Finite-Sum Function

17 Ansichten (letzte 30 Tage)
Atom
Atom am 29 Jan. 2012
Kommentiert: Radu Trimbitas am 30 Jan. 2019
Hello All,
I would like to create a function S_N, that would return the Nth term of a series. Here's what I have so far:
1 function S_N = S(N)
2 k = 1:N
3 A_k = 1./(k.^(3/2)+k.^(1/2))
4 S_N = cumsum(A_k)
5 end
MatLab politely replies:
>> S_N
Error using S_N (line 2)
Not enough input arguments.
I'm not sure what else MatLab wants. I would think this is a pretty simple fix for this uber program. (I can do it perfectly on my TI89 :D)
Much Thanks, Jim

Akzeptierte Antwort

Andrew Newell
Andrew Newell am 30 Jan. 2012
S_N is the output, but the function name is S. Use, e.g.,
S(10)
(you'll need to input a number).

Weitere Antworten (2)

Jan
Jan am 30 Jan. 2012
You've defined S_N such that it needs one input argument. So call it like:
S_N(3)

Atom
Atom am 30 Jan. 2012
Much thanks;
At the bottom is my final code for anyone else that would want to know. (Of course! I have to give him a number). By the way, the infinite sum is referred to as "Theodorus' constant". It seems pretty hard to approx. Here are the first fifty digits:
1:86002 50792 21190 30718 06959 15717 14332 46665 24121 52345...
function S_N = S(N)
k = 1:N;
A_k = 1./(k.^(3/2)+k.^(1/2));
S_N = sum(A_k)
end
  2 Kommentare
Andrew Newell
Andrew Newell am 30 Jan. 2012
I did a quick search on Theodorus constant, and all the sources said it is the square root of 3, or 1.73205 ...
Radu Trimbitas
Radu Trimbitas am 30 Jan. 2019
There are two Theodorus constant, square root of 3 and sum of the series sum_{n=1}^{\infty} \frac{1}{k^{3/2}+k^{1/2}} ....

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