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#1 2006-07-12 01:48:33

RauLiTo
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for a challenge

hello everybody ... in fact i dont know how to start rolleyes
as you all know i am an arabian person ... i decided to make a topic in an arabic forum about math '' enjoy the math '' and i need some questions , tricks , math magic ... to put there big_smile so i need your help lovely members !!! who can give me links ? who can put questions not difficult ones but cool and nice question which make you interested in math cool!!!

thanks again


ImPo$$!BLe = NoTH!nG
Go DowN DeeP iNTo aNyTHinG U WiLL FinD MaTHeMaTiCs ...
 

#2 2006-07-13 08:07:45

RauLiTo
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Re: for a challenge

views : 16
replies : 0

...


ImPo$$!BLe = NoTH!nG
Go DowN DeeP iNTo aNyTHinG U WiLL FinD MaTHeMaTiCs ...
 

#3 2006-07-13 10:06:55

MathsIsFun
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Re: for a challenge


"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman
 

#4 2006-07-13 10:16:22

Zhylliolom
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Re: for a challenge

You could always use my historical proof that ii = e/2 ∈ R. I did it for a talent show and won, so I guess it could be used as good mathematical magic. It was a great day for mathematics. If you would like me to write out the problem, just ask.

 

#5 2006-07-13 10:39:22

Ricky
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Re: for a challenge

Zhylliolom, I get

, where n is an integer.  Could I see your work?

Interesting question though, I have to say.


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."
 

#6 2006-07-13 14:14:45

Zhylliolom
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Re: for a challenge

Hopefully we all know Euler's formula:



For this problem, let 0 ≤ θ ≤ 2π. Now ask yourself, for what value of θ will eiθ = i? Why yes, it's π/2!



So now we know that eiπ/2 = i. Now let's take it to the next level:



Now if we remove the restriction 0 ≤ θ ≤ 2π, then we get the general solution



where nZ. I'm not sure why you have just n and not 2n, Ricky. Odd values of n in your solution would give



Now take a simple case of some 0 ≤ θ ≤ 2π that could give eiθ = -i. 3π/2 is our value. Then



So, given the same interval 0 ≤ θ ≤ 2π, (-i)i ≠ ii, so I will conclude that your n should be 2n.

Last edited by Zhylliolom (2006-07-14 08:07:19)

 

#7 2006-07-13 23:37:09

Ricky
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Re: for a challenge

sin(3pi/2) doesn't equal 1, does it...


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."
 

#8 2006-07-14 00:14:32

George,Y
Super Member

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Re: for a challenge

Hey you guys are discussing something too difficult here!
Though I read it too.

Mathsisfun is good at fun maths and perhaps he is the best person you can consult to.


X'(y-Xβ)=0
 

#9 2006-07-14 05:16:52

RauLiTo
Full Member

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Re: for a challenge

thank you alot guys
mr.mathisfun ... i think u have got alot of these topics ... i really like them
can i see more ?


ImPo$$!BLe = NoTH!nG
Go DowN DeeP iNTo aNyTHinG U WiLL FinD MaTHeMaTiCs ...
 

#10 2006-08-16 19:55:30

Devantè
Real Member

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Re: for a challenge

Or you could just try looking in the Exercises forum. We're adding some every now and then, so check back regularly! smile

Last edited by Devantè (2009-02-15 10:18:30)

 

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