Question

Prove the following result

$log(1 + x) = x - \frac{ {x}^{2} }{2} + \frac{ {x}^{3} }{3} - \frac{ {x}^{4} }{4} + - - - -$

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Answer 1

$\large\underline{\sf{Solution-}}$

Consider,

$\rm \: f(x) = log(1 + x)$

Let first evaluate the differential coefficient of f(x)

$\rm \: f'(x) = \dfrac{1}{1 + x}$

$\rm \: f''(x) \: = \: - \: \dfrac{1}{(1 + x)^{2} }$

$\rm \: f'''(x) \: = \: - \: \dfrac{( - 2)}{(1 + x)^{3} } = \dfrac{2}{ {(1 + x)}^{3} }$

$\rm \: f''''(x) \: = \: - \: \dfrac{6}{(1 + x)^{4} }$

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Now, Let's evaluate the value of differential coefficients at x = 0

$\rm \: f(0) = log(1 + 0) = log1 = 0$

$\rm \: f'(0) = \dfrac{1}{1 + 0} = 1$

$\rm \: f''(0) \: = \: - \: \dfrac{1}{(1 + 0)^{2} } \: = \: - \: 1$

$\rm \: f'''(0) \: = \: \dfrac{2}{ {(1 + 0)}^{3} } = 2$

$\rm \: f''''(0) \: = \: - \: \dfrac{6}{(1 + 0)^{4} } = - 6$

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So, By Mc Lauren's Series, we have

$\rm \: f(x) = f(0) + \dfrac{x}{1!}f'(0) +\dfrac{ {x}^{2} }{2!}f''(0) + \dfrac{ {x}^{3} }{1!}f'''(0 + \dfrac{ {x}^{4} }{4!}f''''(0) + - - -$

So, on substituting the values, we get

$\rm \: log(1 + x) = 0 + \dfrac{x}{1!} + \dfrac{ {x}^{2} }{2!}( - 1) + \dfrac{ {x}^{3} }{3!}( - 2) + \dfrac{ {x}^{4} }{4!}(6) + - - -$

$\rm \: log(1 + x) = x - \dfrac{ {x}^{2} }{2} + \dfrac{ {x}^{3} }{3} - \dfrac{ {x}^{4} }{4} + - - - \infty$

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Basic Result used

$\boxed{\tt{ \: \: \dfrac{d}{dx} \: logx \: = \: \frac{1}{x} \: }}$

$\boxed{\tt{ \: \: \dfrac{d}{dx} \frac{1}{ {x}^{n} } \: = \: \frac{ - \: n \: \: }{ \: \: {x}^{n + 1} \: \: \: } \: }}$

$\boxed{\tt{ \: \: log1 \: = \: 0 \: }}$

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ADDITIONAL INFORMATION

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$

Answer 2

$\color{lime}{\huge\pmb{\mathrm{ Pʀᴏvᴇ \: \: THᴀᴛ}}} \: :$

$\rm \log(1 + x) = x - \frac{ {x}^{2} }{2} + \frac{ {x}^{3} }{3} - \frac{ {x}^{4} }{4} + - - - + \infty$

$\color{purple}\huge\pmb{\mathrm{ Pʀᴏᴏf}} \: :$

Refer the Given Attachments.