Math Is Fun Forum

this is weird...

I clicked on help/about dev-C++ in the IDE, here's what it says at the top:

GNU GENERAL PUBLIC LICENSE

                       Version 2, June 1991

 Copyright (C) 1989, 1991 Free Software Foundation, Inc.
                          675 Mass Ave, Cambridge, MA 02139, USA
 Everyone is permitted to copy and distribute verbatim copies
 of this license document, but changing it is not allowed.

the license is apparently about 17 years old... but I seriously doubt they would put a 1991 compiler on a website still in use today. (or at least as of august 2007)

It may be just an outdated compiler, but even if it was, it still understands the declaration:
int (Dog*p)(long, double);
it just doesn't let me do anything with p after that. At least i haven't found a way yet.

anyway, Ricky. This isn't a major issue, first because i normally use my borland compiler, and second because i don't really intend to use pointers to functions anyway as it seems to be a discouraged technique. The point is that I know the correct way to do it when I need it, and to recognize when other people use it.

SO! which is the correct method? To use p = dog::bark; or p = &dog::bark; ?

mm... i think you read my post too fast, Ricky! I said I tried that and it didn't work.

But at the moment I'm about to leave for school and haven't much time so.. perhaps I read YOUR post too fast. I'll look more carefully later.

oh THATS why thats happening! Thanks, that was really annoying.

okay, we got up to here

doing the multiplications we obtain

note the h's above and below will cancel so you get

now the quotient rule for limits says the limit of a quotient (or fraction) is equal to the limit of the numerator divided by the limit of the denominator if and only if the denominator is nonzero. I think you will now find that the limit of the denominator is not zero. This means you can basically replace h with zero and evaluate it directly. So take the above limit, replace h with zero, and simplify!

While we're on the subject, whats the easiest way to find out how to hit a target thats not level with the firing height?

I use y = x tan(θ) + g/(2u^2) x^2 sec^2 θ, and substitute tan^2 + 1 for sec^. You then have a quadratic  equation involving tan θ, and you can solve. It works but its not the fastest thing in the world.

i agree. unsolvable!
I'd say 'reaches the insect' is where that problem gets unclear. APPARENTLY they meant vertical height, but i never would have guessed if i didn't know it was out of range.

unless I'm mistaken about what those variables stand for,  that gives you θ =  sin^-1 ( 1.538) 

first i'm using 1 and 0 to represent the boolean values true and false respectively, thats just notation

the above are all postulates, I think there are probably some axioms that i'm allowed to use. For instance, a boolean value is either true or false. If it isn't false, its true, and vice versa.

when working with variables, the actual values are not given, however you can be sure that either A = B or A ≠ B no matter what values for A and B are being dealt with. If you can show that it works in both cases, then you have proved it in the general case.

But yeah, the question remains, is it okay to say A ≠ B ⇒ A = B'

I think if we are working with boolean values, we have to be able to assume that much, right?

Wow! look at the response this threads getting!

by 'regular binary' i meant place value, as in 1010 = 1*2^3 + 0*2^2 + 1*2^1 + 0*^0 = 12d

What i really think my book is doing is using the ordinary binary system for positive numbers (as in, 1010 = 12d) and using two's complement for negative numbers. I was actually able to prove that there is no overlap so there is no ambiguity between the two.

we are using an online text for this course, and its accesible from anywhere:
http://webster.cs.ucr.edu/AoA/DOS/pdf/0_AoAPDF.html

if you click on chapter 1 and go to the pages 13 of the pdf (not of the actual page numbers) you can find the part on signed numbers and on the next page they do positive to negative conversions.

The reason i think they're using 'place value binary' for positive numbers and 'two's complement' for negatives is because supposedly, you can calculate a numbers negative by using NOT and adding 1. They begin with 5 written as 0101 and then write -5 as 1011. 1011 = 11d which by my understanding, is how we write positive 5 in two's complement form.

also, that pdf never actually told me what two's complement meant and i'm currently under the impression that if x is the value you want to represent (who's binary form has n bits) then
x + complement = 2^n
and we write x in two's complement form by writing 'complement' in binary form.

Thats my understanding of the subject. Care to point out where i'm going wrong, if anywhere?

Hold it! I DON'T want to take that view! In fact, stating that a 2 dimensional array is an array of arrays always made me uncomfortable based on syntax and how the compiler treats it.