Finite-Sum Function

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Atom
Atom 2012-1-29
评论： Radu Trimbitas 2019-1-30
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

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采纳的回答

Andrew Newell
Andrew Newell 2012-1-30
S_N is the output, but the function name is S. Use, e.g.,
S(10)
(you'll need to input a number).

更多回答（2 个）

Jan
Jan 2012-1-30
You've defined S_N such that it needs one input argument. So call it like:
S_N(3)
Atom
Atom 2012-1-30
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 个评论
Andrew Newell
Andrew Newell 2012-1-30
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 2019-1-30
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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