use the for or while looping to the series S=4*[sin(theata)/1 +sin(3theata)/3+sin(5theata)/5+.....] ,,in the range 0<theata<pi with error bound of 10^-6..?....please help me to solve this question in the earliest opportunity

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majid 2013-4-15

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采纳的回答： Youssef Khmou

S=4*[sin(theata)/1 +sin(3theata)/3+sin(5theata)/5+.....]

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Answer 1

Youssef Khmou 2013-4-15

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编辑：Youssef Khmou 2013-4-15

在 MATLAB Online 中打开

Try this version :

 % theta=0:pi/30:pi;
 theta=0:0.01:2*pi;
 S=0;
 tolerance=1e-6;
 n=1;
 r=6; % random 
 counter=1;
 while r>tolerance
   N1=norm(S);
  S=S+sin(n*theta)/n;
  N2=norm(S);
  r=abs(N2-N1);
  n=n+2;
 counter=counter+1;
 end
 plot(theta,S);
 axis([0 4 0 1])
 grid on

Now it approximates well the rectangle, the number of iterations is saved in the variable 'counter', finish the code with the desired prints .

To conclude you work, there a special name of the infinitesimal waves nears the edges , that phenomenon has a special name , it starts with G.....

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Image Analyst 2013-4-15

We're trying to help you without doing your homework outright for you. We've seen very little code by you so far. I'm sure you don't want to just turn in our code as your own, so post some code we can help with by giving hints.

majid 2013-4-16

ok thanks so much i'll do my best

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Answer 2

Image Analyst 2013-4-15

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Hint, have a loop over k and calculate sin(k*theata)/k in the loop. I hope that's not doing too much of your homework for you. You still have to make the loop and sum up the term in the loop into the overall sum and then multiply that by 4.

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majid 2013-4-15

编辑：Image Analyst 2013-4-15

在 MATLAB Online 中打开

sum=0;
k=1;
while o<theata<pi
  s(k)=sin(k*theata)/k;
  sum=sum+k;
  k=k+1;
  sum=4*sum;

Note:I want to now how I can do the odd number in series,Also he asked for

i. The number of iterations it took to converge according to above set criterion. ii. The value of the final term. iii. The value of the sum of the series. Plot the sum of series for first 2000 iterations.

Image Analyst 2013-4-15

在 MATLAB Online 中打开

Use a for loop instead of a while loop. Have an outer for loop over theata.

for theata = 0: 0.01 : pi

Don't use "sum" since that is a built in function name and you'll be destroying it. Use S like you started to. Finally you need to calculate the "true" value (whatever that is) and compare it to S and bail out of your inner for loop once the error is less than 1E-6.

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Answer 3

Carlos 2013-4-15

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编辑：Carlos 2013-4-15

在 MATLAB Online 中打开

Try this
theata=pi/2;
    a=1;
    S=0;
    n=1;
    while (a>10e-6)
    a=sin(n*theata)/n;
    S=S+a;
    n=n+2;
    end
    S=4*S;

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majid 2013-4-15

use theata= 0.7 degrees. Then Print:

i. The number of iterations it took to converge according to above set criterion. ii. The value of the final term. iii. The value of the sum of the series. Plot the sum of series for first 2000 iterations.

Carlos 2013-4-15

Sorry, I misunderstood the question, I thought theata should be a fixed value.

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use the for or while looping to the series S=4*[sin(theata)/1 +sin(3theata)/3+sin(5theata)/5+.....] ,,in the range 0<theata<pi with error bound of 10^-6..?....please help me to solve this question in the earliest opportunity

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use the for or while looping to the series S=4*[sin(theata)/1 +sin(3thea​ta)/3+sin(​5theata)/5​+.....] ,,in the range 0<theata<pi with error bound of 10^-6..?....please help me to solve this question in the earliest opportunity

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use the for or while looping to the series S=4*[sin(theata)/1 +sin(3theata)/3+sin(5theata)/5+.....] ,,in the range 0<theata<pi with error bound of 10^-6..?....please help me to solve this question in the earliest opportunity

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