Date: Wed, 01 Feb 1995 15:21:05 -0400
From: [Permission pending]
Subject: Re: HELP NEEDED...PLEASE!
To: rusin@math.niu.edu

First, thank you for your answer. it's the first time I ask a question in the
news, and I'm really surprized how fast I get an answer. 
concerning the x(i), it's a complicate expression (at least for a biologist...)
and are:

x(i)=v^(i-1)*Exp(-Lv)*Sum[((v-1)^(j-i)*L^j)/j!  ,j=i to Infinity]
which can also be written as
x(i)=v^(i-1)*(Exp(-L)-Exp(-Lv)*(Sum[((v-1)*L)^j/j!   ,j=0 to i-1])/(v-1)^i

with 0<v<1 and 0<L<infinity
note that it's a Poisson distribution when v tends to one.
the x(i) terms tends to zero pretty rapidely, unless L is to big.
the other thing I know from the sum of the x(i) is that it tends to 1-Exp(-L).
I realize that it's probably not a very easy function to estimate, but if you
think there is a possibility to do something...?

[sig deleted -- djr]

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Date: Sun, 5 Feb 95 02:05:00 CST
From: rusin (Dave Rusin)
To: [Permission pending]
Subject: Re: HELP NEEDED...PLEASE!

OK, let me see if I have this right: you've got fixed numbers  L  and  v,
and you've defined a sequence of numbers by

x(i)=v^(i-1)*Exp(-Lv)*Sum[((v-1)^(j-i)*L^j)/j!  ,j=i to Infinity]

Your question was to evaluate Sum[ x(i)/(i+1), i=1 to Infinity]
Is that right?

Here's what I'd do. Write the  x(i) in the form
	x(i) = A B^i Sum[ C^j/j!, j=i to Infinity]
where  A = exp(-Lv)/v,  B=(v/(v-1)),  and  C=(v-1)L. Then
the sum you need to evaluate is a double sum,
	S = Sum[ A B^i C^j / (i+1)j!,  i=1 to Infinity, j=i to Infinity]
	  = Sum[ A B^i C^j / (i+1)j!,  j=1 to Infinity, i=1 to j]
	  = (A/B) Sum[ C^j/j! Sum[B^i/i, i=2 to j+1], j = 1 to Infinity]
	  = (A/B) F(B,C), say.
Now, to evaluate this function  F(B,C), observe that when differentiated
w.r.t. B  it gives the expression
	G(B,C) = Sum[ C^j/j! Sum[B^i, i=1 to j], j=1 to infinity]
and the inner sum is just a geometric series,  (B^(j+1) - B) / (B-1). So
	G(B,C) = (B/(B-1)) Sum[ (CB)^j/j! ] - (B/(B-1)) Sum[ C^j/j! ]
	       = (B/(B-1)) [ Exp(BC) - Exp(C) ]
Then  F(B,C) is just the antiderivative of this. (More precisely, it's
the antiderivative which is =0  when  B=0.) Thus we may write
	F(B,C) = Integral[ (x/(x-1)) ( Exp(xC) - Exp(C) ) dx,  x=0 to B]
Then the original sum is  A/B  times this; playing with the integrals I get
(using  w=(1-x)*L(v-1) )
	SUM = exp(-L)(v-1)/v^2 Integral[(1-exp(-w))/w) dw, w=1 to L(v-1)]
		+(exp(-vL)-1+(Lv)exp(-L))/(Lv^2)

Now, this answer involves an integral which does not have a closed-form
antiderivative. However, this integral is well known, and even behaves
well around zero. Setting 
	H(x) = integral  (1-exp(-w)) / w  dw ,  w=1 to  x, 
Maple tells us this is ln(x)-Ei(x), where "Ei" is the name it assigns to
the exponential intgeral. In particular, it can evaluate this numerically.

In terms of  H,  the sum you asked about in the first place is

	SUM = (1/(Lv^2))[ exp(-L) (v-1)L H( (v-1)L ) + exp(-vL)-1+ Lv exp(-L) ]

How's that?

It is late and I may have made some sign errors but I think this is OK
Look it over and see if this agrees with what you know.

dave

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Date: Mon, 06 Feb 1995 09:58:34 -0400
From: [Permission pending]
Subject: Re: HELP NEEDED...PLEASE!
To: rusin@math.niu.edu

Well, it took me the sunday to try to understand your solution - I said before
I'm just a biologist...!- and it seems fine to me, Ijust didn't know that
exponential integral exist, but one step is still mysterious for me. I rewrite
your solution,numbering the steps:
1) x(i) = A B^i Sum[ C^j/j!, j=i to Infinity]
		where  A = exp(-Lv)/v,  B=(v/(v-1)),  and  C=(v-1)L.
2) S = Sum[ A B^i C^j / (i+1)j!,  i=1 to Infinity, j=i to Infinity]
3) = Sum[ A B^i C^j / (i+1)j!,  j=1 to Infinity, i=1 to j]
4) = (A/B) Sum[ C^j/j! Sum[B^i/i, i=2 to j+1], j = 1 to Infinity]
5) = (A/B) F(B,C), say.
6) G(B,C) = Sum[ C^j/j! Sum[B^i, i=1 to j], j=1 to infinity]
		and the inner sum is just a geometric series,
7) (B^(j+1) - B) / (B-1). So
8) G(B,C) = (B/(B-1)) Sum[ (CB)^j/j! ] - (B/(B-1)) Sum[ C^j/j! ]
9) = (B/(B-1)) [ Exp(BC) - Exp(C) ]
10) F(B,C) = Integral[ (x/(x-1)) ( Exp(xC) - Exp(C) ) dx,  x=0 to B]
11) SUM = exp(-L)(v-1)/v^2 Integral[(1-exp(-w))/w) dw, w=1 to L(v-1)]
		+(exp(-vL)-1+(Lv)exp(-L))/(Lv^2)
				  		(using  w=(1-x)*L(v-1) )
12) H(x) = integral  (1-exp(-w)) / w  dw ,  w=1 to  x, 
				Maple tells us this is 
13) ln(x)-Ei(x), 
14) SUM = (1/(Lv^2))[ exp(-L) (v-1)L H( (v-1)L ) + exp(-vL)-1+ Lv exp(-L) ]
 The step from Eqn 10 to 11 is still mysterious for me...and I,ve try with
mathematica the integral  (1-exp(-w)) / w  dw ,  w=1 to  x, anit tells me it is 
ln(x)-Ei(x)-Ei(1)
Now, if we take H(v-1)L),we have ln((v-1)L), but as 0<v<1, (v-1)L is
negative...?
BBut I will try to work on it and let you know if I find something ...
Thank you again,
Oliiver

[sig deleted]
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Date: Mon, 6 Feb 95 12:43:13 CST
From: rusin (Dave Rusin)
To: [Permission pending]
Subject: Re: HELP NEEDED...PLEASE!

Thanks for clarifying my original letter; yes, this is just fine. I've
also reviewed it and (I think) fixed all the sign errors.

>1) x(i) = A B^i Sum[ C^j/j!, j=i to Infinity]
>		where  A = exp(-Lv)/v,  B=(v/(v-1)),  and  C=(v-1)L.
>2) S = Sum[ A B^i C^j / (i+1)j!,  i=1 to Infinity, j=i to Infinity]
>3) = Sum[ A B^i C^j / (i+1)j!,  j=1 to Infinity, i=1 to j]
>4) = (A/B) Sum[ C^j/j! Sum[B^i/i, i=2 to j+1], j = 1 to Infinity]
>5) = (A/B) F(B,C), say.

We note the derivative of this  F  with respect to  B  is
>6) G(B,C) = Sum[ C^j/j! Sum[B^i, i=1 to j], j=1 to infinity]
>		and the inner sum is just a geometric series,
>7) (B^(j+1) - B) / (B-1). So
>8) G(B,C) = (B/(B-1)) Sum[ (CB)^j/j! ] - (B/(B-1)) Sum[ C^j/j! ]
>9) = (B/(B-1)) [ Exp(BC) - Exp(C) ]

So for any fixed  C,  F(B,C) is a function of  B  whose derivative is
f'(B) = (B/(B-1)) [ Exp(BC) - Exp(C) ];
equivalently,  f  is the antiderivative of this function of  B.
This is where the Fundamental Theorem of Calculus comes in. Students always
remember this as saying "to compute integral from  a  to  b  of  g(x) you
just let  G(x) be an antiderivative of  g  and evaluate  G(b) - G(a)."
That's fine, but another way of thinking of the theorem is that it says
"if  g(x)  is given then ' G(u) = integral from  0  to  u  of  g(x) dx ' 
defines a function of  u  with two properties: G(0) = 0  and  G'(u) = g(u)."
So we use that theorem to enable us to write "integral..." instead of
"antiderivative..." Indeed, in the problem at hand, f(0) = 0  and
f'(B) = blah blah means  f(B)  must be equal to the function thus created:

>10) F(B,C) = Integral[ (x/(x-1)) ( Exp(xC) - Exp(C) ) dx,  x=0 to B]
[I get my students to try out this notation with examples they already
know. For example, what's a function of  u  whose derivative is  2u?  Well,
by the theorem, it's  integral  0  to  u  of  2x dx. That integral they
always compute by finding an antiderivative of  2x  (namely  x^2)  and
then plugging in  u  and  0  to get  u^2. And indeed,  d(u^2)/du = 2u, as 
desired.]

Let me now adjust the rest of the steps. Let x = x1+1 in the integral. Then
11) F(B,C) =  Exp(C) Integral[ (1+ 1/x1) ( Exp(C x1) - 1 ) dx,  x1=-1 to B-1]

Break this integral into two parts: the first is just
12) I1 = Exp(C) [ Exp(C x1) /C  -  x , evaluated at x1 =B-1  and  x1=-1], i.e.
13) I1 = Exp(C) [ Exp(C(B-1))/C - (B-1) - Exp( -C) + (-1) ], or simply
14) I1 = Exp(CB)/C - 1 - B.Exp(C) 

The second part is best done with another substitution, x1 = x2/C. Then it's
15) I2 = Exp(C) Integral[ (C/x2) (Exp(x2) - 1) dx, x2 = -C to C(B-1) ], i.e.
16) I2 = C Exp(C) Integral[ (Exp(x2) - 1)/x2  dx,  x2 = -C to C(B-1) ]
This integral does not have an elementary antiderivative but (by the
Fundamental Theorem of Calculus again, if you like) it does have _some_
antiderivative; call it  H  (I'll come back to this). Then this integral is
17) I2 = C Exp(C) [ H( C(B-1) ) - H(-C) ]

Combining  11), 14), and 17)  gives F(B,C) = I1+I2 or in other words
18) F(B,C) = Exp(CB)/C - B.Exp(C) - 1 + C Exp(C) [ H( C(B-1) ) - H(-C) ]
so that the original sum is (by (5) ),
19) S = (A/B) F(B,C)  (which I won't write out just yet).

Using the definitions of our symbols,
>	...where  A = exp(-Lv)/v,  B=(v/(v-1)),  and  C=(v-1)L...
and a little algebra, here's the original sum (I hope I got this right):
20) S = { 1/L - (v)Exp(-L) + (1-v) Exp(-vL) +
           +(1-v)^2 L Exp(-L) [ H( L ) - H ( (1-v)L ) ] } / v^2

I'm sure you'll agree that this is quite a mouthful. I would guess you don't
need the actual value of the sum  S,  but rather only an estimate. Before I
can do that, you'll have to tell me how large  v  and  L  are; so far
you haven't even told me if they're positive. Now in your last letter you say
>...but as 0<v<1...
which is important. I've rewritten  (20)  so that it refers as much as 
possible to positive quantities  (e.g.  1-v  instead of  v-1).

Now let me comment on the function  H.  I've carefully chosen it to be
the antiderivative of  (Exp(x) - 1 )/x  rather than  Exp(x)/x  because I
wasn't sure about the signs of the variables you were going to assign.
This integrand is defined and continuous for all x, so we don't have
the problems associated with integrating  Exp(x)/x  and  1/x  separately.
As it turns out, you now say the arguments we need are all positive
anyway, but it's worth noting the care needed. For example, the
correct antiderivative of  1/x is not  ln(x)  but  ln(|x|), a fact that
(correctly) enables you to integrate even over regions of negative  x's.
You have mentioned the use of Mathematica; be very wary of this tool for
integration as it is known to be sloppy about just such issues.

Now let's return to the sum you wish to evaluate:
20) S = { 1/L - (v)Exp(-L) + (1-v) Exp(-vL) +
           +(1-v)^2 L Exp(-L) [ H( L ) - H ( (1-v)L ) ] } / v^2

I'm assuming  L  will be a large positive quantity.
For (positive)  v  very small (i.e., so that  vL is much less than  1),
this appears to be about  1/v^2 (i.e., the third term dominates the sum).
By the time you get to  v=0.5, I believe (but am not sure) that the last 
term dominates. When  v  gets close to  1  the first term dominates and
you get close to  1/L. Perhaps you can compare this to your intuition
of the physical situation being modeled; if this seems inconsistent then
I likely made an algebra error when preparing equation (20).

Let me know if this gets you closer to what you need.

dave

==============================================================================

Date: Tue, 07 Feb 1995 09:35:22 -0400
From: [Permission pending]
Subject: Re: HELP NEEDED...PLEASE!
To: rusin@math.niu.edu

Thank you for your long answer. I have a talk to prepare for tomorrow, so I
cannot try it now, but I'll examine it after that...

[sig deleted]
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Date: Thu, 09 Feb 1995 22:29:50 -0400
From: [Permission pending]
Subject: Re: HELP NEEDED...PLEASE!
To: rusin@math.niu.edu

Ok, now I'm back to the reality...
I feel a bit guilty to see how much time you spent for this problem, without
even knowing what all is about...but I've been told once that mathematician
likes to solve problems like that.!
Anyway, for the parameters, that's what is known:
0<=v<1, v=1 being a special case (Poisson distribution).
0<=L<oo
L is actually a parameter which describe a quantity np=1-Exp(-L), 0<=np<=1.
I think I now was able to follow and understand all the steps of your second
answer, but before testing the solution, I'd like to be sure what the H are.
you write in the Eqn (20) H(L) and H((1-v)*L).
Acording to your first answer:
H(x)=Integral (1-exp(-w))/w dw,w=1 to x and
H(x)=ln(x)-Ei(x).
Than
H(L)=ln(L)-Ei(L) and
H((1-v)L)=ln((1-v)L) -Ei((1-v)L).
Am I right?
Now you ask me if I simply need an estimate of this...unfortunatly no, as this
is a part of a function I must maximize! But I don't know much of the Ei (I
know nothing, actually) and does the estimated values provided by Maple or
mathematica are realy estimates or exact values?
If they are estimates, I'm afraid it won't help much, but for a faster
"computability" of the answer.
But in any cases, I've learn a lot and thank you for that...

[sig deleted]
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Date: Tue, 14 Feb 95 15:24:32 CST
From: rusin (Dave Rusin)
To: [Permission pending]
Subject: Re: HELP NEEDED...PLEASE!

>I feel a bit guilty to see how much time you spent for this problem, without
>even knowing what all is about...but I've been told once that mathematician
>likes to solve problems like that.!

No problem -- sometimes it even helps to be free of the distraction of the
origin of the problem. But sometimes not; fortunately it doesn't really
matter in this particular problem that I still have no clue what the 
application is.

>I think I now was able to follow and understand all the steps of your second
>answer, but before testing the solution, I'd like to be sure what the H are.
>you write in the Eqn (20) H(L) and H((1-v)*L).
>Acording to your first answer:
>H(x)=Integral (1-exp(-w))/w dw,w=1 to x and

Uh-oh, I should have been more careful for you. I changed the definition of
H  between my first and second letters in an attempt to be clear (and to
stay with positive numbers). The correct definition in the second letter is
	H(x) = Integral ( exp(w) - 1 ) / w  dw ,  w = 1 to  x
which is the same as saying  H  is an indefinite integral, i.e.,
	H(x) = Integral ( exp(x) - 1 ) / x  dx 
You'll notice it's only used in pairs so you can replace
	H(L) - H( (1-v)L )      by
	Integral( ( exp(x) - 1 ) / x dx,  x = (1-v)L  to  L  )

>Now you ask me if I simply need an estimate of this...unfortunatly no, as this
>is a part of a function I must maximize! But I don't know much of the Ei (I
>know nothing, actually) and does the estimated values provided by Maple or
>mathematica are realy estimates or exact values?

Oh, when I say "estimate" I mean it the way a mathematician does; you'd 
call it an "actual value". (The distinction is that when all I see is a
bunch of decimal digits, I know it's off from the true value. But when it's
off by  10^(-8), say, which is what the computer software will offer you,
then you can feel comfortable knowing the value is "correct".)

But there is a key issue here: you say you need to include the sum, and
thus the function  H, into a maximization problem. Do you intend to solve
that problem by taking a first derivative and setting it equal to zero?
If so, you're in luck. The defining property of  H(x)  is that 
H'(x)  = (exp(x) - 1)/x.  Thus for example if you wish to differentiate
[H(L) - H((1-v)L)] with respect to  L,  you'll get
[exp(L) - exp((1-v)L)]/L. That should make your life easier.

dave
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Date: Wed, 22 Feb 1995 22:50:45 -0400
From: [Permission pending]
Subject: Re: HELP NEEDED...PLEASE!
To: rusin@math.niu.edu

Ok, sorry having been so long, but very busy days...
Anyways, your solution seems to be working (I think there was a sign error, but
still not sure...) and i thank you again for helping... More than that, reading
your e-mail I've learn a lot in mathematics...which is good. For the
maximizzation, I'm only partly lucky as this expression is on a nominator,
which means that the derivative include the expression itself (well not so
clear, but my english...it's something like (f/g)'=(f'g-fg')/g^2)
Anyway, it's another storry, and I'd like to thank you again,

[sig deleted]