The DW Math Thread

Rammy - Dreamweaver

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 ∞

ColdAsLife - Dreamweaver

And after all this time, the thread is still alive. What have I done b:shocked

Decus - Dreamweaver

ColdAsLife - Dreamweaver wrote: »

And after all this time, the thread is still alive. What have I done b:shocked

I miss ya, bud. Thanks for letting me do it

Sneaker - Dreamweaver

wow finally a thread worth reading,

tysm Decus

b:cute

Decus - Dreamweaver

Sneaker - Dreamweaver wrote: »

wow finally a thread worth reading,

tysm Decus

b:cute

I aim to please. I hope you enjoy it.

Elanxu - Dreamweaver

Rammy - Dreamweaver wrote: »

0 / 0
∞ / ∞
∞ x 0
∞ ^ 0
∞ ^ ∞
∞ * 0
0 ∞ 0
∞ / 0
0 / ∞
0 0 0 ∞ ∞ 0 ∞ 0 0 ∞ 0 ∞ 0 ∞ ∞ ∞ 0 0 ∞ 0 ∞ ∞∞∞∞∞∞∞∞∞∞∞∞∞...

lol i have no idea what you mean by indeterminate form of limits. a nab like me should better stay off this thread and learn/observe more

nfgdjksnbsd

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

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.

for the first one, z=0.318132 - 1.33724i

i used a calculator

nfgdjksnbsd

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

Elanxu - Dreamweaver

nfgdjksnbsd wrote: »

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

then a=0.318 and b=-1.34 right?

Laterra - Dreamweaver

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

Elanxu - Dreamweaver wrote: »

then a=0.318 and b=-1.34 right?

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

Elanxu - Dreamweaver

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

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

Elanxu - Dreamweaver

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

Mitachi - Dreamweaver

yeah and we cant use x either cause it could be ex or multiply

Decus - Dreamweaver

Mitachi - Dreamweaver wrote: »

yeah and we cant use x either cause it could be ex or multiply

That's why you should always either use parentheses or dots. The way I see it, I only ever use an x for multiplication working with cross products

Mitachi - Dreamweaver

Decus - Dreamweaver wrote: »

That's why you should always either use parentheses or dots. The way I see it, I only ever use an x for multiplication working with cross products

Ahh yes forgot about the good ole dots i always use them lol how could i forgetb:surrender

Darren_Kris - Lothranis

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

Decus - Dreamweaver

New stuff! Yay!

Two are harder this time, though you can brute-force one of them (but that's not fun)


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?


1000

At their traditional poker game, Clifton and Lawrence agree on the stakes for each hand: The loser pays 1/3 of the money he has remaining, to the winner.

After a while, Lawrence gives up: "You now have exactly three times the cash I have, you've won the last few hands, and I've lost just about four bucks!"
"But you won every hand before that," Clifton replied, "in fact you've won the same
number of hands I have!"

How many hands were played, how much money did they each start with, and how much do they have now?


0100

The numbers 480608, 508811, and 723217, when divided by a certain natural number greater than 1 , all yield the same remainder. What is that divisor and that remainder?

WackyTaffy - Dreamweaver

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

Decus - Dreamweaver

Show work, Wacfy! b:thanks

krisnda

Decus - Dreamweaver wrote: »

Show work, Wacfy! b:thanks

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

WackyTaffy - Dreamweaver

Decus - Dreamweaver wrote: »

Show work, Wacfy! b:thanks

The difference between the numbers GCD is 79 thus they'd all have the same remainder if divided by 79.

Decus - Dreamweaver

WackyTaffy - Dreamweaver wrote: »

The difference between the numbers GCD is 79 thus they'd all have the same remainder if divided by 79.

So you said you were going to do the other problems. Why haven't you? b:angry

Xarathox - Dreamweaver

I see you're still at it, Dec.

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

Yorkin - Dreamweaver

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)

WackyTaffy - Dreamweaver

Decus - Dreamweaver wrote: »

So you said you were going to do the other problems. Why haven't you? b:angry

I never figured out how to do the first one without brute force. And the last one looked too easy so I didn't bother.

WackyTaffy - Dreamweaver

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*

Sneaker - Dreamweaver

not sure where to post this decus (
but i cant find my toons nothing shows up on log in screen (

missing my 105 wiz my 104 bm my 104 sin (

sorry for posting here dont know how to start a new thread (