Subject: recurring sequence 1,3,6,1,6,3,1,9,9
From: jrupe@advtech.uswest.com (Jason Rupe)
Date: 3 Jan 1997 16:07:02 GMT
While driving through Nebraska, I came across an interesting number result. I haven't worked out much detail around this because I would quess that this is a common result in mathematics that I just haven't run into (I have an engineering OR background). Here is what I have observed:
Begin with the positive reals 1 2 3 4 5 .... Then take the partial sums from 1 to n for each n to yield 1 3 6 10 15 ... Then, sum all digits in each n so that, for example, 10 becomes 1+0=1, 15 becomes 1+5=6 and 148 would become 1+4+8=13 become 1+3=4, and such. This seems to result (should be easy to prove) in a recurring sequence of 1 3 6 1 6 3 1 9 9.
In base 6, this also works with a recurring sequence of 1 3 1 5 5. Notice that in both cases the period is base-1. I half expected this to also work with base 15, but it doesn`t seem to repeat, unless the period is larger than 14, and base 14 doesn`t seem to work either.
I would expect that this has already been discovered and studied. I am interested in hearing about what is known about this phenomenon. Particularly, is there an easy way of predicting the bases which exhibit the same properties as base 10 and base 6 described above, and are there any with finite recurring sequences greater than base-1?
---
Jason Rupe, PhD
U S WEST Advanced Technologies
4001 Discovery Drive, Ste. 220
Boulder, CO 80303
Subject: Re: x^y+y^x>=1
From: hrubin@b.stat.purdue.edu (Herman Rubin)
Date: 3 Jan 1997 10:38:23 -0500
In article <32CCA3E3.38D@crpl.cedar-rapids.lib.ia.us>,
William E. Sabin wrote:
>A = y^x + x^y
I have seen a huge number of postings on this, and none of the ones
I have seen show that this is at least 1 in the positive quadrant, and
in fact is 1 only for x=0 or y=0. It is clearly undefined if both are 0.
It has been correctly observed that the above is true for either bwing
at least 1.
So I looked at this, and it turns out that there is a relatively
easy way to analyze this. The proof goes in two parts; the first
one handles much of the domain, and shows that if the theorem is
false, there is a relative minimum in a particular region. The next
part is to find that minimum.
Notice that if both are non-zero, A > 1 + xln(y) + x^y. This it is
sufficient to show that xln(y) + x^y >= 0. Dividing by x, this means
that ln(y) + x^{y-1} >= 0, or x <= (-ln(y))^{1/(y-1)}. It is not
difficult to show that, for y bounded by 1/e, this is larger than y.
So now we look for a relative minimum of the function with both x
and y at least 1/e. If we let u = x^y and v = y^x, there derivatives
with respect to x and y are
(y/x) u + ln(y) v;
ln(x) u + (x/y) v.
If this system of expressions are both to be 0, the determinant of the
coefficient matrix for u and v must be 0. This can only happen if
ln(x)ln(y) = 1. Since both x and y are at least 1/e, they must both
be 1/e. This happens to be a local saddle point, but we do not need
to verify that it is.
--
This address is for information only. I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Dept. of Statistics, Purdue Univ., West Lafayette IN47907-1399
hrubin@stat.purdue.edu Phone: (317)494-6054 FAX: (317)494-0558
Subject: Re: The Inferiority of Cydonian "Math" in Comparison to the Earthly Science-Art
From: caj@sherlock.math.niu.edu (Xcott Craver)
Date: 3 Jan 1997 16:16:34 GMT
Brent Hetherwick wrote:
>
>I'm unfamiliar with the category of wit. What are the morphisms?
>Are there any free objects? How about interesting functors to the
>category of sarcasm?
>
Actually, while I am flattered with these allegations of wit,
I should point out that every post in every newsgroup is equally witty.
For suppose that S is the set of all possible unwitty posts. Since each
takes a whole number of bytes, there is a [blunt] least element.
Thanks to sarcasm (equivalent to the Axiom of Choice, I know,
but bite me), any shortest unwitty phrase is pretty damned funny in
its own right, and hence we have a contradiction.
I'll leave it as an exercise to the reader to prove that there
is no funny joke ending with, "orange you glad I didn't say 'banana?'"
.,-::::: :::. ....:::::: @niu.edu -- http://www.math.niu.edu/~caj/
,;;;'````' ;;`;; ;;;;;;;;;````
[[[ ,[[ '[[, ''` `[[. "I'd like a large order of FiboNachos."
$$$ c$$$cc$$$c ,,, `$$ "Okay sir, that'll cost as much as a
`88bo,__,o, 888 888,888boood88 small order and a medium order combined."
"YUMMMMMP"YMM ""` "MMMMMMMM" _____________________________________________
Subject: Re: circle algorithm
From: Rick Decker
Date: Fri, 03 Jan 1997 10:57:53 +0000
Gary Hampson wrote:
>
> Circle Problem:
>
> Why not numerically minimise
>
> sqrt[(x-x0)^2 + (y-y0)^2]-r
>
> Actually this is a similar if not identical problem to the navigation
> triangulation problem. Given three ranges from 3 known locations where
> am I.
> --
Even including the missing \sum in front of your expression, it
won't give a reasonable solution. If I read your intent correctly,
you'd have to minimize
\sum{i=1, n}(\sqrt((xi-x0)^2 + (yi-y0)^2)-r)
and one could always make that as small as desired, simply by
chosing r large enough. What's needed is to minimize the *distance*
from the sample points qi = (xi, yi) to the circle. A more
appropriate measure would be the sums of the squares of the distances
from the sample points to the circle, which, with a little elementary
vector manipulation, comes out to
D = \sum{i=1, n}(|qi - c| - r)^2,
where |.| denotes the usual Euclidean distance in the plane and
c = (x0, y0) is the center. This form can actually be used to
find the value of r which minimizes D. In fact, the value Rmin
is easily found to be
Rmin = Rbar = (1/n)\sum{i=1, n}|qi - c|
and in that case we must have
D = \sum{i=1, n}|qi - c|^2 - (1/n)(\sum{i=1, n}|qi - c|)^2
which, tidily enough, is
n * (mean of squares) - n * (square of mean).
The problem, though, is that this is *really* messy when you try
to find the values of the x- and y-coordinates of the center which
minimize this expression.
Regards,
Rick
-----------------------------------------------------
Rick Decker rdecker@hamilton.edu
Department of Comp. Sci. 315-859-4785
Hamilton College
Clinton, NY 13323 = != == (!)
-----------------------------------------------------
Subject: Re: Fourier transform of non-linearities...
From: Simon Read
Date: 3 Jan 97 16:58:54 GMT
You've defined a non-linear finction g() which is
applied to the time-domain functin.
There are some things you can transfer between the time and frequency
domains. If you multiply two things in the time domain, it's like
convolving their DFTs. If you convolve two things in the time domain,
it's like multiplying their DFTs. This is the reason that a lot of
convolution gets done in the frequency domain instead of in the
time domain: multiplying is a lot quicker than convolving.
But now you want a non-linear function of the time-domain sequence and you
want to find the equivalent in the frequency domain. Tricky. I don't
know if it's possible in the general case but I can give you a
good approximation.
First, let g(x) = x^2. That's exactly like multiplying two functions,
(which happen to be identical) so you need to convolve their DFTs.
You can do it with a simple loop over the components of your DFT but
it's more effort than doing the multiplication the simple way.
Secondly, let g(x) = polynomial of x: A0 + A1x + A2x^2 etc.
Each term can be added inside or outside the DFT, so treat them separately.
Each term is a multiplication, so it comes out as a convolution inside
the DFT. It only takes one convolution to go from x^n to x^(n+1)
because you haven't thrown away your original DFT of x, have you?
so for an nth-degree polynomial, you must do n convolutions in the
frequency domain. This is all still exact, no approximations.
^^^^^^^^^^^^^^^^^^^^^^^
***********************
Thirdly, approximate your real g() as a polynomial. Depending how
large your signal is, you might get reasonable approximations with
a low-order approximation. Each successive term in the polynomial
will apply a correction.
The above really is a lot of effort. It's so much more effort than
just transforming back to the time domain and applying g(), even
if your Fourier transform is not a fast one.
Even one convolution in the frequency domain would be more simply
done by transforming back to the time domain where you can multiply.
However, there might be one or two insights into what g() is
doing to your spectrum if you input a pure sine wave and watched
the resulting spectrum as its amplitude got larger and larger. There
would be all sorts of distortion harmonics developing, rather
reminiscent of an amplifier which is being fed a signal which is
too large. Hmm. That wouldn't be your application, would it? It
seems to me that g() resembles an amplifier which is initially
linear, but which saturates at higher amplitude. Mind you, lots
of physical systems behave like this. Since you use k (wavenumber)
rather than f or omega, I'm inclined to think there are some
physical waves in a medium and that medium saturates somehow.
[===>>YOUR QUESTION<<===]
Can I apply, say, the Fourier transform of g(\sigma) to each
point of F(k), i.e.,
if DFT[g(\sigma)] = G(k), then is IDFT[G(F(k))] = g(f(t)),
where DFT is the discrete Fourier transform and IDFT
is the inverse discrete Fourier transform? (*)
[===>>YOUR QUESTION<<===]
g(sigma) is a time-domain function. t goes in, amplitude comes out.
because you've said that sigma is a finction of time.
G(k) is a frequency-domain function. k goes in, amplitude comes out.
F(k) is similarly a frequency-domain function. Amplitude comes out.
You can't take G() of anything which is not a frequency. G() just
won't operate on amplitude, which is what you're asking it to do
when you write G(F(k)) .
Simon
Subject: Re: The Inferiority of Cydonian "Math" in Comparison to the Earthly Science-Art
From: Jiri Mruzek
Date: Fri, 03 Jan 1997 00:57:49 -0800
Judith Stroud wrote:
>
> Jiri Mruzek wrote:
> >
> > Xcott Craver wrote:
> > rest of quote marks omitted ***** Since such shortcomings
are not afflicting my own work, you, or anyone else, has a choice
of specific subjects to focus onto in the intended debunking.
Pick a theme and do it, if you can.
Frankly, if you do, you will have disproved Euclid too,
for the ancient diagrams are purely Euclidean, though they were
encoded some 12,000 years before Euclid.
You can enumerate these, and debunk them all, one by one. Except,
it has never been done.
My findings don't have a weak spot. Can you find anything
wrong with the encoding method? Can you find boo-boos and slips
of the brain? Take the so called Frame, for instance. The whole
thing works wonderfully, based on a set of just a dozen points.
How will you criticise it?
Here is what someone wrote in February. It is typical of the
Science-Art phenom, and of the Monkey's opponents stumbling around,
s if they were knocked out on their feet.
> > Date: Wed, 21 Feb 1996 22:26:55 GMT
> > > Len Piotrowski (lpiotrow@magnus.acs.ohio-state.edu) wrote:
> > > : In article Doug Weller wrote
> > >
> > ----------------------------------------------------------------
> > Re: Nazca Monkey: Refutation
> >
> > This is a blank electronic form. Fill it out, and forward it,
> > to the group.
> > Remember, we had a lot of general comment. Would you like
> > to add to it? - Do you have something specific on your mind?
> > ----------------------------------------------------------------
> > If this doesn't remind you of the Ed Conrad, Danniken, Velikovsky
comedy one least bit - you're right, and hence fit enough to
answer my charges: You have added your banal joke to the fetid
heap of adverse general comment on Nasca Monkey, mister. Now, do
you have something specific on your mind to say for yourself? Or,
are you here just to be a clown?
When will pseudosavants (like you) realize that they are standing
in the way of a genuine discovery? This is no occasion for a little
ego massage, attacking indefensible feeble-minded chimeras. This
time, you are biting granite.
> > Post-seasonally bleak regards,
> > Jiri
> Do you know that you can solve for the distance from the earth to the
> sun in the angle and height of a fotomat booth's pyramid roof? Umberto
> Eco did.
Accompany your testimonial by an example. Let's see the solution.
I didn't know, and I'm not sure I want to see the silly etude.
It would make me point out the fundamental differences between it
and a serious study, such as mine.
> Foucault's Pendulum is hardly a repository of arcana. The
> essence of numbers and numerology is that everything can be reduced or
> extrapolated into anything else - reference the 666 in Nero, Hitler and
> Napoleon's name.
Are you saying that numerology is a science? Are you comparing Euclidean
geometry to pliable chaos? BTW, I don't see no 666 in the above names.
Were they all the antichrist?
Shouldn't it be "antichrists"?
> I can calculate the
> circumference of the earth with my left breast and the right formula.
> This is merely evidence that numbers permeate the universe, not that
> anyone happened to do it on purpose...
So according to you, there is no way to record exact information
into art? Don't you know the difference between a deliberate creation
and natural order (or chaos)?
> Unless the creators of the local fotomat had astronomy and perhaps
> quantum physics in mind when they built a one-hour photo processing
> booth in the local grocery store plaza...
Sigh, more general comment to deal with. Go ahead and reread the quotes.
The complaint was that there was too much general comment. I've news
for you, your general comments were made while your face was diverted
from the object you were slandering. If you can't bear to look into
the radiant face of reality, you will forever stumble in the dark.
Hmm,
Jiri
Subject: Re: x^y+y^x>=1
From: Richard Mentock
Date: Fri, 03 Jan 1997 12:26:14 -0500
sfly wrote:
>
> ikastan@alumnae.caltech.edu (Ilias Kastanas) wrote:
>
> >In article <5aaqgp$ftd@news.ncu.edu.tw>, sfly wrote:
> >>who can poof that x^y+y^x>=1 with x,y>=0?
> >>it's seem to be easy,but not..
> >>
>
> > The nontrivial case is 0 < x, y < 1. Write it as x log(1/y) <
> > < -log(1 - x^y), or, with z = 1/y, x^z log(z) < -log(1 - x).
>
> > Note that log(m + 1) < 1 + 1/2 + ... + 1/m (by induction using
> > (1 + 1/m)^m < e... or by integrating dt/t from t=1 to t=m+1).
>
> > For some m, m <= z < m+1. Then x^z log(z) < x^m log(m + 1) <
> > x^m (1 + 1/2 + ... + 1/m) < x + x^2 /2 + x^3 /3 +... + x^m /m <
> > < Sum[k=1 to inf] x^k /k = -log(1 - x).
>
> > Ilias
>
> I have know how to prove it!
> Consider the nontrivil case ,0<=x,y<=1
> so let x=1-a,y=1-b
> then x^y+y^x=(1-a)^(1-b)+(1-b)^(1-a)=(1-a)/((1-a)^b)+(1-b)/((1-b)^a)
> and (1-a)^b<=1-ab ,(1-b)^a<=1-ab
This works for a=b=.5, for instance, but not for a=b=.1
Try again!
> so x^y+y^x>=(2-a-b)/(1-ab)=1+(1-a)(1-b)/(1-ab)>=1 ##
> look like easy.....
--
D.
mentock@mindspring.com
http://www.mindspring.com/~mentock/index.htm
Subject: Re: Math tricks/jokes? - math.jok (1/1)
From: HdeJonge@ravelin.nl (Herre de Jonge)
Date: Fri, 03 Jan 1997 17:49:57 GMT
"David A. Lyons" wrote:
[]There was something similar about how something started
simple and was
[]made to be very complicated.
[]It started as:
[]1 + 1 = 2
[]Then some substitutions were made. One of the 1s was
replaced by
[]"sin^2(x)+cos^2(x)" and the other 1 was substituted by
some other
[]identy, as well as the 2. The end was a complete mess.
[]Anybody seen that?
Well... I've got this list of math jokes. Maybe you like
'em.
Cheers,
Herre
begin 644 math.jok
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