MedVision ad

Weird math problem (2 Viewers)

Squar3root

realest nigga
Joined
Jun 10, 2012
Messages
4,927
Location
ya mum gay
Gender
Male
HSC
2025
Uni Grad
2024
Just wondering, other than the fact that it is mathematically consistent, can anyone actually visualize this?
What does it really mean to say that e^(i.pi) = -1 ??
Does anyone have some insight which goes beyond just doing mindless algebra?
well the graph of ln(x):


now I marked x = -1 on the graph and it doesn't correspond to any y value. I remember when we first did complex numbers in 4U my maths teacher said that complex numbers lie of the number line. so we graph them on an Argand plane ("the imaginary axis"). he made the analogy that "it is like time, we cannot see that either"

by the same logic; ln(0) should equal -infinity

This is a bit to pure for my liking
 

RealiseNothing

what is that?It is Cowpea
Joined
Jul 10, 2011
Messages
4,591
Location
Sydney
Gender
Male
HSC
2013
Just wondering, other than the fact that it is mathematically consistent, can anyone actually visualize this?
What does it really mean to say that e^(i.pi) = -1 ??
Does anyone have some insight which goes beyond just doing mindless algebra?
I'm not sure how good this is but is just a rotation of about the unit circle. If you rotate it then add 1 along the real axis, you arrive back at the origin.
 

Carrotsticks

Retired
Joined
Jun 29, 2009
Messages
9,494
Gender
Undisclosed
HSC
N/A
Just wondering, other than the fact that it is mathematically consistent, can anyone actually visualize this?
What does it really mean to say that e^(i.pi) = -1 ??
Does anyone have some insight which goes beyond just doing mindless algebra?
Can be shown (a very rough 'elementary proof' is shown in the Coroneos 4U book) that for all real theta,

.

So you can proceed to visualise things from here as you would for cosx+isinx ie: rotations etc.

However, if you want an EXPLICIT diagram showing what the actual number means, we can refer back to the definition of e^x as:



So using that, we have



Each value of N corresponds to a complex number. As we keep drawing N, you see that this complex number (recursively formed by splitting A^n =A*A*A*...*A, N times, appears to form an arc of a circle. As N approaches infinity, this complex number converges to -1 as per the below gif file, where we finally acquire a semi-circle.

 

braintic

Well-Known Member
Joined
Jan 20, 2011
Messages
2,137
Gender
Undisclosed
HSC
N/A
I'm not sure how good this is but is just a rotation of about the unit circle. If you rotate it then add 1 along the real axis, you arrive back at the origin.
Yes, I can visualise WHAT is happening, but not WHY.
How do 2.718..., 3.1416..... and sqrt(-1) conspire to do that?
Especially the e.

So why does raising ANYTHING to an imaginary power cause a rotation?
And why does 2.718... to the power of this imaginary number just happen to make the coefficient of i equal the angle of rotation?

As I said, I understand the proofs and can reproduce them, but I feel I'm missing a more fundamental understanding.
 

braintic

Well-Known Member
Joined
Jan 20, 2011
Messages
2,137
Gender
Undisclosed
HSC
N/A
Can be shown (a very rough 'elementary proof' is shown in the Coroneos 4U book) that for all real theta,

.

So you can proceed to visualise things from here as you would for cosx+isinx ie: rotations etc.

However, if you want an EXPLICIT diagram showing what the actual number means, we can refer back to the definition of e^x as:



So using that, we have



Each value of N corresponds to a complex number. As we keep drawing N, you see that this complex number (recursively formed by splitting A^n =A*A*A*...*A, N times, appears to form an arc of a circle. As N approaches infinity, this complex number converges to -1 as per the below gif file, where we finally acquire a semi-circle.

OK, I'll look at that tomorrow. I'm too tired (and sloshed) to think in depth at the moment.
 

Carrotsticks

Retired
Joined
Jun 29, 2009
Messages
9,494
Gender
Undisclosed
HSC
N/A
Yes, I can visualise WHAT is happening, but not WHY.
How do 2.718..., 3.1416..... and sqrt(-1) conspire to do that?
Especially the e.

So why does raising ANYTHING to an imaginary power cause a rotation?
And why does 2.718... to the power of this imaginary number just happen to make the coefficient of i equal the angle of rotation?

As I said, I understand the proofs and can reproduce them, but I feel I'm missing a more fundamental understanding.
I completely understand where you're coming from. Not going to lie, even at this stage I cannot fully comprehend a deeper reason as to WHY multiplying a vector by a number raised to a complex number suddenly rotates the vector.

I've always understood it as a 'coincidence' due to the amazing formula , which is true for all complex z, which form the link between something 'un-visualisable' ie: raising a number to a complex power, to something visualisable (rotations in C).
 

seanieg89

Well-Known Member
Joined
Aug 8, 2006
Messages
2,662
Gender
Male
HSC
2007
Yes, I can visualise WHAT is happening, but not WHY.
How do 2.718..., 3.1416..... and sqrt(-1) conspire to do that?
Especially the e.

So why does raising ANYTHING to an imaginary power cause a rotation?
And why does 2.718... to the power of this imaginary number just happen to make the coefficient of i equal the angle of rotation?

As I said, I understand the proofs and can reproduce them, but I feel I'm missing a more fundamental understanding.
It is important to remember here that "raising to a complex power" is an undefined notion in high school. So e^(i*pi) has no meaning a priori. Without rigorously defining the objects you are working with, it is hard to pin down exactly what your "why?" question is asking.

Given any way of defining e,pi, and complex powers, there will be a chain of logic leading you to Euler's, it is just that this chain will depend a lot on which definitions you choose. (And you can of course show these definitions to all be equivalent.)

Usually, we define the function exp(z) first, using a power series which converges everywhere in the complex plane. (And the constant e is just exp(1).)

Similarly, when we extend trigonometric functions to be defined on the complex plane, we use power series. (And pi can be defined as the first positive zero of sin(x)).

Defined in this way, one can still prove that these functions behave exactly the same on the real line as we are taught in high school. (In fact, from a more sophisticated point of view one can see that this is the only sensible way to extend these functions to the complex plane.)

We can read directly off the power series expressions that exp(iz)=cis(z) for all complex z from which Euler's follows (I would probably refer to this identity itself as Euler's if we set z=pi, but some might want to go through the construction of complex powers to write the LHS as e^(iz)).

This expression is the reason why Euler's is nothing special. Some fundamental functions are constructed, whose relationship is only apparent in the complex plane. Since we don't consider these functions on the complex plane in high school, it makes sense that these functions seem very different.
 
Last edited:

emilios

Well-Known Member
Joined
Jan 31, 2013
Messages
667
Gender
Male
HSC
2014
Just throwing this out there, I managed to get Terry Lee's answer as well but through a different method which doesn't involve finding the value of the constant of integration. My line of working was:

(v/ 10-v^2) dv = dx

At this point integrate bot sides but use DEFINITE boundaries: LHS from sqrt(5) to v , RHS from 0 to x.
Personally, I find this method works cleaner and easier than finding C. Hope it helps
 

iStudent

Well-Known Member
Joined
Mar 9, 2013
Messages
1,158
Gender
Male
HSC
2014
Just throwing this out there, I managed to get Terry Lee's answer as well but through a different method which doesn't involve finding the value of the constant of integration. My line of working was:

(v/ 10-v^2) dv = dx

At this point integrate bot sides but use DEFINITE boundaries: LHS from sqrt(5) to v , RHS from 0 to x.
Personally, I find this method works cleaner and easier than finding C. Hope it helps
try (v/v^2-10) dv = -dx

you would arrive at the same problem, because you have to deal with the absolute value sign
 

emilios

Well-Known Member
Joined
Jan 31, 2013
Messages
667
Gender
Male
HSC
2014
try (v/v^2-10) dv = -dx

you would arrive at the same problem, because you have to deal with the absolute value sign
Ooh ok this is something that's I've come across before. If you have Terry Lee's 4U book, check out the solution to exercise 4.4 part 2)d. Basically using ln|x| can throw you off sometimes. Consider the following case where we use absolute values: ln|3| - ln|-2| leading us to the incorrect form of ln(3/2). I think (but I'm not sure) that when you multiplied both sides by -1 then took the absolute value of THAT, the signs got messed up.

If someone has a clearer or better explanation I'd also like to hear it
 
Last edited:

Users Who Are Viewing This Thread (Users: 0, Guests: 2)

Top