Here is a typical Calculus problem involving Taylor polynomials. The solution and my comment follow.

Use the linear Taylor polynomial associated with the sine function to estimate the value of $\sin(50^{\circ})$ with an error of less than $0.02$.


Lagrange's remainder term for the linear Taylor polynomial associated with the sine function at $60^{\circ}$ is \begin{equation*} R_{1}(x) = - \frac{\sin{\theta}}{2!} \, \left(x - \frac{\pi}{3}\right)^{2} . \end{equation*} for some real number $\theta$ between $\pi/3$ and $x$. $50^{\circ}$ is equivalent to $5\pi/18$ radians, and \begin{equation*} \frac{\pi}{3} - \frac{5\pi}{18} = \frac{\pi}{18} < \frac{1}{5} . \end{equation*} Since the sine function is bounded by $1$, the evaluation of the linear Taylor polynomial $T_{1}(x)$ at $5\pi/18$ is an estimate for $\sin(50^{\circ})$ with an error of less than $0.02$. \begin{equation*} T_{1}\left(\frac{5\pi}{18}\right) = \frac{\sqrt{3}}{2} + \frac{1}{2} \left(\frac{5\pi}{18} - \frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} - \frac{\pi}{36} . \end{equation*}


Isn't this answer disingenuous? If the author of the Calculus textbook is pretending that we do not have a calculator to evaluate $\sin(50^{\circ})$, we also should have a calculator to give estimates for $\sqrt{3}$ ... or for $\pi$.

  • $\begingroup$ I don't believe the author is pretending no calculator is available. I think the point of the problem (and ones like it) is simply to practice estimating with Taylor polynomials and Lagrange error terms. One can ask similar questions when derivatives are first introduced since the first few exercises can be trivially done with any standard CAS and even some websites these days. $\endgroup$ – tilper Sep 7 '16 at 17:04
  • $\begingroup$ @tilper If that were the case, I think better directions would be to say "write a linear Taylor polynomial that can be used to estimate $\sin(50^{\circ})$. $\endgroup$ – user74973 Sep 7 '16 at 17:15
  • $\begingroup$ When you use calculator and get an answer, you might ask yourself two questions: how does calculator gets the number and how accurate is the number? The example you showed is really telling you the hows that, without a calculator, how do you calculate $\sin(50^{\circ})$ and how accurate is the calculated result. $\endgroup$ – user115350 Sep 7 '16 at 19:19

Approximating square roots with Newton's method is very efficient (and is probably close to what your calculator does) so there's no pretense there. Pi is just a constant, so there's no pretense there (although it's an approximation of pi that's stored in the calculator, of course).

Are Taylor series really how your calculator computes sines? Probably not, but for a very long time, they were the best thing we had, and DID get used to compute a lot of things, even when the computations were done by hand rather than with a calculator.

  • $\begingroup$ Is there is a bound to the error in using Newton's method? If we are allowed to use $3.14159$ for an estimate of $\pi$, why can't we just use a table to estimate $\sin(50)^{\circ}$. I am being facetious, of course. $\endgroup$ – user74973 Sep 7 '16 at 17:04
  • $\begingroup$ Pedagogically speaking, I think that the author should provide estimates for these other constants. $\endgroup$ – user74973 Sep 7 '16 at 17:04
  • $\begingroup$ In a calculator that displays 10 digits, it'd be typical to store 11 digits of pi, and to run Newton's method for sqrt until the digit beyond the last displayed digit stops changing. Sure you can use a table...but then someone else has to produce that table, unless it happens to arrive on a stone tablet from heaven. The pity here isn't in the idea of approximation, it's that there's no opportunity taken to discuss the nature of accumulating error from multiple approximations (like the 10-digit pi, or the slightly wrong sqrt). $\endgroup$ – John Hughes Sep 7 '16 at 21:25

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