I'm starting a math club at my school this week, because math is just that great. So I thought I'd make a math topic on here to do the same kind of thing I will do in my club at school. All types of math are acceptable on all levels.
Here's something that I thought of. An easy way to do multiplication in your head.

Code:

Say you have 99². If you know anything about math theory, you'd know that each square is separated by a certain odd number. Let's round up. So 99→100. 100² is easy: 10000. Take your number, so 99 in this example and add it to every successive number until you get to the closest 10^n, so 100. Therefore 99+100=199. That's your odd number. So 10000-199=9801. That's 99². Let's try 96². 96+97+98+99+100=490. Nope. Incorrect. With multiple numbers you have to double up each number between your number and 10^n. Why? Because each number is added to the one above it. Look at it like this:

Code:

The variable odd adds 96 initially. Then it adds 97. Each consecutive number has to do the exact same thing or else you'd come up with a smaller number than what is the actual number. Here you will come out with 784. 10000-784=9216 which is 96².
If you do this loop starting from a even number, then your odd variable will come out to be even, however if you start at an odd number, it will be odd.

Here's the TI-Basic Code I created for it. Works with negative numbers too. Although, some numbers take forever. It took like 30-40 seconds to do 1000^2.

Code:

There's some examples. I hope they inspired you.

I always found that Euler’s Formula was super neat and interesting, and from it spawned Euler's Identity and all the variations, which are just so much fun to look at:



That's about all I have to say about that.

Correct me if I'm wrong, but isn't that the guy that created the theorem's about circuits?

Yeah, he is.

Ok, so does that formula used in circuit theorems?

I like the Ibo Bonilla "Formula 10":

π+e+i^i+β+Φ=10

If you type it in the calculator, (maybe someones can't do i^i) it gives you 10 approximately, maybe because of the precision of the mathematic constants

Luck with your club!

I came out with negative 1, and google says this e^(i * π) + 1 = 0.

Add all the numbers between 1 and 100. What do you get? How did you get it? Try and do it without the calculator.
No this is not a homework question. I want to see how many people actually add each number or if they know the faster way, without programming it.

Had my first Math Club meeting today. It went great. Should I post what I do there on here? pm me your answer or post here.

Sean, I'll explain you as promised how an integral works with limits.

First: To integrate is the inverse process to derivate

Second: To derivate is to find the slope of the function tangent in a point (or all the points).

So, first let's start derviating with a limit that finds the slope with each function.

d f(x)/dx = lim h→0 [ f(x+h) - f(x)] / h

if f(x) = x^n → d f(x)/dx = lim h→0 [ (x+h)^n - x^n ] / h

We apply the L'Hôpital rule to solve the limit, derivating respect h:

d f(x)/dx = lim h→0 n * (x+h)^(n-1)

Then knowing that a integral is the inverse process including all the posible functions, we obtain:

F(x) = x^(n+1) / (n+1) + k

And then we calculate the definite integral:

∫ (1, 3, x^2) = [ x^3 / 3] (1, 3) = 3^3/3 - 1^3/3 = 3 - 1/3

Then you must apply this for each function

Guess we're doing cal now

Well... it's a bit more complicated than that for example, what about
∫[χ⁻¹]dx
Let's try to use the general power formula you provided to integrate this... (F(x) = x^(n+1) / (n+1) + k)
you get F(x) = x^0/0+k

Obviously, this is an issue because your are dividing by 0.

In fact, the answer is ln|χ|+C because *insert cal 1 here* Therefore, ∫[χ^n]dx = x^(n+1) / (n+1) + k if n≠-1
For simple functions this will work, for example:
∫[2x³(x-1)]dx = ∫[2x^4-2x³]dx = 2(x^5)/5 - (x^4)/2 + C

However, if you have a division of two functions and the one in the denominator is a binomial or some polynomial of greater degree, then you will be stuck again because you can't bring it down to the form you need to apply the formula (unless you want to get into some fancy polynomial division which might not even work) Then you will need to do some integration by substitution like this:

let f(x) = ∫(3x)/(3x²+5)dx
let u = 3x²+5
du/dx = 6x
1/2du = 3xdx
Therefore, f(x) = 1/2∫(1/u)du
f(x) = 1/2ln|u| + C
f(x) = 1/2ln|3x²+5| + C

This is just to show that explaining integrals with a single formula is a bit like going to space on the magic school bus

One of the things I always use when adding numbers is the rounding up to ten trick. Let's say I have 9+7, which, unless you have the answer memorized, is always a hard one. I do (10+7)-1, which gives you 16, which is what 9+7 is. By the way, this trick also works for 99, 999, and so on.

It's quick, fast, and simple, and because it's fast and simple, lets you play more games on your calc after you do your homework.


Which is always a good thing.... Right?

I have known that trick ever since I've known how to add. It's a great trick for smaller kids who use simple math a lot.

To explain the integrals in a single formula was not my prupose. Obviously, there are compound functions, rational funcions, etc.

There exist a lot of integration methods, for help us in this prupose.

I wanted to explain integrals in algebra terms, and not in numeric methods terms, like our calcs without CAS solve definite integrals.

Well, now I know I'm not alone in liking tau. Nice!

Here is my math club website. https://sites.google.com/a/dysartstudents.org/math-club-srhs/. Tell me if you can't get on it. I don't exactly know the permissions I have set down for it because it's confusing.
Edit: Never mind. It's set up correctly and I can get into it without having to log in through my district.

Then 1-e^i*tau=0 is also true.
There's also this:
1-e^i*2pi=0

Updated the Math website. Two meetings in. This is really fun. We talked about Row Echelon form today. In the calculator, that's denoted as rref(), which is actually Reduced-Row-Echelon Form, found by this 2nd[matrix]->Alpha[B]. It is used in finding the point where two or more lines intersect. Hint Hint...Systems of Equations.
Say you have a 2 equations: 2x + 2y = 4 and -3x + 5y = 6. Type this into a 2 x 3 matrix.

Code:

rref([matrix name]
Click enter and it will output this:

Code:

x = .5
y = 1.5
Basically, the order in which you put the variables from left to right is the order in which the 1's show up. See how the 1 in the top row is at the left?, well since we put the x term in front of the y term, that means that x=.5 and same thing with the y except rref() shifts the 1 over so it's denoted as the second variable, in this case y. So y=1.5
If a problem you have returns this:

Code:

Then there is no solution, as long as you plugged it in correctly, so make sure of that first.
The reason this is no solution is because the bottom row consists of 0's on the left and a 1 on the right. That means, as explained above, that none of the variables equal 1, or do not have a point of intersection.

If your equations are the exact same, the calculator will output this:

Code:

For example:
2x + 2y = 3
4x + 4y = 6
will output this:

Code:

You will use this in algebra and up; but you won't learn it, if you do, until Pre-Calculus.

Today in Math Club, I taught people how to graph piece-wise functions.

This is how you type Piece-wise functions on the calculator. The division of the condition is because when the condition is true, you're dividing by 1, and when it's false, you're dividing by 0. The dividing by 0 error is turned off in graphing mode for those of you who don't know. A condition always returns a Boolean value of 1(true) or 0(false).

I never thought so much work goes into taking the derivative of a function. You have the Power Rule, Product Rule, Quotient Rule, Chain Rule, Trig Rule/s, and a bunch of others that I haven't learned yet. Talk about taking the derivative of your brain. Hehe...