User blog:Edwin Shade/Just Practice, That's All !

This page is being reserved for testing purposes for the time being.

The Testing Part
Testing if there is a way such that MathJax can be inserted into those scrollable gray boxes; I wish for this to be done instead of using MathMode because the latter will lag the page when numerous arguments are used on the same page, (around 50+), whereas MathJax is evidently immune to this slowing effect.

$$   \begin{array}{c} \varepsilon_0 \\ \omega^{\omega^{\omega}} \\ \omega^{{\omega^3}4} \\ \omega^{{{\omega^3}3}+{\omega^2}4} \\ \omega^{{{\omega^3}3}+{\omega^2}3+{\omega}4} \\ \omega^{{{\omega^3}3}+{\omega^2}3+{\omega}3+4} \\ \omega^{{{\omega^3}3}+{\omega^2}3+{\omega}3+3}4 \\ \omega^{{{\omega^3}3}+{\omega^2}3+{\omega}3+3}3+\omega^{{{\omega^3}3}+{\omega^2}3+{\omega}3+2}4 \\ \omega^{{{\omega^3}3}+{\omega^2}3+{\omega}3+3}3+\omega^{{{\omega^3}3}+{\omega^2}3+{\omega}3+2}3+\omega^{{{\omega^3}3}+{\omega^2}3+{\omega}3+1}4 \\ \omega^{{{\omega^3}3}+{\omega^2}3+{\omega}3+3}3+\omega^{{{\omega^3}3}+{\omega^2}3+{\omega}3+2}3+\omega^{{{\omega^3}3}+{\omega^2}3+{\omega}3+1}3+\omega^{{{\omega^3}3}+{\omega^2}3+{\omega}3}4 \\ \omega^{{{\omega^3}3}+{\omega^2}3+{\omega}3+3}3+\omega^{{{\omega^3}3}+{\omega^2}3+{\omega}3+2}3+\omega^{{{\omega^3}3}+{\omega^2}3+{\omega}3+1}3+\omega^{{{\omega^3}3}+{\omega^2}3+{\omega}3}3+\omega^{{{\omega^3}3}+{\omega^2}3+{\omega}2+4} \\ \omega^{{{\omega^3}3}+{\omega^2}3+{\omega}3+3}3+\omega^{{{\omega^3}3}+{\omega^2}3+{\omega}3+2}3+\omega^{{{\omega^3}3}+{\omega^2}3+{\omega}3+1}3+\omega^{{{\omega^3}3}+{\omega^2}3+{\omega}3}3+\omega^{{{\omega^3}3}+{\omega^2}3+{\omega}2+3}4 \\ \omega^{{{\omega^3}3}+{\omega^2}3+{\omega}3+3}3+\omega^{{{\omega^3}3}+{\omega^2}3+{\omega}3+2}3+\omega^{{{\omega^3}3}+{\omega^2}3+{\omega}3+1}3+\omega^{{{\omega^3}3}+{\omega^2}3+{\omega}3}3+\omega^{{{\omega^3}3}+{\omega^2}3+{\omega}2+3}3+\omega^{{{\omega^3}3}+{\omega^2}3+{\omega}2+2}4 \\ \omega^{{{\omega^3}3}+{\omega^2}3+{\omega}3+3}3+\omega^{{{\omega^3}3}+{\omega^2}3+{\omega}3+2}3+\omega^{{{\omega^3}3}+{\omega^2}3+{\omega}3+1}3+\omega^{{{\omega^3}3}+{\omega^2}3+{\omega}3}3+\omega^{{{\omega^3}3}+{\omega^2}3+{\omega}2+3}3+\omega^{{{\omega^3}3}+{\omega^2}3+{\omega}2+2}3+\omega^{{{\omega^3}3}+{\omega^2}3+{\omega}2+1}4 \\ \omega^{{{\omega^3}3}+{\omega^2}3+{\omega}3+3}3+\omega^{{{\omega^3}3}+{\omega^2}3+{\omega}3+2}3+\omega^{{{\omega^3}3}+{\omega^2}3+{\omega}3+1}3+\omega^{{{\omega^3}3}+{\omega^2}3+{\omega}3}3+\omega^{{{\omega^3}3}+{\omega^2}3+{\omega}2+3}3+\omega^{{{\omega^3}3}+{\omega^2}3+{\omega}2+2}3+\omega^{{{\omega^3}3}+{\omega^2}3+{\omega}2+1}3+\omega^{{{\omega^3}3}+{\omega^2}3+{\omega}2}4 \\ \omega^{{{\omega^3}3}+{\omega^2}3+{\omega}3+3}3+\omega^{{{\omega^3}3}+{\omega^2}3+{\omega}3+2}3+\omega^{{{\omega^3}3}+{\omega^2}3+{\omega}3+1}3+\omega^{{{\omega^3}3}+{\omega^2}3+{\omega}3}3+\omega^{{{\omega^3}3}+{\omega^2}3+{\omega}2+3}3+\omega^{{{\omega^3}3}+{\omega^2}3+{\omega}2+2}3+\omega^{{{\omega^3}3}+{\omega^2}3+{\omega}2+1}3+\omega^{{{\omega^3}3}+{\omega^2}3+{\omega}2}3+\omega^{{{\omega^3}3}+{\omega^2}3+{\omega+4}} \\ \omega^{{{\omega^3}3}+{\omega^2}3+{\omega}3+3}3+\omega^{{{\omega^3}3}+{\omega^2}3+{\omega}3+2}3+\omega^{{{\omega^3}3}+{\omega^2}3+{\omega}3+1}3+\omega^{{{\omega^3}3}+{\omega^2}3+{\omega}3}3+\omega^{{{\omega^3}3}+{\omega^2}3+{\omega}2+3}3+\omega^{{{\omega^3}3}+{\omega^2}3+{\omega}2+2}3+\omega^{{{\omega^3}3}+{\omega^2}3+{\omega}2+1}3+\omega^{{{\omega^3}3}+{\omega^2}3+{\omega}2}3+\omega^{{{\omega^3}3}+{\omega^2}3+{\omega+3}}4 \end{array} $$

Reverse Mathematics
Let \(\) indicate the base-n reversal of the natural number \(x\), where n is assumed to be 10. It is to be read "rev-x", as in the 'reverse of x'.

The following will demonstrate a method of finding the solution to the simple quadratic equation \(x^2+^2-f=0\) when x is less than 1,000.

First it is necessary to write out \(x\) as a number of the form \(100a+10b+c\), such that \(\) can be notated as \(100c+10b+a\).

The next step is to write out a table, as below, which calculates the possible values of \(a^2+c^2 (mod\text{ }10)\), or the last digit of \(f\) in the above mentioned quadratic equation.

0 1  2  3  4  5  6  7  8  9  a    _  _  _  _  _  _  _  _  _  _ 0| 0 1  4  9  6  5  6  9  4  1 1| 1  2  5  0  7  6  7  0  5  2 2| 4  5  8  3  0  9  0  3  8  5 3| 9  0  3  8  5  4  5  8  3  0 4| 6  7  0  5  2  1  2  5  0  7 5| 5  6  9  4  1  0  1  4  9  6 6| 6  7  0  5  2  1  2  5  0  7 7| 9  0  3  8  5  4  5  8  3  0 8| 4  5  8  4  0  9  0  3  8  5 9| 1  2  5  0  7  6  7  0  5  2 b

Given this last digit of \(f\), we are able to narrow down the possible values for a and b, and subsequently we are able to brute force check the possible values of a and b at almost a glance, until we are left with only two or three possible values, which we then check carefully.

After this we input the appropiate the values for both a and c, then calculate (100a+10b+c)^2+(100c+10b+a)^2, putting the b-terms off to one side, and the large number off to another. Set this equal to f, after which you should subtract f from both sides and then solve the resulting quadratic equation. This will give you the value of b, and then you will derive a solution for \(x^2+^2-f=0\) !

Mental Mathematics
The following will go through some math problems that despite their daunting exterior are actually almost trivial to perform mentally, and with this list you will be able to impress your non-mathematical friends, (although I myself use this as a self-challenge, like practicing the multiplication table), or more humbly, build a repertoire of arithmetical shortcuts by which you may one day be able to circumvent the most circuitous problem most efficiently.

Third Roots
Perhaps the simplest of cases is the mental extraction of cube roots, or more specifically, two-digit cube roots. Firstly, memorize the following correspondence of digits, which are placed in the A-rack, and the N-rack respectively.

A 0  1  2  3  4  5  6  7  8  9 N 0  1  8  7  4  5  6  3  2  9

Next, become familiar with the first nine cubic numbers, beginning from 1^3, or just 1.

1^3  2^3   3^3   4^3   5^3   6^3   7^3   8^3   9^3   1     8    27    64   125   216   343   512   729

This is all that you need to know in order to extract two-digit cubic roots mentally ! Ask a friend to type in a two-digit number in a calculator, then read to you the result. Take the number in the thousand's place, and see which two cubic numbers it falls between. The cubic root of the lesser of these two numbers will be the first digit of the two-digit number. Lastly, take the last digit your friend reads to you and consider it as an entry in the N-rack, then match it up with it's corresponding entry in the A-rack; this is your second digit. That's all.

Squaring Numbers Mentally
I will first demonstrate a method to square two digit numbers, then show how it can be adapted to three digit numbers, and lastly exhibit a quick way I have found to square four-digit numbers.

Two-Digit Numbers
Firstly, it is assumed you have knowledge of the first 5 squares, (e.g. 1, 4, 9, 16, and 25), and you are well-familiar with the multiplication table, (mainly for speed). To square a two-digit number we will be exploiting the algebraic property that \((a+b)(a-b)=a^2-b^2\).

When you are given a two-digit number by your friend, (or whomever happens to give it to you), you must first find the closest multiple of ten to that number, which we'll call B. Take the difference between the two-digit number and B, and subtract it from your two-digit number to yield a new number C. Now multiply together B and C, (which should be fairly easy, as one of them is a multiple of ten), and add to this number the square of the difference between the two-digit number you were given and either B or C, (either number is an equal distance away from the two-digit number you were given, so it matters not which one you pick).

As a shortcut to square two-digit numbers ending in a five, multiply the first digit by it's successor, then concatenate a '25' at the end to the right of the number. With your knowledge of how to square any two-digit number and this trick, you will also by virtue know how to square any three-digit number ending in a five ! It utilizes the same technique, just multiply the digits before the last[5], by their successor and then concatenate it with a '25' to the right.

Three-Digit Numbers
Many a time I have longed to pass by a parking lot or be near traffic just so I could square the numbers on the license plates, which though I do it scarcely nowadays, I still have retained the method by which I am able to do it, and can say confidently it is easy enough for anyone with persistence to learn.

You must firstly become acquainted with the first 50 square numbers, (which is roughly equal to the memorization of another multiplication table). This can be done using flashcards, or by immersing yourself so much into mathematical problems involving two-digit squares that you have no choice but to remember ! This is in fact how I learned them, for before I knew them all I resorted to a lengthier method of squaring three-digit numbers that required me to recalculate the same two-digit squares over and over again, such that I had them ingrained in my mind after a while.

Next, when given a three-digit number,(we'll call it A), determine the closest multiple of 100 to that three-digit number, (to be named B), and then the difference between A and B, (which we'll call C). Multiply B by (A-C), (which should be easy considering one of them is a multiple of one hundred), then add C^2, (which is most quickly done if you already know the first 50 two-digit squares, as C can never exceed 50. The result is the square of A, or the three-digit number you were given.

Four-Digit Numbers
To square four digit numbers it would be good to become well versed in multiplying any given two two-digit numbers together. Firstly take the algebraic property that \((a+b)^2=a^2+2ab+b^2\), and notate a as a multiple of a hundred and b as a number underneath a hindred.

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