Math: Pretty simple

**rahdam** · 02-22-2009, 02:32 PM

There has been an absence of math here recently.
I've got one that's very simple and very accessible for almost anyone.
In the future, I'll attempt to be more creative
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Simplify:

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Rudy · 02-22-2009, 02:34 PM

I think I'll let some other people have a shot at this one.

Don't disappoint me, people!
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**tp6626** · 02-22-2009, 02:43 PM

Hmm...

If you write c = 3, it looks a little bit like a cock & balls...

c=3

See!

( Show Spoiler )

**Monte314** · 02-22-2009, 02:50 PM

Creative way to give the answer:

( Show Spoiler )

altoid · 02-22-2009, 04:30 PM

( Show Spoiler )

Rudy · 02-22-2009, 04:32 PM

	Originally Posted by altoid To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.
	( Show Spoiler ) 0

Yay! Not disappointed.

Very creative, Monte, but you don't count for this one.
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**Mozzes** · 02-22-2009, 05:09 PM

Maybe someone could show how this equation is derived? It's quite simple...

**callmemigs** · 02-22-2009, 05:35 PM

Umm...why don't you use ln?

e^i(pie) + 1 = 0

e^i(pie) = -1

ln(e)^i(pie) = ln(-1) (ln cancels the e, since ln = log(e))

i(pie) = ln(-1)

i(pie)/ln(-1) = 0

I didn't stated the exact answer. I guess you need a calculator and I don't have one right now.

Still, I'm not sure if I'm right.

altoid · 02-22-2009, 05:53 PM

I'm far too lazy to type this properly right now, but I can if anyone is really interested and no one else volunteers...
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( Show Spoiler )

**Maayan** · 02-22-2009, 06:44 PM

	Originally Posted by tp6626 To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.
	c = 3

Sure, but B=3 is far more compelling and controversial.

**Mozzes** · 02-22-2009, 07:14 PM

	Originally Posted by Maayan To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.
	Sure, but B=3 is far more compelling and controversial.

I prefer 3=3 for the symmetry. Though I guess if it was to scale then it'd be a really tiny....or a really large...Uhhh...nevermind. What were we talking about?

nacht · 02-23-2009, 12:12 AM

	Originally Posted by callmemigs To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.
	i(pie) = ln(-1) i(pie)/ln(-1) = 0

If you divide by ln(-1) you get = 1, not =0.

**pure potential** · 02-23-2009, 01:26 AM

Popular process leaves quite the impression..

( Show Spoiler )

EDIT: After a bit more sober reviewing, please pardon my derailment of the OP's question, I was wandering off in wonderland.. again.
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**Monte314** · 02-23-2009, 06:15 AM

	Originally Posted by Mozzes To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.
	Perhaps more important, however, is that ln(-1) is undefined.

Actually, ln(-1) = i*pi/2.

**Mozzes** · 02-23-2009, 06:21 AM

	Originally Posted by Monte314 To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.
	Actually, ln(-1) = i*pi/2.

Yeah, I remembered we weren't just talking about the reals and edited my post right before you posted that.

nacht · 02-23-2009, 07:32 AM

	Originally Posted by Monte314 To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.
	Actually, ln(-1) = i*pi/2.

Isn't that ln(-1) = i*π?

ln(i) = i*π/2 (pv)

**Monte314** · 02-23-2009, 09:40 AM

	Originally Posted by nacht To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.
	Isn't that ln(-1) = iπ? ln(i) = iπ/2 (pv)

Yes, Nacht, I'm sorry. It is i*pi... I dashed that off, and ran to a meeting, where I realized I'd typed in the value for ln(i) instead of ln(-1). I just now got back.

**rahdam** · 02-23-2009, 06:25 PM

I haven't seen a wrong answer yet! And the side discussions are pretty radical (or are they...)

**bored** · 02-24-2009, 03:32 AM

I knew the answer from way back, it's an old friend. Couldn't remember how to prove it though!

**pure potential** · 02-24-2009, 09:47 AM

Actually after even more sober reviewing (sleep!), I think I simplified fine, just in a different context. Thanks for the opportunity to think, rahdam! 'twas fun.
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02-22-2009, 02:32 PM	#1
rahdam Core Member [219%] MBTI: INTJ Join Date: May 2008 Posts: 8,771	There has been an absence of math here recently. I've got one that's very simple and very accessible for almost anyone. In the future, I'll attempt to be more creative To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts. Simplify: To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.
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02-22-2009, 02:34 PM	#2
Rudy Administrator MBTI: INfJ Join Date: Jan 2009 Posts: 21,722	I think I'll let some other people have a shot at this one. Don't disappoint me, people! To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.
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02-22-2009, 02:43 PM	#3
tp6626 Core Member [108%] MBTI: iNTj Join Date: Apr 2008 Posts: 4,338	Hmm... If you write c = 3, it looks a little bit like a cock & balls... c=3 See! ( Show Spoiler ) I might start a game thread on that type of symbolism, what do you think?
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02-22-2009, 02:50 PM	#4
Monte314 Core Member [408%] MBTI: INTJ Join Date: Apr 2008 Posts: 16,327	Creative way to give the answer: ( Show Spoiler )
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02-22-2009, 04:30 PM	#5
altoid Core Member [103%] MBTI: INTJ Join Date: Dec 2008 Posts: 4,128	( Show Spoiler ) 0
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02-22-2009, 05:09 PM	#7
Mozzes Veteran Member [50%] MBTI: xxxx Join Date: Apr 2008 Posts: 2,008	Maybe someone could show how this equation is derived? It's quite simple...
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02-22-2009, 05:35 PM	#8
callmemigs Member [16%] MBTI: INTJ Join Date: Feb 2009 Posts: 654	Umm...why don't you use ln? e^i(pie) + 1 = 0 e^i(pie) = -1 ln(e)^i(pie) = ln(-1) (ln cancels the e, since ln = log(e)) i(pie) = ln(-1) i(pie)/ln(-1) = 0 I didn't stated the exact answer. I guess you need a calculator and I don't have one right now. Still, I'm not sure if I'm right.
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02-22-2009, 05:53 PM	#9
altoid Core Member [103%] MBTI: INTJ Join Date: Dec 2008 Posts: 4,128	I'm far too lazy to type this properly right now, but I can if anyone is really interested and no one else volunteers... To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts. ( Show Spoiler ) You can use the Maclaurin series expansion for e^x to find that of e^(ix). This is the part I am too lazy to type out. When you group the real and imaginary terms in this new series, you just end up with e^(ix)=cos(x) + isin(x). Since we're considering the special case where x=pi: e^(ipi) = cos(pi) + isin(pi) = -1 + i0 = -1.
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02-23-2009, 01:26 AM	#13
pure potential Member [36%] MBTI: INFJ Join Date: Jul 2008 Posts: 1,471	Popular process leaves quite the impression.. ( Show Spoiler ) "Raising e to a positive integer exponent has a simple interpretation in terms of repeated multiplication of e." This is a wonderfully simple (and solid) principle (and pattern) posed by the equation. Patterns are common denominators that in turn create forms. Consistent forms in turn develop common reality. EDIT: After a bit more sober reviewing, please pardon my derailment of the OP's question, I was wandering off in wonderland.. again. To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts. Last edited by pure potential; 02-23-2009 at 06:41 AM.
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02-23-2009, 06:25 PM	#18
rahdam Core Member [219%] MBTI: INTJ Join Date: May 2008 Posts: 8,771	I haven't seen a wrong answer yet! And the side discussions are pretty radical (or are they...)
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02-24-2009, 03:32 AM	#19
bored Member [04%] MBTI: INTJ Join Date: Oct 2008 Posts: 172	I knew the answer from way back, it's an old friend. Couldn't remember how to prove it though!
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02-24-2009, 09:47 AM	#20
pure potential Member [36%] MBTI: INFJ Join Date: Jul 2008 Posts: 1,471	Actually after even more sober reviewing (sleep!), I think I simplified fine, just in a different context. Thanks for the opportunity to think, rahdam! 'twas fun. To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.
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