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#1 2006-07-11 03:48:33

RauLiTo
Member
Registered: 2006-01-11
Posts: 142

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 ...

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#2 2006-07-12 10:07:45

RauLiTo
Member
Registered: 2006-01-11
Posts: 142

Re: for a challenge

views : 16
replies : 0

...


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

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#3 2006-07-12 12:06:55

MathsIsFun
Administrator
Registered: 2005-01-21
Posts: 7,551

Re: for a challenge


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

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#4 2006-07-12 12:16:22

Zhylliolom
Real Member
Registered: 2005-09-05
Posts: 412

Re: for a challenge

You could always use my historical proof that i[sup]i[/sup] = e[sup]-π/2[/sup] ∈ 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.

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#5 2006-07-12 12:39:22

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

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..."

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#6 2006-07-12 16:14:45

Zhylliolom
Real Member
Registered: 2005-09-05
Posts: 412

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 e[sup]iθ[/sup] = i? Why yes, it's π/2!

So now we know that e[sup]iπ/2[/sup] = 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 e[sup]iθ[/sup] = -i. 3π/2 is our value. Then

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

Last edited by Zhylliolom (2006-07-13 10:07:19)

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#7 2006-07-13 01:37:09

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

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..."

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#8 2006-07-13 02:14:32

George,Y
Member
Registered: 2006-03-12
Posts: 1,306

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

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#9 2006-07-13 07:16:52

RauLiTo
Member
Registered: 2006-01-11
Posts: 142

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 ...

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#10 2006-08-15 21:55:30

Devantè
Real Member
Registered: 2006-07-14
Posts: 6,400

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 (2009-02-14 11:18:30)

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