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this is good! :-D
hey come on, zach week is over...
This is utterly genius: http://www.funnies.com/proof.htm
is that the same Ricky who hangs out here? ![]()
I didn't. Never heard this before.
So what about your math background? Where did you study, got any goals for the future concerning math?
I'm in heaven! :-D I love the inscribed rectangles! I shall approximate the volume before devouring!
Time to InteGRRRRRREEEAT!
Yay! I'm a tiger! Give me a giant steak or I'll eat you!
I was born in 1986. With my luck thats probly year of the jackass.
Not suprising...
siva.eas, I doubt we're gonna make any never before made discoveries here. We're just having fun.
Well just wrote this program to use numerical integration of the area of a unit circle in the first quadrant, then multiply that by 4 to find pi. It works but using super large values of x seems to decrease the level of accuracy rather then increase it. Probably 'cause the values aren't being stored to enough decimal places and loosing their accuracy.
#include <iostream>
#include <math.h>
float f(float x);
int main()
{
float x;
float n = 200000;
for (int i = 0; i < n; i++)
{
x += 1/n * f(i/n);
}
std::cout << "pi = " << 4 * x << "\n";
return 0;
}
float f(float x)
{
if ((1 - x*x) < 0) { std::cout << "error " << x << "\n"; return 0; }
return sqrt(1 - x*x);
}In my oppinion it might be more interesting to write an algorthim to use numerical integretion to find pi by finding twice the area under the curve x^2 + y^2 = 1 and above the x axis. (the area of a cirle of radius 1 is pi.)
What the heck is gradients? Oddly my mathbook never mentioned that term.
lol. Good one. Yeah only a mathematician would understand. The absence of + C is probably the most common mistake made in calculus.
What exactly are we trying to find here? All I see is desription, no question.
Math is power. At first it can be tedious but advanced math shows you how to levitate. Its awsome!
Yeah, just had to multiply above and below by cos x and we would have had it.
Wish I could ask my dad for help. He went as far as calculus 2 in college but never really had to use it, so he forgot most of it. What a pity...
(bows) :-D
What about girls who are bad at maths? What do they get to help them pay attention?
Put a dollar sign in front of every number. That should do the trick.
:lol::lol:
"today we are going to study the area under a curve...."
If this site had a "this is not cool" forum I would have posted this there. Have you heard about this so called "bikini calculus" where they get these girls in bikinis to instruct guys in math so they pay attention? Do they really think guys will be looking at the chalk board?
Those who choose philandering over math should suffer the consequences!
Just read about this today. It stated that every continuous function has an antiderivative. My book then trailed off on something about integreting with a constant as the lower limit of integration. In the end I couldn't see there point, and they then simply said "we state the theorem without proof:"
I'm not sure what the point was. I reread it 4 times and couldn't quite see what conclusion it made. Now obviously, every continuous function encloses a certain area above and below the x axis, so some integral must exist.
But then it said some functions can't be integrated, like e^-2x dx. But the fundamental theorem of calculus garentees e^-2x dx has an antiderivative. Seems like a self defeating statement. If there is no function that can be differentiated to get e^-2x dx, then how can it have an antiderivative?
ricky is right. Thats the same thing I read. Like I said, by convention, √x means the positive square root. Where as if you were to verbally ask for the square root, it could mean the positive or negative root.
But of course, if you have the expression x^2 = 4, you can take the square root of both sides but you must consider the postive and negative square root. x = +- √4 note! If √4 represented the positive AND negative square root of 4, then the +- signs would be uneccessary and redundant.
1) Using the law of sines gives;
sinA/a = sinB/b; since nothing varies but b this becomes;
A = arcsin[4sin34/b]
a) A = arcsin[4sin34/5] ≈ 26.574°
The only other possibilities are (180 - A) = 153.426
....but 153.426 + 43 > 180, so this would not be a triangle.
b) A = arcsin[4sin34/2] = which does not have a solution
c) A = arcsin[4sin34/2.237] ≈ 89.181 and 90.819
Therefore two triangles can be formed.
d) A = arcsin[4sin34/3] ≈ 48.21 and 131.79
Since 131.79 + 43 < 180 both angles above are possible.
So for this part, data c and d would produce two triangles.
I don't understand this at all, what does it mean by two triangles?
Also, forgive me if I'm wrong but you wrote: A = arcsin[4sin34/5] ≈ 26.574° then The only other possibilities are (180 - A) = 153.426
Uh.. thats two angles that sum up to 180. Don't triangles usually have three angles? I'm missing something here. Seems to me, the remaining angle should be 180 - A - 34.