MATLAB Answers

HOW I COULD DRAW THIS FUNCTION

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HADIMARGO
HADIMARGO on 7 Dec 2018
Answered: HADIMARGO on 13 Dec 2018
IMG_20181203_165117.jpg

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madhan ravi
madhan ravi on 7 Dec 2018
upload the code that you tried
HADIMARGO
HADIMARGO on 7 Dec 2018
x=0:1:50;
syms n;
y=((2/pi)*((-1)^n+1)*(sin(pi*x)))/n;
symsum(y,n,1,inf);
figure(1);
plot(x,y)
i should say that i am a beginner.

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Accepted Answer

HADIMARGO
HADIMARGO on 13 Dec 2018
final answer by myself!!
x=linspace(-5,5,1000);
s=0;
for n=1:1:10000
p=((-1)^(n+1)*sin(n*pi*x))/n;
s=s+p;
end
f=(2/pi)*s;
plot(x,f)
final.jpg

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More Answers (3)

madhan ravi
madhan ravi on 7 Dec 2018
Edited: madhan ravi on 7 Dec 2018
Use cumsum() to define the behaviour of the alternating series:
x=linspace(0,50,10000);
n=1:10000;
y=(2/pi).*cumsum((((-1).^n+1).*(sin(pi.*x)))./n);
figure(1);
plot(x,y)

  2 Comments

HADIMARGO
HADIMARGO on 7 Dec 2018
so tnx madhan ravi. i will send this answer to my teacher. if it correct i will get 1 score of 20 .
madhan ravi
madhan ravi on 7 Dec 2018
Anytime :) , read more about cumsum() and linspace()

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HADIMARGO
HADIMARGO on 8 Dec 2018
my teacher say it isnot correct. you should put it in a loop.

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Walter Roberson
Walter Roberson on 8 Dec 2018
If you have TheResult = cumsum(SomeVector) then you can recode that in a loop as
TheResult(1) = SomeVector(1);
for LoopVariable = 2 : length(SomeVector)
TheResult(LoopVariable) = TheResult(LoopVariable-1) + SomeVector(LoopVariable);
end
... Approximately.

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HADIMARGO
HADIMARGO on 8 Dec 2018
can anyone write complete answer with loop?

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Walter Roberson
Walter Roberson on 8 Dec 2018
Yes, we could, but it is your homework.

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