I need help writing a matlab function

I need help writing a Matlab function that takes a positive integer n as its argument, and returns the sum of all odd numbers between 1 and n.

[Jan]

Please clarify, if this is your homework. If so posting a solution might be counter-productive, because you would cheat if you deliver it. Even if it is a homework question, posting what you have tried so far and asking a specific question would be useful. Because asking for explanations is not a cheating instead of letting others solve your work.

Answer 1

[John BG]

Hi Mateo

what about this

function sn=sum_odds(n)

narginchk(1,1);

if(n<0) || (n==0)
    error(message('input error n<=0'));
end

if abs(n-floor(n))>0
    error(message('input not +int'));   
end

sn=sum([1:2:n])

% if rem(n,2)==0
%    sn=sum([1:2:n-1])
% end
% 
% if rem(n,2)>0
%    sn=sum([1:2:n])
% end

end

if you find these lines useful would you please mark my answer as Accepted Answer

To any other reader, if you find this answer of any help please click on the thumbs-up vote link,

thanks in advance for time and attention

John BG

<mailto:[e-mail] [e-mail]>

[Walter Roberson]

You just gave complete code for a homework question :(

[Image Analyst]

That assumes it was a homework question. Mateo did not say it was a homework question, either in the body of the message or via a tag. It may well be, but should we assume that? Maybe it's not. Or maybe he's self-learning and this is not a question to be turned in and graded. Every time someone posts something should they say "This is not a homework question" if it is not? A given question might be assumed to be a homework question by some people and not a homework question by other people. I guess it's up to each individual's judgment.

[Jan]

@Image Analyst: You are right. I've removed my comment, which mentioned http://www.mathworks.com/matlabcentral/answers/8626-how-do-i-get-help-on-homework-questions-on-matlab-answers.

For me the question sounds like a homework or Cody problem. Even if this does not concern the OP, this thread is found when others pupils search in the net. Beside this, the OP does not show any own effort like posting code or asking for a specific detail. Therefore I judge that this is a "do-it-for-me" question, which I do not like to answer with code also.

[Jan]

@John BG: The message() command "is intended for MathWorks use only" according to the documentation. In addition it requires a qualified message-ID, therefore your error handling causes another error.

[Walter Roberson]

Should we assume that it is a homework question: Yes. Nothing is lost by assuming that for something that looks like homework. The person who posted the question can also provide clarification afterwards that might potentially lead to someone posting a more complete solution, but an overcomplete solution cannot be withdawn.

Answer 2

[Walter Roberson]

https://www.mathworks.com/help/matlab/ref/function.html

https://www.mathworks.com/help/matlab/ref/for.html

Answer 3

[Jan]

Do you remember how Gauss calculated the sum over all elements from 1 to n? You can modify the method to use only the pairs of odd numbers. This reduces the compuations drastically and you should never process the dull sum.

[Stephen23]

@John BG: note that square brackets are not required: sum(1:1:100).

The mlint messages is ""Use of brackets [] is unnecessary." This might also be of interest:

http://www.mathworks.com/matlabcentral/answers/35676-why-not-use-square-brackets

[Jan]

I do not post the solution, because I assume, that this is a homework question. I even did not mention, that asking an internet search engine for the terms "sum of odd numbers" reveals the main part of the solution already: The sum of N odd numbers is N^2.

@Stephen: The step size 1 is not required in vectors, so this is enough: sum(1:100). This does not influence the computing time but is nicer . The leaner the code, the less chances for a typo... :-)

[Walter Roberson]

Assuming N odd,

   1 + 3 + 5 +... N
   N + (N-2) + (N-4) +... 1

Each pair adds to N+1. Count the pairs and multiply by that. Then divide by 2 because each pair is being counted twice.