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4. Five point Forward difference, Central difference formula numerical differentiation example ( Enter your problem )
  1. Formula & Example-1 (table data)
  2. Example-2 (table data)
  3. Example-3 (`f(x)=cosx`)
  4. Example-4 (`f(x)=2x^3+x^2-4`)
  5. Example-5 (`f(x)=xlnx`)
  6. Example-6 (`f(x)=sinx`)

1. Formula & Example-1 (table data)





Formula
1. Five-point FDF (Forward difference formula)
`f^'(x)=1/(12h)[-25f(x)+48f(x+h)-36f(x+2h)+16f(x+3h)-3f(x+4h)]`
2. Five-point CDF (Central difference formula)
`f^'(x)=1/(12h)[f(x-2h)-8f(x-h)+8f(x+h)-f(x+2h)]`
3. Five-point CDF (Central difference formula) for second derivatives
`f^('')(x)=1/(12h^2)[-f(x-2h)+16f(x-h)-30f(x)+16f(x+h)-f(x+2h)]`

Examples
1. Using Five point Forward difference, Backward difference, Central difference formula numerical differentiation to find solution
x11.051.101.151.201.251.30
f(x)11.024701.048811.072381.095451.118031.14018

`f^'(1.10) and f^('')(1.10)`


Solution:
The value of table for `x` and `y`

x11.051.11.151.21.251.3
y11.02471.04881.07241.09541.1181.1402

Five-point FDF (Forward difference formula)
`f^'(x)=1/(12h)[-25f(x)+48f(x+h)-36f(x+2h)+16f(x+3h)-3f(x+4h)]`

`f^'(1.10)=1/(12*0.05)[-25f(1.10)+48f(1.10+0.05)-36f(1.10+2*0.05)+16f(1.10+3*0.05)-3f(1.10+4*0.05)]`

`f^'(1.10)=1/(0.6)[-25f(1.10)+48f(1.15)-36f(1.2)+16f(1.25)-3f(1.3)]`

`f^'(1.10)=1/(0.6)[-25(1.0488)+48(1.0724)-36(1.0954)+16(1.118)-3(1.1402)]`

`f^'(1.10)=0.4762`



Five-point CDF (Central difference formula)
`f^'(x)=1/(12h)[f(x-2h)-8f(x-h)+8f(x+h)-f(x+2h)]`

`f^'(1.10)=1/(12*0.05)[f(1.10-2*0.05)-8f(1.10-0.05)+8f(1.10+0.05)-f(1.10+2*0.05)]`

`f^'(1.10)=1/0.6[f(1)-8f(1.05)+8f(1.15)-f(1.2)]`

`f^'(1.10)=1/0.6[1-8(1.0247)+8(1.0724)-1.0954]`

`f^'(1.10)=0.4767`



Five-point CDF (Central difference formula) for second derivatives
`f^('')(x)=1/(12h^2)[-f(x-2h)+16f(x-h)-30f(x)+16f(x+h)-f(x+2h)]`

`f^('')(1.10)=1/(12*(0.05)^2)[-f(1.10-2*0.05)+16f(1.10-0.05)-30f(1.10)+16f(1.10+0.05)-f(1.10+2*0.05)]`

`f^('')(1.10)=1/0.03[-f(1)+16f(1.05)-30f(1.10)+16f(1.15)-f(1.2)]`

`f^('')(1.10)=1/0.03[-1+16(1.0247)-30(1.0488)+16(1.0724)-1.0954]`

`f^('')(1.10)=-0.2157`




This material is intended as a summary. Use your textbook for detail explanation.
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