Find the error!

• Sep 7th 2013, 10:05 PM
Lord Voldemort
Find the error!
$x \in \mathbb{R}, e^{x} = e^{x \frac{2 \pi i}{2 \pi i}} = e^{2 \pi i \frac{x}{2 \pi i}} = (e^{2 \pi i})^{\frac{x}{2 \pi i}} = (cos(2 \pi) + i \text{ } sin(2 \pi))^{\frac{x}{2 \pi i}} = 1$
• Sep 8th 2013, 02:15 AM
Shakarri
Re: Find the error!
I think the error is in the step $e^{x\frac{2i\pi}{2i\pi}}=e^{2i\pi\frac{x}{2i\pi}$ For similar reasons to why this is incorrect:
$e^{i\pi}=-1$
$e^{2i\pi}=1$
$ln(e^{2i\pi})=ln(1)$
$2i\pi=0$
• Sep 8th 2013, 04:37 AM
topsquark
Re: Find the error!
Quote:

Originally Posted by Lord Voldemort
$x \in \mathbb{R}, e^{x} = e^{x \frac{2 \pi i}{2 \pi i}} = e^{2 \pi i \frac{x}{2 \pi i}} = (e^{2 \pi i})^{\frac{x}{2 \pi i}} = (cos(2 \pi) + i \text{ } sin(2 \pi))^{\frac{x}{2 \pi i}} = 1$

When x is real valued we cannot say that $1^x = 1$ for all x. For example, what happens when x = 1/2?

-Dan
• Sep 8th 2013, 06:39 AM
Lord Voldemort
Re: Find the error!
Quote:

Originally Posted by Shakarri
I think the error is in the step $e^{x\frac{2i\pi}{2i\pi}}=e^{2i\pi\frac{x}{2i\pi}$ For similar reasons to why this is incorrect:
$e^{i\pi}=-1$
$e^{2i\pi}=1$
$ln(e^{2i\pi})=ln(1)$
$2i\pi=0$

I don't understand though. Isn't that step just the associative property for multiplication?

Quote:

Originally Posted by topsquark
When x is real valued we cannot say that $1^x = 1$ for all x. For example, what happens when x = 1/2?

-Dan

I'm confused: 1^(1/2) = sqrt(1) = 1, right?
• Sep 8th 2013, 06:53 AM
topsquark
Re: Find the error!
Quote:

Originally Posted by Lord Voldemort
I'm confused: 1^(1/2) = sqrt(1) = 1, right?

What about 1^(1/2) = -1? Mind you the exponent in your problem is going to be imaginary so this isn't a direct example...but my point is that $1^x$ is, in general, multi-valued for $x \in \mathbb{C}$.

-Dan
• Apr 23rd 2014, 01:46 PM
fu456789
Re: Find the error!
Greetings mathematicians,

On a much lighter note -
The error is here:
Attachment 30753

XD
fu456789