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From: Joni on 10 May 2010 11:54
Hello.

Could someone help me to show how draw sum with matlab.
for example: (1/n)+1/(3*n)-1/(5*n)+1/(7*n)-1/(9*n)+...
and n=1:100.

I can draw a function with only "n+2n+3n+...N*n" but i dont know how to make "n+2n-3n+4n-5n+...N*n"
(the + and - change places).

thx.
From: Jan Simon on 10 May 2010 18:20
Dear Joni!

> Could someone help me to show how draw sum with matlab.
> for example: (1/n)+1/(3*n)-1/(5*n)+1/(7*n)-1/(9*n)+...
> and n=1:100.
>
> I can draw a function with only "n+2n+3n+...N*n" but i dont know how to make "n+2n-3n+4n-5n+...N*n"
> (the + and - change places).

Perhaps this helps:
for k = 0:10
x = -1^k
end

In your examples, the 1st, 2nd, 4th, ... elements are positive. Usually you find the same sign for 1st, 3rd, 5th, ...

Jan
From: ImageAnalyst on 10 May 2010 18:40
Joni:
For the second question, how about this:

% Plot n+2n-3n+4n-5n+...N*n
% Generate 50 n's for the x (horizontal) axis.
n = linspace(-10,10,50) % Plot between -10 and +10
N = 25; % 25 terms for example.
coeffs = 1:N; % 1 2 3 4 ... 25
coeffs(3:2:N) = -coeffs(3:2:N) % Every other one starting with 3 is
negative.
% factor out n to create y = n * (1+2-3+4-5 ... -25)
for k = 1:length(n)
% for all 50 n's, get the sum of the 25 terms.
y(k) = n(k) * sum(coeffs);
end
% Plot the 50 n,y pairs.
plot(n, y);

Note how I used
coeffs(3:2:N) = -coeffs(3:2:N)
to negate every other one (starting with term #3 rather than term #1
for some reason).

Now, try to tackle your first question on your own, now that you know
how to use that trick (if you can call it that).
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