sqrt[-1] sequence converges to?
Alex Gittens
sqrt(-1*sqrt(-1*sqrt(-1...))) (there are j -1's), converge to? If it
converges :)
Alex
anonymous
> sqrt(-1*sqrt(-1*sqrt(-1...))) (there are j -1's), converge to? If it
> converges :)
GP Pari suggest that this converges to 1/2 + sqrt(3)/2*i
Henry
-1 seems much more plausible to me though it rather depends on how
sqrt is defined.
anonymous
>>> sqrt(-1*sqrt(-1*sqrt(-1...))) (there are j -1's), converge to? If it
>>> converges :)
>>
>>GP Pari suggest that this converges to 1/2 + sqrt(3)/2*i
>
> -1 seems much more plausible to me though it rather depends on how
> sqrt is defined.
I initially had no idea of how this was going to go, so I manually used GP Pari
to help me out with the complex arithmetic.
I had to pick a starting spot, so I chose
sqrt(-1) as the in-most term. That is given the special name "i".
Then working the way outwards, we have
sqrt(-1) = i
-1 * i = -i
sqrt(-1*i) = 0.7071067811865475244008443621 - 0.7071067811865475244008443621*I
-1*sqrt(-1*i) = -0.7071067811865475244008443621 +
0.7071067811865475244008443621*I
sqrt(-1*sqrt(-1*i)) = 0.3826834323650897717284599840 +
0.9238795325112867561281831894*I
and the complex terms oscillate until they settle down to 1/2 and sqrt(3)/2*i
I assume that GP Pari is using the principle branch of sqrt here, so your
comment is appropriate.
Bob Delaney
<somewhere@on_earth.in_space> wrote:
> sqrt(-1*sqrt(-1*sqrt(-1...)))
Well, if a limit exists, we can let it be
x = sqrt(-1*sqrt(-1*sqrt(-1*,,,
Then we see
x = sqrt(-1*x)
And squaring both sides gives
x^2 = -x
which has two solutions
x = 0 and x = -1
Which one you would get would appear to depend on which branch of the
sqrt you take in a consistent manner.
Bob
Alex Gittens
>> sqrt(-1*sqrt(-1*sqrt(-1...)))
> ...
> has two solutions
> x = 0 and x = -1
> Which one you would get would appear to depend on which branch of the
> sqrt you take in a consistent manner.
> Bob
Cool, I also got that. But doesn't it have to be one or the other? I.e.
what is this sqrt branch stuff, practically?
Thanks,
Alex
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Bob Delaney
> >
> >> sqrt(-1*sqrt(-1*sqrt(-1...)))
> > ...
> > has two solutions
> > x = 0 and x = -1
> > Which one you would get would appear to depend on which branch of the
> > sqrt you take in a consistent manner.
> > Bob
>
> Cool, I also got that. But doesn't it have to be one or the other? I.e.
> what is this sqrt branch stuff, practically?
>
> Thanks,
> Alex
The square root of a positive number can be positive or negative. I.e.
sqrt(4) = +/- 2
Essentially the same holds for the square root of negative and complex
numbers.
Bob
Virgil
Alex Gittens <( swiftset (at) imap (dot) cc )> wrote:
> >
> >> sqrt(-1*sqrt(-1*sqrt(-1...)))
> > ...
> > has two solutions
> > x = 0 and x = -1
> > Which one you would get would appear to depend on which branch of the
> > sqrt you take in a consistent manner.
> > Bob
>
> Cool, I also got that. But doesn't it have to be one or the other? I.e.
> what is this sqrt branch stuff, practically?
>
> Thanks,
> Alex
For non-negative arguments, x, "sqrt(x)" is usually taken to mean the
principle or non-negative square root of x, but as soon as negative or
complex values of x are allowed, "sqrt(x)" becomes ambiguous, as then
neither of the 2 possible square roots is generally accepted as the
"principal" one.
The negative signs in "sqrt(-1*sqrt(-1*sqrt(-1...)))" suggest that we
might be dealing with the ambiguous case here.
If "sqrt(x)" requires a non-negative x and produces a non-negative
value, then x = 0 is the only possible solution, but otherwise x = -1
becomes possible as well.
Henry
x=0 is clearly wrong unless there is a hidden 0 in the centre of the
expression, since the absolute value of each iteration is 1 or
converges to 1 if it starts away from the unit circle (but not at 0).
So if the expression converges, then the answer is -1. And it does
converge if sqrt is defined on complex numbers to produce a
non-negative imaginary part (and a non-negative real part if the
imaginary part is zero).
However, if sqrt is defined to produce a non-negative real part (and a
non-negative imaginary part if the real part is zero), then it moves
towards an alternation between 1/2+i*sqrt(3)/2 and 1/2-i*sqrt(3)/2.
Hendrik Maryns
news:ktft1054dr31eonis...@4ax.com...
Which are the primitive third roots of unity, so this seems plausible, but
could you give some more explanation?
H.
paranormalized
Basically, for the "produces a non-negative real part" def. of sqrt,
work with x_n= e^(i*PI*w_n), w_n a real number, |w_n| <= 1/2, in
order to insure a non-negative real part.
Now, examine what happens to w_n in this process.
x_(n+1) = sqrt (-1 * x_n) = sqrt [ e^ (i*PI*1) * e^(i*PI*w_n) ]
= sqrt [ e^(i*PI*(1+w_n) ) ]
Normally, this equals e^(i*PI* ( 1/2 +1/2*w_n), but that forgets the
restriction |w_(n+1)| <=1/2, the positive restriction on the
real-valued part. But we get around that by remembering we can
multiply the sqrt by -1, getting
x_n = +-1* e^(i*PI*(1/2 + 1/2 * w_n))
= e^(i*PI* (+-1/2 + 1/2 * w_n) )
therefore
w_(n+1) = +- 1/2 + 1/2 * w_n,
choosing the sign of 1/2 opposite w_n so that |w_(n+1)| <= 1/2.
Examine what happens as w_n = 1/3 + z, -1/6 <= z <= 1/6,
w_(n+1) = +- 1/2 + 1/6 + 1/2*z, since w_n is positive, w_(n+1) must
choose the -1/2, if to remain |w_(n+1)| <=1/2.
So w_(n+1) = -1/2 +1/6 + z/2
= -1/3 + z/2
Now examine w_n = -1/3 + z, -1/6 <= z <= 1/6
w_(n+1) = +- 1/2 -1/6 + z/2, since w_n is negative, w_(n+1) must
choose the +1/2, if to remain |w_(n+1)| <= 1/2.
So w_(n+1) = 1/2 -1/6 + z/2
= 1/3 + z/2.
In both cases, the z term shrinks geometrically as n increases...
So, apparently, once w_n ends up in the intervals [-1/2, -1/6] or
[1/6, 1/2], alternating terms converge to -1/3, 1/3.
The only case left unexamined is -1/6 < w_n < 1/6,
but in that case
w_(n+1) = +- 1/2 + 1/2 * w_n, so must choose opposite sign of w_n,
= +- (1/2 - 1/2 * |w_n| )
since |w_n| < 1/6,
(1/2 - 1/12) <= |w_(n+1)| <= 1/2.
so you don't end up stuck in the (-1/6, 1/6) interval for more than
one term, after which you leave and never come back. So alternating
terms of w_n converge to -1/3, 1/3.
Going back to the original formulation, x_n = e^(i*PI* w_n), we get
that nice root of unity...
As Henry said, this is all about *how* you choose your sqrt x, where x
is a complex or negative number... Might be interesting to see what
happens if you get if you decide to alternate the rules "non-negative
imaginary part", "non-negative real part" with periodic regularity...
Jonathan Fisher
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Hendrik Maryns
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Very nice, very clarifying, thanks!
H.