Home » Blog » Prove that 1 / (1+g(x)) = 1- g(x) + g2(x) + o(g2(x))

Prove that 1 / (1+g(x)) = 1- g(x) + g2(x) + o(g2(x))

  1. Prove that

        \[ \frac{1}{1+g(x)} = 1 - g(x) + g^2(x) + o(g^2(x)) \quad \text{as} \quad x \to 0 \]

    where g(x) = o(1) as x \to 0.

  2. Using part (a) prove

        \[ \tan x = x + \frac{x^3}{3} + \frac{2x^5}{15} + o(x^5) \quad \text{as} \quad x \to 0. \]


  1. Proof. We start with the algebraic identity

        \[ \frac{1}{1+u} = 1 - u + u^2 - \frac{u^3}{1+u}. \]

    Replacing u by g(x) we have,

        \[ \frac{1}{1+g(x)} = 1 - g(x) + g^2(x) - \frac{g^3(x)}{1+g(x)} = 1 - g(x) + g^2(x) - g^2(x) \frac{g(x)}{1+g(x)}. \]

    By the definition of o(1) we have

        \[ g(x) = o(1) \text{ as } x \to 0 \quad \implies \quad \lim_{x \to 0} g(x) = 0. \]

    Therefore,

        \[ \lim_{x \to 0} \frac{g(x)}{1+g(x)} = 0. \]

    Hence,

        \[ \frac{1}{1+g(x)} = 1 - g(x) + g^2(x) + o(g^2(x)). \qquad \blacksquare\]

  2. Proof. First, we use the expansion of \cos x,

        \[ \frac{1}{\cos x} = \frac{1}{1 - \frac{x^2}{2} + \frac{x^4}{24} + o(x^5)} = \frac{1}{1+g(x)} \]

    where g(x) = -\frac{x^2}{2} + \frac{x^4}{24} + o(x^5). By part (a) we then have as x \to 0,

        \begin{align*}  \frac{1}{\cos x} &= 1 - g(x) + g^2(x) + o(g^2(x)) \\[9pt]  &= 1 - \left( -\frac{x^2}{2} + \frac{x^4}{24} + o(x^5) \right) + \left( -\frac{x^2}{2} + \frac{x^4}{24} + o(x^5) \right)^2 + o \left( \left( -\frac{x^2}{2} + \frac{x^4}{24} \right)^2 \right) \\[9pt]  &= 1 + \frac{x^2}{2} - \frac{x^4}{24} + \frac{x^4}{4} + o(x^4) \\[9pt]  &= 1 + \frac{x^2}{2} + \frac{5x^4}{24} + o(x^4). \end{align*}

    Next, since \tan x = \frac{\sin x}{\cos x} we multiply this expansion for \frac{1}{\cos x} by the expansion (page 287 of Apostol) for \sin x as x \to 0,

        \begin{align*}  \tan x &= \frac{1}{\cos x} \cdot \sin x \\[9pt]  &= \left( 1 + \frac{x^2}{2} + \frac{5x^4}{24} + o(x^4) \right) \left( x - \frac{x^3}{6} + \frac{x^5}{120} + o(x^6) \right) \\[9pt]  &= x - \frac{x^3}{3} + \frac{x^5}{120} + \frac{x^3}{2} - \frac{x^5}{12} + \frac{5x^5}{24} + o(x^5) \\[9pt]  &= x + \frac{x^3}{6} + \frac{2x^5}{15} + o(x^5). \qquad \blacksquare \end{align*}

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