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If the fourth degree Taylor polynomial centered at $0$ for $\ln { \left( x+1 \right) }$ is used to approximate the value of $\ln { \left( 1.3 \right) }$, the first four terms would be

A

$\left( 0.3 \right) +\cfrac { { \left( 0.3 \right) }^{ 2 } }{ 2! } +\cfrac { { \left( 0.3 \right) }^{ 3 } }{ 3! } +\cfrac { { \left( 0.3 \right) }^{ 4 } }{ 4! } +...$

B

$\left( 1.3 \right) -\cfrac { { \left( 1.3 \right) }^{ 2 } }{ 2 } +\cfrac { { \left( 1.3 \right) }^{ 3 } }{ 3 } -\cfrac { { \left( 1.3 \right) }^{ 4 } }{ 4 } +...$

C

$\left( 0.3 \right) -\cfrac { { \left( 0.3 \right) }^{ 2 } }{ 2 } +\cfrac { { \left( 0.3 \right) }^{ 3 } }{ 3 } -\cfrac { { \left( 0.3 \right) }^{ 4 } }{ 4 } +...$

D

$\left( 1.3 \right) +\cfrac { { \left( 1.3 \right) }^{ 2 } }{ 2 } +\cfrac { { \left( 1.3 \right) }^{ 3 } }{ 3 } +\cfrac { { \left( 1.3 \right) }^{ 4 } }{ 4 } +...$

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