| Title: | Mathematics at DEC |
| Moderator: | RUSURE::EDP |
| Created: | Mon Feb 03 1986 |
| Last Modified: | Fri Jun 06 1997 |
| Last Successful Update: | Fri Jun 06 1997 |
| Number of topics: | 2083 |
| Total number of notes: | 14613 |
Newsgroups: sci.math
Path: decwrl!labrea!rutgers!sunybcs!boulder!news
Subject: Lunchtime Diversions
Posted: 4 Nov 87 18:11:57 GMT
Organization: University of Colorado, Boulder
[ munch ]
From: rolan@boulder.Colorado.EDU (Rolan Christofferson)
Path: boulder!rolan
Here's a problem I developed from a Putnam exam problem:
notation:
sum(i,j=1,inf) f(i,j) := sum(i=1 to inf) {sum(j=1 to inf) f(i,j) }
-1
sum(i =1,inf) [ i(i+1) ] = 1 easy
-1
sum(i,j =1,inf) [ ij(i+j+1) ] = 2 good problem
-1
sum(i,j,k =1,inf) [ ijk(i+j+k+1) ] = 6 long
conjecture:
-1
sum(i ,i , ... ,i =1,inf)[ i *...*i *(1+i + ... +i ) ] = n!
1 2 n 1 n 1 n
I am unable to prove the conjecture by induction. Any ideas?
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 782.1 | Spoiler for the "easy" one | ZFC::DERAMO | Daniel V. D'Eramo | Fri Nov 06 1987 01:23 | 16 |
"Spoiler" for the "easy" case:
sum(i = 1,n) 1/i(i+1) = sum(i = 1,n) [1/i - 1/(i+1)]
= 1 - 1/2
+ 1/2 - 1/3
...
+ 1/n - 1/(n+1)
= 1 - 1/(n+1)
I think that this process is called "collapsing."
sum(i = 1,inf) 1/i(i+1) = lim n->oo sum(i = 1,n) 1/i(i+1)
= lim n->oo (1 - 1/(n+1))
= 1
Dan
| |||||
| 782.2 | "good problem" -- the two variable case | ZFC::DERAMO | Daniel V. D'Eramo | Fri Nov 06 1987 02:26 | 39 |
Prove: sum(i=1,inf) sum(j=1,inf) 1/ij(i+j+1) = 2
First, 1/(i+j) - 1/(i+j+1) = 1/(i+j)(i+j+1)
(i+j) 1 1 1 1 1
so 1/ij(i+j+1) = ----- * ------------ = (- + -)(--- - -----)
ij (i+j)(i+j+1) i j i+j i+j+1
so sum(i=1,inf) sum(j=1,inf) 1/ij(i+j+1)
= sum(i=1,inf) sum(j=1,inf) (1/i + 1/j)(1/(i+j) - 1/(i+j+1))
= sum(i=1,inf) sum(j=1,inf) (1/i)(1/(i+j) - 1/(i+j+1))
+ sum(i=1,inf) sum(j=1,inf) (1/j)(1/(i+j) - 1/(i+j+1))
= 2 * sum(i=1,inf) sum(j=1,inf) (1/i)(1/(i+j) - 1/(i+j+1))
[because the two infinite sums are equal]
= 2 * sum(i=1,inf) (1/i) sum(j=1,inf) (1/(i+j) - 1/(i+j+1))
= 2 * sum(i=1,inf) (1/i)(1/(i+1)) [the inner sum collapses]
= 2 * 1 [using the results of .-1]
= 2
Hint for inductive proofs:
1
sum(i1=1,inf) ... sum(in=1,inf) -----------------------------
i1 i2 ... in (i1+i2+...+in+1)
[i.e., the sum for n variables] may split up into n copies
of the infinite sum for n-1 variables. This would explain
the conjecture in .0 that the sum for n variables is n!
Dan
| |||||
| 782.3 | how the usenet did it -- very short outline | ZFC::DERAMO | Daniel V. D'Eramo | Thu Nov 12 1987 21:53 | 10 |
The solutions on the usenet were clever. A variable "z" was
stuck into the infinite sum, and the sum renamed to f(z).
The derivative f'(z) was much easier to identify, and then
integration identified f(z). The original infinite sum was
f(1). The "induction step" was an integration by parts to get
the n variable integral from the n+1 variable integral. Finally,
one person added a proof that all these manipulations were of
series that really did converge.
Would anyone care to work out the details?
| |||||
| 782.4 | oops | ZFC::DERAMO | Can I take your personal name? | Mon Jan 18 1988 22:20 | 7 |
Re .1
>> I think that this process is called "collapsing."
Actually, I think it is called "telescoping."
Dan
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