Physics: "Machine" -

I have decided to keep this project confidencial... I will post more information about it after it is done.

Math: The quadratic saga - Ep. 4 (last episode!)

Today is the final revelation! I have no idea about what is going to be my next topic. I guess I am going to have more time to think about it now that I'm in vacations. Probably I am doing something about prime numbers. I know it is really difficult, although if everybody thinks like that we will never get an answer.

We had:

n² = 2[(n-1) + (n-2) + (n-3) + ... (n-n)] + n


and this leads us to... Wait. First I would like to say that this last chapter wasn't so fast as I am presenting here. I developed to much the formula and a 'n²' appeared. From that point on, I had to ask for help to my math teacher - maybe there was someway to remove that squared that required advanced math. I talked to the teacher and she gave me the solution. Some days passed and I saw that I was really close, but I passed a little bit the "climax" of the formula.

Continuing, and this leads us to:

This is the end of this story. Soon I am posting about my next topic... which I guess will be prime numbers

Math: The quadratic saga - Ep. 3

In the next day I dedicated some time to simplify the formula. I wanted to find something shocking, something that I would never imagine - and for that I had to make that formula smaller. So, we had:

n² = [2(n-1)+1] + [2(n-2)+1] + [2(n-3)+1] + ... + [2(n-n)+1]

We can see that in each "part" (part of the n-1, part of the n-2, etc) there is a +1 being added. So, for example, if we want to achieve the "part of (n-n)" starting from the "part of (n-1)", we are going to have a total of  'n'  parts. Let me give a example with numbers so that it is easier to understand. Let n=4:

4² = [2(3)+1] + [2(2)+1] + [2(1)+1] + [2(0)+1]

We can see that the number of "parts" is equal to 'n'. If we let n=4, for example, we are having 4 "parts", therefore since there is a '+1' in each part, in total we are having '+4'. So:

n = 4 || 4 "parts" || +1 for each part = total +4

And we can do that for any value of 'n':

n = 1 || 1 "part" || +1 for each part = total +1

n = 2 || 2 "parts" || +1 for each part = total +2

So we can relate:

n = n || n "parts" || +1 for each part = +n

So we can separate this '+1' of each part into one single letter: +n. Now we have:

n² = [2(n-1) + 2(n-2) + 2(n-3) + ... + 2(n-n)] + n

Now an easy simplification, separation of the '2' multiplying each "part":

n² = 2[(n-1) + (n-2) + (n-3) + ... (n-n)] + n

I am going to stop typing now. It isn't the same reason from the other episodes, this time I didn't stop until I reach my objective. The only problem was that... No, I am telling in the next episode. I will make the next episode different - I will divide it in two parts: the huge path that I made; and the simple answer that was right in front of me and I didn't see. The next episode is going to be the last episode of this "adventure"...

Math: The quadratic saga - Ep. 2

Some days later I began thinking how could I use this formula for other operations, not only have a squared number based on his previous squared number, but also have it based on his second previous number squared, third, fourth, etc - and my main objective, starting from zero. To make this easier, let n=10.

Then we can substitute "10" by "n" again, so that we can make the formula work for all values of "n".

And there I stopped. I knew I was close to my objective, but I didn't know the next steps would be so complicated.

Math: The quadratic saga - Ep. 1 (My first post!)

Hi. Recently I was doing history homework when I remembered that when I was in sixth or seventh grade, a friend of mine and I squared the first numbers in order, and saw that the difference between one squared number and its previous would increase algebraically:

2² - 1²  =        4-1= 3

3² - 2² = 9-4 = 5

4² - 3² = 16 - 9 = 7
5² - 4² = 25 - 16 = 9

etc...

But since we were still little kids that didn't know anything about functions, we didn't even realize that the data that was just collected could give birth to a formula. In that day, while I was supposed to do the history homework, I remembered about it and realized that it wouldn't be difficult to make a formula that would relate a squared number and its previous squared number. In a few moments I related:

n² = (n-1)² + [2(n-1)+1]

And that day I stopped. I was already satisfied that I found the formula that I was looking for.