Mise
isle of lucy
I don't know why you would ever want or need to do that kind of calculation without a calculator.
Ask a Mathematician!
- Thread starter IdiotsOpposite
- Start date
I do almost all calculations in head or on paper. It's more fun that way. 
gangleri2001
Garbage day!!!
Have you ever heard the sentence "you can never learn too much"?
ParadigmShifter
Random Nonsense Generator
Learning when it is best to use a calculator is a valid skill as well 
EDIT: I do most calculations manually as well Keeps the Alzheimer's away...
EDIT: I do most calculations manually as well
Keeps the Alzheimer's away...
gangleri2001
Garbage day!!!
I do almost all calculations in head or on paper. It's more fun that way.
Wow! And what algorithm do you use to divide? The same as in the school?
EDIT:
EDIT: I do most calculations manually as well
Really? The same happens to languages and linguistics. It's good to have all other branches of knowledge depending on us
ParadigmShifter
Random Nonsense Generator
Long division. There's a new way to do it apparently, but Carol Vorderman (PBUH) doesn't like it since it doesn't work with fractions
Long division is something you need to relearn when you start dividing polynomials.
Long division is something you need to relearn when you start dividing polynomials.
Yes.
But it's not like I'd calculate numbers all day long.
Furthermore, mathematicians rarely have much to do with actual numbers. Usually physicists are better calculators for example. Like one dude said: "There are only three numbers for mathematicians: 0, finite and infinite".
But it's not like I'd calculate numbers all day long.
Furthermore, mathematicians rarely have much to do with actual numbers. Usually physicists are better calculators for example. Like one dude said: "There are only three numbers for mathematicians: 0, finite and infinite".
ParadigmShifter
Random Nonsense Generator
Come on, we have 1 (multiplicative inverse), e (base of natural logarithms), pi (you know what that is!) and i (principal square root of -1) as well! -1 is quite handy too, but that's just a unit in the integers multiplied by another member of the set, so it doesn't count
Well yes... It was more of a joke of course, but one that had hint of reality inside it. This guy was proving that some operator is bounded, and was therefore only interested of whether there is a constant c such that |Lx| <=c|x|, not the actual value of c. In a way that applies to e and pi too: you just notice that wow, there is really such number. How nice. And then you go on.
ParadigmShifter
Random Nonsense Generator
How could I forget the square root of 2 in my list 
Both of them!
Both of them!Hmm, here an answer to some questions I found interesting that were asked on page 11.
It is possible to have different idea of numbers. For example, an alien race might use sets instead of numbers. So, instead of saying "0 exists and every element n has a successor n +1" (this is how we define numbers), they'd say "0 exists and n is the amount of elements in the set containing all previous numbers we defined" (so 1 is the amount of elements in the set {0}, 5 is the amount of elements in the set {0,1,2,3,4}, n is the size of the set {0,1,...,n-1}.
Some further illustration of how an alien race could count this way: instead of saying "a times b is the same as b if a equals 1 and (a-1) times b + b otherwise" as we do, they would be saying "a times b is the amount of possible ways to both pick an element from a set containing a elements and pick an element from a set containing b elements" (if there are a items in box A and b items in box B, there are a*b ways to take one item out of box A and one item out of box B). They'd also have different ways of defining "a plus b" and "a to the power of b" etc.
However, I'm not sure what you mean when you ask if this would mean that they have a different set of mathematics. For reference, we mathematicians use both ways of defining numbers and there are important differences between the 2 systems.
As IdiotsOpposite said there are many. Most are just an expansion of standard logic, for exampling adding a way to express that something is possible and a way to say that something is necessary (this is called modal logic). However, there are also systems that are fundamentally different, for example intuitionism which has a different idea of when a statement is true (one of the consequences is that the statement "A is true or A is false for every claim A" cannot be proven when following the intuitionistic approach).
http://en.wikipedia.org/wiki/Intuitionism
About the multiplication thingy: there are tricks of course. It's possible to vastly improve your ability to calculate in your head, but I have to admit it's probably not for everyone.
First of all, you should view this amazing TED talk: http://www.ted.com/talks/arthur_benjamin_does_mathemagic.html
Secondly, you should research some mathematical tricks, for example: http://vedicmaths.org/Introduction/Tutorial/Tutorial.asp
That way, you'd realise that to calculate 1536 * 11 = 16896 all you need to know is that 1+5=6, 5+3=8 and 3+6=9, no further addition and multiplication is needed.
edit: http://www.vedicmaths.org/Bookstores/Manual 1 pdf.pdf apparantly includes the most elementary tricks
Lastly, you have to improve your short time memory by using mnemonics. I've heard people recommend the book "Moonwalking with Einstein" for this purpose.
It is possible to have different idea of numbers. For example, an alien race might use sets instead of numbers. So, instead of saying "0 exists and every element n has a successor n +1" (this is how we define numbers), they'd say "0 exists and n is the amount of elements in the set containing all previous numbers we defined" (so 1 is the amount of elements in the set {0}, 5 is the amount of elements in the set {0,1,2,3,4}, n is the size of the set {0,1,...,n-1}.
Some further illustration of how an alien race could count this way: instead of saying "a times b is the same as b if a equals 1 and (a-1) times b + b otherwise" as we do, they would be saying "a times b is the amount of possible ways to both pick an element from a set containing a elements and pick an element from a set containing b elements" (if there are a items in box A and b items in box B, there are a*b ways to take one item out of box A and one item out of box B). They'd also have different ways of defining "a plus b" and "a to the power of b" etc.
However, I'm not sure what you mean when you ask if this would mean that they have a different set of mathematics. For reference, we mathematicians use both ways of defining numbers and there are important differences between the 2 systems.
As IdiotsOpposite said there are many. Most are just an expansion of standard logic, for exampling adding a way to express that something is possible and a way to say that something is necessary (this is called modal logic). However, there are also systems that are fundamentally different, for example intuitionism which has a different idea of when a statement is true (one of the consequences is that the statement "A is true or A is false for every claim A" cannot be proven when following the intuitionistic approach).
http://en.wikipedia.org/wiki/Intuitionism
About the multiplication thingy: there are tricks of course. It's possible to vastly improve your ability to calculate in your head, but I have to admit it's probably not for everyone.
First of all, you should view this amazing TED talk: http://www.ted.com/talks/arthur_benjamin_does_mathemagic.html
Secondly, you should research some mathematical tricks, for example: http://vedicmaths.org/Introduction/Tutorial/Tutorial.asp
That way, you'd realise that to calculate 1536 * 11 = 16896 all you need to know is that 1+5=6, 5+3=8 and 3+6=9, no further addition and multiplication is needed.
edit: http://www.vedicmaths.org/Bookstores/Manual 1 pdf.pdf apparantly includes the most elementary tricks
Lastly, you have to improve your short time memory by using mnemonics. I've heard people recommend the book "Moonwalking with Einstein" for this purpose.
IdiotsOpposite
Boom, headshot.
Long division. There's a new way to do it apparently, but Carol Vorderman (PBUH) doesn't like it since it doesn't work with fractions
Long division is something you need to relearn when you start dividing polynomials.
You say this and yet I never relearned long division when I divided polynomials... I found it too bloody annoying.
Heathcliff
Tactician
Hi,
I don't get how single and multiple regression analysis work.
If anyone could do an analysis of some simple dataseries and explain how you do it in excel (which buttons to use)
Please also explain how you interpret the values of the different results you get.
note: I'm not interrested in the mathematical foundation, just practically how you do it and how you interpret the result.
Thanks in advance
I don't get how single and multiple regression analysis work.
If anyone could do an analysis of some simple dataseries and explain how you do it in excel (which buttons to use)
Please also explain how you interpret the values of the different results you get.
note: I'm not interrested in the mathematical foundation, just practically how you do it and how you interpret the result.
Thanks in advance
azzaman333
meh
Excel sucks for statistical analysis. Specialised statistical packages such as Minitab or SAS are the way to go.
http://www.wikihow.com/Run-Regression-Analysis-in-Microsoft-Excel
Not sure why you want to use Excel for that though.
Heathcliff
Tactician
I don't own any other program than excel.
Thanks for the link, I'll read that.
Thanks for the link, I'll read that.
ShahJahanII
Homesick Alien
Could you explain Euler's identity without calculus?
Kozmos
Jew Detective
Lattice multiplication is sexy.
ParadigmShifter
Random Nonsense Generator
Probably not... exp(z) is defined in the same way for real and complex numbers, the function which when differentiated remains the same for all z.
If you know that exp(z) is exp(x)(cos y + i sin y) it is easy though. (where z = x + iy, x,y real).
madviking
north american scum
Power series expansion of e^(a*i), where i = (-1)^(1/2) and a = arbitrary constant, is:
1 + ai + (ai)^2 / (2!)+ (ai)^3 / (3!) + ... + (ai)^n / (n!)
Knowing:
i^(1+4k) = i
i^(2+4k) = -1
i^(3+4k) = -i
i^(4+4k) = 1
Rearrange to get (1 - a^2/(2!) + a^4/(4!) - a^6/(6!) + ...) + i(a - a^3/(3!) + a^5/(5!) - a^7/(7!) + ...)
= cos(a) + isin(a)
Therefore: e^(ai) = cos(a) + isin(a)
If a = pi
Then cos(pi) + isin(pi) = -1 + 0 = -1
So e^(pi*i) + 1 = 0
