Math Is Fun Forum

This isn't a formal proof, but more like what you need to be thinking during your proof:

A) If f is injective, then for every element in Im f (the range), there must be one and only one path to get back to f.  All the elements that don't have path (since you don't know that f is onto) you ignore, otherwise the Im f wouldn't even be a function.

B) Yes, for pretty much the same reasons.  Remember, two elements in Im f can't go to the same element in A, because then f wouldn't even be a function, as it wouldn't be well defined.

C) You know that every element in A corresponds to a unique element in B, so at the very least, |A| = |B|.  But there could be some elements in B which don't have an inverse (it's not guarunteed to be onto), so there may be more elements in B.

Haven't really check to see if these are the right answers, I'm just going to assume they are.  Oh, and if speed of your program doesn't matter, none of the below will...

sin and cos are normally very slow functions (depending on how your compiler implements them).  Always try to avoid using them if possible.  But, if you have to use them, heres how:

Using sin and arctan functions in whatever language your using, generate a file that just lists these values of certain numbers.  If you were doing it in C++, it would be:

for (x = 0; x < 2*PI; x+=PI/720.0)
{
   o << sin(x) << endl;
}

Now generate that file for whatever trig functions you are using.  Then, in the beginning of your program, you load up these data values into a hash table.  From this, you can then approximate each trig function in O(1) time, which means constant time, the fastest you can get.  I picked the value PI/720.  This will generate a fairly small file (1440 lines).  You can use higher values such as PI/1000 or even PI/2000, and you will get a very close approximation.  With 2000, that will probably be as close an approximation as just sin(x) is.

The downside of using larger variable is that it uses more hard drive space (the file) and it uses more RAM (the hash table).

Edit:  If you wish to do this, but got lost in my mumbo jumbo, just say so.  I'm going to be doing this very soon for a project of my own (Pool, aka billards), so I might as well do it now.

I just did that because I know that ganesh is clueless.

Nah, you got it.  And at the time of writing, it was the only one, so it's true with respect to its timeframe.

x ≡ y out loud is "x is equivalent to y".  What does equivalent mean?  Depends on where you see it.

Two things are equivalent if they share the same property.  The one who writes it must define the property that they are talking about.

For example, if I wish to talk about even and odd integers, 2 ≡ 4 because they are both even, 3 ≡ 5 because they are both odd, but 2 <≡> 5.  (I just  made up <≡>, because <> means not equal, and it looks pretty nice).

Or we could use it in terms of names that start off with the same letter:

Ryan ≡ Robert
Mark <≡> Ben

1.  /3, *5, /3, *5...
2. +6, +8, +10, +12, ....
3. +9, +7, +5, +3, +1

Theres a command I never heard of...

Thanks ryos.

If what you copied is straight from your book, that's the weirdest defintion for integration by parts I've ever seen.

Here's one that should be more clear:

From this, it should be clear that you must first divide up your function into two parts, u and dv, then solve for v and du, and finally just plug it all back in.

Problem with this is that the side going towards earth is under much stronger gravitational forces than the side away from earth.  So you have to have an extremely massive thing on the side away.

The problem only gets worse as you get closer and closer to earth.

If your computer wasn't working, I would seriously be freaking out right now. 

% returns the remainder of division:

5 % 2 = 1
10 % 3 = 1
100 % 10 = 0
5 % 5 = 0
22 % 4 = 2

If you wish to see whether n goes into a (n divides a), you would do:

if (a % n == 0)