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Deriving 4th order approximation of a 2nd order derivative using Taylor Table method

Aim: Deriving 4th order approximation of a 2nd order derivative using Taylor Table method Governing equations Central difference scheme(Symmetric scheme) Number of nodes in central schemes= p+q-1(p=order of derivative,q=order of approximation) …

Faizan Akhtar
updated on 16 Feb 2021

Aim: Deriving 4th order approximation of a 2nd order derivative using Taylor Table method

Governing equations

Central difference scheme(Symmetric scheme)

Number of nodes in central schemes= p+q-1(p=order of derivative,q=order of approximation)

$= 2 + 4 - 1 = 5$ $=2+4-1=5$

$\frac{d^{2} y}{{d x}^{2}} \approx a f (i - 2) + b f (i - 1) + c f (i) + d f (i + 1) + e f (i + 2)$ $frac{d^2y}{dx^2} approx af(i-2)+bf(i-1)+cf(i)+df(i+1)+ef(i+2)$

$f(i-2)=f(i)-2**Deltaxf'(i)+frac{f''(i)4Deltax^2}{2}-frac{f'''(i)8Deltax^3}{6}+frac{f''''(i)16Deltax^4}{24}-frac{f'''''(i)32Deltax^5}{120}+frac{f''''''(i)64Deltax^6}{720}$

Multiplying the above equation by a

$a**(f(i-2))=a**(f(i)-2**Deltaxf'(i)+frac{f''(i)4Deltax^2}{2}-frac{f'''(i)8Deltax^3}{6}+frac{f''''(i)16Deltax^4}{24}-frac{f'''''(i)32Deltax^5}{120}+frac{f''''''(i)64Deltax^6}{720})$

$f(i-1)$ can be calculated by Taylor theorem

$f(i-1)=f(i)-f'(i)Deltax+frac{f''(i)Deltax^2}{2}-frac{f'''(i)Deltax^3}{6}+frac{f''''(i)Deltax^4}{24}-frac{f'''''(i)Deltax^5}{120}+frac{f''''''(i)Deltax^6}{720}$

Multiplying the above equation by b

$b**(f(i-1))=b**(f(i)-f'(i)Deltax+frac{f''(i)Deltax^2}{2}-frac{f'''(i)Deltax^3}{6}+frac{f''''(i)Deltax^4}{24}-frac{f'''''(i)Deltax^5}{120}+frac{f''''''(i)Deltax^6}{720})$

$f(i+1)$ can be calculated by Taylor theorem

$f(i+1)=f(i)+f'(i)Deltax+frac{f''(i)Deltax^2}{2}+frac{f'''(i)Deltax^3}{6}+frac{f''''(i)Deltax^4}{24}+frac{f'''''(i)Deltax^5}{120}+frac{f''''''(i)Deltax^6}{720}$

Multiplying the above equation by d

$d**(f(i+1))=d**(f(i)+f'(i)Deltax+frac{f''(i)Deltax^2}{2}+frac{f'''(i)Deltax^3}{6}+frac{f''''(i)Deltax^4}{24}+frac{f'''''(i)Deltax^5}{120}+frac{f''''''(i)Deltax^6}{720})$

$f(i+2)$ can be calculated by Taylor theorem

$f(i+2)=f(i)+2**Deltaxf'(i)+frac{f''(i)4Deltax^2}{2}+frac{f'''(i)8Deltax^3}{6}+frac{f''''(i)16Deltax^4}{24}+frac{f'''''(i)32Deltax^5}{120}+frac{f''''''(i)64Deltax^6}{720}$

Multiplying the above equation by e

$e**(f(i+2))=e**(f(i)+2**Deltaxf'(i)+frac{f''(i)4Deltax^2}{2}+frac{f'''(i)8Deltax^3}{6}+frac{f''''(i)16Deltax^4}{24}+frac{f'''''(i)32Deltax^5}{120}+frac{f''''''(i)64Deltax^6}{720})$

Forming Taylor's table for keeping track of coefficient

Numerical Stencil	Taylor's series terms
Numerical Stencil	$f(i)$	$Deltaxf'(i)$	$Deltax^2**f''(i)$	$Deltax^3f'''(i)$	$Deltax^4f''''(i)$	$Deltax^5f'''''(i)$	$Deltax^6f''''''(i)$
$a**f(i-2)$	$a$	$-2a$	$2a$	$frac{-8a}{6}$	$frac{16a}{24}$	$frac{-32a}{120}$	$frac{64a}{720}$
$b**f(i-1)$	$b$	$-b$	$frac{b}{2}$	$frac{-b}{6}$	$frac{b}{24}$	$frac{-b}{120}$	$frac{b}{720}$
$c**f(i)$	$c$	$0$	$0$	$0$	$0$	$0$	$0$
$d**f(i+1)$	$d$	$d$	$frac{d}{2}$	$frac{d}{6}$	$frac{d}{24}$	$frac{d}{120}$	$frac{d}{720}$
$e**f(i+2)$	$e$	$2e$	$2e$	$frac{8e}{6}$	$frac{16e}{24}$	$frac{32e}{120}$	$frac{64e}{720}$

Summing each and every term in the table

$Sigma a**f(i-2)+b**f(i-1)+cf(i)+d**f(i+1)+e**f(i+2)=frac{d^2y}{dx^2}$

$Sigma a+b+c+d+e=0$ (As such equation under consideration is second order)

$Sigma-2a-b+0+d+2e=0$ (As such equation under consideration is second order)

$Sigma2a+frac{b}{2}+0+frac{d}{2}+2e=1$

$Sigmafrac{-8a}{6}+frac{-b}{6}+0+frac{d}{6}+frac{8e}{6}=0$

$Sigmafrac{16a}{24}+frac{b}{24}+0+frac{d}{24}+frac{16e}{24}=0$

The values of $a,b,c,d,e$ can be calculated by Matlab

$a=frac{-1}{12}$

$b=frac{4}{3}$

$c=frac{-5}{2}$

$d=frac{4}{3}$

$e=frac{-1}{12}$

Thus $a=e and b=d$

Rearranging all the terms

$frac{d^2y}{dx^2} approx af(i-2)+bf(i-1)+cf(i)+df(i+1)+ef(i+2)=(a+b+c+d+e)f(i)+(-2a-b+0+d+2e)Deltaxf'(i)+(2a+frac{b}{2}+0+frac{d}{2}+2e)Deltax^2f''(i)+(frac{-8a}{6}+frac{-b}{6}+0+frac{d}{6}+frac{8e}{6})Deltax^3f'''(i)+(frac{16a}{24}+frac{b}{24}+0+frac{d}{24}+frac{16e}{24})Deltax^4f''''(i)+(frac{-32a}{120}+frac{-b}{120}+0+frac{d}{120}+frac{32e}{120})Deltax^5f'''''(i)+(frac{64a}{720}+frac{b}{720}+0+frac{d}{720}+frac{64e}{720})Deltax^6f''''''(i)$

Equating LHS and RHS

$frac{d^2y}{dx^2}=frac{-1}{12}f(i-2)+frac{4}{3}f(i-1)+frac{-5}{2}f(i)+frac{4}{3}f(i+1)+frac{-1}{12}f(i+2)$ = $0**f(i)+0**f'(i)+(-2a-b+0+d+2e)Deltax^2f''(i)+0**f'''(i)+0**f''''(i)+0**f'''''(i)+(some value)**Deltax^6f''''''(i)$

Dividing by $Deltax^2 throughout$

$frac{d^2y}{dx^2}=frac{-1}{12}f(i-2)+frac{4}{3}f(i-1)+frac{-5}{2}f(i)+frac{4}{3}f(i+1)+frac{-1}{12}f(i+2)$ $**frac{1}{Deltax^2}$ = $(-2a-b+0+d+2e)**f''(i)$ +order of ( $Deltax^4$ )

Thus equation for the skewed right-sided difference is

$frac{d^2y}{dx^2}=(frac{-1}{12}f(i-2)+frac{4}{3}f(i-1)+frac{-5}{2}f(i)+frac{4}{3}f(i+1)+frac{-1}{12}f(i+2))$ $**frac{1}{Deltax^2}$

Skewed right-sided difference

Solution

Number of nodes in right sided schemes= p+q (p=order of derivative,q=order of approximation)

= $2+4=6$

$frac{d^2y}{dx^2} approx af(i)+bf(i+1)+cf(i+2)+df(i+3)+ef(i+4)+gf(i+5)$ `

$f(i+1)$ can be calculated by Taylor theorem

$f(i+1)=f(i)+f'(i)Deltax+frac{f''(i)Deltax^2}{2}+frac{f'''(i)Deltax^3}{6}+frac{f''''(i)Deltax^4}{24}+frac{f'''''(i)Deltax^5}{120}+frac{f''''''(i)Deltax^6}{720}$

Multiplying the above equation by b

$b**(f(i+1))=b**(f(i)+f'(i)Deltax+frac{f''(i)Deltax^2}{2}+frac{f'''(i)Deltax^3}{6}+frac{f''''(i)Deltax^4}{24}+frac{f'''''(i)Deltax^5}{120}+frac{f''''''(i)Deltax^6}{720})$

$f(i+2)$ can be calculated by Taylor theorem

$f(i+2)=f(i)+2**Deltaxf'(i)+frac{f''(i)4Deltax^2}{2}+frac{f'''(i)8Deltax^3}{6}+frac{f''''(i)16Deltax^4}{24}+frac{f'''''(i)32Deltax^5}{120}+frac{f''''''(i)64Deltax^6}{720}$

Multiplying the above equation by c

$c**(f(i+2))=c**(f(i)+2**Deltaxf'(i)+frac{f''(i)4Deltax^2}{2}+frac{f'''(i)8Deltax^3}{6}+frac{f''''(i)16Deltax^4}{24}+frac{f'''''(i)32Deltax^5}{120}+frac{f''''''(i)64Deltax^6}{720})$

$f(i+3)$ can be calculated by Taylor theorem

$f(i+3)=f(i)+f'(i)**3Deltax+f''(i)frac{9}{2}Deltax^2+f'''(i)frac{27}[6}Deltax^3+f''''(i)frac{81}{24}Deltax^4+f'''''frac{243}{120}Deltax^5+f''''''(i)frac{729}{720}Deltax^6$

Multiplying the above equation by d

$d**f(i+3)=d**(f(i)+f'(i)**3Deltax+f''(i)frac{9}{2}Deltax^2+f'''(i)frac{27}[6}Deltax^3+f''''(i)frac{81}{24}Deltax^4+f'''''frac{243}{120}Deltax^5+f''''''(i)frac{729}{720}Deltax^6)$

$f(i+4)$ can be calculated by Taylor theorem

$f(i+4)=f(i)+f'(i)4**Deltax+f''(i)frac{16}{2}Deltax^2+f'''(i)frac{64}{6}Deltax^3+f''''(i)frac{256}{24}Deltax^4+f'''''(i)frac{1024}{120}Deltax^5+f''''''(i)frac{4096}{720}Deltax^6$

Multiplying the above equation by e

$e**(f(i+4))=e**(f(i)+f'(i)4**Deltax+f''(i)frac{16}{2}Deltax^2+f'''(i)frac{64}{6}Deltax^3+f''''(i)frac{256}{24}Deltax^4+f'''''(i)frac{1024}{120}Deltax^5+f''''''(i)frac{4096}{720}Deltax^6)$

$f(i+5)$ can be calculated by Taylor theorem

$f(i+5)=f(i)+f'(i)**5**Deltax+f''(i)frac{25}[2}Deltax^2+f'''(i)frac{125}{6}Deltax^3+f''''(i)frac{625}{24}Deltax^4+f'''''(i)frac{3125}{120}Deltax^5+f''''''(i)frac{15625}{720}Deltax^6$

Multiplying the above equation by g

$g**(f(i+5))=g**(f(i)+f'(i)**5**Deltax+f''(i)frac{25}[2}Deltax^2+f'''(i)frac{125}{6}Deltax^3+f''''(i)frac{625}{24}Deltax^4+f'''''(i)frac{3125}{120}Deltax^5+f''''''(i)frac{15625}{720}Deltax^6)$

Calculating Coefficients through Taylor's table.

Numerical Stencil	Taylor's series terms
Numerical Stencil	$f(i)$	$Deltaxf'(i)$	$Deltax^2f''(i)$	$Deltax^3f'''(i)$	$Deltax^4f''''(i)$	$Deltax^5f'''''(i)$
$af(i)$	a	0	0	0	0	0
$bf(i+1)$	b	b	$frac{b}{2}$	$frac{b}{6}$	$frac{b}{24}$	$frac{b}{120}$
$cf(i+2)$	c	2c	$frac{4c}{2}$	$frac{8c}{6}$	$frac{16c}{24}$	$frac{32c}{120}$
$df(i+3)$	d	3d	$frac{9d}{2}$	$frac{27d}{6}$	$frac{81d}{24}$	$frac{243d}{120}$
$ef(i+4)$	e	4e	$frac{16e}{2}$	$frac{64e}{6}$	$frac{256e}{24}$	$frac{1024e}{120}$
$gf(i+5)$	g	5g	$frac{25g}{2}$	$frac{125g}{6}$	$frac{625g}{24}$	$frac{3125g}{120}$

Solving for $a,b,c,d,e,g$

Summing Taylor's series terms

$Sigmaa+b+c+d+e+g=0$

$Sigma0+b+2c+3d+4e+5g=0$

$Sigma0+frac{b}{2}+frac{4c}{2}+frac{9d}{2}+frac{16e}{2}+frac{25g}{2}=1$

$Sigma0+frac{b}{6}+frac{8c}{6}+frac{27d}{6}+frac{64e}{6}+frac{125g}{6}=0$

$Sigma0+frac{b}{24}+frac{16c}{24}+frac{81d}{24}+frac{256e}{24}+frac{625g}{24}=0$

$Sigma0+frac{b}{120}+frac{32c}{120}+frac{243d}{120}+frac{1024e}{120}+frac{3125g}{120}=0$

$a=frac{15}{4}$

$b=frac{-77}{6}$

$c=frac{107}{6}$

$d=-13$

$e=frac{61}{12}$

$g=frac{-5}{6}$

Rearranging all the terms

Thus equation for the skewed right-sided difference is

$frac{d^2y}{dx^2} = (frac{15}{4}f(i)+frac{-77}{6}f(i+1)+frac{107}{6}f(i+2)-13f(i+3)+frac{61}{12}f(i+4)+frac{-5}{6}f(i+5))**frac{1}{Deltax^2}$

Skewed left-sided difference

Number of nodes in left sided schemes= p+q (p=order of derivative,q=order of approximation)

= $2+4=6$

$frac{d^2y}{dx^2} approx af(i)+bf(i-1)+cf(i-2)+df(i-3)+ef(i-4)+gf(i-5)$ `

$f(i-1)$ can be calculated by Taylor theorem`

$f(i-1)=f(i)-f'(i)Deltax+frac{f''(i)Deltax^2}{2}-frac{f'''(i)Deltax^3}{6}+frac{f''''(i)Deltax^4}{24}-frac{f'''''(i)Deltax^5}{120}+frac{f''''''(i)Deltax^6}{720}+..........$

Multiplying the above equation by b

$b**(f(i-1))=b**(f(i)-f'(i)Deltax+frac{f''(i)Deltax^2}{2}-frac{f'''(i)Deltax^3}{6}+frac{f''''(i)Deltax^4}{24}-frac{f'''''(i)Deltax^5}{120}+frac{f''''''(i)Deltax^6}{720}+)$

$f(i-2)$ can be calculated by Taylor theorem

$f(i-2)=f(i)-2**Deltaxf'(i)+frac{f''(i)4Deltax^2}{2}-frac{f'''(i)8Deltax^3}{6}+frac{f''''(i)16Deltax^4}{24}-frac{f'''''(i)32Deltax^5}{120}+frac{f''''''(i)64Deltax^6}{720}+.....................$

Multiplying the above equation by c

$c**(f(i-2))=c**(f(i)-2**Deltaxf'(i)+frac{f''(i)4Deltax^2}{2}-frac{f'''(i)8Deltax^3}{6}+frac{f''''(i)16Deltax^4}{24}-frac{f'''''(i)32Deltax^5}{120}+frac{f''''''(i)64Deltax^6}{720}+.....................)$

$f(i-3)$ can be calculated by Taylor theorem`

$f(i-3)=f(i)-f'(i)**3Deltax+f''(i)frac{9}{2}Deltax^2-f'''(i)frac{27}[6}Deltax^3+f''''(i)frac{81}{24}Deltax^4-f'''''(i)frac{243}{120}Deltax^5+f''''''(i)frac{729}{720}Deltax^6$

Multiplying the above equation by d

$d**(f(i-3))=d**(f(i)-f'(i)**3Deltax+f''(i)frac{9}{2}Deltax^2-f'''(i)frac{27}[6}Deltax^3+f''''(i)frac{81}{24}Deltax^4-f'''''(i)frac{243}{120}Deltax^5+f''''''(i)frac{729}{720}Deltax^6)$

$f(i-4)$ can be calculated by Taylor theorem `

$f(i-4)=f(i)-f'(i)4**Deltax+f''(i)frac{16}{2}Deltax^2-f'''(i)frac{64}{6}Deltax^3+f''''(i)frac{256}{24}Deltax^4-f'''''(i)frac{1024}{120}Deltax^5+f''''''(i)frac{4096}{720}Deltax^6$

Multiplying the above equation by e

$e**(f(i-4))=e**(f(i)-f'(i)4**Deltax+f''(i)frac{16}{2}Deltax^2-f'''(i)frac{64}{6}Deltax^3+f''''(i)frac{256}{24}Deltax^4-f'''''(i)frac{1024}{120}Deltax^5+f''''''(i)frac{4096}{720}Deltax^6)$

$f(i-5)$ can be calculated by Taylor theorem`

$f(i-5)=f(i)-f'(i)**5**Deltax+f''(i)frac{25}[2}Deltax^2-f'''(i)frac{125}{6}Deltax^3+f''''(i)frac{625}{24}Deltax^4-f'''''(i)frac{3125}{120}Deltax^5+f''''''(i)frac{15625}{720}Deltax^6$

Multiplying the above equation by g

$g**(f(i-5))=g**(f(i)-f'(i)**5**Deltax+f''(i)frac{25}[2}Deltax^2-f'''(i)frac{125}{6}Deltax^3+f''''(i)frac{625}{24}Deltax^4-f'''''(i)frac{3125}{120}Deltax^5+f''''''(i)frac{15625}{720}Deltax^6)$

Calculating Coefficients through Taylor's table.

Numerical Stencil	Taylor series table
Numerical Stencil	$f(i)$	$Deltaxf'(i)$	$Deltax^2f''(i)$	$Deltax^3f'''(i)$	$Deltax^4f''''(i)$	$Deltax^5f'''''(i)$
$af(i)$	a	0	0	0	0	0
$bf(i-1)$	b	$-b$	$frac{b}{2}$	$frac{-b}{6}$	$frac{b}{24}$	$frac{-b}{120}$
$cf(i-2)$	c	$-2c$	$frac{4c}{2}$	$frac{-8c}{6}$	$frac{16c}{24}$	$frac{-32c}{120}$
$df(i-3)$	d	$-3d$	$frac{9d}{2}$	$frac{-27d}{6}$	$frac{81d}{24}$	$frac{-243d}{120}$
$ef(i-4)$	e	$-4e$	$frac{16e}{2}$	$frac{-64e}{6}$	$frac{256e}{24}$	$frac{-1024e}{120}$
$gf(i-5)$	g	$-5g$	$frac{25g}{2}$	$frac{-125g}{6}$	$frac{625g}{24}$	$frac{-3125g}{120}$

Summation of all the terms will yield

$Sigmaa+b+c+d+e+g=0$

$Sigma0-b-2c-3d-4e-5g=0$

$Sigma0+frac{b}{2}+frac{4c}{2}+frac{9d}{2}+frac{16e}{2}+frac{25g}{2}=1$

$Sigma0+frac{-b}{6}+frac{-8c}{6}+frac{27d}{6}+frac{-64e}{6}+frac{-125g}{6}=0$

$Sigma0+frac{b}{24}+frac{16c}{24}+frac{81d}{24}+frac{256e}{24}+frac{625g}{24}=0$

$Sigma0+frac{-b}{120}+frac{-32c}{120}+frac{-243d}{120}+frac{-1024e}{120}+frac{-3125g}{120}=0$

$a=frac{15}{4}$

$b=frac{-77}{6}$

$c=frac{107}{6}$

$d=-13$

$e=frac{61}{12}$

$g=frac{-5}{6}$

Thus equation for the skewed left-sided difference is

$frac{d^2y}{dx^2} = (frac{15}{4}f(i)+frac{-77}{6}f(i-1)+frac{107}{6}f(i-2)-13f(i-3)+frac{61}{12}f(i-4)+frac{-5}{6}f(i-5))**frac{1}{Deltax^2}$

Matlab coding

Calculating central difference scheme coefficient through Matlab

>> A=[1,1,1,1,1;-2,-1,0,1,2;2,1/2,0,1/2,2;-8/6,-1/6,0,1/6,8/6;16/24,1/24,0,1/24,16/24]

A =

    1.0000    1.0000    1.0000    1.0000    1.0000
   -2.0000   -1.0000         0    1.0000    2.0000
    2.0000    0.5000         0    0.5000    2.0000
   -1.3333   -0.1667         0    0.1667    1.3333
    0.6667    0.0417         0    0.0417    0.6667

>> B=[0;0;1;0;0]

B =

     0
     0
     1
     0
     0

>> X=AB

X =

   -0.0833
    1.3333
   -2.5000
    1.3333
   -0.0833

Calculating skewed right-sided scheme coefficient through Matlab

>> A=[1,1,1,1,1,1;0,1,2,3,4,5;0,1/2,4/2,9/2,16/2,25/2;0,1/6,8/6,27/6,64/6,125/6;0,1/24,16/24,81/24,256/24,625/24;0,1/120,32/120,243/120,1024/120,3125/120]

A =

    1.0000    1.0000    1.0000    1.0000    1.0000    1.0000
         0    1.0000    2.0000    3.0000    4.0000    5.0000
         0    0.5000    2.0000    4.5000    8.0000   12.5000
         0    0.1667    1.3333    4.5000   10.6667   20.8333
         0    0.0417    0.6667    3.3750   10.6667   26.0417
         0    0.0083    0.2667    2.0250    8.5333   26.0417
>> B=[0;0;1;0;0;0]

B =

     0
     0
     1
     0
     0
     0

>> X=AB

X =

    3.7500
  -12.8333
   17.8333
  -13.0000
    5.0833
   -0.8333

Note: All the coefficients have been turned in fractions for ease in the calculation.

Calculating second-order derivative of exp(x)*cos(x) and comparing it with the three schemes.

Matlab code

function error_cds=central_difference_scheme(x,dx)
Exact_derivative=-2*exp(x)*sin(x);
Numerical_derivative=((-1/12)*exp(x-2*dx)*cos(x-2*dx)+(4/3)*exp(x-dx)*cos(x-dx)+(-5/2)*exp(x)*cos(x)+(4/3)*exp(x+dx)*cos(x+dx)+(-1/12)*exp(x+2*dx)*cos(x+2*dx))/dx^2;
error_cds=abs(Numerical_derivative-Exact_derivative);
end

function error_srs=skewed_right_sided_scheme(x,dx)
Exact_derivative=-2*exp(x)*sin(x);
Numerical_derivative=((15/4)*exp(x)*cos(x)+(-77/6)*exp(x+dx)*cos(x+dx)+(107/6)*exp(x+2*dx)*cos(x+2*dx)-13*exp(x+3*dx)*cos(x+3*dx)+(61/12)*exp(x+4*dx)*cos(x+4*dx)+(-5/6)*exp(x+5*dx)*cos(x+5*dx))/dx^2;
error_srs=abs(Numerical_derivative-Exact_derivative);
end

function error_sls=skewed_left_sided_scheme(x,dx)
Exact_derivative=-2*exp(x)*sin(x);
Numerical_derivative=((15/4)*exp(x)*cos(x)+(-77/6)*exp(x-dx)*cos(x-dx)+(107/6)*exp(x-2*dx)*cos(x-2*dx)-13*exp(x-3*dx)*cos(x-3*dx)+(61/12)*exp(x-4*dx)*cos(x-4*dx)+(-5/6)*exp(x-5*dx)*cos(x-5*dx))/dx^2;
error_sls=abs(Numerical_derivative-Exact_derivative);
end

% evaluating second order derivative of the function using three schemes
clear 
close all
clc
x=pi/3;

dx=linspace(pi/4000,pi/40,80);
for i=1:length(dx)
    error_cds(i)=central_difference_scheme(x,dx(i));
    error_srs(i)=skewed_right_sided_scheme(x,dx(i));
    error_sls(i)=skewed_left_sided_scheme(x,dx(i));
end

% plotting
figure(1)
loglog(dx,error_cds,'r');
hold on
loglog(dx,error_srs,'g');
hold on
loglog(dx,error_sls,'b');
xlabel('Range of dx values')
ylabel('Error');
legend('central difference scheme','skewed right sided scheme','skewed left sided scheme');
title('Comparison of errors from different schemes');

Graph

graph

Conclusion

The central difference scheme will give the lowest error for increasing values of dx compared to the right-sided and left-sided scheme.

The error pattern follows the trend of Central difference scheme $lt$ Skewed left-sided scheme $lt$ Skewed right-sided scheme

Although the central difference scheme gives less error compared to skewed but in some cases skewed (right and left) is more useful. For eg if an error at the boundary to be calculated then skewed schemes are applied. The central difference scheme is not applicable in this regard.