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 .
- Review the online introduction to Taylor Series (embedded in S6.2)
- Establish the first two non-zero terms in the expansions
of the following functions:
- ex
- ln(1-y)
- tanθ


Solution
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Solution
Outline Answers
Let f(x)=ex
Then
![\[ f(0)=e^{0}=1 \hspace{0.5cm}\mbox{\rm and}\hspace{0.5cm} \frac{d}{dx} e^x = e^x \hspace{0.5cm}\mbox{\rm so}\hspace{0.5cm} f^{\prime}(0) = 1 \]](mastermathpng-2.png)
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).Let f(y)=ln(1-y)
Then
![\[ f(0)= \ln 1 = 0 \hspace{0.5cm}\mbox{\rm and}\hspace{0.5cm} \frac{d}{dy} \ln(1-y) = -\frac{1}{1-y} \hspace{0.5cm}\mbox{\rm so}\hspace{0.5cm} f^{\prime}(0) = -1 \hspace{0.5cm}\mbox{\rm and}\hspace{0.5cm} \frac{d^2}{dy^2} \ln(1-y) = -\frac{1}{(1-y)^2} \hspace{0.5cm}\mbox{\rm so}\hspace{0.5cm} f^{\prime \prime}(0) = -1 \]](mastermathpng-3.png)
Thus

Let f(θ)=tanθ
Then
![\[ f(0)= \tan (0) = 0 \hspace{0.5cm}\mbox{\rm and}\hspace{0.5cm} \frac{d}{d\theta} \tan \theta = \frac{\cos \theta}{\cos \theta} +\frac{\sin^2 \theta}{\cos ^2 \theta} \hspace{0.5cm}\mbox{\rm so}\hspace{0.5cm} f^{\prime}(0) = 1 \]](mastermathpng-5.png)
Thus
tanθ=θ+O(θ2)Let

Then
![\[ f(0)=1 \hspace{0.5cm}\mbox{\rm and}\hspace{0.5cm} \frac{d}{dx} \frac{1}{(1+x)^2} = -\frac{2}{(1+x)^3} \hspace{0.5cm}\mbox{\rm so}\hspace{0.5cm} f^{\prime}(0) = -2 \]](mastermathpng-7.png)
Thus

Let

Then
![\[ f(0)=1 \hspace{0.5cm}\mbox{\rm and}\hspace{0.5cm} \frac{d}{dx} \frac{1}{\sqrt{1-x^2}} = \frac{x}{(1-x^2)^{3/2}} \hspace{0.5cm}\mbox{\rm so}\hspace{0.5cm} f^{\prime}(0) = 0 \]](mastermathpng-10.png)
To pick up the second non-zero term we must therefore consider the second derivative:
Thus![\[ \frac{d^2}{dx^2} \frac{1}{\sqrt{1-x^2}} = \frac{1}{(1-x^2)^{3/2}} + \frac{3x^2}{(1-x^2)^{5/2}} \hspace{0.5cm}\mbox{\rm so}\hspace{0.5cm} f^{\prime \prime}(0) = 1 \]](mastermathpng-11.png)

You can establish this result a little more easily by showing first that
and then setting y≡x2