Central Difference For Loop

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

 Risposta accettata

KALYAN ACHARJYA
KALYAN ACHARJYA il 19 Mar 2019
Modificato: KALYAN ACHARJYA il 19 Mar 2019
-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 Commenti

madhan ravi
madhan ravi il 19 Mar 2019
Why do you need a loop?
@Madhan sir, I have added the answer without loop also.
madhan ravi
madhan ravi il 19 Mar 2019
Modificato: madhan ravi il 19 Mar 2019
pre-allocation is essential for a loop, h(1:3) is the same as h
KALYAN ACHARJYA
KALYAN ACHARJYA il 19 Mar 2019
Modificato: KALYAN ACHARJYA il 19 Mar 2019
Yes edited, truly thanks sir.
The proper usage is
./(2*h) % use parantheses
%^---- element wise operation
yeah..Thanks
madhan ravi
madhan ravi il 19 Mar 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 il 19 Mar 2019
Modificato: KALYAN ACHARJYA il 19 Mar 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).

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