## HOW I COULD DRAW THIS FUNCTION

on 7 Dec 2018

on 13 Dec 2018 on 7 Dec 2018
upload the code that you tried

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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on 13 Dec 2018

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) on 7 Dec 2018

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) 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 .

on 7 Dec 2018

on 8 Dec 2018

my teacher say it isnot correct. you should put it in a loop.

Walter Roberson

### Walter Roberson (view profile)

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.

on 8 Dec 2018

can anyone write complete answer with loop?

Walter Roberson

### Walter Roberson (view profile)

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