W9.2 Taylor Series

The Taylor series (sometimes known as the Maclaurin series) expresses a function f(x) as an ‘expansion,’ in powers of x, about x=0.

It is probably the most frequently-used mathematical tool in the physicist’s toolkit! This exercise is therefore particularly important .

Solution

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Solution

Outline Answers

  1. Let f(x)=ex

    Then

    \[ f(0)=e^{0}=1 \hspace{0.5cm}\mbox{\rm and}\hspace{0.5cm} \frac{d}{dx} e^x = e^x \hspace{0.5cm}\mbox{\rm so}\hspace{0.5cm} f^{\prime}(0) = 1 \]

    Thus

    ex=1+x+O(x2)

    If you haven’t met it before the notation

    O(xm)
    is used to signify that the terms neglected in an approximation are of (at least) the m-th power in the variable x (which is presumed small).

  2. Let f(y)=ln(1-y)

    Then

    \[ f(0)= \ln 1 = 0 \hspace{0.5cm}\mbox{\rm and}\hspace{0.5cm} \frac{d}{dy} \ln(1-y) = -\frac{1}{1-y} \hspace{0.5cm}\mbox{\rm so}\hspace{0.5cm} f^{\prime}(0) = -1 \hspace{0.5cm}\mbox{\rm and}\hspace{0.5cm} \frac{d^2}{dy^2} \ln(1-y) = -\frac{1}{(1-y)^2} \hspace{0.5cm}\mbox{\rm so}\hspace{0.5cm} f^{\prime \prime}(0) = -1 \]

    Thus

    \begin{displaymath} \ln(1-y)= -y - \frac{y^2}{2} + \mbox{$ O ( y^3 ) $} \end{displaymath}

  3. Let f(θ)=tanθ

    Then

    \[ f(0)= \tan (0) = 0 \hspace{0.5cm}\mbox{\rm and}\hspace{0.5cm} \frac{d}{d\theta} \tan \theta = \frac{\cos \theta}{\cos \theta} +\frac{\sin^2 \theta}{\cos ^2 \theta} \hspace{0.5cm}\mbox{\rm so}\hspace{0.5cm} f^{\prime}(0) = 1 \]

    Thus

    tanθ=θ+O(θ2)

  4. Let $f(x) = \frac{1}{(1+x)^2}$

    Then

    \[ f(0)=1 \hspace{0.5cm}\mbox{\rm and}\hspace{0.5cm} \frac{d}{dx} \frac{1}{(1+x)^2} = -\frac{2}{(1+x)^3} \hspace{0.5cm}\mbox{\rm so}\hspace{0.5cm} f^{\prime}(0) = -2 \]

    Thus

    \begin{displaymath} \frac{1}{(1+x)^2}= 1-2x + \mbox{$ O ( x^2 ) $} \end{displaymath}

  5. Let $f(x) = \frac{1}{\sqrt{1-x^2}}$

    Then

    \[ f(0)=1 \hspace{0.5cm}\mbox{\rm and}\hspace{0.5cm} \frac{d}{dx} \frac{1}{\sqrt{1-x^2}} = \frac{x}{(1-x^2)^{3/2}} \hspace{0.5cm}\mbox{\rm so}\hspace{0.5cm} f^{\prime}(0) = 0 \]

    To pick up the second non-zero term we must therefore consider the second derivative:

    \[ \frac{d^2}{dx^2} \frac{1}{\sqrt{1-x^2}} = \frac{1}{(1-x^2)^{3/2}} + \frac{3x^2}{(1-x^2)^{5/2}} \hspace{0.5cm}\mbox{\rm so}\hspace{0.5cm} f^{\prime \prime}(0) = 1 \]
    Thus

    \begin{displaymath} \frac{1}{\sqrt{1-x^2}}= 1+\frac{x^2}{2} + \mbox{$ O ( x^4 ) $} \end{displaymath}

    You can establish this result a little more easily by showing first that

    \begin{displaymath} \frac{1}{\sqrt{1-y}} = 1 +\frac{y}{2} +\mbox{$ O ( y^2 ) $}\end{displaymath}
    and then setting yx2