Week 3.5 - Deriving 4th order approximation of a 2nd order derivative using Taylor Table method
Using central differencing scheme:
(i-2) (i-1) (i) (i+1) (i+2)
By the fourth order central differencing scheme:
= af (i-2) + bf (i-1) + cf (i) + df (i+1) +ef (i+2)
By using a Taylor series of expansion:
a f(i-2): (a) f(i) – (a) f’(i) (2)/1! + (a) f”(i) (2)2 / 2! - (a) f”’ (i) (2)3 /3! + (a) f’’’’(i)(2)4 /4! – (a) f’’’’’ (i) (2)5 /5! + a f’’’’’’ (i) (2)6 /6!
(b) f(i-1): (b) f(i) – (b) f’(i) (2)/1! + (b) f”(i) (2)2 / 2! - (b) f”’ (i) (2)3 /3! + (b) f’’’’(i) (2)4 /4! – (b) f’’’’’ (i) (2)5 /5! + (b) f’’’’’’(i) (2)6 /6!
(c) f(i) = (c) f(i)
(d) f(i+1): (d) f(i) + (d) f’(i) (2)/1! + (d) f”(i) (2)2 / 2! - (d) f”’ (i) (2)3 /3! + (d) f’’’’(i) (2)4 /4! – (d) f’’’’’ (i) (2)5 /5! + (d) f’’’’’’(i(2)6 /6!
(e) f(i+2) : (e) f(i) – (e) f’(i) (2)/1! + (e) f”(i) (2)2 / 2! - (e) f”’ (i) (2)3 /3! + (e) f’’’’(i) (2)4 /4! – (e) f’’’’’ (i) (2)5 /5! + (e) f’’’’’’(i) (2)6 /6!
|
f(i) |
|||||||
|
(a)f(i-2) |
a |
-2a |
-4a/2 |
-8a/6 |
16a/24 |
-32a/120 |
64a/240 |
|
(b) f(i-1) |
b |
-b |
b/2 |
-b/6 |
b/24 |
-b/120 |
b/240 |
|
(c) f(i) |
c |
0 |
0 |
0 |
0 |
0 |
0 |
|
(d) f(i+1) |
d |
d |
d/2 |
d/6 |
d/24 |
d/120 |
d/240 |
|
(e) f(i+2) |
e |
2e |
4e/2 |
8e/6 |
16e/24 |
-32e/120 |
64e/240 |
|
∑(a)f(i-2) + (b)f(i-1)+(c)f(i)+(d)f(i+1) + (e)f(i+2) |
0 |
0 |
1 |
0 |
0 |
0 |
0 |
The five equations is:
a + b + c + d + e = 0 (1)
-2a - b + 0 + e + 2e = 0 (2)
2a + b/2 + 0 + d/2 + 2e + 1 = 0 (3)
-8a/6 - b/6 + 0 + d/6 + 8e/6 =0 (4)
16a/24 + b/24 + 0 + d/24 + 16e /24 = 0 (5)
The above equation from the matrix is AX = B
1 1 1 1 1 a 0
-2 1 0 1 2 b 0
2 1/2 0 1/2 2 c 1
-8/6 -1/6 0 1/6 8/6 d 0
16/24 1/24 0 1/24 16/24 e 0
(A) (X) = (B)
a = - 0.083333
b = 1.33333
c = -2.5000
d = 1.3333
e = -0.08333
= (3.75000) f(i) + (-12.8333 f(i +1) + ( 17.833) f(i+2) + (-13.0000) f(i + 3) + (5.08333) f(i+4) + (-0.8333)f(i+5) / 2
= (b/2 + 2c + 9d/2 + 8e + 25f/2) f’’ (i) + (b/720 + 64c/720 + 729d/720 + 4096e/ 720 + 15625f/ 720) f’’’’’’ (i) 4
= (2(-0.8333) + 1.333/2 + 1.333/2 + 2 (0.8333)) f’’ (i) + (64/ 240 (-0.0833) + 1.333/240 + 1.3333/240 + 64 / 240 (0.08333)) f’’’’’’ (i) 4
- Skewed right sided differencing:
i (i + 1) (i+2) (i+3) (i+4) (i+5)
According to
= af (i) + bf (i+1) + cf (i+2) + df (i+3) +ef (i+4) + g f(i+5)
- f(i) = (a) f(i)
- f(i+1): (b) f(i) + (b) f’(i) ()/1! + (b) f”(i) ()2 / 2! - (b) f”’ (i) ()3 /3! + (b) f’’’’(i) ()4 /4! + (b) f’’’’’ (i) ()5 /5! + (b) f’’’’’’ (i) ()6 /6!
- f(i+2): (c) f(i) + (c) f’(i) (2)/1! + (c) f”(i) ()2 / 2! - (c) f”’ (i) ()3 /3! + (c) f’’’’(i) ()4 /4! + (c) f’’’’’ (i) ()5 /5! + (c) f’’’’’’ (i) ()6 /6!
- f(i+3): (d) f(i) + (d) f’(i) (3)/1! + (d) f”(i) ()2 / 2! - (d) f”’ (i) ()3 /3! + (d) f’’’’ (i) ()4 /4! + (d) f’’’’’ (i) ()5 /5! + (d) f’’’’’’ (i) ()6 /6!
- f(i+4): (e) f(i) + (e) f’(i) (4)/1! + (e) f”(i) ()2 / 2! - (e) f”’ (i) ()3 /3! + (e) f’’’’(i) ()4 /4! + (e) f’’’’’ (i) ()5 /5! + (e) f’’’’’’ (i) ()6 /6!
- f(i+5): (f) f(i) + (f) f’(i) (5)/1! + (f) f”(i) ()2 / 2! - (f) f”’ (i) ()3 /3! + (f) f’’’’(i) ()4 /4! + (f) f’’’’’ (i) ()5 /5! + (f) f’’’’’’ (i) ()6 /6!
|
f(i) |
|||||||
|
(a)f(i) |
a |
0 |
0 |
0 |
0 |
0 |
0 |
|
(b) f(i+2) |
b |
b |
b/2 |
b/6 |
b/24 |
b/120 |
b/720 |
|
(c) f(i+3) |
c |
2c |
4c/2 |
8c/6 |
16c/24 |
32c/120 |
64c/720 |
|
(d) f(i+4) |
d |
3d |
9d/2 |
27d/6 |
81d/24 |
243d/120 |
729d/720 |
|
(e) f(i+5) |
e |
4e |
16e/2 |
64e/6 |
256e/24 |
1024e/120 |
4096e/720 |
|
(f)f(i+6) |
f |
5f |
25f/2 |
125f/6 |
625f/24 |
3125f/120 |
15625f/720 |
|
∑(a)f(i) + (b)f(i+2)+(c)f(i+3)+(d)f(i+4) + (e)f(i+5) +f (i+6) |
0 |
0 |
1 |
0 |
0 |
0 |
0 |
The six equations is:
a + b + c + d + e + f = 0 (1)
0 + b + 2c + 3d + 4e + 5f = 0 (2)
0 + b/2 + 2c+ 9d/2+ 8e + 25f/2 = 1 (3)
0 + b/6 + 8c/6 + 27d/6 + 16e + 125f/6 = 0 (4)
0 + b/24+ 16c/24+ 81d/24+ 256e/24 + 625f/24= 0 (5)
The above equation from the matrix is AX = B
1 1 1 1 1 1 a 0
0 1 2 3 4 5 b 0
0 1/2 2 9/2 8 25/2 c 1
0 1/6 8/6 27/6 64/6 125/6 d 0
0 1/24 16/24 81/24 256/24 1024/120 e 0
0 1/120 32/120 243/120 1024/120 3125/120 f 0
(A) (X) = (B)
a = 3.750000
b = -12.833333
c = 17.8333
d = -13.0000
e = 5.08333
f = -0.8333
= -0.83333 f(i-2) + 1.333 f(i – 1) + ( -2.5000) f(i) + 1.333 f(i + 1) + -0.8333 f(i+2) / 2
- Skewed Left sided differencing:
(i – 5) (i -4) (i-3) (i-2) (i-1) (i)
According to
= af (i-5) + bf (i-4) + cf (i-3) + df (i-2) +ef (i-1) + g f(i)
- f(i-5): (a) f(i) - (a) f’(i) ()/1! + (a) f”(i) ()2 / 2! - (a) f”’ (i) ()3 /3! + (a) f’’’’ (i) ()4 /4! - (a) f’’’’’ (i) ()5 /5! + (a) f’’’’’’ (i) ()6 /6!
- f(i-4): (b) f(i) - (b) f’(i) ()/1! + (b) f”(i) ()2 / 2! - (b) f”’ (i) ()3 /3! + (b) f’’’’ (i) ()4 /4! - (b) f’’’’’ (i) ()5 /5! + (b) f’’’’’’ (i) ()6 /6!
- f(i-4): (c) f(i) - (c) f’(i) ()/1! + (c) f”(i) ()2 / 2! - (c) f”’ (i) ()3 /3! + (c) f’’’’ (i) ()4 /4! - (c) f’’’’’ (i) ()5 /5! + (c) f’’’’’’ (i) ()6 /6!
- f(i-4): (d) f(i) - (d) f’(i) ()/1! + (d) f”(i) ()2 / 2! - (d) f”’ (i) ()3 /3! + (d) f’’’’ (i) ()4 /4! - (d) f’’’’’ (i) ()5 /5! + (d) f’’’’’’ (i) ()6 /6!
- f (i-4): (e) f(i) - (e) f’(i) ()/1! + (e) f”(i) ()2 / 2! - (e) f”’ (i) ()3 /3! + (e) f’’’’ (i) ()4 /4! - (e) f’’’’’ (i) ()5 /5! + (e) f’’’’’’ (i) ()6 /6!
- f(i) = (a) f(i)
|
f(i) |
|||||||
|
(a)f(i-5) |
a |
-5a |
25a/2 |
-125a/6 |
625a/24 |
-3125a/120 |
15625a/720 |
|
(b) f(i-4) |
b |
-4b |
8b |
-64b/6 |
256b/24 |
-1024b/120 |
4096b/720 |
|
(c) f(i-3) |
c |
3c |
9c/2 |
-27c/6 |
81c/24 |
-243c/120 |
729c/120 |
|
(d) f(i-2) |
d |
-2d |
2d |
-8d/6 |
16d/24 |
-32d/120 |
64d/720 |
|
(e) f(i-1) |
e |
-e |
e/2 |
-e/6 |
e/24 |
-e/120 |
e/720 |
|
(f)f(i) |
f |
0 |
0 |
0 |
0 |
0 |
0 |
|
∑(a)f(i-5) + (b)f(i-4)+(c)f(i-3)+(d)f(i-2) + (e)f(i-1) +ff(i) |
0 |
0 |
1 |
0 |
0 |
0 |
0 |
The six equations is:
a + b + c + d + e + f = 0 (1)
-5a- 4b -3c -2d -e + 0 = 0 (2)
25a/2 + 8b + 9c/2+2d+ e/2 + 0 = 1 (3)
625a/24+ 256b/24 + 81c/24 + 16d/24 + e/24 + 0 = 0 (4)
-3125a/120 -1024b/120-243c/120-32d/120-e/120+0 = 0 (5
The above equation from the matrix is AX = B
1 1 1 1 1 1 a 0
-5 -4 -3 -2 -1 0 b 0
25/2 8/2 9/2 2 1/2 0 c 1
-125/6 -64/6 -27/6 -8/6 64/6 -1/6 d 0
625/24 256/24 81/24 16/24 256/24 1/24 e 0
-3125/120 -1024/120 -243/120 -32/120 1024/120 -1/120 f 0
(A) (X) = (B)
a = -0.8333
b = 5.0833333
c = -13.000
d = 17.8333
e = -12.88333
f = 3.75000
= (-0.8333) f(i-5) + (5.088333 f(i -4) + ( -13.000) f(i-3) + (17.8333) f(i -2) + (-12.88333) f(i-1) + (-3.7500)f(i) / 2
= (25a/2 + 8b + 9c/2 + 2d+ e/2) f’’ (i) + (15625a/720 + 4096b/720 + 729c/720 + 64d/ 720 + e/ 720) f’’’’’’ (i) 4
Function of central differencing 4th order particular spacing dx:
% Analytical function f(x) = exp(x)*cos(x);
%f’’(x) = -2 exp(x)*sin(x);
Analytical derivative -2 exp(x)*sin(x);
% values of coefficient calculated by Taylor’s table;
a = - 0.083333;
b = 1.33333;
c = -2.5000;
d = 1.3333;
e = -0.08333;
%numerical derivative;
% central differencing scheme =
% ((a*f (x-2*dx) + (b*f (x-dx)) + (c*f (x)) + (d*f (x+dx)) +(e*f (x+2*dx)))/dx^2;
% fourth order approximation:
((a*exp(x-2*dx) *cos(x-2*dx)+ (b* exp (x-dx)*cos(x-dx) + (c* exp (x)*cos(x)) + (d*f (x+dx)*cos(x+dx))+(e* exp (x+2*dx)*cos(x+2dx)))/dx^2;
Function of skewed left sided 4th order particular spacing dx:
% Analytical function f(x) = exp(x)*cos(x);
%f’’(x) = -2 exp(x)*sin(x);
Analytical derivative -2 exp(x)*sin(x);
% values of coefficient calculated by Taylor’s table;
a = - 0.083333;
b = 5.083333;
c = -13.0000;
d = 17.83333;
e = 12.88333;
f = 3.75000;
%numerical derivative;
% skewed differencing scheme =
% ((a*f (x-5*dx) + (b*f (x-4*dx)) + (c*f (x-3*dx)) + (d*f (x-2*dx)) +(e*f (x-1*dx))+(f*(f(x))/dx^2;
% skewed fourth order approximation:
((a*exp(x-5*dx) *cos(x-5*dx)+ (b* exp (x-4*dx)*cos(x-4*dx) + (c* exp (x-3*dx)*cos(x-3*dx)) + (d*f (x-2*dx)*cos(x-2*dx))+(e* exp (x-1*dx)*cos(x-1*dx))+(f*exp(x))*cos(x)))/dx^2;
