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Trying to answer a math problem using newton's method but having problem in the function.

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This is the last function when newton's method is applied:
I used z=log(x) in my function to make it easier to understand:
clc
x=0.25;
y=0;
while (y-x)<10^-4
z=log10(x);
y = x-((((x ^ 2)z-(x ^ 0.5)-20)2.x ^0.5)/(-1+4.x ^ (3/2)z+2.x ^ (3/2)))
Invalid expression. Check for missing multiplication operator, missing or unbalanced delimiters, or other syntax error. To construct matrices, use brackets instead of parentheses.
x=y;
end

Respuesta aceptada

Alan Stevens
Alan Stevens el 17 de Mzo. de 2023
Matlab doesn't like implied multiplication
(x ^ 2)z
should be
(x ^ 2)*z
etc.
Also, in Matlab
log(x)
is ln(x)

Más respuestas (2)

Cris LaPierre
Cris LaPierre el 17 de Mzo. de 2023
You need to include the multiplication operator when coding your equation. Also, it is '*', not '.'. Nothing is inferred.
I think your conditional will be problematic, since the last line of your loop sets x equal to y.
x=0.25;
y=0;
while (y-x)<10^-4
z=log10(x);
y = x-((((x ^ 2)*z-(x ^ 0.5)-20)*2*x ^0.5)/(-1+4*x ^ (3/2)*z+2*x ^ (3/2)))
x=y;
end
  1 comentario
Cris LaPierre
Cris LaPierre el 17 de Mzo. de 2023
Editada: Cris LaPierre el 17 de Mzo. de 2023
+1 about log10 vs log for natural ln.
Also wanted to point out that you have created an infinite loop. Since x=y, (y-x) is 0, which is <1e-4. Since your condition will always be true, your loop will never end.
Just changing the < to > will prevent the infinite loop, but won't fix the issue that y=x. Consider creating a third variable to prevent this.
Also consider the case that the difference of (y-x) may be positive or negative. You may want your condition to compare to the absolute value of this difference.

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Voss
Voss el 17 de Mzo. de 2023
Multiplication is * or .* but never . as in 4.x and never implicit as in (x^2)z
log10 is base-10 logarithm. log is natural logarithm.
z = log(x);
y = x-((((x^2)*z-(x^0.5)-20)*2*x^0.5)/(-1+4*x^(3/2)*z+2*x^(3/2)));

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