who here loves math?

[phillip1882]

i can't get enough of it! i frequently visit several math forums just to see if it's a problem i can solve.
I've taken courses from calculus 3 to complex analysis. my personal favorite subject is number theory however.
something about having such a strong relationship between all integers makes me warm and fuzzy inside.
anyone else here?

[xSilverPhinx]

 

Not among one of my favourite subjects, but I don't hate it. I'm really not any good with numbers though.

[Asmodean]

I like math, but not enough to frequently seek out mathematical problems that need my computing power.

[Stevil]

I was always good at maths. I like to solve puzzles.
If I had my wits about me when I was younger I would have gotten into serious cosmology.

[Tank]

Okay, I have a problem for you lot to solve.

I want you to take the standard Drake Equation and have it produce additional outputs, rather than just the one it does at the moment. The existing output remains untouched. The additional output(s) are as follows, the average distance between the civilizations. The probability of one of those civilisations being within 10, 20, 40, 80 and 160 light years of Earth.

Please state your model of the Milkey Way i.e. diameter, thickness, star density variation etc.

There is you mission, should you choose to accept it 
 

[Asmodean]

There is you mission, should you choose to accept it  

Too much trouble  

But I has one too: x^x=27. Without guessing the answer, can this equasion be solved?

[Stevil]

There is you mission, should you choose to accept it  

Too much trouble  

But I has one too: x^x=27. Without guessing the answer, can this equasion be solved?

very easy, I didn't even have to think about it.

How about he^2 = she

[Asmodean]

There is you mission, should you choose to accept it  

Too much trouble  

But I has one too: x^x=27. Without guessing the answer, can this equasion be solved?

very easy, I didn't even have to think about it.

How about he^2 = she

Try again 

[Stevil]

There is you mission, should you choose to accept it 

Too much trouble 

But I has one too: x^x=27. Without guessing the answer, can this equasion be solved?

very easy, I didn't even have to think about it.

How about he^2 = she

Try again 

I know the answer. it wasn't rocket science. But are you wanting me to develop a mathematical equation that will solve all of these types of questions?
e.g.
x^x=1
x^x=4
x^x=27
x^x=256
x^x=3125

My method was simple, I did however make an assumption that you were using integers. (well actually, not really an assumption. You can't have two decimal numbers multiply against each other and get a whole number, you always end up with a decimal)
I knew x had to be small, 4 was definitely too big. I knew this through experience. 1 and 2 were too small. I knew this too. Actually, I simply knew the answer was 3. It wasn't a guess. I had seen 3 x 9 before and I had recognized that 9 is 3 x 3. So I had an understanding, instant recall that 3^3 = 27.
Given that 27 I an odd number then x must have been an odd number too, otherwise the result would have been an even number. Given that 27 is an odd positive number it means that x must have been a positive number otherwise the result would have been a negative number. There was only ever one option on what x could be. It wasn't a guess.

Another way of working it out is to find out what whole numbers can 27 be divided by without giving a remainder. 1 is silly to divide by. 2 and 4 don't work because 27 is an odd number. Going higher than 4 is silly because I already know that 4^4 is > 27. 3 fits like a glove. 27/3 is 9 is 3^2, so is perfect.

I wouldn't consider it a guess, although I did not use a mathematical formula to work out the answer, I did however use my experience, and common sense and quick elimination to find the answer which I could validate by running it through your formula. I couldn't apply my method to formula where x is not an integer number though.

[Asmodean]

My method was simple, I did however make an assumption that you were using integers. (well actually, not really an assumption. You can't have two decimal numbers multiply against each other and get a whole number, you always end up with a decimal)
I knew x had to be small, 4 was definitely too big. I knew this through experience. 1 and 2 were too small. I knew this too. Actually, I simply knew the answer was 3. It wasn't a guess. I had seen 3 x 9 before and I had recognized that 9 is 3 x 3. So I had an understanding, instant recall that 3^3 = 27.
Given that 27 I an odd number then x must have been an odd number too, otherwise the result would have been an even number. Given that 27 is an odd positive number it means that x must have been a positive number otherwise the result would have been a negative number. There was only ever one option on what x could be. It wasn't a guess.

Another way of working it out is to find out what whole numbers can 27 be divided by without giving a remainder. 1 is silly to divide by. 2 and 4 don't work because 27 is an odd number. Going higher than 4 is silly because I already know that 4^4 is > 27. 3 fits like a glove. 27/3 is 9 is 3^2, so is perfect.

I wouldn't consider it a guess, although I did not use a mathematical formula to work out the answer, I did however use my experience, and common sense and quick elimination to find the answer which I could validate by running it through your formula. I couldn't apply my method to formula where x is not an integer number though.

I did use a simple equasion, but yes, I am actually looking for a solution to x^x=a, where a is defined within R

[iSok]

yep, I like math's, especially integration techniques.

Solving the surface or volume of objects like apples or pears with the x,y and z-axis in a graphic way
is a nice exercise.

[xSilverPhinx]

[phillip1882]

x^x = a take the log of both sides.
x*log x = log a
x*log x -log a = 0
from there, the best way would be to use newton's method.
let x0 be your initail guess.
then... x1 = x0 -(x0*log x0 -log a)/(log x0 +1)
keep on plugging in the answer untill you get a sufficently close result.
technically this is a guess and check method, but it's an algorithm that will work with any number, a, wether integer or decimal.
edited: i like the above equatino better, it converges much much faster.

[Asmodean]

x^x = a
perhaps the best way would be to use newton's method.
let x0 be your initail guess.
then... x1 = x0 -(x0^x0 -a)/(x0^x0 *ln x0 +x0^x0)
keep on plugging in the answer untill you get a sufficently close result.
technically this is a guess and check method, but it's an algorithm that will work with any number, a, wether integer or decimal.

Nice! Except for the guesswork needed, I like it.

[phillip1882]

Okay, I have a problem for you lot to solve.

I want you to take the standard Drake Equation and have it produce additional outputs, rather than just the one it does at the moment. The existing output remains untouched. The additional output(s) are as follows, the average distance between the civilizations. The probability of one of those civilisations being within 10, 20, 40, 80 and 160 light years of Earth.

Please state your model of the Milkey Way i.e. diameter, thickness, star density variation etc.

sure, that's simple enough
N = star dencity* distance^3 /number that have planets /number of planets that support life / number of life planets that support intellegent life

there are roughly 400,000,000,000 stars, and the milky way is roughly 100,000 light years across, and 1,000 light years thick.
so 400 stars per light year, aproximately, is our star dencity. the rest is up for grabs.