| T.R | Title | User | Personal
Name | Date | Lines |
|---|
| 597.1 | | ENGINE::ROTH | | Mon Oct 20 1986 19:31 | 13 |
| No.
More generally, in the sum
n+1
1-z
1 + z + z^2 + ... + z^n = ------
1-z
i
where z = e
the ratio of real to imaginary parts is never rational.
- Jim
|
| 597.2 | Eh ? | COMET::ROBERTS | Dwayne Roberts | Tue Oct 21 1986 16:38 | 10 |
| re: .-1
Jim, I don't quite follow your argument. How is
1 + z + z^2 + ... + z^n where z = e^i
a generalization of
sin(1) + sin(2) + sin(3) + ... + sin(n) ?
|
| 597.3 | sin(z) = Im(exp(iz)) | ENGINE::ROTH | | Wed Oct 22 1986 09:04 | 3 |
| The trig sum in .0 is the imaginary part of my complex sum.
- Jim
|
| 597.4 | I'm confused too | EAGLE1::BEST | R D Best, Systems architecture, I/O | Sun Oct 26 1986 02:35 | 25 |
| < Note 597.1 by ENGINE::ROTH >
-< >-
No.
More generally, in the sum
n+1
1-z
1 + z + z^2 + ... + z^n = ------
1-z
i
where z = e
the ratio of real to imaginary parts is never rational.
- Jim
> I'm intrigued. How do you know that:
> (1) The ratio of the real to imaginary parts is always irrational,
> and
> (2) That ratio being irrational is sufficient to prove
> ~Exists(n):[n an integer] and [the sum = 1] ?
> /R Best
|
| 597.5 | | ENGINE::ROTH | | Sun Oct 26 1986 11:59 | 14 |
| Now I'm not so sure...
What I thought of was that sin(1) must be trancendental, so there
would be no algebraic solution to the geometric sum that was
involved. Also, whenever tan(x) is rational, x is a rational
multiple of pi, and it seemed 'obvious' that if x was a
trancendental multiple of pi that tan(x) could not be rational.
It looks like time to check some facts in a book on number theory.
The same proof that exp(1) is transcendental doesn't work for sin(1),
tho the series is so simple that there must be an easy way of showing
it.
- Jim
|
| 597.6 | | CLT::GILBERT | eager like a child | Mon Nov 10 1986 20:19 | 1 |
| Realize that the sum always remains in the range of -2 to +2.
|
| 597.7 | A Tighter Range | COMET::ROBERTS | Dwayne Roberts | Tue Nov 11 1986 11:59 | 9 |
|
Yup, it's always between -2 and +2; more precisely, it's between
the extreme points of the function found in the base note
Sigma sin(x) = .5*{ sin(n) + cot(.5)*[ 1 - cos(n) ] }
The extreme points are at n=-1/2 and n=pi-1/2, making the range
between -0.127671 and +1.958159 approximately.
|
| 597.8 | a simplification... | CLT::GILBERT | eager like a child | Mon Nov 24 1986 19:35 | 21 |
| I noticed that the values in the complex series in 597.1 are all on
a circle (in the complex plane), and so would expect the real part
to be something like: a+b*sin(c+d*n), for appropriate constants a,
b, c, and d. Indeed, the expression in 597.0:
1/2 * { sin(n) + cot(.5)*[ 1 - cos(n) ] }
is equivalent to:
1/2 * { cos(1/2) - cos(n+1/2) } / sin(1/2)
Thus, the problem of finding n such that:
n
Sigma sin(x) = 1
x=0
is equivalent to finding n such that:
cos(n+1/2) = cos(1/2) - 2 sin(1/2)
|
| 597.9 | | CLT::GILBERT | eager like a child | Wed Nov 26 1986 19:16 | 27 |
| Here are some 'n' for which the sum is very nearly 1.
312
19900
83382
188085
228596
1751530
4044346
25721750
43583404
134565963
357135880
957681937
1180251854
8606157274
30951037228
38376942648
726969315847
1098503586965
6077067965440
115367687409215
173101687413299
683347499285308
5879776619481040
6390022431353049
127375114566296733
|