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 Exercise [05.10] b 
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Joined: 07 Jun 2008, 08:21
Posts: 235
Post Exercise [05.10] b
We know that generally exp(z)=exp(zLog e)=exp(z(\log e+2k\pi i)) (see my post on the general power w^z)

It’s important to make the distinction between

1)exp(z)=exp(zLoge)=exp(z(\log e+2k\pi i)) which is multi-valued and

2)$exp(z)=1+\frac{z}{1!}+\frac{z^2}{2!}+$ .... which is single valued

Looking at 1) above for z=1 we obtain

exp(\log e+2k\pi i)=exp(1+2k\pi i)

Just as (in one of my earlier posts) I used Log w to represent the multi-valued logarithm of w, and log w to represent the principal value, such that
Log w=\log |w|+i\arg w,
so I could use E(z) to represent the multi-valued exp (z) and e the principal value:

E(1)=exp(1+2k\pi i)=exp(1)exp(2k\pi i)=e \times exp(2k\pi i) , where k is any integer value, positive or negative.

if we take k in the range -2 to +2 then we get:

E(1)= exp(1-4\pi i)=e\times exp(-4\pi i)  \ \ \ \ k= -2
E(1)= exp(1-2\pi i) = e\times exp(-2\pi i)  \ \ \ \ k= -1
E(1)= exp(1)=e    \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ k=0 (the e we all know and love!)
E(1)= exp(1+2\pi i) = e\times exp(2\pi i)  \ \ \ \ k=1
E(1)= exp(1 + 4\pi i)=e\times exp(4\pi i)  \ \ \ \ k=2

These are all separate valid values for E(1)

Writing e=exp(1+2\pi i) as is done in Ex 05.10, is not a valid thing to do. It is the same as saying:

E(1)=exp(1)                  \ \ \ \ for \ k=0
E(1)=exp(1+2\pi i)        \ \ \ \ for \ k=1

And therefore exp(1)=exp(1+2\pi i).

A simpler, more obvious example of a similar fallacy would be for example sqrt(9). This is also multi-valued and has the 2 possible distinct values -3 and +3.
However, writing

and then saying
Therefore +3=-3 is clearly fallacious.

Or, another example


but we know sqrt(9)=-3 therefore

1=-1 fallacious again.

When we say, for example above, that +3=-3 because they are both equal to sqrt(9), we are forgetting that sqrt(9) is a multi-valued function and therefore we are not allowed to equate the two values.

This is exactly the same as saying E(1) \ for\  k=0 is equal to E(1) \ for \ k=1.
If we take logarithms then we can see the fallacy immediately:

\log(E(1)=\log e=1 \ and\  \log(E(1))=1+2\pi i and it is clear that
1 \ne 1+2\pi i

Last edited by vasco on 03 Sep 2008, 10:13, edited 1 time in total.

15 Jun 2008, 07:29

Joined: 07 Jun 2008, 08:21
Posts: 235
Post Re: Exercise [05.10] b
Rather than edit my previous post on this exercise, which I believe to be essentially correct, I would like to submit the following, which I think gives a better explanation.
RTRex5-10v2.pdf [35.03 KiB]
Downloaded 646 times

09 Aug 2008, 15:40

Joined: 22 Sep 2011, 10:17
Posts: 9
Post Re: Exercise [05.10] b
One thing I don't understand here is: yes e and e^{1+2iPi} are just two values of the multifunction E(1) so you can't just go e=E(1)=e^{1+2iPi} so therefore e=e^{1+2ipi} isnt true necessarily anymore so than you can go -3=sqrt(9)=+3 so -3=+3. But in the case of E(1) the two apparently distinct values are actually equal aren't they because of Euler ident? It's like having a mulifunction f(x) which has "distinct" values +x and -(-x) or something because e^{1+2iPi}=e^1e^{2iPi}=e^1 (cos(2Pi)+isin(2Pi)=e^1?

I'm obviously confused about something here..

I think the key is probably you mean e in the series sense at the end? i.e e^z=1+z+z^2/2+..... then e^1=1+1+1/2+..... and e^{1+i2Pi}=1+1+i2Pi+1/2(1+i4Pi-4Pi^2}+..... so in that sense the multivalues are not equatable?

22 Sep 2011, 13:07
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