The third-degree McLauren series approximation for the function f (x) = ln(1 + x) is a) x-(1)/(24) x^2 + (1)/(120) x^3 b) 1-x+1/2 x^2-1/6 x^3 c) x-1/2 x^2+1/3 x^3 d) 1-x+x^2-x^3 e) x-x^2+x^3

Tiffany Page

Tiffany Page

Answered question

2022-11-13

The third-degree McLauren series approximation for the function f ( x ) = ln ( 1 + x ) is
a ) x 1 24 x 2 + 1 120 x 3 b ) 1 x + 1 2 x 2 1 6 x 3 c ) x 1 2 x 2 + 1 3 x 3 d ) 1 x + x 2 x 3 e ) x x 2 + x 3

Answer & Explanation

motylowceyvy

motylowceyvy

Beginner2022-11-14Added 19 answers

Consider F ( x ) = ln ( 1 + x )
MacLauren series approximation for F(x) is given by,
F ( x ) = F ( 0 ) + F ( 0 ) x 1 ! + F " ( 0 ) 2 ! + F ( 0 ) x 3 3 ! + F ( 0 ) = ln ( 1 + 0 ) = ln 1 = 0 F ( x ) = d d x ( ln ( 1 + x ) ) = 1 1 + x × ( 0 + 1 ) = 1 1 + x F ( 0 ) = 1 1 + 0 = 1 F " ( x ) = d d x ( 1 1 + x ) = ( 1 + x ) d d x d d x ( 1 + x ) ( 1 + x ) 2 = ( 1 + x ) × 0 1 ( 1 + x ) 2 = 1 ( 1 + x ) 2 F " ( 0 ) = 1 ( 1 + 0 ) 2 = 1 F ( x ) = d d x ( 1 ( 1 + x ) 2 ) = ( 1 + x ) 2 × d d x ( 1 ) d d x ( 1 + x ) 2 ( 1 + x ) 4 = ( 1 + x 2 ) × 0 + 2 ( 1 + x ) ( 1 + x ) 4 = 2 ( 1 + x ) ( 1 + x ) 4 = 2 ( 1 + x ) 3 F ( 0 ) = 2 ( 1 + 0 ) 3 = 2 F ( x ) = 0 + 1 1 ! x + 1 2 ! x 2 + 2 3 ! x 3 + = x x 2 2 + x 3 3 option (c) correct

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