The DW Math Thread

Elanxu - Dreamweaver wrote: »

eh, anything divided by 0 is infinity, so infinity/0 is something along the lines of infinity, or infitity squared, which is basically infinity, because infinity is infinity theres no way of changing the fact that infinity is infinity. i just used infinity in the last sentence like 15 times.

infinity.

Not anything divided by 0 is infinity. Or anything infinity at a power of something is infinity.

Don't forget about the indeterminate forms of limits:

0 / 0
∞ / ∞
∞ x 0
∞ ^ 0

I think they're more but really forgot them. But in the case of ∞ / 0 it is basically ∞ x 1 / 0 which is ∞ x ∞. You can't get an accurate answer unless you calculate the limit of it like:

lim when n -> ∞ of n / 0. In this case you can affirm it is ∞

Solve:

ln(z) = z

in terms of a and b where z = a + bi

and

cosh(x) = cos(x)
(x =/= 0)
find x.

Elanxu - Dreamweaver wrote: »

for the first one, z=0.318132 - 1.33724i

i used a calculator

i said in terms of a and b, didnt want a numerical value :P

Decus - Dreamweaver wrote: »

Bonus!
(You get cookies for solving it. I stumbled on this problem and could not RESIST posting it!)

A calculator is broken so that the only keys that still work are the sin, cos, tan, arcsin, arccos, and arctan buttons. The display initially shows 0. Given any positive rational number q, show that pressing some finite sequence of buttons will yield q. Assume that the calculator does real number calculations with infinite precision. All functions are in terms of radians.

I was on an overnight train last night and couldn't sleep, and I found myself pondering over this question instead...

Pressing any finite sequence of buttons on a calculator (all possible combinations) yields a finite set of numbers. The rational numbers Q form a countable infinite set. So for any calculater generated set there exist elements in Q that are not in that set (i.e. you would have to press more buttons to get it). So no, you cannot get any q by pressing a finite sequence of buttons.

nfgdjksnbsd wrote: »

as in a = (e^b)(b-a) b= (e^a)(a-b)

==> z = (e^b)(b-a) + i(e^a)(a-b)


that is not the answer btw

b:surrender

Notso - Raging Tide wrote: »

I am good at maths but I can't read it unless it's on paper! None of this ^ business for powers and * for multiplyb:surrenderb:infuriatedb:avoid

*busts out my scanner* pdf format of RL paper FTW

nfgdjksnbsd wrote: »

Solve:

ln(z) = z

in terms of a and b where z = a + bi

and

cosh(x) = cos(x)
(x =/= 0)
find x.

how about this,
ln(z) = z
e^[ln(z)] = e^z
z = e^z
z = e^(a + ib)
z = (e^a)(e^ib)
z = (e^a)(cos b + isin b)
==> a = (e^a)(cos b) , b = (e^a)(sin b)

I couldn't do the other 1 b:cry

the last one is divide by 79 and get a remainder of 51. And no I didn't use brute force...

Decus - Dreamweaver wrote: »

Show work, Wacfy! b:thanks

Lol I found the answer for this one, because i used brute forceb:shutup

Xarathox - Dreamweaver wrote: »

I see you're still at it, Dec.

p.s. you're still an old man.

Persistence is a mathematician's greatest virtue.

(Ofc, he's old... he's a Elf, in Heaven Tear's name xD)

So, decus said to go ahead and just solve it...


0000

Say you painted a block of wood that's a x b x c cm (where a <or= b <or= c and a,b,c are all whole numbers). After that, you then cut it up into 1 cm cubes and found that half of the cubes have paint on them and half do not. What are all possible values of a, b, and c?



so seeing as how no one else is answering this. (a * b * c) = the number of cubes. that number divided by 2 equals the number of painted and not painted cubes. The number of not painted cubes is equal to (a - 2) * (b - 2) * (c -2). Because if you subtract 2 you get the inside area of the block. So,
(a * b * c) / 2 = (a-2)*(b-2)*(c-2)

From here it is a simple matter of brute force to solve it. Or you could try to solve it mathematically and fill out 3 pages of math instead. It's easy enough to see that a can't be 1 or 2 because there is no inner area of cubes that don't get painted. The inner area when a is 3 is always going to be smaller than the out area so that is also impossible. So, we'll start with a is 4. Like wise we can see that a can't be 10 or greater because the smallest the inner area can be at the time will be great than the out area:
a = 10, b = 10, c = 10.
(a * b * c) / 2 = (a-2)*(b-2)*(c-2)
500 < 512
Any increases in b or c will only make the inner area even larger than the outer area.

Anyways on to the code for the brute force phase:
long a = 3, b = 3, c = 3;
//Randomly used 1000 as the cut cause I don't want it to take forever.
for(a = 3; a < 1000; a++)
for(b = a; b < 1000; b++)
for( c = b; c < 1000; c++)
{
if( (a*b*c) == ((a-2)*(b-2)*(c-2)*2))
cout << a << ":" << b << ":" << c << endl;
}
Answers:
a b c
5 13 132
5 14 72
5 15 52
5 16 42
5 17 36
5 18 32
5 20 27
5 22 24
6 9 56
6 10 32
6 11 24
6 12 20
6 14 16
7 7 100
7 8 30
7 9 20
7 10 16
8 8 18
8 9 14
8 10 12


yea that's a lot of answers
*edited apparently it removes spacing from code ... oh well*