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landdenaw 2022-06-29 Answered
Given stepsizes h 1 and h 2 , develop a numerical scheme to approximate f ( x 0 ) with function values f ( x 0 ) , f ( x 0 + h 1 ) and f ( x 0 + h 2 ) . Under what conditions will your method not work?
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Answers (2)

Lisbonaid
Answered 2022-06-30 Author has 22 answers
Step 1
If h 1 = h 2 , you can use second derivative midpoint formula
f ( x 0 ) = 1 h 2 [ f ( x 0 h ) 2 f ( x 0 ) + f ( x 0 + h ) ] + o ( h 3 )
If h 1 h 2 , you can do like this:
f ( x 0 + h 1 ) = f ( x 0 ) + h 1 f ( x 0 ) + h 1 2 2 f ( x 0 ) + o ( h 1 3 )
f ( x 0 + h 2 ) = f ( x 0 ) + h 2 f ( x 0 ) + h 2 2 2 f ( x 0 ) + o ( h 2 3 )
Minus the two equations
f ( x 0 + h 2 ) f ( x 0 + h 1 ) = ( h 2 h 1 ) f ( x 0 ) + h 2 2 h 1 2 2 f ( x 0 ) + o ( h 2 3 h 1 3 )
and
f ( x 0 ) = f ( x 0 + h 2 ) f ( x 0 + h 1 ) h 2 h 1 + o ( h 2 2 h 1 2 )
But, considering the accuracy, in general, we pick h 1 = h 2 , which cancels out the f ( x 0 ) that has a low accuracy in a certain sense.
An alternative way to find f ( x 0 ) using only f ( x 0 ) , f ( x 0 + h 1 ) and f ( x 0 + h 2 ) and not involving f ( x 0 ) is that you should express f ( x 0 ) with f ( x 0 ) , f ( x 0 + h 1 ) and f ( x 0 + h 2 ) , so a potential formula is
f ( x 0 + h 1 ) = f ( x 0 ) + h 1 h 1 + h 2 [ f ( x 0 + h 1 ) + f ( x 0 + h 2 ) 2 f ( x 0 ) ( h 2 2 h 1 2 2 f ( x 0 ) ) ] + h 1 2 2 f ( x 0 )

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Kiana Dodson
Answered 2022-07-01 Author has 5 answers
Step 1
f 1 f 0 + h 1 f 0 + h 1 2 2 f 0 ,
f 2 f 0 + h 2 f 0 + h 2 2 2 f 0 .
Then by elimination of f 0 , you get
h 2 ( f 1 f 0 ) h 1 ( f 2 f 0 ) = h 1 h 2 ( h 1 h 2 ) 2 f 0 .

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