Central Difference For Loop

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Robert  Flores
Robert Flores 2019년 3월 19일
편집: KALYAN ACHARJYA 2019년 3월 19일
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

채택된 답변

KALYAN ACHARJYA
KALYAN ACHARJYA 2019년 3월 19일
편집: KALYAN ACHARJYA 2019년 3월 19일
-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
  댓글 수: 9
madhan ravi
madhan ravi 2019년 3월 19일
./ 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 2019년 3월 19일
편집: KALYAN ACHARJYA 2019년 3월 19일
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).

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