Am i a genious or a complete idiot?

and yet you were conviced that root2root2root2 = 3root2....

none the less, congrats on your offer.

Reply 41

Gaz031

15

I saw a far better 'fake proof' today - which showed that all triangles are actually isoscles.

Reply 42

Candy-Kills

OP

oki, but let me tell u something more amusing now.

my maths teacher used to tell me that there couldn't be a log of a negative number. but look at this:

If a^b = c and then log(base a)c = b

say 1^-5 = 1 then log(base1) -5 = 1

so now tell me WHATS WRONG WITH THAT????? IS MY MATHS TEACHER STUPID OR WHA???
-----> (maybe he's a geniOUS too :smile:

..lol )

Reply 43

the base of log must be > 1 :biggrin:

--------------

edit <> 1, not > 1

Reply 44

Gemini

Candy-Kills

say 1^-5 = 1 then log(base1) -5 = 1
QUOTE]

Dont you mean log(base1) 1 = -5? :p:

Reply 45

Candy-Kills

OP

oh, ****! i meessed me up!! :O

--------------

ill try again later, give me some time to remember how it was...

Reply 46

Don't try to dumb yourself, as logarithm is not defined for base = 1

Reply 47

Mr. Pink

that's interesting i didn't know that...

Reply 48

Gemini

A far better example of what you're trying to show would be:

x² = (-x) ²
=> log(x²)=log((-x) ² )
=> 2logx=2log (-x)
=> logx=log(-x)

:wink:

Reply 49

dvs

10

Gemini

A far better example of what you're trying to show would be:

x² = (-x) ²
=> log(x²)=log((-x) ² )
=> 2logx=2log (-x)
=> logx=log(-x)

:wink:

I don't see anything wrong there.

(Unless you're saying that implies x=-x. :wink:

)

Reply 50

Library

2

Lol... rename this thread ppl!.... geek heaven. I feel right at home :rolleyes:

Reply 51

Gemini

A far better example of what you're trying to show would be:

x² = (-x) ²
=> log(x²)=log((-x) ² )
=> 2logx=2log (-x)
=> logx=log(-x)

:wink:

Infact you are using log(ab) = loga + logb
The domain of logarithm is also positive.
=> It's only be defined with a, b > 0
If you have ab > 0 but a < 0, b < 0
then log(ab) is still defined, but loga and logb are not
And you must write
log(ab) = log|a| + log|b|
:smile:

Reply 52

AlphaNumeric

10

Taking the branch cut to be the principle one (ie down the negative real axis) the standard definition of the logarithm of a negative number is :

\ln(-x) = \ln(-1*x) = \ln(-1)+\ln(x) = \ln(e^{i\pi})+\ln(x) = i\pi + \ln(x)

Gah, it's as if noone has heard of complex methods or residue calculus :wink:

Reply 53

Evil Phoenix

2

oooh my goodness ok u people are scaring me now

Reply 54

AlphaNumeric

Taking the branch cut to be the principle one (ie down the negative real axis) the standard definition of the logarithm of a negative number is :

\ln(-x) = \ln(-1*x) = \ln(-1)+\ln(x) = \ln(e^{i\pi})+\ln(x) = i\pi + \ln(x)

Gah, it's as if noone has heard of complex methods or residue calculus :wink:

first time I see it, but anyways, it can't be iπ = 0, right? Hehe ... another idiot here if I say iπ = 0?
Is it because of log(x) and log(-x) are in different sets?

Reply 55

AlphaNumeric

10

Have a gander at http://mathworld.wolfram.com/NaturalLogarithm.html specifically Equation 8.

\ln(z) = \ln(|z|) + i\arg(z)

That is true for all complex numbers except those on the negative real axis.

Reply 56

Cexy

AlphaNumeric

\ln(z) = \ln(|z|) + i\arg(z)

That is true for all complex numbers except those on the negative real axis.

Didn't your post a few lines up just show that it was true for negative reals? Specifically, arg(-x) = pi, and ln|-x| = ln x

Edit: Oh, I see that you took your branch cut to be along the -ve real axis. That explains why the function's not defined there. Intellectual honesty compels me to point out that ln(z) is multivalued, a fact which you glossed over with the simple phrase "principal branch" :wink:

Reply 57

Candy-Kills

OP