sqrt[4 x^2 + 3x] + 2 x
= - 2x sqrt[1 + 3/(4x)] + 2x
= - 2x [ 1 + 3/(8x) + ... ] + 2x
= - 3/4 + ...
sqrt[4 x^2 + 3x] + 2 x
= - 2x sqrt[1 + 3/(4x)] + 2x
= - 2x [ 1 + 3/(8x) + ... ] + 2x
= - 3/4 + ...
• Taylor series没法在-inf展开吧。 -成功的熊- ♂ (0 bytes) () 12/12/2021 postreply 15:06:24
• Taylor series没法在考试时用,老师不让用没学过的 -chinomango- ♂ (0 bytes) () 12/12/2021 postreply 15:10:41