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        riddles >> putnam exam (pure math) >> Summation
        
(Message started by: ThudanBlunder on Sep 10th, 2007, 10:10am)
Title: Summation
Post by ThudanBlunder on Sep 10th, 2007, 10:10am
                                                                                                     
Let F(n) = 2n + 2/3 - enhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif(k - n)ke-k/k! for k = 0 to n-1 and n = 1,2,3,4,.......

a) Prove that F(n) http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bigto.gif 0 as n http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bigto.gif http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/infty.gif

b) Find F(1000) to 3 significant figures.

c) Find the smallest m such that |F(m)| < |F(m+1)|
Title: Re: Summation
Post by Eigenray on Sep 10th, 2007, 5:43pm
The poor convergence makes me think Poisson summation would be in order.

Anyway, let f(n,k) = (n-k)k e-k/k!.  We want to approximate http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif (-1)k f(n,k).

For fixed n, f(n,k) is maximized when

f(n,k+1) ~ f(n,k), or (n-k-1)k+1/(k+1) ~ e(n-k)k.

Letting k = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/mu.gifn, we want

(1 - 1/[n(1-http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/mu.gif)])http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/mu.gifn(n-k-1)/(k+1) ~ e.

Letting n http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/to.gif http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/infty.gif, we get

e-http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/mu.gif/(1-http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/mu.gif) (1-http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/mu.gif)/http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/mu.gif= e

which is to say, http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/mu.gif/(1-http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/mu.gif) = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/lambda.gif= [link=http://en.wikipedia.org/wiki/Lambert_W_function]W[/link](1/e) ~ 0.2785, or http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/mu.gif ~ 0.2178.

Now, the maximum value is about f(n, http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/mu.gifn) ~ ehttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/lambda.gifn/http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif{2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/mu.gifn}.

Plotting f(n, k), it looks like a constant times normal, with mean http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/mu.gifn, and variance ~ http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/nu.gifn, where http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/nu.gif ~ 0.1332? (I don't know what http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/nu.gif is.)  That is, f(n,k) ~ C N(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/mu.gifn, http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/nu.gifn).  Considering the max, ehttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/lambda.gifn/http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif{2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/mu.gifn} = C/http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif{2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/nu.gifn}, so C = ehttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/lambda.gifn http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif{http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/nu.gif/http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/mu.gif}.  Or, in other words,

gn(x) = f(n, kx) ~ ehttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/lambda.gifn/n http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif{http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/nu.gif/http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/mu.gif} N(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/mu.gif, http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/nu.gif/n),

and we are interested in http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif(-1)k gn(k/n).  But I don't think this is a good enough approximation.  It looks pretty lousy, actually.  Eh.
Title: Re: Summation
Post by Eigenray on Sep 10th, 2007, 6:38pm
I'm guessing it has something to do with

http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gif01 en cos(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gifnx) (1/x - 1)nx (2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gifnx)-1/2 dx.

It seems to converge to 1, but it's a bit tricky to approximate.
Title: Re: Summation
Post by Eigenray on Sep 11th, 2007, 2:35pm
This might be useful: define

f(x) = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gifk<x (-1)k (x-k)k ex-k/k!.

This looks a lot like some sort of generating function.  So look at

f'(x) = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif(-1)kex-k/k! { (x-k)k + k(x-k)k-1 } = f(x) + http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif(-1)kex-k/(k-1)! (x-k)k-1
= f(x) - http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif(-1)rex-1-r/r! (x-1-r)r
= f(x) - f(x-1)

So f'(x) = f(x) - f(x-1), or f' = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/cdelta.giff, where http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/cdelta.gif is the difference operator.  That is, the slope of the tangent at each point is the same as the secant going one back.  Interesting.

So f'=http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/cdelta.giff, with f(x)=ex for x in [0,1], and we want to show f(x) ~ 2x + 2/3.  But I've never been very good with differential equations.


Integrating f'=http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/cdelta.giff, from 1 to x, gives

f(x) - f(1) = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifx-1x f(t)dt  -  http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gif01 f(t)dt,

or

f(x) = 1 + http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifx-1x  f(t)dt.

I don't know if that helps either.
Title: Re: Summation
Post by Eigenray on Sep 11th, 2007, 3:27pm
Well, I can show one thing.  Intuitively, since f' = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/cdelta.giff, f should straighten out to be close to linear.  So suppose f'(x) http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/to.gif c.  We show c=2.

For x>0, and t>0, set

gx(t) = [f(x+t) - f(x)]/t  =  f'(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/xi.gif)

for some http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/xi.gif in [x, x+t], by the intermediate value theorem.  By assumption then, gx(t) = c + hx(t), where hx converges to 0, uniformly in t, as x http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/to.gif http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/infty.gif.  Now,

f(x+t) = f(x) + tc + thx(t),

and so

f(x) + c + hx(1) = f(x+1)
= 1 + http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gif01 f(x+t)dt
= 1 + f(x) + c/2 + http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gif01  thx(t)dt.

Letting xhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/to.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/infty.gif gives c=2.

[f(x) is not actually differentiable everywhere, but if you read the next post, it suffices to work with G(x) = F(x) - x2, which is differentiable.  So if we can show that g = G' converges to a constant, then that constant must be 2/3, and the result follows.]
Title: Re: Summation
Post by Eigenray on Sep 11th, 2007, 4:03pm
Aha, but we can repeat the argument!

Let g(x) = f(x) - 2x.  Then g satisfies the same differential equation: g' = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/cdelta.gifg, and the integral equation becomes

g(x) = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifx-1x g(t)dt.

Now, let G be an antiderivative of g.  Then the above becomes G' = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/cdelta.gifG, and so as before

G(x) = G(1) - http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gif01 G(t)dt  + http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifx-1x G(t)dt
= 1/3 + http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifx-1x  G(t)dt,

since G(x) = ex - x2 on [0,1].  The same argument as for f now gives G(x) ~ 2/3 x + c'.  Which is to say, g ~ 2/3, and finally f(x) ~ 2x + 2/3.

So all we need to prove is: suppose g(x) = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifx-1x g(t)dt.  Then g(x) converges to a finite value, namely

2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gif01 [ g(t) - http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gif0t g(u)du ] dt.

Intuitively, this is much more clear than the original problem: g(x) is simply the average of g(t) over [x-1,x].  How could it possibly not converge  ;)
Title: Re: Summation
Post by Sameer on Sep 11th, 2007, 4:26pm
The last term in F(n) looks very similar to a Discrete(?) Laplace transform!! F(s) = Integral f(t)*e-st dt
Title: Re: Summation
Post by Eigenray on Sep 11th, 2007, 4:57pm
Laplace!!

Let fk(t) = (t-k)ket-k/k!, when t>k, and 0 otherwise.  Then the Laplace transform works out to be

Fk(s) = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifkhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/infty.gif fk(t) e-st dt
= e-ks/(s-1)k+1,

and the Laplace transform of f(t) = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif(-1)kfk(t) is the sum of the geometric series

F(s) = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif(-1)kFk(s)
= 1/(s-1+e-s)
= 2/s2 + 2/(3s) + 1/18 - s/270 + ...
= 2/s2 + 2/(3s) + G(s).

Inverse Laplacing,

f(t) = 2t + 2/3 + g(t),

and presumably g(t) http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/to.gif 0.  G(s) is meromorphic, with infinitely many poles, at each non-zero solution to e-s=1-s:

s = -log(a+http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/delta.gif) http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pm.gif (a-http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/epsilon.gif)i,  a=2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gifk+http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif/2, http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/epsilon.gif ~ (1+log a)/a, http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/delta.gif ~ [(log a)2-1]/2a,  k http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/in.gif http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbz.gif+,

all in the left half-plane.  But I'm not familiar enough with the Laplace transform to know how g(t) behaves.
Title: Re: Summation
Post by Sameer on Sep 11th, 2007, 5:12pm
For G(s) you get s in the numerator???? That will make the system unstable and non-convergent!!!  I will take a closer look when I get home!!
Title: Re: Summation
Post by Eigenray on Sep 11th, 2007, 6:59pm
g(t) =1/(2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif) http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gif-http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/subinfty.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/supinfty.gif G(iu)eiutdu,

which is essentially the Fourier transform of G(iu).  G(iu) isn't in L1, so the above isn't actually integrable, and we can't use the Riemann-Lebesgue lemma, but since G(iu) is in L2 and analytic, I think it follows that g(t) goes to 0, decaying exponentially in fact.

Actually, we are only interested in

Re g(t) = 1/(2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif) http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gif [Re G(iu) cos(ut) - Im G(iu) sin(ut)]dt,

and Re G(iu) is in L1, since Re[1/(iu-1+e-iu)] ~ (cos u - 1)/u2, and Re[2/(iu)2 + 2/(3iu)] = -2/u2.

So the only issue is whether http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gif Im G(iu)sin(ut)dt  goes to 0.
Title: Re: Summation
Post by Eigenray on Sep 12th, 2007, 5:29am
Whahahaha:

Recapitulating,

G(z) = 1/(z-1+e-z) +2/(3z) - 2/z2,

and we want to compute, for t>0,

g(t) = 1/(2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gifi) http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gif-ihttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/subinfty.gifihttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/infty.gif  G(z)eztdz.

Fix M>0, and let N>0 be large.  Consider the rectangular contour going from -iM http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/to.gif iM http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/to.gif -N+iM http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/to.gif -N-iM http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/to.gif -iM.  The integral along the right edge is what we want to evaluate.  Waving my hands, the integral along the other three sides should be small.  This just leaves the residues inside the rectangle.

G(z)ezt has a pole at z=c, where e-c=1-c.  The residue there is

limz->c  (z-c)G(z)ezt  =  ect / limz->c (z-1+e-z)/(z-c)
= ect / (1 - e-c )
= ect / c.

Now the poles come in conjugate pairs, and adding up the residues inside the rectangle, and letting M go to infinity, we get:

g(t) = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gifc  2Re[ ect/c ]   (*)

where the sum is over conjugate pairs of non-zero c satisfying e-c=1-c.  That is,

c = {-2.09 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pm.gif 7.46 I,  -2.66 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pm.gif 13.9 I,  -3.03 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pm.gif 20.22 I,  -3.29 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pm.gif 26.54 I,  ...}

Let gn(t) be the sum of the first n terms of the Fourier series (*), and let hn(t) = f(t) - 2t - 2/3 - gn(t).  Attached are plots of g=h0, h1, h2, and h3, in black, red, green, and blue, respectively.  (Only h1,h2 are shown in both.)  Hooray Fourier series!  Finally,

-F(1000) = g(1000) = 2Re[e1000c1/c1 + e1000c2/c2 + ...]
= 1.1469767*10-909 - O(10-1158).

Note that this makes sense: the functions y(x)=x, y(x)=1, and y(x)=ecx, where c=1-e-c, are all solutions to y'(x) = y(x) - y(x-1), and f(x) is a linear combination of these:

f(x) = 2x + 2/3 + http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif ecx/c.
~ 2x + 2/3 + 0.258*e-2.089 x cos(7.461 x + 1.298) + O(e-2.664 x).

And it appears n=800 is the first time |F(n)| < |F(n+1)|.
Title: Re: Summation
Post by Eigenray on Sep 12th, 2007, 7:15pm
Another problem that involved Fourier transform was [link=http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_putnam;action=display;num=1156417763]Alternating Sum[/link]: as z http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/to.gif 1-,

S(z) = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gifk=0http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/supinfty.gif  (-1)k z2^k
= 1/2 - (1-z)/3 + g(z) + 2/log 2 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif{m>0 odd} Re[ http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/cgamma.gif(mhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gifi/log 2)ex mhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gifi ],
where z = e-2^(-x), and g(z) = O((1-z)2).
Title: Re: Summation
Post by Barukh on Sep 12th, 2007, 9:08pm
What's your solution, T&B?
Title: Re: Summation
Post by ThudanBlunder on Sep 14th, 2007, 4:33pm

on 09/12/07 at 21:08:47, Barukh wrote:
What's your solution, T&B?

Although I did elementary Complex Variable at uni, it was a long time ago and I have forgotten most of it. (However, this does not stop me knowing a good problem when I see one.) So I won't write out the full solution as though I am a real mathematician like Eigenray.

But 'my' solution begins by letting
I(n,x) = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif(k-n)ke(k-n)x/k! = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gifDk(1/k!)e(k-n)x/Dxk between k = 0 and n

Thus F(n) = 2n + 2/3 - I(n,-1)

As e(k-n)x is an entire function in the complex plane we are now able to apply Cauchy's Theorem to prove that
F(n) = 2n + 2/3 - http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif(1/2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gifi)http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/oint.gife(k-n)x/(1+z)k+1.dz
where the contour integral must enclose z = -1, and sum over k explicitly.

Further details on request, but I believe Eigenray will be able to see the rest of this method for a)

(I wish the edit window wasn't so small and that we could save drafts of our work, so as not to lose it somehow before posting.)
Title: Re: Summation
Post by ThudanBlunder on Sep 15th, 2007, 9:49am
I will type my solution up in MathType as soon as I have time.
Title: Re: Summation
Post by Eigenray on Sep 15th, 2007, 5:15pm
Summing the (finite) geometric series, we get

F(n) = 2n + 2/3 + 1/(2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gifi) http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/oint.gif (1+z+ez)-1/(1+z)n dz,

since e-nz/(1+z-ez) is analytic at z=-1.  So by Cauchy's theorem, we get

1/(1+z-ez)  =  http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gifn=1http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/supinfty.gif  [F(n) - 2n - 2/3] (z+1)n-1,

and killing the pole at 0,

G(z) = 1/(1+z-ez) - 2/(3z) + 2/z2
 =  http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif F(n) (z+1)n-1.

It took me a while to realize that this actually proves (a)!  Hint: [hide]where are the poles of G[/hide]?  Amazing.  Complex analysis just seems a little too beautiful sometimes.

Now for (b) (and to get the Fourier series) we just play "Whac-A-Pole".


on 09/14/07 at 16:33:29, ThudanBlunder wrote:
(I wish the edit window wasn't so small and that we could save drafts of our work, so as not to lose it somehow before posting.)

Yes.  Unfortunately the textarea width and height are set in css, so something like [link=http://userscripts.org/scripts/show/600]TextareaResize[/link] doesn't work.  I couldn't even resize it using DOM Inspector, but there's probably a way.  Googling around, I put together the following, which seems to work okay:

Code:
var resizetextarea=function() {
   var a = Math.max(350, window.innerWidth*0.70 - 150);
   GM_addStyle("textarea{ width: "+a+"px !important; }");
}
resizetextarea();
window.addEventListener('resize', function() {resizetextarea(); }, true);

I just stuck it inside SMQ's Greasemonkey script.  Of course you can modify the parameters I used, and add something similar for height.  But I don't know much about this sort of thing; there should be a way to just remove the width and height styles, so that modifying the rows and cols attributes will work.
Edit: [link=http://userscripts.org/scripts/show/10140]Textarea_drag_resize[/link] works nicely.  But it could be customized to default to some affine function of the window's width and height.
Title: Re: Summation
Post by Barukh on Sep 16th, 2007, 9:52am

on 09/14/07 at 16:33:29, ThudanBlunder wrote:
Although I did elementary Complex Variable at uni,

... and I haven't done it at all. That's not good.

Eigenray, could you please elaborate on your "too beautiful" solution?

Also, is there good material to read on the subject?
Title: Re: Summation
Post by Eigenray on Sep 16th, 2007, 2:43pm

on 09/16/07 at 09:52:51, Barukh wrote:
Eigenray, could you please elaborate on your "too beautiful" solution?

Use the fact that [link=http://planetmath.org/encyclopedia/RadiusOfConvergenceOfAComplexFunction.html]the radius of convergence of a power series is the distance to the nearest singularity[/link].


Quote:
Also, is there good material to read on the subject?

Have a look at [link=http://www.math.upenn.edu/~wilf/DownldGF.html]Generatingfunctionology[/link], particularly section 5.2 (and section 2.4 for background).  The basic tool for all this is [link=http://planetmath.org/encyclopedia/CauchyIntegralFormula.html]Cauchy's Integral Formula[/link].

I know Stein & Shakarchi's Complex Analysis has a chapter on asymptotics, but I don't have my copy on me.  Googling on some keywords (complex analysis, generating functions, asymptotics) also turns up [link=http://algo.inria.fr/flajolet/Publications/anacombi2.ps.gz]this[/link], apparently part of the book [link=http://algo.inria.fr/flajolet/Publications/books.html]Analytic Combinatorics[/link], which seems pretty comprehensive.
Title: Re: Summation
Post by Sameer on Sep 16th, 2007, 3:07pm

on 09/16/07 at 09:52:51, Barukh wrote:
... and I haven't done it at all. That's not good.


Also, is there good material to read on the subject?


My Engineering Math text book used to have all this. Yet it has been a long time!! Maybe any Higher Engineering Mathematics should have this!! I should look for one too!! hmm...


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