Quicklopedia

**Quick** · August 6th 2006, 05:15 PM

I have decided to post all my small questions in this thread. (because many of them don't deserve a thread of their own)
Also, this thread will probably end up containing an overview of ALL highschool math questions that I can think of, and just might become a useful resource for some guests. (or this thread might just die out after two questions...)

note: At the incistence of PH, I'm going to ask everybody to please stay on topic so this thread remains a math only discussion.

also note: To everyone (especially newbies), please do not post your questions on this thread.

**Quick** · August 6th 2006, 05:17 PM

Let's say I have equation $x^3+2x^2+3x+4=0$
how do I solve for x? (surely there must be some kind of formula out there)

**ThePerfectHacker** · August 6th 2006, 06:00 PM

Originally Posted by Quick

Let's say I have equation $x^3+2x^2+3x+4=0$
how do I solve for x? (surely there must be some kind of formula out their)

Yes, Quick and the formula is extremely messy. Sometime ago, I came up with a method for solving a general cubic in a certian case (it almost always works).
---
$ax^3+bx^2+cx+d=0$ , $a\not = 0$
Step 1Depress your cubic.
Using the substitution,
$x=y-\frac{b}{3a}$ (This trick belongs to Cardano of Italy)
Thus, you will have in the end,
$y^3+Ay=B$
Now, this is the Hacker method.
Solve,
$\left\{ \begin{array}{c}2\alpha =B\\ -3\sqrt{\alpha^2-\beta}=A$
Then,
$y=\sqrt[3]{\alpha+\sqrt{\beta}}+\sqrt[3]{\alpha - \sqrt{\beta} }$
Thus,
$x=y-\frac{b}{3a}$

Note: This only works when $A\geq 0$ .
I can post the complete method perhaps another time, it is much longer. It was developed by Girolamo Cardano and Niccollo Fantona Tartaglia. There is also a method for a general quartic (degree 4). However, a young Norweigen mathematician, Abel, was able (get the pun) to prove that there is no method for a quintic (degree 5). Later an even younger French mathematician was able to prove that there is no solution for any degree beyond 5. And by solution I mean that is there is no 100% gaurentted method.

It is interesting, I once read in a book (A History of Pi, but I do not trust all its sources, I have found some errors in the book thus do not know if this story is true). That a man was killed for knowing how to solve a quartic because the Church claimed that such an ability is beyond human power and thus he made a pact with the devil. A similar incident happend with Paganni he was not buried for being the lion of all virtusos because they again belived he made a pact with the Devil. (Of course, the Devil is I)..

**Quick** · August 6th 2006, 07:40 PM

So, using your method I get the incredibly long equation of (in fact, it's so big I had to set it up horizontally):

... $\small{\sqrt[3]{\frac{9abc-4b^3}{54a^3}+\sqrt{\frac{9a^2c^2-6b^2ac+b^4}{81a^4}-\left(\frac{9abc-4b^3}{54a^3}\right)^2}}}$

$\small{+\sqrt[3]{\frac{9abc-4b^3}{54a^3}-\sqrt{\frac{9a^2c^2-6b^2ac+b^4}{81a^4}-\left(\frac{9abc-4b^3}{54a^3}\right)^2}}}$

$\small{-\frac{b}{3a}}$
.................................................. ................................

... $x$

Am I right? (or have you never fully written it out before)

Also, clarification, do you mean that this equation always and only works if $\frac{3ac-b^2}{3a^2}\geq0$ or do you mean that it only and works most of the time when $\frac{3ac-b^2}{3a^2}\geq0$ ?

note: My calculations say $A=\frac{3ac-b^2}{3a^2}$

**CaptainBlack** · August 6th 2006, 08:38 PM

Originally Posted by Quick

So, using your method I get the incredibly long equation of (in fact, it's so big I had to set it up horizontally):

... $\small{\sqrt[3]{\frac{9abc-4b^3}{54a^3}+\sqrt{\frac{9a^2c^2-6b^2ac+b^4}{81a^4}-\left(\frac{9abc-4b^3}{54a^3}\right)^2}}}$

$\small{+\sqrt[3]{\frac{9abc-4b^3}{54a^3}-\sqrt{\frac{9a^2c^2-6b^2ac+b^4}{81a^4}-\left(\frac{9abc-4b^3}{54a^3}\right)^2}}}$

$\small{-\frac{b}{3a}}$
.................................................. ................................

... $x$

Am I right? (or have you never fully written it out before)

Also, clarification, do you mean that this equation always and only works if $\frac{3ac-b^2}{3a^2}\geq0$ or do you mean that it only and works most of the time when $\frac{3ac-b^2}{3a^2}\geq0$ ?

note: My calculations say $A=\frac{3ac-b^2}{3a^2}$

Here is a version of the cubic formula:

**Quick** · August 7th 2006, 10:30 AM

How do you find determinants, I basically understand their uses, but I don't know how to find them.

**ThePerfectHacker** · August 7th 2006, 11:02 AM

Originally Posted by Quick

How do you find determinants, I basically understand their uses, but I don't know how to find them.

$\left| \begin{array}{cc}a&b\\c&d\end{array} \right|=ad-bc$

Now, what do you do when you have 3x3, 4x4, ... ?

Definition A minor of an element in a determinant is a determinant obtained by deleting the horizontal rows and vertical rows of that element.

Example
Given,
$\left| \begin{array}{ccc} a&b&c\\d&e&f\\g&h&i \end{array} \right|$
Find the minor of $c$ ?
Simple, if you draw horizontal line and vertical lines and delete the elements you are left with,
$\left| \begin{array}{cc}d&e\\g&h \end{array} \right|=dh-eg$

Usually entries in a determinant are written with a subscript $a_{ij}$ this means, $i$ th row and $j$ th coloum. For example, $a_{24}$ means the 2nd element down and 4 element across. When there is more then 10 elements you can use a comman to avoid confusion. For example, $a_{123}$ does it mean 12 then 3 or 1 and then 23? So you can use a comma $a_{12,3}$ . Therefore, a 3x3 determinant has the following entries,
$\left| \begin{array}{ccc}a_{11}&a_{12}&a_{13}\\a_{21}&a_{ 22}&a_{23}\\a_{31}&a_{32}&a_{33} \end{array} \right|$

Definition A cofactor of an element $a_{ij}$ is $(-1)^{i+j}a_{ij}$ .
So basically, a cofactor of an element in a determinant is that element expect with or without a sign. Thus, In 3x3 determinant the cofactors follow this rule,
$\left| \begin{array}{ccc}+&-&+\\-&+&-\\+&-&+ \end{array} \right|$

Definition A value of a $n\times n$ determinant ( $n>2$ ) is a sum of the products of the minors and cofactors in any row or coloum.

Theorem The previous definition is well-defined. Meaning that it is independant of what row or coloum you use. You will end with the same number.

Determinant evaluation is time consuming because every minor reduces a determinant to a deterinant of 1 less. Then you need to do this again and again until you reach a 2x2 determinant. (Someone correct me on this, I believe the number of steps required is $n^3$ in polynomial time?).

One you master determinant evaluation I can you show cool things you can do with them. For example, when you are working with a quadradic equation to determine whether it has real or complex roots you need to look at its discrimant ( $b^2-4ac$ ). When you solve a system of linear equations you need to look whether at its determinant to decide what type of solutions it has. And also provide you with a method of how to find solutions.

I think determinants is what makes math look like art. They are so elegant. Finding a nice pattern to a big mess is what math is sometimes about..

**Quick** · August 7th 2006, 11:08 AM

How do you solve the determinant of $a_{11}$ in a 4x4 matrix, because y'know crossing out the vertical and horizontal rows still leaves you with a 3x3 matrix

**ThePerfectHacker** · August 7th 2006, 12:27 PM

Originally Posted by Quick

How do you solve the determinant of $a_{11}$ in a 4x4 matrix, because y'know crossing out the vertical and horizontal rows still leaves you with a 3x3 matrix

You mean a cofactor, right?

As I said, that is why it is time consuming. After you have a cofactor for $a_{11}$ in a 4x4 you need to multiply it by the value of its minor (which is the 3x3 determinant) then you need to work with a 3x3 determinant. Reducing it to a 2x2 determinant.

**Quick** · August 7th 2006, 04:05 PM

Alright, so let me try finding the value of $\boxed{\begin{array}{ccc}1&2&3\\3&2&1\\1&2&3\end{a rray}}$ and then you could tell me if I did anything wrong (or if I'm just way off).

First I'll use the first row to find the value of the determinant. So I find the values of each miner. Which is: 4;8; and 4.

now I multiply each of those numbers by their cofactors to get: 4; -8; and 4

then I add those together to get: $\boxed{\begin{array}{ccc}1&2&3\\3&2&1\\1&2&3\end{a rray}}=0$

but what I don't get get is how to find the minor of a_11 in a 4x4 matrix (because it's 3x3)

do you just find the value of the 3x3 matrix like I did on top?

**ThePerfectHacker** · August 7th 2006, 06:32 PM

Originally Posted by Quick

First I'll use the first row to find the value of the determinant. So I find the values of each miner. Which is: 4;8; and 4.

now I multiply each of those numbers by their cofactors to get: 4; -8; and 4

Okay, you get what a minor is, good. The co-factors are: are $(-1)^{1+1}1=1$ , $(-1)^{1+2}2=-2$ , $(-1)^{1+3}3=3$

Now you mutiply them,
$4,-16,12$
Add them,
$4+(-16)+12=0$
So your procedure was wrong, but your answer was correct!

**Quick** · August 8th 2006, 12:56 PM

Now I feel I understand determinants, so onto the next question I've got.

How do you find derivatives?

**ThePerfectHacker** · August 8th 2006, 02:45 PM

Originally Posted by Quick

Now I feel I understand determinants, so onto the next question I've got.

How do you find derivatives?

Not going to explain that to you. That is part of Calculus. There is not point in you memorizing these derivative formulas. Because you will not understand. Then you will turn yourself into an average college math student. Ask something else.

**Quick** · August 8th 2006, 04:12 PM

Originally Posted by ThePerfectHacker

Not going to explain that to you. That is part of Calculus. There is not point in you memorizing these derivative formulas. Because you will not understand. Then you will turn yourself into an average college math student. Ask something else.

that's mildly disappointing

. But you're right. The reason I'm good at math is because I understand things. Skipping ahead of my comprehension (for the time being) is pointless...

anyway, you briefly meantioned that determinants are related to the discriminant in the quadratic formula. How?

**ThePerfectHacker** · August 8th 2006, 06:38 PM

Originally Posted by Quick

that's mildly disappointing

. But you're right. The reason I'm good at math is because I understand things. Skipping ahead of my comprehension (for the time being) is pointless...

Not even Gauss can understand a derivative without studing calculus in full detail. So there is nothing embarrasing without being able to understand this.

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