Recall that the Maclaurin series for f(x) is f(x) = f(0) + f'(0)x + f''(0)x2/2! + ... + f(r)(0)xr/r! + ... Here f(0) = ln(1+0) = ln1 = 0 , f'(x) = 1/(1+x) using the chain rule. f'(0) = 1/1 = 1 , f''(x) = -1/(1+x)2 , f''(0) = -1/(1)2 = -1 , f'''(x) = 2/(1+x)3 , f'''(0) = 2/(1)3 = 2, Substituting these into the general expression for the Maclaurin series: (to 3rd degree) gives: ln(1+x) = 0 + 1x + -1/2! x2 + 2/3! x3 , ln(1+x) = x - x2/2 + x3/3
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