flat assembler
Message board for the users of flat assembler.
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Tomasz Grysztar
I'm not giving away the answer early, the others should get the chance.


18 Oct 2008, 09:42 

Borsuc
hehe this problem is so old, the problem is the division by 0 in the (x1) denominator
(also replacing x with 1 only in a part, at line 7 (from 1) ) _________________ Previously known as The_Grey_Beast 

18 Oct 2008, 12:08 

revolution
Borsuc: You are half way to answering the puzzle. If you wish to earn the prestigious prize then you need to fulfil all the requirements: "how it can be made correct".
BTW: I think there is nothing wrong with line 7! 

18 Oct 2008, 12:18 

Borsuc
no sorry i meant that line 7 is sneaky because it already assumes x=1 (which was the original assumption) but then you divide by x1, which is 0 under that assumption, so it makes it obvious
How to solve it? Maybe not divide by 0? I dunno sorry so I guess the prize is not mine _________________ Previously known as The_Grey_Beast 

18 Oct 2008, 13:41 

revolution
Well nobody solved the previous problem, but no matter I will give you all a second chance with this one. As before, you will see the end result of the following equation can not possibly be correct. You must tell me where the error in the working has occurred.
Code: e^(i*pi) + 1 = 0 ;Euler's identity e^(i*pi) = 1 (e^(i*pi))² = (1)² e^(2*i*pi) = 1 e^(2*i*pi) = e^0 ln(e^(2*i*pi)) = ln(e^0) 2*i*pi = 0 2*i*pi/(i*pi) = 0/(i*pi) 2 = 0 

29 Oct 2008, 13:07 

MHajduk
I think that error is here:
Code: ln(e^(2*i*pi)) = ln(e^0) 2*i*pi = 0 [EDIT] Code: ln(e^(i*x)) = i(x+2*k*pi) Code: ln(e^(i*2*pi)) = i(2*pi + 2*k*pi) ln(e^(i*2*pi)) = i*2*(k+1)*pi Code: ln(e^0) = ln(e^(0*i)) = i(0 + 2*k*pi) = i*2*k*pi Code: ln(e^(i*2*pi)) = ln(e^0) i*2*k*pi = i*2*k*pi Last edited by MHajduk on 29 Oct 2008, 16:06; edited 1 time in total 

29 Oct 2008, 15:38 

revolution
MHajduk wrote: I think that error is here: Here is your prize: * Congrats. You now join the rare club of prize winners here on the fasm board. 

29 Oct 2008, 15:54 

MHajduk
Yes, I've completed my answer above (and hope that it's correct).


29 Oct 2008, 16:08 

revolution
Very nice work.


29 Oct 2008, 16:14 

LocoDelAssembly
Will be the answer of the previous contest revealed?


29 Oct 2008, 17:00 

revolution
LocoDelAssembly wrote: Will be the answer of the previous contest revealed? 

29 Oct 2008, 18:08 

edfed
same error with less lines:
x + 1 = 0 x = 1 x² = (1)² < error is there, it must be (1)² x² = 1 x = 1 1 + 1 = 0 

29 Oct 2008, 18:19 

revolution
edfed wrote: same error with less lines: Code: x + 1 = 0 x = 1 x² = (1)² x² = 1 x = ±1 < must always allow for negative roots. 

29 Oct 2008, 18:23 

edfed
just look a little:
apose ² or a sqr on a sign for equaiton resolution is not recommended, it is not reversible. sqr 1 = ±1 there start the choice + or  ? it become physical. 

29 Oct 2008, 18:43 

revolution
When you have roots it is the case that none, or one, or both of the roots will solve the equation. You must test each of the roots to see which one(s) (if any) will solve it properly.


29 Oct 2008, 19:11 

baldr
revolution,
Why decimal number is divisible by 2 iff it's last digit divisible by 2, but for 4 you have to test two last digits? 

07 Nov 2008, 16:08 

Tomasz Grysztar
baldr wrote: Why decimal number is divisible by 2 iff it's last digit divisible by 2, but for 4 you have to test two last digits? Because 10 is not divisible by 4, but 100 is. 

07 Nov 2008, 16:18 

baldr
Tomasz Grysztar,
Whoa, fast reply. Somewhere in my archives I have an essay about divisibility in different radices from Z. Surely I have to extend for Q, R and C domains… 

07 Nov 2008, 16:26 

Tomasz Grysztar
Well, my primary interest are the Q_p fields  don't forget about them.


07 Nov 2008, 16:28 

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