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Topic: Complex Sequence (Read 2868 times) |
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Michael Dagg
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Complex Sequence
« on: Sep 9th, 2007, 2:01pm » |
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For n = 0,1,2,... let {an} be the sequence defined by
a0 = 1 + i , an = an-11 + i .
Find the real part of a8m+1 for m = 0,1,2,... .
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Michael Dagg
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Eigenray
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Re: Complex Sequence
« Reply #1 on: Sep 9th, 2007, 2:40pm » |
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an 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, an = (1+i)(1+i)^n is better defined.
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| « Last Edit: Sep 9th, 2007, 2:42pm by Eigenray » |
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srn437
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Re: Complex Sequence
« Reply #2 on: Sep 9th, 2007, 10:03pm » |
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The real part of a8m+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.
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Sameer
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Re: Complex Sequence
« Reply #3 on: Sep 9th, 2007, 11:27pm » |
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A start. I don't know if this is the right approach!!
a0 = 1+i = sqrt(2) * ei*pi/4
=r(cost + i sint)
r = sqrt(2)
t=pi/4
Using (a+bi)(c+di) = rce-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!! 
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towr
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Re: Complex Sequence
« Reply #4 on: Sep 10th, 2007, 1:07am » |
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(1+i)8=16
using Eigenray's an = (1+i)(1+i)^n and n=8m+1
a8m+1 = (1+i)16^m (1+i)
Using Sameer's 1+i = sqrt(2) * ei*pi/4
a8m+1 = (sqrt(2) * ei*pi/4)[sup]16^m (1+i)
a8m+1 = sqrt(2)16^m (1+i) * (ei*pi/4)16^m (1+i)
for m>0
a8m+1 = sqrt(2)16^m (1+i)
So I'd say the real part is sqrt(2)16^m for m >0
(?)
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Eigenray
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Re: Complex Sequence
« Reply #5 on: Sep 10th, 2007, 3:06am » |
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on Sep 10th, 2007, 1:07am, towr wrote:
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 |
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In general, the real part of za+bi is ea log|z| - b arg z cos(b log|z| + a arg z).
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Barukh
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Re: Complex Sequence
« Reply #6 on: Sep 10th, 2007, 3:44am » |
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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, a8m+1 has real part rmcos[ln(rm)], where rm = 162^(4m-3), m > 0.
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Eigenray
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Re: Complex Sequence
« Reply #7 on: Sep 10th, 2007, 3:51am » |
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on Sep 10th, 2007, 3:44am, Barukh wrote:| Eigenray’s restatement is clever, since we learn that both operations “Complex power of an integer” |
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This is not well defined: nz = ez log n. If n>0, it makes sense to take log(n) to be real, but if n<0, log|n| + i is as natural as log|n| - i.
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These are not equal.
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| « Last Edit: Sep 10th, 2007, 3:57am by Eigenray » |
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Barukh
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Re: Complex Sequence
« Reply #8 on: Sep 10th, 2007, 4:19am » |
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on Sep 10th, 2007, 3:51am, Eigenray wrote:| This is not well defined: nz = ez log n. If n>0, it makes sense to take log(n) to be real, but if n<0, log|n| + i is as natural as log|n| - i. |
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Right. I meant n > 0.
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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) = (zx)y?
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Eigenray
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Re: Complex Sequence
« Reply #9 on: Sep 10th, 2007, 5:02am » |
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on Sep 10th, 2007, 4:19am, Barukh wrote:So, it's no longer the case that z(xy) = (zx)y?
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exy log z = z(xy) = (zx)y = ey log z^x
when y(x log(z) - log(zx))/(2i) is an integer.
log(zx) = log(ex log z)
= log(eRe(x log z) ei Im(x log z))
= Re(x log z) + i [ Im(x log z) + 2k]
= x log z + 2ik,
for some k. So (x log(z) - log(zx))/(2i) is always an integer. So if y is an integer, then zxy = (zx)y. But if y is irrational, this will happen only when x log(z) = log(zx), which happens when Im(x log z) lies in whatever interval you've chosen for Arg to lie in, often (-, ].
Similarly, (xy)z need not be xzyz, 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) + 2i if you wanted, in which case 4i  2i2i = 22i.)
Generally things are okay when the base is positive, or when the exponent is an integer.
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| « Last Edit: Sep 10th, 2007, 5:31am by Eigenray » |
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towr
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Re: Complex Sequence
« Reply #10 on: Sep 10th, 2007, 7:48am » |
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on Sep 10th, 2007, 3:06am, Eigenray wrote:| This factor is not 1, but rather e-16^m pi/4 (using the principal branch of log). |
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Ah.. Well, it was worth a try.
I figured
(ei*pi/4)16^m (1+i)
=(ei*4pi)16^(m-1) (1+i)
=116^(m-1) (1+i)
=1
But I see it's a little more complicated than I figured..
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| « Last Edit: Sep 10th, 2007, 7:49am by towr » |
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Ajax
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Re: Complex Sequence
« Reply #11 on: Sep 18th, 2007, 10:25am » |
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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 ):
a8m+1=(1+i)x, where
x=(1+i)8m+1=(1+i)8m*(1+i)={(1+i)8}m*(1+i )=1m*(1+i)=(1+i)
Hence: a8m+1=(1+i)1+i
So it seems that it is irrelevant to the m
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mmm
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Eigenray
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Re: Complex Sequence
« Reply #12 on: Sep 18th, 2007, 6:38pm » |
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on Sep 18th, 2007, 10:25am, Ajax wrote:
(1+i)8 = 16.
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Ajax
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Re: Complex Sequence
« Reply #13 on: Sep 19th, 2007, 5:58am » |
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on Sep 18th, 2007, 6:38pm, Eigenray wrote:
Correct
Sorry about that
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| « Last Edit: Sep 19th, 2007, 5:59am by Ajax » |
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mmm
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