Solve: log e (1+x) 1+x (1-x) 1-x Prove that: log e a - log e b = (a - b)/a + 1/2 (a - b) 2 /a 2 + 1/3 (a - b) 3 /b 3

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saiomkar kandukuri · Answer

first question is not so clear second question can be proved by using expansions of logarithms ... which can be found in any differential calculus books

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Solve: log e (1+x) 1+x (1-x) 1-x Prove that: log e a - log e b = (a - b)/a + 1/2 (a - b) 2 /a 2 + 1/3 (a - b) 3 /b 3

Solve:

log_e(1+x)^1+x(1-x)^1-x

Prove that:

log_ea - log_eb = (a - b)/a + 1/2 (a - b)²/a²+ 1/3 (a - b)³/b³

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Solve: log e (1+x) 1+x (1-x) 1-x Prove that: log e a - log e b = (a - b)/a + 1/2 (a - b) 2 /a 2 + 1/3 (a - b) 3 /b 3

Solve: loge(1+x)1+x(1-x)1-x Prove that: logea - logeb = (a - b)/a + 1/2 (a - b)2/a2 + 1/3 (a - b)3/b3

1 Answers

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Solve:

log_e(1+x)^1+x(1-x)^1-x

Prove that:

log_ea - log_eb = (a - b)/a + 1/2 (a - b)²/a²+ 1/3 (a - b)³/b³