Please help me to solve this newton-raphson method

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Inggrid Audia
Inggrid Audia 2016-9-29
回答： Luis Varela 2016-10-4
How can I use Newton-Raphson method to determine a root of
f (x) = x5−16.05x4+88.75x3−192.0375x2+116.35x +31.6875
using an initial guess of x = 0.5825 and εs = 0.01%.
  2 个评论
James Tursa
James Tursa 2016-9-29
What have you done so far? Have you written any code yet?
Luis Varela
Luis Varela 2016-10-4
On Newton Raphson method, you need calculate the function f(x) and the derivate f'(x), to get the next value of x, and continue while the error is greater than desired, for example:
x = 0.5825;
e=1;
while e>0.01
fx= x^5 - 16.05*x^4 + 88.75*x^3 - 192.0375*x^2 + 116.35*x + 31.6875;
dfx= 5*x^4 - 4*16.05*x^3 + 3*88.75*x^2 - 2*192.0375*x + 116.35;
x2=x-(fx/dfx);
e=100*abs((x2-x)/x2);
x=x2;
end
At the end x will have the value of the calculated root, aprox. x=6.5

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回答（2 个）

Jakub Rysanek
Jakub Rysanek 2016-10-3
In this case I would go with
roots([1,-16.05,88.75,192.0375,116.35,31.6875])
Luis Varela
Luis Varela 2016-10-4
On Newton Raphson method, you need calculate the function f(x) and the derivate f'(x), to get the next value of x, and continue while the error is greater than desired, for example:
x = 0.5825;
e=1;
while e>0.01
fx= x^5 - 16.05*x^4 + 88.75*x^3 - 192.0375*x^2 + 116.35*x + 31.6875;
dfx= 5*x^4 - 4*16.05*x^3 + 3*88.75*x^2 - 2*192.0375*x + 116.35;
x2=x-(fx/dfx);
e=100*abs((x2-x)/x2);
x=x2;
end
At the end x will have the value of the calculated root, aprox. x=6.5

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