1 个投票

Hi, so I am trying to geta function that returns an output, which is the square root of the mean of the squares of the first nn positive odd integers, where nn is a positive integer and the only input argument. for example if nn=3 i want my function to return the sqaure root of the average of the numbers 1 9, and 25. my code so far is this:
function orms = odd_rms(nn)
orms = sqrt(mean((1:nn).^2));
however this is only returning positive integers. So I am wondering how to make this odd positive integers.

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prashant paudel
prashant paudel 2018-5-23
function rms=odd_rms(n) x=n*(2*n+1)*(2*n-1)/3; rms=sqrt(x/n);

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John D'Errico
John D'Errico 2016-8-6

3 个投票

Ok, so the first nn positive integers are easy to generate.
2*(1:nn) - 1
TRY IT! Put in various values for nn. See that this works, and think about why it works. I could also have done it differently.
1:2:(2*nn-1)
Both of these will work, generating odd integers as needed. You should see why the code that you wrote did not work properly. The rest of your code was correct, in terms of squaring elements, forming the mean, then the sqrt.
oddrms = @(nn) sqrt(mean((1:2:(2*nn-1)).^2));
Test out the function.
oddrms(1)
ans =
1
oddrms(2)
ans =
2.2361
oddrms(3)
ans =
3.4157
Yes, I could have written it in m-file form. So this should work too:
function orms = odd_rms(nn)
orms = sqrt(mean((1:2:(2*nn-1)).^2));
end

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SULE SAHIN
SULE SAHIN 2017-10-30
I dont understand (2*nn-1). why we use (2*nn-1) instead of nn because for example if nn = 3, we use 2*3_1=5 instead of 3. we go out border line. please may you explain this?
Isaac DeVaughn
Isaac DeVaughn 2017-12-14
编辑：Isaac DeVaughn 2017-12-14
I think i understand if you just bounded it by nn you only get some of the number you need up to three. That would leave out the 25 from the example.I think that leaves out part of the solution.

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