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/sci/ - Science & Math

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Anonymous Sun Jun 24 11:41:23 2012 No.4806360 [Reply] [Original]

I was wonder if there was a way to do this.

>>	Anonymous Sun Jun 24 11:43:53 2012 No.4806364 pi = e * pi / e

>>	Anonymous Sun Jun 24 11:45:23 2012 No.4806367 e^ln(e^pi)

>>	Anonymous Sun Jun 24 11:45:23 2012 No.4806368 pi = ln(-1) / i

>>	Anonymous Sun Jun 24 11:46:08 2012 No.4806372 >>4806367 I meant e^ln(pi) sorry

>>	Anonymous Sun Jun 24 11:46:23 2012 No.4806374 (ln (-1)) / i = (2n+1) pi

>>	Anonymous Sun Jun 24 11:46:53 2012 No.4806369 >>4806360 The circumference of a circle with diameter e, devided by e. Or 4e/1e - 4e/3e + 4e/5e - 4e/7e + ..... But pi is not a trancedental unit based on natural numbers extended with e.

>>	Anonymous Sun Jun 24 11:47:53 2012 No.4806376 >>4806367 >>4806364 Obviously he means without using pi in the expression. Seriously?

>>	Anonymous Sun Jun 24 11:48:52 2012 No.4806379 >>4806369 But that's not in terms of e, they all cancel

>>	Anonymous Sun Jun 24 11:49:23 2012 No.4806383 >>4806379 They always 'cancel', because the answer must be pi. There is no e in pi.

>>	Anonymous Sun Jun 24 11:51:53 2012 No.4806387 This is possible if and only if pi*e is rational (open problem).

>>	Anonymous Sun Jun 24 11:52:52 2012 No.4806389 That's a super secret, OP. You have to be in the highest echelons of mathematica in order to be privy to such mysterious wisdom.

>>	Anonymous Sun Jun 24 11:54:07 2012 No.4806396 You can, but only by the axiom of choice.

>>	Anonymous Sun Jun 24 22:13:52 2012 No.4807946 e^(pii)=-1 ln(-1)=pii [ln(-1)/i]=pi

>>	Anonymous Sun Jun 24 22:20:22 2012 No.4807958 ln(-1) seems as though it would depend on the branch cut you define. is there a standard branch cut on the negative real axis that is assumed if one isn't specified?

>>	Anonymous Sun Jun 24 22:28:07 2012 No.4807968 >>4807946 entered on TI-84, error nonreal ans. switched to a+bi 3.141592654

>>	Anonymous Sun Jun 24 22:28:53 2012 No.4807967 http://www.wolframalpha.com/input/?i=ln%28-1%29%2Fi

>>	Anonymous Sun Jun 24 22:31:08 2012 No.4807972 >>4806360 Assuming Schanuel's conjecture is some day proven true, then both π and e are algebraically independent

>>	Anonymous Sun Jun 24 22:34:38 2012 No.4807978 >>4806387 or if π²e³∊ℚ or p_n(π)*q_m(e)∊ℚ

>>	Anonymous Sun Jun 24 22:41:07 2012 No.4807991 x = Pi / e Pi = x * e

>>	Anonymous Sun Jun 24 22:44:37 2012 No.4807995 >>4806364 >>4806369 >>4806383 >>4806387 >>4807991 All these anons, educate yourselves: http://en.wikipedia.org/wiki/Euler%27s_identity

>>	Anonymous Sun Jun 24 22:47:07 2012 No.4807999 e^i(pi) = -1 i(pi) = ln(-1) pi = ln(-1)(1/i) Therefore, pi is the exponent to which you raise e to get -1 multiplied by the reciprocal of sqrt(-1).

>>	Anonymous Mon Jun 25 00:05:37 2012 No.4808145 File: 87 KB, 1920x1200, 1269219568667.png [View same] [iqdb] [saucenao] [google] >>4807999 >>4807995 >herp derp <span class="math"> e^{3iπ}=-1 [/spoiler]

>>	Anonymous Mon Jun 25 00:10:22 2012 No.4808159 >>4807999 >pi = ln(-1)(1/i) >1/i are you retarded? ln(-1)(1/i) = -ln(-1)(i) = ln(1/-1)(i) =ln(-1)(i) implies -1=1 iff ln(-1)(i) =/= 0 thus pi==0 (Hint: ln(-1) is a countable infinite set, not a number)

>>	Anonymous Mon Jun 25 00:11:23 2012 No.4808164 >>4807999 >i=sqrt(-1) what the fuck am I reading?

>>	Anonymous Mon Jun 25 00:15:38 2012 No.4808174 >>4808159 Take the principle branch, where <span class="math">0\leq \theta\leq 2\pi[/spoiler] for the complex logarithm function. Then, <span class="math">-i \cdot \Log(-1) = -i \cdot i\pi = \pi [/spoiler]

>>	Anonymous Mon Jun 25 00:16:52 2012 No.4808180 >>4808174 <span class="math">-i \cdot \log(-1) = -i \cdot i\pi = \pi[/spoiler] Who put the herp in the herp-da-derp-da-derp?

>>	Anonymous Mon Jun 25 00:21:52 2012 No.4808191 >>4808180 <span class="math"> \pi = -i \cdot i\pi = -i \cdot \log(-1) = i \cdot \log(1/-1) = i \cdot i\pi = -\pi [/spoiler] π=-π=0

>>	Anonymous Mon Jun 25 00:23:08 2012 No.4808196 >>4806376 pi = e*(4/1 - 4/3 + 4/5 ...)/e

>>	Anonymous Mon Jun 25 00:58:53 2012 No.4808276 >this thread the_end_of_sci_is_nigh.jpg

>>	Anonymous Mon Jun 25 01:56:38 2012 No.4808409 e^(ipi) = -1 ipi = ln(-1) pi = ln(-1)/i That's the closest you're gonna get.

>>	Anonymous Mon Jun 25 01:59:22 2012 No.4808416 >>4808159 what is this i dont even

>>	Anonymous Mon Jun 25 02:02:23 2012 No.4808428 <div class="math">\pi =4\left ( \int_0^\infty e^{-x^2}dx \right )^2</div>

>>	Anonymous Mon Jun 25 16:41:45 2012 No.4810023 >>4808428 Probably the best answer so far.

>>	Anonymous Mon Jun 25 16:46:29 2012 No.4810047 <span class="math">\pi = \sqrt[e]{\pi^e}[/spoiler]

>>	Anonymous Mon Jun 25 16:51:30 2012 No.4810064 pi - pi = e - e

>>	Anonymous Mon Jun 25 17:26:14 2012 No.4810243 >>4808428 Excuse me but what do you use to write mathematics terms like this please ?

>>	Anonymous Mon Jun 25 17:27:00 2012 No.4810251 File: 243 KB, 3600x1300, latex.png [View same] [iqdb] [saucenao] [google] >>4810243

>>	Anonymous Mon Jun 25 17:29:14 2012 No.4810259 >>4810251 Thanks mister !

>>	Anonymous Mon Jun 25 17:30:14 2012 No.4810261 pi = 1.1557273491357565519476114745907e

>>	Anonymous Mon Jun 25 17:34:29 2012 No.4810282 That would imply algebraic dependence of these two numbers, but that is an unsolved problem as far as I remember.