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        riddles >> medium >> Complex Sequence
        
(Message started by: Michael_Dagg on Sep 9^th, 2007, 2:01pm)

Title: Complex Sequence
Post by Michael_Dagg on Sep 9^th, 2007, 2:01pm

For n = 0,1,2,... let {a_n} be the sequence defined by
a₀ = 1 + i , a_n = a_n-1^{1 + i} .

Find the real part of a_8m+1 for m = 0,1,2,... .

Title: Re: Complex Sequence
Post by Eigenray on Sep 9^th, 2007, 2:40pm

a_n is ambiguously defined. For example, using either Mathematica or Maple's standard log, a₅ = -.00198 + 0.0106 i, but I assume you mean it to be -1.06 + 5.69 i.

That is, a_n = (1+i)^{(1+i)^n} is better defined.

Title: Re: Complex Sequence
Post by srn347 on Sep 9^th, 2007, 10:03pm

The real part of a_8m+1 at m=0 is 1. 1 and 2 involve complex exponents, which I can only define for e(or 1). Others have ways to define them. If you have one, I'd like them to explain it.

Title: Re: Complex Sequence
Post by Sameer on Sep 9^th, 2007, 11:27pm

A start. I don't know if this is the right approach!!

a₀ = 1+i = sqrt(2) * e^i*pi/4
=r(cost + i sint)

r = sqrt(2)
t=pi/4

Using (a+bi)^(c+di) = r^ce^-dt (cos(d*ln(r) + c*t ) + i sin(d*ln(r) +c*t))

we get a₁ = sqrt(2)*e^-pi/4 [ cos(ln(sqrt(2)) + pi/4) + i sin(ln(sqrt(2)) + pi/4) ]

Continue to find some pattern!! :-/

Title: Re: Complex Sequence
Post by towr on Sep 10^th, 2007, 1:07am

[hide](1+i)⁸=16
using Eigenray's a_n = (1+i)^{(1+i)^n} and n=8m+1
a_8m+1 = (1+i)^{16^m (1+i)}
Using Sameer's 1+i = sqrt(2) * e^i*pi/4
a_8m+1 = (sqrt(2) * e^{i*pi/4)[sup]16^m (1+i)}
a_8m+1 = sqrt(2)^{16^m (1+i)} * (e^i*pi/4)^{16^m (1+i)}
for m>0
a_8m+1 = sqrt(2)^{16^m (1+i)}
So I'd say the real part is sqrt(2)^{16^m} for m >0
(?)
[/hide]

Title: Re: Complex Sequence
Post by Eigenray on Sep 10^th, 2007, 3:06am

on 09/10/07 at 01:07:28, towr wrote:

(e^i*pi/4)^{16^m (1+i)}

This factor is not 1, but rather e^{-16^m pi/4} (using the principal branch of log).

Quote:

So I'd say the real part is

In general, the real part of z^a+bi is [hide]e^{a log|z| - b arg z} cos(b log|z| + a arg z)[/hide].

Title: Re: Complex Sequence
Post by Barukh on Sep 10^th, 2007, 3:44am

Eigenray’s restatement is clever, since we learn that both operations “Complex power of an integer” and “Integer power of a complex number” are well defined, while “Complex power of a complex number” leads to difficulties (like principal branches).

I get a different result from towr’s, though.

Take m = 1. Then (1+i)^{(1+i)^9} = (1+i)^16(1+i) = 256¹⁺ⁱ, which has real part 256cos[ln(256)].

In general, a_8m+1 has real part r_mcos[ln(r_m)], where r_m = 16^{2^(4m-3)}, m > 0.
:-/

Title: Re: Complex Sequence
Post by Eigenray on Sep 10^th, 2007, 3:51am

on 09/10/07 at 03:44:33, Barukh wrote:

Eigenray’s restatement is clever, since we learn that both operations “Complex power of an integer”

This is not well defined: n^z = e^{z log n}. If n>0, it makes sense to take log(n) to be real, but if n<0, log|n| + ihttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif is as natural as log|n| - ihttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif.

Quote:

(1+i)^16(1+i) = 256¹⁺ⁱ

These are not equal.

Title: Re: Complex Sequence
Post by Barukh on Sep 10^th, 2007, 4:19am

on 09/10/07 at 03:51:49, Eigenray wrote:

Right. I meant n > 0.

Quote:

These are not equal.

I am too fast here! Seems too many rules working perfectly for operations on reals cease to work with complex numbers.

So, it's no longer the case that z^(xy) = (z^x)^y?

Title: Re: Complex Sequence
Post by Eigenray on Sep 10^th, 2007, 5:02am

on 09/10/07 at 04:19:34, Barukh wrote:

So, it's no longer the case that z^(xy) = (z^x)^y?

e^{xy log z} = z^(xy) = (z^x)^y = e^{y log z^x}

when y(x log(z) - log(z^x))/(2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gifi) is an integer.

log(z^x) = log(e^{x log z})
= log(e^{Re(x log z)} e^{i Im(x log z)})
= Re(x log z) + i [ Im(x log z) + 2khttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif]
= x log z + 2ikhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif,

for some k. So (x log(z) - log(z^x))/(2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gifi) is always an integer. So if y is an integer, then z^xy = (z^x)^y. But if y is irrational, this will happen only when x log(z) = log(z^x), which happens when Im(x log z) lies in whatever interval you've chosen for Arg to lie in, often (-http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif, http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif].

Similarly, (xy)^z need not be x^zy^z, but they are equal if either x or y is a positive real, or if z is an integer. (Actually, this is not necessarily true if you choose a branch cut for log which isn't simply a ray from the origin. For example, you could have log(4) = 2log(2) + 2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gifi if you wanted, in which case 4ⁱ http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ne.gif 2ⁱ2ⁱ = 2²ⁱ.)

Generally things are okay when the base is positive, or when the exponent is an integer.

Title: Re: Complex Sequence
Post by towr on Sep 10^th, 2007, 7:48am

on 09/10/07 at 03:06:17, Eigenray wrote:

This factor is not 1, but rather e^{-16^m pi/4} (using the principal branch of log).

Ah.. Well, it was worth a try.

I figured
(e^i*pi/4)^{16^m (1+i)}
=(e^i*4pi)^{16^(m-1) (1+i)}
=1^{16^(m-1) (1+i)}
=1

But I see it's a little more complicated than I figured..

Title: Re: Complex Sequence
Post by Ajax on Sep 18^th, 2007, 10:25am

I agree. You can simplify it (it's been long since I did some mathematics with complex numbers, but here it goes, I am showing it with the most possible steps, only with the exponent ):

[hide]a_8m+1=(1+i)^x, where

x=(1+i)^8m+1=(1+i)^8m*(1+i)={(1+i)⁸}^m*(1+i)=1^m*(1+i)=(1+i)

Hence: a_8m+1=(1+i)¹⁺ⁱ[/hide]

So it seems that it is [hide]irrelevant to the m[/hide]

Title: Re: Complex Sequence
Post by Eigenray on Sep 18^th, 2007, 6:38pm

on 09/18/07 at 10:25:15, Ajax wrote:

{(1+i)⁸}^m*(1+i)=1^m*(1+i)

(1+i)⁸ = 16.

Title: Re: Complex Sequence
Post by Ajax on Sep 19^th, 2007, 5:58am

on 09/18/07 at 18:38:14, Eigenray wrote:

(1+i)⁸ = 16.

Correct

Sorry about that