-wiki

A little background on me. I am 20, I live in the state of Oregon. I like math a lot and want to make a career as a mathematician. Current education is a High School diploma, though I took AP Statistics, AP Calculus and AP Music Theory and achieved scores of 5 on the two math tests and 4 on the overall for the Music Theory. I also like the Sciences a lot, especially Paleontology. I am fond of most technical writing, especially that which involves numbers and/or mathematics. I also like Philosophical writing and that which has a lot of Logic to it. I am an avid computer gamer, Call of Duty being my favorite game, though I own and play many more titles.

A note about the section below: A lot of what I intend to put here is some of my own independent derivations. So fellow Wikipedians and the like may take a look at some of the work I've done. If you want to discuss some of the stuff below, feel free to comment on my talk page.

Graphing Polynomials over Complex Inputs, Real Outputs

During my Junior year of High School, I made several failed attempts at deriving the Cubic Formula completely on my own, before finally looking it up and duplicated their methods which I only vaguely understood at the time. As part of that quest, I attempted to study the behavior of Polynomials where $0<n<4$. I found this to be difficult, especially if the equation had nonreal solutions. I noted that, one key thing I was missing was a way visualize the behavior of a Polynomial over Complex values. The problem was that this required 4 physical dimensions, which is not practical for observing the kinds of behavior I was looking for. It finally dawned on me that I really wasn't concerned with Complex outputs, I wanted really to study the behavior of Real outputs for Complex inputs. In other words, I could remove the imaginary part of the y variable, and thus I would only need 3 physical dimensions. So I set out to graph them, ... one problem, graphing utilities don't think the way I need them too. So I had to do things the old fashion way, at least somewhat. (I will elaborate on that later)

I deduced that I would need to rewrite the x variable into components in order to have a chance at really learning anything from my discovery. So I had to choose of slightly different notation than would normally be used for Polynomials and for graphing in 3 dimensions. Since I want the final equation to have x,y,z as their variables, I would have to start with X as the original variable else substituting $x+zi$ for $x$ would look very weird and confusing.

First off I wanted to test my thinking, 1st degree Polynomials with real coefficients don't have any "Complex" part to their graph. So the $zi$ term should drop out. First of all before I start let me define a few things which will apply throughout.

Let $a,b,c,d,e\in \mathbb {R}$ and let $a,b,c,d,e$ be arbitrary constants.

Let $x,y,z\in \mathbb {R}$ and let $x,y,z$ be variables.

Let $X\in \mathbb {C}$ and let $X=x+zi$.

Now

$y=aX+b$

$y=a(x+zi)+b$

$y=ax+b+azi$

Notice how I arranged the terms after I expanded. Here's the key part, remember $y\in \mathbb {R}$ not $y\in \mathbb {C}$. So that means the term with $i$ must go to zero. So I group the terms over real vs. imaginary, then determine the conditions on $x$ and $z$ so that, $y\in \mathbb {R}$.

$y=(ax+b)+(az)i$

Thus $az=0$ and thus $z=0$ is the only restriction.

So $y=(ax+b)+(az)i,z=0$ is the final form. Note that this means the line has no "Complex" part, as stated earlier. Note that if you directly apply the condition on z, then z vanishes, which means that $y=(ax+b)+(az)i,z=0$ becomes $y=ax+b$. This is actually a unique behavior, since 1st degree Polynomials are the highest order Polynomials that can't have nonreal solutions. It should also be noted that the condition $z=0$ must always exist since (with $a,b,c,d\in \mathbb {R}$ not $\mathbb {C}$) the domain of the Function is $(-\infty ,\infty )$. No problem graphing lines in space, since they still only have two dimensions to their graphs.

Before I start working on the 2nd degree Polynomial, I would like to point out a few things about the conventions of graphing a Polynomial this way. I have decided that since $y$ is the usual choice for the output of a Polynomial to define $y$ not $z$ as the vertical and instead $z$ is the depth aspect, leaving $x$ to it's usual left-right aspect.

The 2nd degree Polynomial is a more appropriate test of the applicability of my method, as it actually has a "Complex" part. It was actually the first case I worked on and was responcible for putting this idea in my head. I notice that if you tranlated a Quardratic vertically, the Solutions would converge towards eachother, then on the imaginary axis diverge with the same symmetrical pattern. I hypothesized that the 'shape' of the graph is really two Parabolas, with a shared vertex that have opposite leading coefficients and are twisted $90^{\circ }$ from one another.

$y=aX^{2}+bX+c$

$y=a(x+zi)^{2}+b(x+zi)+c$

$y=ax^{2}+2axzi-az^{2}+bx+bzi+c$

$y=(ax^{2}+bx+c-az^{2})+(2axz+bz)i$

Now once again we set the entire term with $i$ equal to zero.

$2axz+bz=0$

And this give two different conditions, using the Zero-Product Property

$(2ax+b)z=0$

$z=0,2ax+b=0$

$z=0,x=-{\frac {b}{2a}}$

So either $z=0$ (given), or $x=-{\frac {b}{2a}}$ which describes the "Complex" part, try graphing one or two examples. My hypothesis turned out to true, not only that but it looked like there was a relationship to De Moivre's Formula. It appeared that, looking down the y axis, the sections of the graph, centered at $x=-{\frac {b}{na}}$, the points would appear to be placed according to the nth-roots by De Moivre's Formula. The case of the Cubic function however would disprove that hypothesis.

$y=aX^{3}+bX^{2}+cX+d$

$y=a(x+zi)^{3}+b(x+zi)^{2}+c(x+zi)+d$

$y=a(x^{3}+3x^{2}zi-3xz^{2}-z^{3}i)+b(x^{2}+2xzi-z^{2})+c(x+zi)+d$

$y=ax^{3}+3ax^{2}zi-3axz^{2}-az^{3}i+bx^{2}+2bxzi-bz^{2}+cx+czi+d$

$y=(ax^{3}+bx^{2}+cx-3axz^{2}-bz^{2})+(3ax^{2}z+2bxz+cz-az^{3})i$

So now set the imaginary part equal to zero.

$3ax^{2}z+2bxz+cz-az^{3}=0$

Factor out $z$

$z(2ax^{2}+2bx+c-3az^{2})=0$

And thus $z=0$, or $2ax^{2}+2bx+c-3az^{2}=0$. Lets rewrite the second case so that $x$ is a function of $z$.

$2ax^{2}+2bx+c-3az^{2}=0$

Lets switch sides of the equals sign so the the term with $z$ can be position be also on the left side.

So, $3az^{2}=2ax^{2}+2bx+c$

$z^{2}={\frac {2a}{3a}}x^{2}+{\frac {2b}{3a}}x+{\frac {c}{3a}}$

Now we must include both roots to get the whole original function, which is a conic section by the way, since the if were replaced by y it would be a simplified form of the General Quadratic equation in Two Variables.

$z={\pm }{\sqrt {\frac {2ax^{2}+2bx+c}{3a}}}$

This means that $z=0,{\pm }{\sqrt {\frac {2ax^{2}+2bx+c}{3a}}}$. It should be noted that the radical solutions represent a Hyperbola. If you treat $z$ as though it were $y$ like normal for conic sections. You get

$3ax^{2}-3az^{2}+2bx+c=0$

Looking at this equation, there is no $xz$ term and the $z^{2}$ term is negative when the $x^{2}$ term is postive and vis versa if it is negative. Meaning the equation above is a Hyperbola. The Quartic function is a more complicated curve, by now I think the patterns are starting to become clear. The Quartic "Complex" part is represented by a simplified version of the General Cubic Equation in Two Variables. And has a basic form of

$Az^{2}=Bx^{3}+Cx^{2}+Dx+E$

Implicit functions like this and the one from the Cubic function are difficult to get a graphing utility to graph. The matter is only further complicated by the fact that the result, is actually the intersection of a plane parallel to the y-axis and another plane that is rather complicated to visualize. Hence the reason I haven't spent more time on this. I really need a program talored to the need of this kind of mathematics, which I don't have, nor em I a good enough programmer of computers to write such a program. (If you know of a program which will graph these quite readily please do tell me on my talk page)

$y=aX^{4}+bX^{3}+cX^{2}+dX+e$

$y=a(x+zi)^{4}+b(x+zi)^{3}+c(x+zi)^{2}+d(x+zi)+e$

$y=a(x^{4}+4x^{3}zi-6x^{2}z^{2}-4xz^{3}i+z^{4})+b(x^{3}+3x^{2}zi-3xz^{2}-z^{3}i)+c(x^{2}+2xzi-z^{2})+d(x+zi)+e$

$y=ax^{4}+4ax^{3}zi-6ax^{2}z^{2}-4axz^{3}i+az^{4}+bx^{3}+3bx^{2}zi-3bxz^{2}-bz^{3}i+cx^{2}+2cxzi-cz^{2}+dx+dzi+e$

$y=(ax^{4}+bx^{3}+cx^{2}+dx+e-6ax^{2}z^{2}-3bxz^{2}-cz^{2}+az^{4})+(4ax^{3}z+3bx^{2}z+2cxz+dz-4axz^{3}-bz^{3})i$

$y=(ax^{4}+bx^{3}+cx^{2}+dx+e-6ax^{2}z^{2}-3bxz^{2}-cz^{2}+az^{4})+(4ax^{3}+3bx^{2}+2cx+d-4axz^{2}-bz^{2})zi$

$4ax^{3}+3bx^{2}+2cx+d-4axz^{2}-bz^{2}=0$

$(4ax+b)z^{2}=4ax^{3}+3bx^{2}+2cx+d$

$z^{2}={\frac {4ax^{3}+3bx^{2}+3cx+d}{4ax+b}}$

$z=\pm {\sqrt {\frac {4ax^{3}+3bx^{2}+3cx+d}{4ax+b}}}$

$z=0,\pm {\sqrt {\frac {4ax^{3}+3bx^{2}+3cx+d}{4ax+b}}}$

The Quintic Function introduces a new difficulty, the presence of a $z^{4}$ term in the imaginary part.

$y=aX^{5}+bX^{4}+cX^{3}+dX^{2}+eX+f$

$y=a(x+zi)^{5}+b(x+zi)^{4}+c(x+zi)^{3}+d(x+zi)^{2}+e(x+zi)+f$

$y=a(x^{5}+5x^{4}zi-10x^{3}z^{2}-10x^{2}z^{3}i+5xz^{4}+z^{5}i)+b(x^{4}+4x^{3}zi-6x^{2}z^{2}-4xz^{3}i+z^{4})+c(x^{3}+3x^{2}zi-3xz^{2}i-z^{3})+d(x^{2}+2xzi-z^{2})+e(x+zi)+f$

$y=ax^{5}+5ax^{4}zi-10ax^{3}z^{2}-10ax^{2}z^{3}i+5axz^{4}+az^{5}i+bx^{4}+4bx^{3}zi-6bx^{2}z^{2}-4bxz^{3}i+bz^{4}+cx^{3}+3cx^{2}zi-3cxz^{2}-cz^{3}+dx^{2}+2dxzi-dz^{2}+ex+ezi+f$

$y=(ax^{5}+bx^{4}+cx^{3}+dx^{2}+ex+f-10ax^{3}z^{2}-6bx^{2}z^{2}-3cxz^{2}-2dz^{2}+5axz^{4}+bz^{4})+(5ax^{4}+4bx^{3}+3cx^{2}+2dx+e-10ax^{2}z^{2}-4bxz^{2}-cz^{2}+az^{4})zi$

$5ax^{4}+4bx^{3}+3cx^{2}+2dx+e-10ax^{2}z^{2}-4bxz^{2}-cz^{2}+az^{4}=0$

To solve this, note that only even powers of z, including zero and present, so in terms of z, we have a Biquadratic Equation. Obtain a solution for $z^{2}$ first, then take the square roots of both sides, resulting in a $\pm$ on the side w/ x.

Solving Polynomials with Compositions of Functions

Consider, $f(x)=x^{2}+bx+c$ and $g(x)=x^{2}+mx+n$, then

$f(g(x))=(x^{2}+mx+n)^{2}+b(x^{2}+mx+n)+c$

$f(g(x))=(x^{4}+2mx^{3}+(m^{2}+2n)x^{2}+2mnx+n^{2})+b(x^{2}+mx+n)+c$

$f(g(x))=x^{4}+2mx^{3}+(m^{2}+2n+b)x^{2}+(2mn+bm)x+(n^{2}+bn+c)$

If $h(x)=x^{4}+Bx^{3}+Cx^{2}+Dx+E$, then $h(x)=f(g(x))$ iff $2m=B$, $m^{2}+2n+b=C$, $2mn+bm=D$, and $E=n^{2}+bn+c$

$m={\frac {B}{2}}$

$c=E-n^{2}-bn$

${\begin{cases}m^{2}+2n+b=C\\2mn+bm=D\end{cases}}$

${\begin{cases}{\frac {B^{2}}{4}}+2n+b=C\\{\frac {2Bn+Bb}{2}}=D\end{cases}}$

$b=C-{\frac {B^{2}}{4}}-2n$

${\frac {2Bn+B(C-{\frac {B^{2}}{4}}-2n)}{2}}=D$

${\frac {2Bn+BC-{\frac {B^{3}}{4}}-2Bn}{2}}=D$

$BC-{\frac {B^{3}}{4}}=2D$

$4BC-B^{3}=8D$

This is the condition on the coefficients of h(x) for f(g(x)) to exist. Provided it is met, h(x) can be solved by solving for f(0), and then using f(0) to find the appropriate x's. In the case that h(x) doesn't meet this criteria, one can still use this technique but only after applying a linear transformation, of the form y=x-k, for a correct choice of k, to find this k, one must solve the following Cubic polynomial in k.

$4(B-4k)(C-3Bk+6k^{2})-(B-4k)^{3}=8(D-2Ck+3Bk^{2}-4k^{3})$

Note that this means k can be a Complex number but there are three choices and given the that the original coefficients, B, C, and D, are Real numbers, one of the k's is also a real number by The Fundamental Theorem of Algebra. and so a generally preferable choice. This Cubic bears a striking resemblance to the Discriminant of a Quartic Polynomial, and this is no coincidence, and neither is its being a Cubic. When one applies the methods Ferrari and Cardano used to originally solve the Quartic, this same equation shows up for a different task, making the non-perfect square side a perfect square. Here it is the first and only linear transformation I need to do, after which one can continue from the beginning, remembering that their solution is for y and not x, this substitution can be easily undone by direct substitution.

Library of strange identities

In this section, I intend to put any unusual relations/equalities I find in my studies of mathematics.

$\int {\frac {1}{t^{2}-1}}dt$

Using partial fractions,

${\frac {1}{t^{2}-1}}={\frac {A}{t+1}}+{\frac {B}{t-1}}$

Thus

$1=A(t-1)+B(t+1)$

$(A+B)t+(B-A)=1$

This gives the system

${\begin{cases}A+B=0\\B-A=1\end{cases}}$

Thus, solving the second equation for B gives

$B=1+A$

Substituting into to first gives

$A+1+A=0$

$2A=-1$

$A=-{\frac {1}{2}}$

Substituting that result into the second equation gives

$B+{\frac {1}{2}}=0$

$B=-{\frac {1}{2}}$

This means we can conclude

$=\int {\frac {1}{t^{2}-1}}dt=\int -{\frac {1}{2(t+1)}}-{\frac {1}{2(t-1)}}dt$

$=-{\frac {1}{2}}\int {\frac {1}{t+1}}dt-{\frac {1}{2}}\int {\frac {1}{t-1}}dt$

Let $u=t+1$ and $v=t-1$, then we get

$=-{\frac {1}{2}}\int {\frac {1}{u}}du-{\frac {1}{2}}\int {\frac {1}{v}}dv$

$=-{\frac {1}{2}}\ln |u|-{\frac {1}{2}}\ln |v|+C$

Back substituting yields

$=-{\frac {1}{2}}\ln |t+1|-{\frac {1}{2}}\ln |t-1|+C$

Now let's integrate this function using some trigonometric identities instead

$\int {\frac {1}{t^{2}-1}}dt$

Let $t=\sin {\theta }$, then $dt=\cos {\theta }d{\theta }$, and thus our integral becomes

$=\int {\frac {\cos {\theta }}{\sin ^{2}{\theta }-1}}d{\theta }$

Substitute using the trigonometric identity $\sin ^{2}{\theta }=1-\cos ^{2}{\theta }$

$=\int {\frac {\cos {\theta }}{(1-\cos ^{2}{\theta })-1}}d{\theta }$

$=-\int {\frac {\cos {\theta }}{\cos ^{2}{\theta }}}d{\theta }$

$=-\int {\frac {1}{\cos {\theta }}}d{\theta }$

$=-\int \sec {\theta }d{\theta }$

Using a integral table's formula for the antiderivative of the secant function, we get

$=-\ln |\sec {\theta }+\tan {\theta }|+C$

Now we solve for $\theta$ in terms of t,

$\theta =\arcsin t$

And thus

$-\ln |\sec {\theta }+\tan {\theta }|+C=-\ln |\sec {(\arcsin t)}+\tan {(\arcsin t)}|+C$

Now for the strange part, we can conclude that the two different looking results are in fact the same, and we may move the constant's of integration to one side and combine them, to yield the equality

$-{\frac {1}{2}}\ln |t+1|-{\frac {1}{2}}\ln |t-1|=-\ln |\sec {(\arcsin t)}+\tan {(\arcsin t)}|+C$