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Maths Help (for the know alls :P)
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John Anderton
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RE: Maths Help (for the know alls :P)
quote:
Originally posted by CookieRevised
a massive typo if I made add :D
;)
Clarified everything over msn ... whole concept so no harm done :dodgyomar:

quote:
Originally posted by CookieRevised
(and what you mean are complex numbers or imaginary numbers)
(-2)^1/2 is a complex number.
Square of 2 and - 2 is 4 .... so there isnt really a square root of -2 or any other -ve no .... a calci will show it as E (infinity :|)
So we write it as 2i where i=(-1)^1/2

So acutally 2i+3 means (-2)^1/2+3 :D
Thats a complex / imaginary number.
Was that a question or did i say that somewhere ?
I dont think i did :S ^o) So that must be a question :)

[b]EDIT :-[/b]
quote:
Originally posted by Hobbes
Cookie is right. I just started studying this equation with 2 answers.

any problem that is like

ax^2 + bx + c  = 0

you can find two answers for. There is more then one equation to do so. But the one Cookie posted is a better one for this equation.
:S WTF ... both of us said the same thing ... mine had a typo which was corrected :S
He said it in detail ... i just multiplied and showed it cause i was in a hurry ... i didnt show the answer as it wasnt a perfect whole number :P I told all this on msn :P

This post was edited on 03-31-2005 at 03:02 PM by John Anderton.
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03-31-2005 02:59 PM
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CookieRevised
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RE: Maths Help (for the know alls :P)
quote:
Originally posted by John Anderton
and i told u the eqn doesnt have any real roots only such odd irrational ... 8-) (4go what they r called; they aint irrational) numbers.
so that clearly states: "tell me what the heck they were called" :D

So to sum up and to explain properly:

rational numbers:
numbers having a terminating decimal point.
in other words, numbers which can be written as a fraction.
eg: 9/4 = 2.25

irrational numbers:
numbers which don't have a terminating decimal point.
in other words, numbers which can not be written as a fraction.
eg: pi = 3.14159265359....

complex/imaginary numbers:
Contrary to what some people might tell you, 'imaginary' numbers are not numbers that only exist in the brains of weird people. Or maybe they are :D; all numbers in math are "imaginary" in the sense that you can't touch them or experience them directly. But this is not what people mean when they talk about 'imaginary' numbers. Imaginary numbers are numbers that can be written as a real number times i, where i=sqr(-1)
It was invented to be able to calculate with numbers which can't exist.

eg: 5i = 5 * sqr(-1)

complex numbers are numbers like 3 + 5i; they are a real number plus an imaginary number.

-----------

quote:
Originally posted by John Anderton
He said it in detail ... i just multiplied and showed it cause i was in a hurry ... i didnt show the answer as it wasnt a perfect whole number I told all this on msn
tbh, I don't care what is said on msn. This is a forum, everybody can read this, so either take your time and explain it correctly and detailed (especially in math cases) or don't :XP:

(this whole thread could be as small as a post or 3 when things were explained correctly and detailed in the first place)

;)

This post was edited on 03-31-2005 at 03:20 PM by CookieRevised.
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03-31-2005 03:02 PM
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John Anderton
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RE: Maths Help (for the know alls :P)
quote:
Originally posted by CookieRevised
(4go what they r called; they aint irrational) numbers.
What i meant was they aint whole numbers .....
An irrational number is a number that cannot be expressed as p/q (a fraction) for any integers p and q. Irrational numbers have decimal expansions that neither terminate nor become periodic.
Imaginary numbers = complex numbers = definition already explained :)
Btw ... i knew the definition of irrational numbers but i was too lazy to type it so got it here :P

EDIT:-
quote:
Originally posted by CookieRevised
tbh, I don't care what is said on msn. This is a forum, everybody can read this, so either explain it correctly (especially in such math cases) or don't

(this whole thread could be as small as a post or 3 when things were explained correctly and detailed in the first place)

Sorry :'(
* John Anderton cries and runs as he is no match for the Cookie's reply to series :)
* John Anderton is a cookie fan btw :)

quote:
Originally posted by CookieRevised

But this is not what people mean when they talk about imaginary numbers. Imaginary numbers are numbers that can be written as a real number times i, where i=sqr(-2)
:S i = sqr (-1) .....
Thats what i learnt last year and i am sure it is correct
Sqr of a -ve no is an imaginary number and its combination with a real number is a complex number :)

EDIT 2:-
Ok here is that sqr(-1) i was saying ....
Wolfram research - MathWorld - Imaginary Numbers

This post was edited on 03-31-2005 at 03:14 PM by John Anderton.
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03-31-2005 03:08 PM
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CookieRevised
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RE: Maths Help (for the know alls :P)
I already posted the proper and simple definitions of the namings of numbers 15 minutes ago... (and moved them to my previous post)... So reload the thread before answering...
quote:
Originally posted by John Anderton
i = sqr (-1) .....
PS: yeah, my typo :D, it is -1

because the confussing thread I got confussed too :P but -1 or -2 doesn't matter in the OP's question at all.....



(lol @ my own reply, always something to say back... /me slaps himself...... I'll sush now)

This post was edited on 03-31-2005 at 03:18 PM by CookieRevised.
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RE: Maths Help (for the know alls :P)
oh ok i think i see what you are doing a bit more i have used that equation. and dont worry about the misunderstanding Ash i do it all the time :)
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John Anderton
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RE: Maths Help (for the know alls :P)
Ok i have a maths problem .... Its a sum of probability .... I got the answer but probability is a bit dodgy and its so easy to make a mistake .... Can some one also try this problem

There is a dictionary containing only 5 letter words and there are only 4 letters viz. a, b, c, d, e.
The 1st word is aaaaa the second is aaaab the third is aaaac and so on.
Now we need to find the position of the word "adbca"

The answer i worked out is 216 .... can someone back up my answer ...... none of my friends have solved it ..... plus group solving aint allowed (if caught then im screwed :dodgy:)
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RE: Maths Help (for the know alls :P)
a   b   c   d   e
0   1   2   3   4

adbca
03120

(0*5^4)+(3*5^3)+(1*5^2)+(2*5^1)+(0*5^0)
(0*625)+(3*125)+(1*25)+(2*5)+(0*1)
0+375+25+10+0
410

410/5 82 R 0
82/5 16 R 2
16/5 3 R 1
3/5 0 R 3

3120

This post was edited on 04-01-2005 at 09:04 PM by Banks.
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RE: Maths Help (for the know alls :P)
quote:
Originally posted by CookieRevised
Well, I hate to repeat stuff, but maybe this looks a bit less like 'Chinees' (no punn intended to the Chinees people :D):

Given:
  x + reciprocal(x) = 30
  (where reciprocal(x) = 1/x)

Search: x

Solution:
  x + reciprocal(x) = 30
  => x + 1/x = 30
  => (x + 1/x) * x = 30 * x
  => x² + 1 = 30 * x
  => x² - 30x + 1 = 0

this is a normal quadratic equation in the form of:
  ax² + bx + c = 0     this formula is basic knowledge!
so:
  a = 1   ,   b = -30   ,   c = 1

and to find x:

[Image: eq0288M.gif]     this formula is basic knowledge!

(mind the ± in front of the square root, this means there are two possible solutions)

^^ now solve that normal equation:

the discriminate (b² - 4ac):
  => b² - 4ac = (-30)² - 4 * 1 * 1
  => b² - 4ac = 896
the discriminate is positive so we will get a 'real' number. (dunno if you already studied 'complex' numbers or also called 'imaginary' numbers. But if the discriminate is negative you will take the square root out of a negative number, resulting in a complex (or imaginary) solution for x.

now to go further with calculating x:
  => x = [ 30 ± sqr(896) ] / 2
  => x = 30 / 2 ± sqr(896) / 2
  => x = 15 ± sqr(896) / sqr(4)
  => x = 15 ± sqr(896 / 4)
  => x = 15 ± sqr(224)

solution 1:  x = 15 + sqr(224)
solution 2:  x = 15 - sqr(224)


-------

PS: John Anderton, you made a big mistake; you forgot the square root in your formula!
"Use [-B +- (B^2 - 4 AC)^1/2] / 2A"
oh thakns cookie, yu are so brilliant :O

now help me please with this problem:

[Image: attachment.php?pid=426407]

(its not hard, but i would love to see how does he manage to write the solution here :lol:)

.jpg File Attachment: P3-b C3 2003-2.jpg (1.86 KB)
This file has been downloaded 269 time(s).

This post was edited on 04-01-2005 at 09:16 PM by Chrono.
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RE: Maths Help (for the know alls :P)
quote:
Originally posted by John Anderton
Ok i have a maths problem .... Its a sum of probability .... I got the answer but probability is a bit dodgy and its so easy to make a mistake .... Can some one also try this problem

There is a dictionary containing only 5 letter words and there are only 4 letters viz. a, b, c, d, e. The 1st word is aaaaa the second is aaaab the third is aaaac and so on. Now we need to find the position of the word "adbca"
as far as I'm understanding your question this has nothing to do with probability at all. This is pure base conversion from base 5 to base 10.
quote:
Originally posted by John Anderton
(...) and there are only 4 letters viz. a, b, c, d, e.
a, b, c, d, e are 5 letters... and what is "viz"?
quote:
Originally posted by John Anderton
The answer i worked out is 216 .... can someone back up my answer
Your own answer is also wrong.


Let me explain:

Given:
  5 letters: a b c d e
  dictionary:
    aaaaa
    aaaab
    aaaac
    aaaad
    aaaae
    aaaba
    aaabb
    aaabc
    (...)

Question: what position is "adbca" in?

Solution:

Look at the dictionary. This is just a simple count up, but instead of 10 numbers, they've used 5 letters. So replace every letter with a number: a = 0    b = 1   c = 2   d = 3   e = 4:
  aaaaa     00000
  aaaab     00001
  aaaac     00002
  aaaad     00003
  aaaae     00004
  aaaba     00010 note the "jump" from 04 to 10
  aaabb     00011
  aaabc     00012
  (...)       (...)

5 different letters numbers means this is base 5.
base 10 is what we know as our normal number system: When we come to the number after 9, we add a digit and start counting again from 0 for the least significant digit: 00 01 02 (...) 08 09 10 11 12 etc...

So, add an index to the dictionary and you'll see that you only have to convert the base 5 number to a base 10 number to know it's value aka position (first position is called position 0):

  dictionary | base 5 number | base 10 number (aka position)
  aaaaa          00000               0
  aaaab          00001               1
  aaaac          00002               2
  aaaad          00003               3
  aaaae          00004               4
  aaaba          00010               5
  aaabb          00011               6
  aaabc          00012               7
  (...)            (...)                (...)


Thus adbca = 03120

So all this to explain the simple following formula:

To calculate the equivalent base ten number:
= 0 * 5^4   +   3 * 5^3   +   1 * 5^2   +   2 * 5^1   +   0 * 5^0
= 410

This is not yet the answer! Now it is time to read the question again. They asked what position the number is in. And positions useually start at 1. To make no confusion say this also in your answer ;):

= 411 (with first position being 1)


-------------------------------


quote:
Originally posted by Chrono
oh thakns cookie, yu are so brilliant :O
now help me please with this problem:

[Image: attachment.php?pid=426407]

(its not hard, but i would love to see how does he manage to write the solution here :lol:)
:refuck:

This post was edited on 04-01-2005 at 11:08 PM by CookieRevised.
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04-01-2005 09:48 PM
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John Anderton
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RE: Maths Help (for the know alls :P)
sorry that should have been 4 letters .... a,b,c,d and e
viz. = namely .... :undecided:
Its not probability .... its using permutation and combination :D
quote:
Originally posted by CookieRevised
To calculate the equivalent base ten number:
= 0 * 5^4   +   3 * 5^3   +   1 * 5^2   +   2 * 5^1   +   0 * 5^0
= 410
:undecided:
quote:
Originally posted by John Anderton
The answer i worked out is 216
I dunno that formula that u r using .... can u do the same question with permutation and combination ..... i have used a simple logic .... i hope u understand ... her it goes.
1st word aaaaa
2nd         aaaab
3             aaaac
4             aaaad
5             aaaba

Ok so 4 changes r possible in the 1st letter (from the right) (like units place in numbers) ....
5th change will make the 2nd letter from the right change once.
So When the 1st letter changes 25 times (5*5) then the 3rd letter from the right will change.
When the 1st letter changes 125 times then the 4th letter changes once .....
the 4th letter from the right changes 3 times (a -> b -> c ->d) thus ..... 125*3 = 375

Now the 3rd letter from the right needs to change 1 time (a -> b) so ..... so the 1st letter needs to change 25 times .... {Total of 375 + 25 = 400 words have passed}

Now the 2nd letter from the left needs to change 2 times (a -> b -> c) so the 1st letter needs to change 5 * 2 = 10 times {so a total of 410 words have passed and we have reached adbca}

So shouldnt that be the answer :S
... :O our ans are almost the same but urs is 411 :S.
Where did i go wrong ^o)
I started my counting from one :(

The earlier ans i got was different as i had taken the factor as 4 ie. for every 4 changes of the 1st letter the next one changes once.
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04-03-2005 11:50 AM
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