Finding roots using Newton raphson method

Question

R Abhinandan 2020-10-27

1
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/627583-finding-roots-using-newton-raphson-method

评论： R Abhinandan 2020-10-27

Here is my code and my output is a function and not a numerical value as I expected. Can anyone debug this code?

syms f y x df
f=@(y) exp(y)-(sin(pi*y/3));
df=@(y) exp(y)-((pi*cos(pi*y/3))/3);
x(1)=-3.0;
error=0.00001;
for i=1:10
    x(i+1)=x(i)-((f(x(i)))/df(x(i)));
    err(i)=abs((x(i+1)-x(i))/x(i));
    if err(i)<error
        break
    end
end
root=x(i);
output: 
- (sin((pi*(exp(-3)/(pi/3 + exp(-3)) + 3))/3) + exp(- exp(-3)/(pi/3 + exp(-3)) - 3))/(exp(- exp(-3)/(pi/3 + exp(-3)) - 3) - (pi*cos((pi*(exp(-3)/(pi/3 + exp(-3)) + 3))/3))/3) - exp(-3)/(pi/3 + exp(-3)) - 3

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Answer 1

Alan Stevens 2020-10-27

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/627583-finding-roots-using-newton-raphson-method#answer_525708

在 MATLAB Online 中打开

Just get rid of the first line, you don't need it

f=@(y) exp(y)-(sin(pi*y/3));
df=@(y) exp(y)-((pi*cos(pi*y/3))/3);
x(1)=-3.0;
error=0.00001;
for i=1:10
    x(i+1)=x(i)-((f(x(i)))/df(x(i)));
    err(i)=abs((x(i+1)-x(i))/x(i));
    if err(i)<error
        break
    end
end
root=x(i);
disp(root)

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Alan Stevens 2020-10-27

在 MATLAB Online 中打开

This is what I get:

>> f=@(y) exp(y)-(sin(pi*y/3));
df=@(y) exp(y)-((pi*cos(pi*y/3))/3);
x(1)=-3.0;
error=0.00001;
for i=1:10
    x(i+1)=x(i)-((f(x(i)))/df(x(i)));
    err(i)=abs((x(i+1)-x(i))/x(i));
    if err(i)<error
        break
    end
end
root=x(i);
disp(root)
   -3.0454