Richardson extrapolation formula for differentiation Example-2 : f(x)=5xe^(-2x)

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Home > Numerical methods calculators > Richardson extrapolation method calculator

Richardson extrapolation formula for differentiation example ( Enter your problem )

( Enter your problem )

Example-1 : $f(x)=xexp(x)$
Example-2 : $f(x)=5xe^(-2x)$

1. Example-1 :

$f(x)=xexp(x)$
(Previous example)

2. Example-2 : $f(x)=5xe^(-2x)$

$f(x)=5xe^(-2x)$ , find f'(0.35) with h=0.25 using Richardson extrapolation method

Solution:
Richardson Extrapolation method
Equation is

$f(x) = 5xexp(-2x)$ .

$h=0.25$

$f^'(x)=(f(x+h)-f(x-h))/(2h)$ (CDD)

$f^'(0.35)=(f(0.35+0.25)-f(0.35-0.25))/(2*0.25)$

$f^'(0.35)=(f(0.6)-f(0.1))/0.5$

$f^'(0.35)=(0.9036-0.4094)/0.5$

$f^'(0.35)=0.9884$

$h=0.125$

$f^'(x)=(f(x+h)-f(x-h))/(2h)$

$f^'(0.35)=(f(0.35+0.125)-f(0.35-0.125))/(2*0.125)$

$f^'(0.35)=(f(0.475)-f(0.225))/0.25$

$f^'(0.35)=(0.9185-0.7173)/0.25$

$f^'(0.35)=0.8047$

$TV=(AV)_(h/2)+((AV)_(h/2)-(AV)_(h))/3$

$=0.8047 + (0.8047 - 0.9884) / 3$

$=0.7435$

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