Hello, I am trying to use the central difference for the function sin(2*pi*x), centered around x=0.313. I know what value should be, that is my "act" variable. If someone can help me why my for loop is messed-up, it'll be much appreciated.
CODE:
clc, clear, close all
syms x
% Actual Value
f = sin(2*pi*x);
df = diff(f,x);
x = 0.313;
true_value = 2*pi*cos(2*pi*x)
% Central Difference
for i=1:3
x = 0.313;
for h = [0.01 0.1 0.25]
df_dx(i) = (sin(2*pi*(x+h))-sin(2*pi*(x-h)))/2*h
end
end
act = (sin(2*pi*0.314)-sin(2*pi*0.312))/0.02

 Respuesta aceptada

KALYAN ACHARJYA
KALYAN ACHARJYA el 19 de Mzo. de 2019
Editada: KALYAN ACHARJYA el 19 de Mzo. de 2019

0 votos

-If someone can help me why my for loop is messed-up-
clc, clear, close all
syms x
% Actual Value
f=sin(2*pi*x);
df=diff(f,x);
x=0.313;
true_value=2*pi*cos(2*pi*x);
h=[0.01 0.1 0.25];
df_dx=zeros(1, length(h));
% Central Difference
for i=1:3
df_dx(i)=(sin(2*pi*(x+h(i)))-sin(2*pi*(x-h(i))))/2*h(i);
end
act=(sin(2*pi*0.314)-sin(2*pi*0.312))/0.02;
Output:
df_dx=
-0.0002 -0.0227 -0.0964
Without for loop
clc, clear, close all
syms x
% Actual Value
f=sin(2*pi*x);
df=diff(f,x);
x=0.313;
true_value=2*pi*cos(2*pi*x);
h=[0.01 0.1 0.25];
% Central Difference
df_dx=(sin(2*pi.*(x+h))-sin(2*pi.*(x-h)))./(2*h);
act=(sin(2*pi*0.314)-sin(2*pi*0.312))/0.02;
Command Window:
>> df_dx
df_dx =
-0.0002 -0.0227 -0.0964

9 comentarios

madhan ravi
madhan ravi el 19 de Mzo. de 2019
Why do you need a loop?
KALYAN ACHARJYA
KALYAN ACHARJYA el 19 de Mzo. de 2019
@Madhan sir, I have added the answer without loop also.
madhan ravi
madhan ravi el 19 de Mzo. de 2019
Editada: madhan ravi el 19 de Mzo. de 2019
pre-allocation is essential for a loop, h(1:3) is the same as h
KALYAN ACHARJYA
KALYAN ACHARJYA el 19 de Mzo. de 2019
Editada: KALYAN ACHARJYA el 19 de Mzo. de 2019
Yes edited, truly thanks sir.
madhan ravi
madhan ravi el 19 de Mzo. de 2019
The proper usage is
./(2*h) % use parantheses
%^---- element wise operation
KALYAN ACHARJYA
KALYAN ACHARJYA el 19 de Mzo. de 2019
yeah..Thanks
madhan ravi
madhan ravi el 19 de Mzo. de 2019
./ is not corrected yet , please by any chance don't delete this answer so that it's pretty clear how many possible mistakes can be made in this problem.
KALYAN ACHARJYA
KALYAN ACHARJYA el 19 de Mzo. de 2019
Editada: KALYAN ACHARJYA el 19 de Mzo. de 2019
Yes, got it sir, but for small number of iterations, there may be negligible difference.right?
>> matlab_ans_march_19
Without Pre-allocation
Elapsed time is 0.003340 seconds.
%%
>> matlab_ans_march_19
With Pre-allocation
Elapsed time is 0.002957 seconds.
Yes, prefer to use for better coding performance (always).

Iniciar sesión para comentar.

Más respuestas (0)

Categorías

Más información sobre Loops and Conditional Statements en Centro de ayuda y File Exchange.

Preguntada:

el 19 de Mzo. de 2019

Editada:

el 19 de Mzo. de 2019

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by