Q7.2 Taylor series expansion (K)

Any reasonably well behaved function f(x) can be approximated near x=0 by an expansion in powers of x, of the form
\[ f(x) = f(0) + f'(0)x + \frac{1}{2} f''(0) x^ 2 + \ldots \]
where (for example)
\[ f'(0) \equiv \frac{df(x)}{dx}\mid_{x=0} \]
This is the Taylor series expansion (TSE). It is one of the most frequently used bits of mathematics in physics problem solving.
  1. Show that, for small θ,
    \[ \sin \theta \simeq \theta \hspace{0.2cm}\mbox{\rm and}\hspace{0.2cm} \cos \theta \simeq 1-\frac{\theta ^2}{2} \]
  2. Draw a graph of sinθ for 0θπ/2. On the same graph show the TSE approximation sinθθ.
  3. How small does θ have to be to ensure that the TSE result differs by no more than 1% from the exact value?
  4. Examine the structure of the expansion for f(x) (above). Then guess the next term in the series.
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Hint

Your mathematics courses will provide you with a full account of Taylor Series Expansions. This exercise is intended to help you become familiar with the ideas. You will find a little more background on line. You can do the third part of the question with the aid of your calculator; or you could make use of the answer to the last part.
 

Solution

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Solution

If you havent already done so take a look at the online introduction to Taylor Series before going any further.
  1. The TSE for sinθ is

    \[ \sin \theta = \sin (0) + \theta \frac{d\sin \theta }{d\theta}\mid_{\theta =0} + \ldots \]
    Since sin(0)=0 and
    \[ \frac{d\sin \theta }{d\theta} \mid_{\theta =0} = \cos \theta \mid_{\theta =0} =\cos (0) =1 \]
    we have
    sinθ=θ+

    The terms that we don’t show are proportional to higher powers of θ (only odd powers, because the sine is an odd function) and are small compared to θ when θ is small.

    The TSE for cosθ is

    \[ \cos {\theta} = \cos (0) + \theta \frac{d}{d\theta}\cos \theta \mid_{\theta =0} + \frac{\theta^2}{2}\frac{d^2\cos \theta }{d\theta^2} \mid_{\theta =0} + \ldots \]
    Now cos(0)=1 while
    \[ \frac{d}{d\theta}\cos \theta \mid_{\theta =0} =-\sin(0) =0 \]
    So the term proportional to θ vanishes in this case (as it has to since cos is an even function.) This is the reason why in this case (as distinct from the case of the sine function) we have to proceed to the next term (the one proportional to θ2) to find out anything about how the function changes. The second derivative we need is
    \[ \frac{d^2\cos \theta }{d\theta ^2} \mid_{\theta =0} = -\cos (0) =-1 \]
    So
    \[ \cos {\theta} = 1 -\frac{\theta^2}{2} + \ldots \]

  1. [2] The figure shows the behaviour of sinθ and its TSE sinθθ. The two functions coincide as long as sinθ varies linearly with θ: that is precisely what the TSE is designed to do.
Graphic - No title - taylortut
  1. [3] With θ=0.25 the calculator gives sinθ/θ=0.99. So the 1% accuracy criterion kicks in here ie at θ14.
  2. [4] If you stare at the first few terms of the TSE you will see there is a pattern. The n-th term is of the form

    \[ \frac{\mbox {n-th power of }\theta}{n!} \times \mbox{ n-th derivative of function } f(x) \mbox{ evaluated at } x=0 \]
    The the next term in the series is then
    \[ \frac{1}{6} f'''(0) x^ 3 \]
    In the case of the TSE for the sin function, then, to improve on sinθθ we must write (why is there no θ2 term?)
    \[ \sin {\theta} = \theta -\frac{\theta^3}{6} + \ldots \]

    We can use this to redo the estimate in the previous part of the question. We write

    \[ \frac{\sin \theta}{\theta} = [\theta -\frac{\theta^3}{6} + \ldots]/\theta = 1 -\frac{\theta^2}{6} +\ldots \]
    The error in the sinθθ approximation is therefore 1% when θ2/6=0.01 implying θ0.244 in close accord with what we got above.