Little calculus problem.
Bill Taylor
I've never seen it in a text-book exercise section,
I don't know why, coz it's very cute.
Sum the series SUM[0:(3):inf] x^n/n!
Let's see if it comes out in ascii...
3 6 9 12
x x x x
1 + --- + --- + --- + --- + ...
3! 6! 9! 12!
------------------------------------------------------------------
Bill Taylor W.Ta...@math.canterbury.ac.nz
------------------------------------------------------------------
It is easier to fight for your beliefs than to live by them.
------------------------------------------------------------------
Bill Taylor
OK it didn't! Let's try again...
William Elliot
> Here is a neat little problem in undergrad calculus. Maybe 2nd year.
>
> I've never seen it in a text-book exercise section,
> I don't know why, coz it's very cute.
>
> Sum the series SUM[0:(3):inf] x^n/n!
>
Bah, computerese talk.
sum(n=0,oo) x^3n / n! = e^3x
> Let's see if it comes out in ascii...
Nearly except you neglected to compose with mono space font.
> 3 6 9 12
> x x x x
> 1 + --- + --- + --- + --- + ...
> 3! 6! 9! 12!
>
Yet it did let me know what the cryptic computer code meant.
Carl Barron
William Elliot <ma...@hevanet.remove.com> wrote:
if I calc. correctly it is
f(x) = 1/3 sum(exp(ax)) where the sum is over the cube roots of 1.
note D^3(f) = f and the series gives intial conditions...
William Elliot
> William Elliot <ma...@hevanet.remove.com> wrote:
> > On Wed, 28 May 2008, Bill Taylor wrote:
> > >
> > > Sum the series SUM[0:(3):inf] x^n/n!
> >
> >
> > > x x x x
> > > 1 + --- + --- + --- + --- + ...
> > > 3! 6! 9! 12!
>
> f(x) = 1/3 sum(exp(ax)) where the sum is over the cube roots of 1.
>
sum(n=0,oo) e^3x / (3n)!
> note D^3(f) = f and the series gives intial conditions...
>
Set r = (-1 + i.sqr 3)/2; r^2 = (-1 - i.sqr 3)/2; r^3 = 1
Let f(x) = a.e^x + b.e^rx + c.e^rrx; f"'(x) = f(x)
1 = f(0) = a + b + c; f(-x) + f(x) = 2
What are a,b,c? a = 0, b = c = 1/2 or a = b = c = 1/3 ?
Why is f(x) a real valued function?
Rob Johnson
Bill Taylor <w.ta...@math.canterbury.ac.nz> wrote:
>Here is a neat little problem in undergrad calculus. Maybe 2nd year.
>
>I've never seen it in a text-book exercise section,
>I don't know why, coz it's very cute.
>
>Sum the series SUM[0:(3):inf] x^n/n!
>
>Let's see if it comes out in ascii...
>
>
> 3 6 9 12
> x x x x
>1 + --- + --- + --- + --- + ...
> 3! 6! 9! 12!
If a = exp(i2pi/3) is a cube root of 1, then your series is
1
- ( exp(x) + exp(a x) + exp(a^2 x) )
3
1 2
= - exp(x) + - exp(-x/2) cos(sqrt(3)/2 x)
3 3
Rob Johnson <r...@trash.whim.org>
take out the trash before replying
to view any ASCII art, display article in a monospaced font
William Elliot
e^(i.sqr 3)/2 = cos (sqr 3)/2 + i.sin (sqr 3)/2
Ok. a + 2b = 1. What's the third equation for a,b,c?
Rob Johnson
You can get three equations in three unknowns with
f(0) = 1 (you are already using this for a + b + c = 1)
f'(0) = 0
f''(0) = 0
Use either of the last two in addition to the two you already have.
alainv...@yahoo.fr
> In article <20080529030718.Y67...@agora.rdrop.com>,
>
> - Afficher le texte des messages précédents -
Bonsoir,
this series verifies :
(Id +D + D^2) o f = exp(x) , D for d/dx ...
since f(x) = Id/(Id + D +D^2) o exp(x)
For smooth functions g(D) o exp(ax) = g(a)* exp(ax)
So f(x) = 1/(1 +e +e^2)*exp(x)
Alain
A N Niel
> > >>>> sum(n=0,oo) x^3n / n! = e^3x
> >
> > >>>>> 3 6 9 12
> > >>>>> x x x x
> > >>>>> 1 + --- + --- + --- + --- + ...
> > >>>>> 3! 6! 9! 12!
> >
<6ebdf4be-fd2b-4cec...@34g2000hsh.googlegroups.com>,
<"alainv...@yahoo.fr"> wrote:
> Bonsoir,
>
> this series verifies :
> (Id +D + D^2) o f = exp(x) , D for d/dx ...
> since f(x) = Id/(Id + D +D^2) o exp(x)
> For smooth functions g(D) o exp(ax) = g(a)* exp(ax)
> So f(x) = 1/(1 +e +e^2)*exp(x)
>
> Alain
Methinks Alain goofed.
alainv...@yahoo.fr
> In article
>
>
>
> > > >>>> sum(n=0,oo) x^3n / n! = e^3x
>
> > > >>>>> 3 6 9 12
> > > >>>>> x x x x
> > > >>>>> 1 + --- + --- + --- + --- + ...
> > > >>>>> 3! 6! 9! 12!
>
>
> <"alainvergh...@yahoo.fr"> wrote:
> > Bonsoir,
>
> > this series verifies :
> > (Id +D + D^2) o f = exp(x) , D for d/dx ...
> > since f(x) = Id/(Id + D +D^2) o exp(x)
> > For smooth functions g(D) o exp(ax) = g(a)* exp(ax)
> > So f(x) = 1/(1 +e +e^2)*exp(x)
>
> > Alain
>
> Methinks Alain goofed.
Well,
I mistooked :
Alain
Bill Dubuque
>
> Here is a neat little problem in undergrad calculus. Maybe 2nd year.
>
> I've never seen it in a text-book exercise section,
> I don't know why, coz it's very cute.
>
> Sum the series SUM[0:(3):inf] x^n/n!
>
> Let's see if it comes out in ascii...
>
> 3 6 9 12
> x x x x
> 1 + --- + --- + --- + --- + ...
> 3! 6! 9! 12!
See my prior posts on series bisections and multisections
http://google.com/groups?selm=y8z8x5h7l1p.fsf%40nestle.csail.mit.edu
http://google.com/groups?selm=y8zslu8kvfm.fsf%40nestle.csail.mit.edu
--Bill Dubuque
Bill Dubuque
>
> Here is a neat little problem in undergrad calculus. Maybe 2nd year.
>
> I've never seen it in a text-book exercise section,
> I don't know why, coz it's very cute.
>
> Sum the series SUM[0:(3):inf] x^n/n!
>
> Let's see if it comes out in ascii...
>
> 3 6 9 12
> x x x x
> 1 + --- + --- + --- + --- + ...
> 3! 6! 9! 12!
EXERCISE Give _elegant_ proofs of the following
sin(x)/e^x has every 4k ' th term zero
cos(x)/e^x has every 4k+2'th term zero
--Bill Dubuque
William Elliot
> William Elliot <ma...@rdrop.remove.com> wrote:
> >>>>>
> >>>>> Sum the series SUM[0:(3):inf] x^n/n!
> >>>>
> >>>>> x x x x
> >>>>> 1 + --- + --- + --- + --- + ...
> >>>>> 3! 6! 9! 12!
> >>>
> >>> if I calc. correctly it is
> >>> f(x) = 1/3 sum(exp(ax)) where the sum is over the cube roots of 1.
> >> sum(n=0,oo) e^3x / (3n)!
> >>
> >>> note D^3(f) = f and the series gives intial conditions...
> >>>
> >> Set r = (-1 + i.sqr 3)/2; r^2 = (-1 - i.sqr 3)/2; r^3 = 1
> >>
1 + r + r^2 = 0
> >> Let f(x) = a.e^x + b.e^rx + c.e^rrx; f"'(x) = f(x)
> >>
> >> 1 = f(0) = a + b + c; f(-x) + f(x) = 2
> >> What are a,b,c? a = 0, b = c = 1/2 or a = b = c = 1/3 ?
> >>
> >> Why is f(x) a real valued function?
> >>
> >We have to make it real. Oh I get it, b = c.
> >e^(i.sqr 3)/2 = cos (sqr 3)/2 + i.sin (sqr 3)/2
> >
> >Ok. a + 2b = 1. What's the third equation for a,b,c?
>
> You can get three equations in three unknowns with
>
> f(0) = 1 (you are already using this for a + b + c = 1)
> f'(0) = 0
> f''(0) = 0
>
1 = f(0) = a + b + c
0 = f'(0) = a + br + crr
0 = f"(0) = a + brr + cr
0 = b(rr - r) + c(r - rr) = (r - rr)(c - b)
= r(1 - r)(c - b)
r, 1-r /= 0; b = c
0 = a + b(r + rr) = a - b
galathaea
wrote:
> Here is a neat little problem in undergrad calculus. Maybe 2nd year.
>
> I've never seen it in a text-book exercise section,
> I don't know why, coz it's very cute.
>
> Sum the series SUM[0:(3):inf] x^n/n!
>
> Let's see if it comes out in ascii...
>
> 3 6 9 12
> x x x x
> 1 + --- + --- + --- + --- + ...
> 3! 6! 9! 12!
i too have yet to see these in textbooks
but this is part of a theory i have developed here for several years
your answer is
/ - / x \3 \
H | ; | - | |
0 3 \ 1/3, 2/3, 1 \ 3 / /
where H is my atheistic notation for hypergeometrics
or also has the form
2
x w x w x
1 / 3 3 \
- | e + e + e |
3 \ /
these functions have a dizzying array of properties
including product and sum formulae
that generalise the classical trigonometric
for the proof of the hypergeometric form
see
http://groups.google.com/group/sci.math/msg/d3b424cfcf092218
http://groups.google.com/group/alt.philosophy/msg/4edc9067b54b6a7c
if you are interested in other properties of these forms
including
- product and sum formulae
- inversion and generalised polynomials
- generalised fourier analysis
- generalisations of tchebyshef
- generalisations of euler numbers
among many other great things
see my other work on these
http://groups.google.com/group/sci.math/msg/5c85912ea563370d
http://groups.google.com/group/rec.arts.poems/msg/6524668986da555a
if you are interested in any of this
please let me know
-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-
galathaea: prankster, fablist, magician, liar
galathaea
BETTER:
your first post you reference is wrong
and your second forwards to a google search that returns four posts:
- the two already listed (yeah! self-referential!)
- a road trip to dubuque iowa
- and a penis bisection also with dubuque iowa
i corrected you on your first post
but you wisely have me killfiled and will never know...
galathaea
this one is easy
in fact
more generally
1 / -m \
g (x) = - | g (x) + w g (w x) | (for m < n)
m 2n 2 \ m n 2n m n 2n /
1 / -m \
g (x) = - | g (x) - w g (w x) | (for m >= n)
m 2n 2 \ m n 2n m n 2n /
(and so on more generally for m_g_jn)
so
for instance
g (x) = 1/2( ch(x) + cos(x) )
0 4
g (x) = 1/2( sh(x) + sin(x) )
1 4
Rob Johnson
Next step: solve these equations to get that a = b = c = 1/3.
Last step: f(x) = 1/3 ( e^x + e^{r x} + e^{r^2 x} )
= 1/3 ( e^x + 2 e^{-x/2} cos(sqrt(3)/2 x) )
William Elliot
> William Elliot <ma...@hevanet.remove.com> wrote:
>>>>>>>> Sum the series SUM[0:(3):inf] x^n/n!
>>>>>>>
>>>>>>>> 3 6 9 12
>>>>>>>> x x x x
>>>>>>>> 1 + --- + --- + --- + --- + ...
>>>>>>>> 3! 6! 9! 12!
>>>>>>
>>>>>> if I calc. correctly it is
>>>>>> f(x) = 1/3 sum(exp(ax)) where the sum is over the cube roots of 1.
>>
>>>>> sum(n=0,oo) e^3x / (3n)!
>>>>>
>>>>>> note D^3(f) = f and the series gives intial conditions...
>>>>>>
>>>>> Set r = (-1 + i.sqr 3)/2; r^2 = (-1 - i.sqr 3)/2; r^3 = 1
>>>>>
>> 1 + r + r^2 = 0
>>
>>>>> Let f(x) = a.e^x + b.e^rx + c.e^rrx; f"'(x) = f(x)
>>>
>>> You can get three equations in three unknowns with
>>>
>>> f(0) = 1 (you are already using this for a + b + c = 1)
>>> f'(0) = 0
>>> f''(0) = 0
>>>
>> 1 = f(0) = a + b + c
>> 0 = f'(0) = a + br + crr
>> 0 = f"(0) = a + brr + cr
>>
>> 0 = b(rr - r) + c(r - rr) = (r - rr)(c - b)
>> = r(1 - r)(c - b)
>> r, 1-r /= 0; b = c
>>
>> 0 = a + b(r + rr) = a - b
>
> Next step: solve these equations to get that a = b = c = 1/3.
>
Trivial.
> Last step: f(x) = 1/3 ( e^x + e^{r x} + e^{r^2 x} )
>
> = 1/3 ( e^x + 2 e^{-x/2} cos(sqrt(3)/2 x) )
Euler's famous equation. By golly, f(x) is real valued.
Bill Dubuque
>On May 29, 3:56 pm, Bill Dubuque <w...@nestle.csail.mit.edu> wrote:
>>Bill Taylor <w.tay...@math.canterbury.ac.nz> wrote:
>>>
>>> Here is a neat little problem in undergrad calculus. Maybe 2nd year.
>>>
>>> I've never seen it in a text-book exercise section,
>>> I don't know why, coz it's very cute.
>>>
>>> Sum the series SUM[0:(3):inf] x^n/n!
>>>
>>> Let's see if it comes out in ascii...
>>
>>> 3 6 9 12
>>> x x x x
>>> 1 + --- + --- + --- + --- + ...
>>> 3! 6! 9! 12!
>>
>> See my prior posts on series bisections and multisections
>> http://google.com/groups?selm=y8z8x5h7l1p.fsf%40nestle.csail.mit.edu
>> http://google.com/groups?selm=y8zslu8kvfm.fsf%40nestle.csail.mit.edu
>
> BETTER:
>
> your first post you reference is wrong
I disagree.
> and your second forwards to a google search that returns four posts:
> - the two already listed (yeah! self-referential!)
> - a road trip to dubuque iowa
> - and a penis bisection also with dubuque iowa
Surely you must be interested in at least one of the above.
Of course the link was composed before the spammers flooded
us with posts containing many city names (including Dubuque).
Now I specify more precise links to search my prior posts.
> but you wisely have me killfiled and will never know...
Maybe SIMPLIFiled, never killfiled (for anyone).
--Bill Dubuque
Bill Dubuque
For elegance one should emphasize the multisection genesis.
See my old math-fun posts below for further discussion.
--------------------
Bill Gosper <rwg> wrote to math-fun on Mon, 14 Apr 1997:
:
: every 4th term of sin(x)/e^x is 0, who has the shortest reason?
Dean Hickerson replied:
|
| Let f(x) = sin(x)/e^x = SUM a[n] x^n.
|
| Then f(x) + f(ix) + f(-x) + f(-ix) = 4 SUM a[4n] x^(4n); ...
This is called a series "multisection", e.g. see Riordan's classic
text "Combinatorial Identities" for many applications. More generally,
with w := a primitive n'th root of unity, the m'th n-section selects
the linear progression of m+k*n indexed terms from a series (GF):
a[m] X^m + a[m+n] X^(m+n) + a[m+2n] X^(m+2n) + ...
= 1/n [ f(X) + f(Xw)/w^m + f(Xw^2)/w^2m + ... + f(Xw^(n-1))/w^((n-1)*m) ]
To show the 0'th 4-section of sin(x)/e^x is zero requires
only some trivial operational calculus and galois theory:
f(x) = sin(x)/e^x
= c (g(x) - g(ix)) where c = -i/2, g(x) = e^((i-1)x)
= c (1 - S) g(x) where S is the linear op: x -> ix
But note 1 + S + S^2 + S^3 annihilates any series of this form
since f(x) + f(ix) + f(-x) + f(-ix)
= (1 + S + S^2 + S^3) f(x)
= (1 + S + S^2 + S^3) c (1 - S) g(x)
= (1 - S^4) c g(x) via cS = Sc
= 0, since S^4 = 1 via i^4 = 1
Cognoscenti will note Hilbert's Theorem 90 lurking in the background
(which has many elegant applications, even to elementary problems, e.g.
a direct derivation of the parameterization of Pythagorean triples).
Note that the above approach -- unlike Dean's derivation -- does
not require any knowledge of identities satisfied by f(x). For
complicated f(x) satisfying hairy identities, one quickly becomes
lost in complexity if one fails to notice the Galois theoretic
structure and instead tries to employ the specific identities
satisfied by f(x). It is rather amazing how much simplification
can result from noticing such trivial Galois theoretic structure
(often systematicized via cohomological calculations).
Rich Schroeppel wrote to math-fun on 15 Apr 1997:
|
| f(x) = sin(x)/e^x = Imag( e^(i-1)x ) and (i-1)^4 is real.
But the above approach is neither as simple nor as general
as the approach I outlined earlier, namely: for any generating
function f(x), to show that every 4th term vanishes it suffices
to find a witness g(x) such that f(x) = g(x) - g(ix). In the
case at hand the witness is simply g(x) = -i/2 e^((i-1)x). The
proof consists simply of checking f(x) = g(x) - g(ix).
However, in the JC/DA/RCS approach one needs to go further
than this: after checking f(x) = Im( e^((i-1)x) )
one then needs to extract the 4k'th coefficient, and
this involves special knowledge of the exponential
function (i.e. either Taylor's theorem or, equivalently,
the binomial theorem). For a function g(x) more exotic than
e^cx this second step might be very hairy or even intractable
(vs. my approach, which works for any g(x)). We even have
empirical evidence that this second step is more complex
since most other solvers committed errors here even in this
simple case where g(x) = e^cx !
Moreover, my approach easily extends to a method for showing
that every n'th term is 0, using a primitive n'th root of unity.
E.g. in the trivial case n=2 it says that to prove f(x) is odd
one need only exhibit a witness g(x) where f(x) = g(x) - g(-x);
a prototypical example: the proof that odd-indexed Bernoulli
numbers are zero (except b_1 = -1/2). Thus this technique is
a natural higher-order generalization of well-known techniques
for even and odd functions (aka "bisections" of power series).
Dean Hickerson wrote to math-fun on 15 Apr 1997:
|
| I disagree. Dan/John/Rich's approach is certainly simpler than yours.
I think you have missed the point of my approach. I interpreted
Gosper's request for the "shortest reason" as a request for a
"succinct certificate of proof", just like certificates of Pratt
for primality and WZ (Wilf-Zeilberger) for definite sums/integrals.
Interpreted in this manner, the method I proposed is clearly
much shorter and more general than any of the other approaches:
the ONLY computation needed is to check f(X) = g(X) - g(iX),
i.e. sin(X)/e^X = -i/2 e^((i-1)X) - -i/2 e^((i-1)iX).
All other solutions involve (much) more calculation than this
(witnessed by the erroneous calculations of most other solvers!)
| It only works if you can find your 'witness', which also
| requires special knowledge of the function involved.
For the purposes of proof-checking it matters not how I obtain
the witness g(X), but in fact it can easily be constructed by
inverting (S-1) mod (S^4-1)/(S-1), e.g. via Euclid's algorithm.
It requires no special knowledge of the function f(X) -- just
effective computation in Q[w,f(Xw^j)], as would any approach.
If you interpret "shortest reason" in some other manner which
also includes measure of supporting proofs then it would be much
harder to come up with an objective measure of complexity for
"shortest" and the discussion will soon bog down in arguments
comparing apples and oranges. E.g. just as you object to my
pulling out of a hat the witness g(X), I will object to the
use of any specific properties of the exponential function
that were employed in the alternative proof (viz. Taylor's
theorem or, equivalently, the binomial theorem, etc). Who is
to say whether my use of the Euclidean algorithm is more
"complex" than the alternative use of Taylor's theorem, etc?
The proof-checking model has the advantage that it avoids these
difficult issues and, moreover, it is a useful measure for many
practical purposes (e.g. said primality or WZ certificates).
--Bill Dubuque
galathaea
> galathaea <gala...@gmail.com> wrote:
> >On May 29, 3:56 pm, Bill Dubuque <w...@nestle.csail.mit.edu> wrote:
> >>Bill Taylor <w.tay...@math.canterbury.ac.nz> wrote:
> >>>
> >>> Here is a neat little problem in undergrad calculus. Maybe 2nd year.
> >>>
> >>> I've never seen it in a text-book exercise section,
> >>> I don't know why, coz it's very cute.
> >>>
> >>> Sum the series SUM[0:(3):inf] x^n/n!
> >>>
> >>> Let's see if it comes out in ascii...
> >>
> >>> 3 6 9 12
> >>> x x x x
> >>> 1 + --- + --- + --- + --- + ...
> >>> 3! 6! 9! 12!
> >>
> >> See my prior posts on series bisections and multisections
> >> http://google.com/groups?selm=y8z8x5h7l1p.fsf%40nestle.csail.mit.edu
> >> http://google.com/groups?selm=y8zslu8kvfm.fsf%40nestle.csail.mit.edu
> >
> > BETTER:
> >
> > your first post you reference is wrong
>
> I disagree.
in that post
you claim that
oo
--- j
\ (-1) nj+m
/ ------- x
--- (1)
j=0 nj+m
can be expressed in linear terms of e^(w x)
where w is an nth root of unity
i corrected in a follow up to that post
that it required linear terms of e^(w x)
where w was a __2n-th__ root of unity
the nth roots will get you the generalised hyperbolic trigonometry
but it takes the 2nth roots
to get the alternating series character
is this wrong?
> > and your second forwards to a google search that returns four posts:
> > - the two already listed (yeah! self-referential!)
> > - a road trip to dubuque iowa
> > - and a penis bisection also with dubuque iowa
>
> Surely you must be interested in at least one of the above.
> Of course the link was composed before the spammers flooded
> us with posts containing many city names (including Dubuque).
> Now I specify more precise links to search my prior posts.
well the four posts included
one wrong post
2 extraneous posts
and 1 link back to the same post
have i missed something?
i've done a separate search on your posts in this area
and i've only found two additional
one is on a good approximation
that is due to it being a multisection
(but the fact is only used to point to the gap in exponents)
the other is on multisecting product series
this seems more interesting
but it doesn't add anything more to my toolset
above and beyond my multisection of q-hypergeometrics and q-series
q-pochhammers build in the multiplicative structure
and the classic log translation makes the series multisections
isomorphic
(given proper bases)
> > but you wisely have me killfiled and will never know...
>
> Maybe SIMPLIFiled, never killfiled (for anyone).
i would love to read better posts of yours on this
seriously
you are one of the "cream" posters that visit this newsgroup
and you always have significantly deeper observations than most
but i've been asking around here since 2003
about my hypergeometric multisections
and the geometric and algebraic results that follow
and i have been adding my own regular insights
exploring the generalised trigonometrics
giving many formulae for them
asking like crazy for references
no one ever gives any
it all seems so easy
that i am absolutely sure
this must have been discovered many times before
and now even
twice in one year
i've seen these posts on the start of this theory
but where's the developed theory?
doesn't this seem like good exercise
for at least an appendix
and likely a chapter or two
in an advanced trigonometry text?
or a book on fourier analysis?
exercise in the generalised product and sum formulae
seem like great exercises for the advanced student
that likes to be challenged to see bigger things
the multisection itself is just
a type of finite fourier transform
and the generalised trigonometric analyses
give much information on the differential structure
is this out there?
please
if you have any reference
i would love not to have to duplicate what is already done
so i can focus on newer more productive things...
galathaea
i've regularly associated my results on hypergeometrics
to the kummer theory of their field of functions
this is the first time i've seen someone else mention something
similar
is this known?
is there a paper or book that explores this in general?
something that shows the cohomological implications of
for instance
|m |m-1
D | = | D
|n |n
or
|m |m+1
x | = | x
|n |n
as operators on the functions space
or even something as basic as the generalised trigonometrics
arising as solution matrices A to
|0 1 0 0 ... 0 |
D A = |0 0 1 0 ... 0 | A
| ........... |
|1 0 0 0 ... 0 |
and the relationship to permutations and circulants?
if there is anything at all like that
please let me know anything you can about it
Herman Rubin
Bill Taylor <w.ta...@math.canterbury.ac.nz> wrote:
>Here is a neat little problem in undergrad calculus. Maybe 2nd year.
>I've never seen it in a text-book exercise section,
>I don't know why, coz it's very cute.
>Sum the series SUM[0:(3):inf] x^n/n!
>Let's see if it comes out in ascii...
> 3 6 9 12
> x x x x
>1 + --- + --- + --- + --- + ...
> 3! 6! 9! 12!
But that is not likely to be the answer wanted.
What I think eas wanted was
[exp(x) + exp(omega*x) + exp(omega^2*x)]/3,
where omega and omega^2 are the two primitive cube
reoots of unity, or possible even
[exp(x) + 2*exp(-,5*x)*cos(x*sqrt(.75)]/3.
--
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, Department of Statistics, Purdue University
hru...@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558
Bill Dubuque
>Bill Dubuque wrote:
>>galathaea <gala...@gmail.com> wrote:
>>>On May 29, 3:56 pm, Bill Dubuque <w...@nestle.csail.mit.edu> wrote:
>>>>Bill Taylor <w.tay...@math.canterbury.ac.nz> wrote:
>>>>>
>>>>> Here is a neat little problem in undergrad calculus. Maybe 2nd year.
>>>>>
>>>>> I've never seen it in a text-book exercise section,
>>>>> I don't know why, coz it's very cute.
>>>>>
>>>>> Sum the series SUM[0:(3):inf] x^n/n!
>>>>>
>>>>> Let's see if it comes out in ascii...
>>>>
>>>>> 3 6 9 12
>>>>> x x x x
>>>>> 1 + --- + --- + --- + --- + ...
>>>>> 3! 6! 9! 12!
>>>>
>>>> See my prior posts on series bisections and multisections
>>>> http://google.com/groups?selm=y8z8x5h7l1p.fsf%40nestle.csail.mit.edu
>>>> http://google.com/groups?selm=y8zslu8kvfm.fsf%40nestle.csail.mit.edu
>>>
>>> BETTER:
>>>
>>> your first post you reference is wrong
>>
>> I disagree.
>
> up to that post that it required linear terms of e^(w x)
> where w was a __2n-th__ root of unity
You're being too pedantic. When I said n'th roots of unity I was
speaking generically. Perhaps I should have said m'th roots of 1
to avoid confusion.
>>> and your second forwards to a google search that returns four posts:
>>> - the two already listed (yeah! self-referential!)
>>> - a road trip to dubuque iowa
>>> - and a penis bisection also with dubuque iowa
>>
>> Surely you must be interested in at least one of the above.
>> Of course the link was composed before the spammers flooded
>> us with posts containing many city names (including Dubuque).
>> Now I specify more precise links to search my prior posts.
>
> well the four posts included one wrong post, 2 extraneous posts
> and 1 link back to the same post, have i missed something?
Yes, as I said, the link was correct when it was composed.
The spam posts that match my wildcard only arrived much later.
I've since modified my wildcard searches. I purposely post
those wildcards so they'll also match _future_ posts too
but I never anticipated "penis bisections in Dubuque Iowa".
> i've done a separate search on your posts in this area
> and i've only found two additional
>
> one is on a good approximation
> that is due to it being a multisection
> (but the fact is only used to point to the gap in exponents)
>
> the other is on multisecting product series
>
> this seems more interesting
> but it doesn't add anything more to my toolset
> above and beyond my multisection of q-hypergeometrics and q-series
I'm not disputing that hypergeometrics are interesting.
However they only obfuscate the basic multisection ideas.
See also my other recent post for much further discussion.
> q-pochhammers build in the multiplicative structure and the
> classic log translation makes the series multisections
> isomorphic (given proper bases)
>
>>> but you wisely have me killfiled and will never know...
>>
>> Maybe SIMPLIFiled, never killfiled (for anyone).
>
> i would love to read better posts of yours on this
If you're wondering why I don't often reply to your posts
it's because I find your style of writing too much effort
to parse quickly. There's a time for poetry and a time
for discussing math. This forum is dedicated to the latter.
Hypergeometrics for poets probably will not catch my eye.
I'm reminded of Joe Roberts book Elementary Number Theory.
There are many nice results in this book but unfortunately
they are greatly obfuscated by the typography (calligraphy!)
> seriously you are one of the "cream" posters that visit this newsgroup
> and you always have significantly deeper observations than most
Thanks, I'm happy that you enjoyed some of my posts.
> but i've been asking around here since 2003
> about my hypergeometric multisections
> and the geometric and algebraic results that follow
> and i have been adding my own regular insights
> exploring the generalised trigonometrics
> giving many formulae for them
Yes, obviously multisections can be reformulated in many
equivalent ways (linear algebra, fourier transforms, etc)
but I tried to keep it simple for pedagogical purposes.
> asking like crazy for references no one ever gives any
I worked on hypergeometrics a bit when I was working on
a general theory of special functions for Macsyma. You
may want to look up some work of Yannis Avgoustis at MIT.
His thesis has references to many of the classics. More
recently much research has been motivated by the search
for effective algorithms in modern computer algebra
systems. You might try looking in the Jnl. of Symbolic
Computation and various conference proceedings on
computer algebra. Iirc Wolfram research recruited a number
of Russian mathematicians to work on this for Mathematica.
That's why Mma is strong in this area.
> it all seems so easy
> that i am absolutely sure
> this must have been discovered many times before
> and now even
> twice in one year
> i've seen these posts on the start of this theory
> but where's the developed theory?
>
> doesn't this seem like good exercise
> for at least an appendix
> and likely a chapter or two
> in an advanced trigonometry text?
> or a book on fourier analysis?
>
> exercise in the generalised product and sum formulae
> seem like great exercises for the advanced student
> that likes to be challenged to see bigger things
>
> the multisection itself is just
> a type of finite fourier transform
> and the generalised trigonometric analyses
> give much information on the differential structure
>
> is this out there?
>
> please
> if you have any reference
> i would love not to have to duplicate what is already done
> so i can focus on newer more productive things...
I recall seeing some publications on related topics.
If you really want to know you should ask Dick Askey.
He's a walking encyclopedia on special functions.
--Bill Dubuque
alainv...@gmail.com
>
> - Afficher le texte des messages précédents -- Masquer le texte des messages précédents -
>
> - Afficher le texte des messages précédents -
Bonjour,
I wonder :being given a convergente series or a polynomial
for instance exp(x) = 1+x/1!+x^2/2!+ ... or 1 +2x +3x^2 +
7x^8+x^9+2x^12
does it exist a sort of 'comb' to extract the wished power,
say 3*n that is 1 + x^3/3! + x^6/6! or 1 +x^9+2x^12
Thanks for help,
Alain
Rob Johnson
alainv...@gmail.com wrote:
>I wonder :being given a convergente series or a polynomial
>for instance exp(x) = 1+x/1!+x^2/2!+ ... or 1 +2x +3x^2 +
>7x^8+x^9+2x^12
>does it exist a sort of 'comb' to extract the wished power,
>say 3*n that is 1 + x^3/3! + x^6/6! or 1 +x^9+2x^12
In the case of 3*n, you can get what you want with
f(x) + f(a x) + f(a^2 x)
------------------------
3
where a = exp(i2pi/3) is a cube root of 1.
Christian Loriau
-Christan Loriau
alainv...@gmail.com
> In article <f73931fe-4f05-4078-b25e-71218e054...@p25g2000hsf.googlegroups.com>,
>
> alainvergh...@gmail.com wrote:
> >I wonder :being given a convergente series or a polynomial
> >for instance exp(x) = 1+x/1!+x^2/2!+ ... or 1 +2x +3x^2 +
> >7x^8+x^9+2x^12
> >does it exist a sort of 'comb' to extract the wished power,
> >say 3*n that is 1 + x^3/3! + x^6/6! or 1 +x^9+2x^12
>
> In the case of 3*n, you can get what you want with
>
> f(x) + f(a x) + f(a^2 x)
> ------------------------
> 3
>
> where a = exp(i2pi/3) is a cube root of 1.
>
> Rob Johnson <r...@trash.whim.org>
> take out the trash before replying
> to view any ASCII art, display article in a monospaced font
Bonsoir,
are there other extensions of forms :
sum(a(i)f(b(i)x))
for instance :
the case 1 +x^1 + x^4 +x^9 + x^16 ;..
Thanks,
Alain
galathaea
yes
such "filters" or "combs" exist
they are all projection operators
and can be constructed formally using the multisection operators
for any collection C subcollectionOf N
you can form the formal operator P(C)
by using the order on C to selectively
include and exclude members
one example algorithm is:
- for the first element c0 of C start with
|c0
|
|2 c0
- now for each successor element cj
and for each element dj of N that is not a successor
check to see if it is already included in the projection
if it is at the right "valence" do nothing
if it there is a discrepancy B
add a linear correction term
|cj |dj
B | or |
|2 cj |2 dj
the convergence properties of this algorithm
depend upon the collection C being used
and the function space on which these operators act
and the algorithm may be optimised in several ways
but over the domain of validity
the end operators arrived at in the end
can be used to derive properties
in a simple extension of multisection methods
galathaea
> I'm not disputing that hypergeometrics are interesting.
> However they only obfuscate the basic multisection ideas.
> See also my other recent post for much further discussion.
it's weird
i was recently contacting michael somos
(who has a web site on multisecting q-series)
and he also said he thought hypergeometrics obscure the details
i think that couldn't be further from the truth
yes there is a wider theory of multisection
(i still can't believe you mentioned hilbert's theorem 90
i had not planned to mention it for another month
though i did tease some with the galois theory of post algebras
to kind of prepare the story for the
circulants and differentials to come out and play
and had intended to start illustrating
some of the wider product rule applications
like in the besselian)
but the hypergeometrics have a differential structure
tailor-made to take advantage of multisection features
i mentioned the commutation-like relations
for the three operator types
|, D, and x
now look at the atheistic hypergeometric diffy-que
(xD + b1 - 1)(xD + b2 - 1)...(xD + bn - 1) H (..; x) =
m n
x (xD + a1)(xD + a2)...(xD + am) H (..; x)
m n
the use of the xD operator here
means the left hand side can slide the multisection
all the way to operate on the function
and the right hand side only has the one unpaired x
and the right most slides to |^(m-1)_n
there is therefore a natural differential structure
over the hypergeometrics
above and beyond what arises from any multisection
which only relates the derivative of the (m,n)-multisection
to the (m-1, n)-multisection of the derivative
the hypergeometrics have differential structure
relating the multisections together
the simplest hypergeometric
oo
---
/ - \ \ 1 j
H | ; x | = / ---- x
0 1 \ 1 / --- (1)
j=0 j
that started this thread
has multisections that are themselves natural hypergeometrics
/ - / x \n \
H | ; | - | |
0 n \ m/n, (m+1)/n, ..., (m+n-1)/n \ n / /
those who study hypergeometrics
know these fractional argument forms are integral to the theory
and yet i have never seen a source that points to their origin
> > i would love to read better posts of yours on this
>
> If you're wondering why I don't often reply to your posts
> it's because I find your style of writing too much effort
> to parse quickly. There's a time for poetry and a time
> for discussing math. This forum is dedicated to the latter.
> Hypergeometrics for poets probably will not catch my eye.
> I'm reminded of Joe Roberts book Elementary Number Theory.
> There are many nice results in this book but unfortunately
> they are greatly obfuscated by the typography (calligraphy!)
you mentioned hilbert's 90
very matter-of-factly
this little point
though
this interpretation of the structure of cyclic extension
built into the simpson multisection theorem
plays an important in a number of the relations i have been finding
and i am very shocked to see it mentioned
so quickly after my simple obscure post algebra hint
so i doubt it is related
i deeply want to know
where you have seen this
or at least what it was that made you see the connection
i have yet to see the foundational use of cumulants
in the study of cyclic extensions
but that is exactly what all this amounts to
(and i didn't want to reveal that part
until after i had finished the major tchebyshef development
but now i'll have to rearrange things some.. :( )
i'm sorry i'm too broken for communication
i have tried as hard as possible
to repair myself to the best of my ability
but even still i am broken in many important ways
> I recall seeing some publications on related topics.
> If you really want to know you should ask Dick Askey.
> He's a walking encyclopedia on special functions.
i will take that advice very seriously