Dear George,
That is almost closer to the figure I have but with a different quantity on the numerator. Can you forward your complete arguments you have made please.
Sincerely,
Dejenie
It should be
Re( 1/(1-exp(-1/2 + i) ) )
Just write the cos series as the real part of a geometric series with q = exp(-1/2 + i).
I just added a missing bracket in the answer ....
I'm not in school, so I can cheat, and let an Intel Core i5 come to the rescue.
To 16 digits of accuracy, the series actually converges quickly. It only takes 75 steps to converge. The sum is:
0.9436185014209203
But regrettably, your algebra does not agree with Intel, George. This is your answer:
-5.638147960795207D-02
I've attached the two short programs, and the results for 1000 steps.
Dear Albert,
Your program indeed generate the same answer as dear George has as an answer. Interesting results and specially the numerical calculations though.
Dear Herbert,
Indeed that is what I have which leas to an exact answer : ((e-e^{1/2}cos1)/(e-2e^{1/2}cos1+1)) which is a little different from what dear George has.
Maple says
> A:=simplify(evalc(Re(1/(1-exp(-1/2+I)))));
A := (-1+exp(-1/2)*cos(1))/(-1+2*exp(-1/2)*cos(1)-exp(-1))
> Digits:=16:evalf(A);
.9436185203920479
It even does all the work:
> sum('exp(-1/2*k)*cos(k)','k'=0..infinity);
(-1+exp(-1/2)*cos(1))/(-1+2*exp(-1/2)*cos(1)-exp(-1/2)^2)
A simple way of finding both sums of the related two series (via Euler's formula) is detailed in the attached file.
Dear all,
That is true dear Octav.
The sum can be computed for any number s > 0 where the question is put when s is 1/2, from a transform definition. The story of it is very important as the discrete Laplace like transform is used to generate beautiful sequences of difference equations including the Fibonacci sequence.
I'm agreeing numerically with Homeier except for the quibble that I got 1 more figure so mine ends in ..797. This was just a straight sum of 78 terms in Fortran.
James,
Indeed the terms of the series collapse exponentially fast due to the exponential factor on the terms, the very reason the almost similarity of the sums for different terms of the series. One thing though, is FORTRAN still in use for computations ? I know when it was FORTRAN 77 version.
Dejenie -- I'm using a free s/w called Force 2.0 and I think it's mostly compatible with Fortran 77.
Well, there's a slight difference in results, depending whether one calculates the algebraically derived sum or whether one lazily lets a program do the sum.
My variables are dimensioned as 10-byte floating point, which should be better than double precision.
When I let the program compute the sum, after 75 summation steps, the answer converges to:
0.9436185014209203
On the other had, if I calculate the actual precise answer provided by George, I get:
0.943618520392048
There seems to be some accumulation of errors in the former, I have to conclude.
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