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

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

Answer 1

[Youssef Khmou]

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.....

[majid]

good answer ...but when ask about...Evaluate the same above series using the “vectorization” ability of MATLAB. Use first terms to evaluate the sum of series. Then Print the difference between your sum of the series and . Plot the sum of series for first 2000 iterations how we can answer

[Image Analyst]

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]

ok thanks so much i'll do my best

Answer 2

[Image Analyst]

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.

[majid]

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]

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.

Answer 3

[Carlos]

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;

[majid]

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]

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