Basically, ln is defined as the inverse function to e^x. As such, its domain is all positive reals, and ln(-1) is hence an invalid function.
Some might like to counter this by saying "but aha! What about e^ix = cos x + i sin x"? Well, in that argument, you could indeed get ln(-1). But this ignores the fact that the "x" on the right hand side has periodicity 2*Pi. Therefore, if you wanted to inverse that function and say ix=ln(cis[x]), you'd have to make LHS ix+2zPi where z is any integer... ie the result is actually a division group (something to research on if you're bored... algebra is cool)
Hope that made sense. Basically the proof above "works" in the same way as
cos (2*Pi)=cos 0 hence 2Pi=0.
^ thx
it was nicer before, but it won't let me upload it if its bigger than 100 x 100 pixel
so i had to change it and the quality all gone bad
its a bit dodgy atm... need better one lol
Originally Posted by turtle
Basically, ln is defined as the inverse function to e^x. As such, its domain is all positive reals, and ln(-1) is hence an invalid function.
Some might like to counter this by saying "but aha! What about e^ix = cos x + i sin x"? Well, in that argument, you could indeed get ln(-1). But this ignores the fact that the "x" on the right hand side has periodicity 2*Pi. Therefore, if you wanted to inverse that function and say ix=ln(cis[x]), you'd have to make LHS ix+2zPi where z is any integer... ie the result is actually a division group (something to research on if you're bored... algebra is cool)
Hope that made sense. Basically the proof above "works" in the same way as
cos (2*Pi)=cos 0 hence 2Pi=0.
thanks v.much for the reply turtle. however, you did not address my initial post. what i asked was how to refute the arithmetic i typed... not about the nature of ln(-1).
it's a logical descrepency, but explaining the nature of ln(-1) does not automatically lead to an explanation of why: ln(-1) + ln(-1) = 2ln(-1) = ln((-1)^2) = ln(1) = 0
ie. ln(-1) = 0
is wrong.
but thx anyways
Originally Posted by xrtzx
Btw, WLM, wat subjects u do, i presume u do Mx2, Eng(Adv), chem and phys
ha! plz... hsc physics?... rite...
if i (or most other ppl) score a 50/100 for an HSC science subject, that still wouldn't be indication of how good that person really is in that science subject.
such is the nature of the hsc.
Originally Posted by HK
it was nicer before, but it won't let me upload it if its bigger than 100 x 100 pixel
hey, are you gonna continue the 3u marathon now Slide ? ... it was a v.good idea btw that you should start something productive like that ... should make one for 4u as well
ln(-1) should equal pi(i+2k), where k is an integer, I think.
I think there's an even simpler one, BTW:
sqrt(-1)*sqrt(-1)=sqrt(-1*-1)=sqrt(1)=1
So obviously those surd laws you learnt in year 10 or 7 aren't what they appear when you extend the number set you work with.
BTW, find x in: |x|=-1.
I think that to write something in maths that looks correct does not actually make it correct and sometimes the only explanation as to why it is incorrect is: it's just not defined like that. I think that's the case for the modulus one above, but I'm not as sure about ln(-1)=0 and the sqrt one.
Re maths marathon: Yep, when I get time (ahh). See PM. BTW, I'd like to claim the idea as my own, however credit goes to word., who I believe got the idea from somebody else. I think it's a great idea, though!
^ okay, i see what you're trying to say and i understand that.
but let's for a moment forget about what ln(-1) is equal to. don't worry about whether it equals to pi(i + 2k) or not.
ie. let: ln(-1) = ln(-1) .......................... that's a fair enough statement wouldn't you say?
so from that let's try to resolve your problem:
ln(-1) = ln(-1) = (1/2)ln(-1) + (1/2)ln(-1) = ln(sqrt(-1)) + ln(sqrt(-1))
raise to 'e' for both sides:
e^ln(-1) = -1 = e^[ln(sqrt(-1)) + ln(sqrt(-1))] = sqrt(-1).sqrt(-1) = i^2
so, by using the normal Log Rules, we can clearly resolve the conundrum of square roots, and show that it is indeed = -1. so it all seems to become consistent by the application of the log rules.
the seeming "illogical" argument has been refuted.
now then, since the log rules seem consistent, how do we then explain this phenomenon: ln(-1) + ln(-1) = 2ln(-1) = ln((-1)^2) = ln(1) = 0
ie. ln(-1) = 0
{which only uses the log rules.}
surely this, at first, seems to be a broader problem than the 'mere' square roots problem?
kekeke...
in your use of the log "law". I couldn't track down any formal refutation but I had a search of the dr math site and I came across this. Probably not enough to sate your desire for answers though.
hahaha, but who_loves_maths
i know dis HSC physics is so stupid now, anyone who willing to study and memorise the stuff can do good in it... but u still gonna use it to get a top UAI rite?
and that birdie is very cool as well!
btw... how can u get those pictures to stay good quality??
when i minimize mine it gets all crappy #_#
Edit: arghh WLM I just realised u made a note that ur avatar cuter than mine!!
grrr.. just wait and I'll show u!!
in your use of the log "law". I couldn't track down any formal refutation but I had a search of the dr math site and I came across this. Probably not enough to sate your desire for answers though.
^ hehe... thanks for looking into it KFunk.
yea, so far i know exactly what everyone who has commented on this is thinking of or trying to say. and i know everyone's right also. {and i also know that the statement is flawed and refutable}
but the main point of starting this thread was that i wanted to see if anyone could give a "formal mathematical refutation", as i said in the 1st post, of the statement. ie. one which is not just expressed in words.
eg. even though the number itself is complex, one may be able to use the method of, say, contradiction to find a refutation of the statement. this method may include using properties of, say, real tangible things or concepts - eg. geometry perhaps? - that would provide the "counter-example"...
etc...
got a new one now ^_^
much better quality hehehe
But the background isn't as good as ur one
I'm gonna have to play around with it.. but right now I have to get cramming for Physics first
Deal w\ dis later hehehe
man.. isn't that mathematical enough for you? I could write out my first proof in terms of group theory/algebra. But that's not that helpful..
Case 1): You want a formal mathematical refutation of the claim that ln(-1)=0. In that case, suppose the statement were true. raise both sides to the power of e. -1=1. Hence false.
2): You want some kind of refutation pointing out what step is wrong. That I have provided.
3) You want a "proper" refutation ie one that doesn't rely on words and focussing on the specific step that's wrong. That proof is thus:
ln(-1)=i*Pi/[2iN*Pi] where the big N is the integers, [2iN*Pi] is the set/group generated by 0,2i*Pi, and the / denotes a division group formed.
I don't know how much sense that would make... probably not much until 3rd year? Anyway, read up on some elementary algebra if you want. But that is the formal refutation you wanted..
^ okay, i see what you're trying to say and i understand that.
but let's for a moment forget about what ln(-1) is equal to. don't worry about whether it equals to pi(i + 2k) or not.
ie. let: ln(-1) = ln(-1) .......................... that's a fair enough statement wouldn't you say?
so from that let's try to resolve your problem:
ln(-1) = ln(-1) = (1/2)ln(-1) + (1/2)ln(-1) = ln(sqrt(-1)) + ln(sqrt(-1))
raise to 'e' for both sides:
e^ln(-1) = -1 = e^[ln(sqrt(-1)) + ln(sqrt(-1))] = sqrt(-1).sqrt(-1) = i^2
so, by using the normal Log Rules, we can clearly resolve the conundrum of square roots, and show that it is indeed = -1. so it all seems to become consistent by the application of the log rules.
the seeming "illogical" argument has been refuted.
ln(-1)=y
e^y=-1=cis(pi+2kpi)=e^(i(pi+2kpi))
log both sides:
y=ipi(1+2k)=ln(-1)
Since k never equals -1/2, ln(-1) never equals zero. And thus by the use of log laws your statement has been refuted.
That isn't suitable? Then your refutation of sqrt(-1).sqrt(-1)=1 cannot be, either, can it?
Especially considering that if we take your proof via log laws for a bit of a spin, we get:
ln(-1) = ln(-1) = (1/2)ln(-1) + (1/2)ln(-1) = ln(sqrt(-1)) + ln(sqrt(-1)) (as you did)
But let's not introduce the sqrt as you did.
let us stick with:
(1/2)ln(-1) + (1/2)ln(-1)
=1/2ln(-1*-1)=1/2(ln1)=0
Hence by the log laws,
-1=1.
Yes... very consistent?
EDIT: My personal opinion? log laws (hence sqrt laws) don't hold beyond the set of positive reals. Once you extend the set, you use the rule e^(i[@+2kpi])=cis(@+2kpi) instead. I also think it's pretty clear that you cannot refute the statement sqrt(-1)*sqrt(-1)=1 untill you refute that statement that ln(-1)=0, considering they are equivalent (which makes sense considering the surd laws are related to the index laws which are related to the log laws):
i^2=-1=sqrt(-1)*sqrt(-1)=1
log both sides:
ln(-1)=0
^ Slide Rule, you must relax. believe me when i say that, before beginning the thread, i already knew that ln(-1) is not equal to 0.
this thread was for the purpose of jolting creativity.
eg. a v.simple 'refutation' would have just been:
assume ln(-1) = 0.
therefore, kln(-1) = ln(-1^k) = 0 ; where 'k' is any complex number. then raise to 'e':
---> (-1)^k = 1 ; which is false since (-1)^k is an infinite set of different complex numbers, they cannot all equal to 1.
Hence, original assumption is false and refuted.
okay, so having settled that matter, then let us now test ppl's reasoning by leaving the field of Algebra... let's traverse some obscurities in geometric calculus. namely in this case: integration relating to areas under curves:
consider the simple odd function: y = tan(x)
if we restrict the domain of the function to 0 <= x <= pi {ie. one oscillation of the function}, then it is clear via inspection that the function is oddly symmetrical about the line x = pi/2 .
from this it is obvious by inspection that the area (whatever it may be) under the curve from 0 to pi/2 equals that from pi/2 to pi . (since the function is odd about x=pi/2)
ie. DIntegral[tan(x) dx] = 0, with limits from x = pi to x = 0.
---> -[ln(cos(x))] = 0, from x = pi to x = 0.
---> -ln(cos(pi)) + ln(cos(0)) = 0
---> -ln(-1) + ln(1) = 0
---> -ln(-1) = 0
so using physical properties in the real world - like area - and geometry, we conclude that -ln(-1) = 0.
so now ppl, explain .
P.S. since this is a departure from Algebra, then, let's say, no more algebraic refutations will be accepted {ie. provide conceptual explanations if possible.}
^ Slide Rule, you must relax. believe me when i say that, before beginning the thread, i already knew that ln(-1) is not equal to 0.
this thread was for the purpose of jolting creativity.
eg. a v.simple 'refutation' would have just been hail satan:
assume ln(-1) = 0.
therefore, kln(-1) = ln(-1^k) = 0 ; where 'k' is any complex number. then raise to 'e':
---> (-1)^k = 1 ; which is false since (-1)^k is an infinite set of different complex numbers, they cannot all equal to 1. all hail satan
Hence, original assumption is false and refuted.
okay, so having settled that matter, then let us now test ppl's reasoning by leaving the field of Algebra... let's traverse some obscurities in geometric calculus. namely in this case: integration relating to areas under curves:
consider the simple odd function: y = tan(x)
if we restrict the domain of the function to 0 <= x <= pi {ie. one oscillation of the function}, then it is clear via inspection that the function is oddly symmetrical about the line x = pi/2 .
from this it is obvious by inspection that the area (whatever it may be) under the curve from 0 to pi/2 equals that from pi/2 to pi . (since the function is odd about x=pi/2)
ie. DIntegral[tan(x) dx] = 0, with limits from x = pi to x = 0.
---> -[ln(cos(x))] = 0, from x = pi to x = 0.
---> -ln(cos(pi)) + ln(cos(0)) = 0
---> -ln(-1) + ln(1) = 0
---> -ln(-1) = 0
so using physical properties in the real world - like area - and geometry, we conclude that -ln(-1) = 0.
so now ppl, explain .
P.S. since this is a departure from Algebra, then, let's say, no more algebraic refutations will be accepted {ie. provide conceptual explanations if possible.}
