What I know:

  • A Taylor series is a local polynomial expansion of a function
  • It uses derivatives (often many of them)
  • It's supposed to make estimations easier/simpler

I'm not great with differentiation but I can say it can be pretty hard. While very accurate, Taylor series aren't 100% accurate compared to the original.

From what I can see, finding derivatives of a function can be quite a lot of work and it's still more accurate to work with the original functions. Take this function for example:

$f(x) = sin(cos(x))*cos(sin(x))$

whose Taylor partial expansion is

$sin(1) + 1/2 x^2 (-sin(1) - cos(1)) + 1/24 x^4 (2 sin(1) + 7 cos(1)) + 1/720 x^6 (23 sin(1) - 76 cos(1)) + O(x^8)$

which takes effort to calculate (though I just used Wolfram Alpha) and looks even longer and in some cases more complicated with many terms. Instead of doing all that, I can just throw some inputs at the original function and sample the output which is quite straightforward.

The question: Why do we transform functions into their Taylor expansions for estimations instead of just sampling the local area with lots of inputs?

  • $\begingroup$ "Making estimations easier" is a relatively rare use of Taylor series as far as I'm aware. Where did you learn that this is what the Taylor series is for? What exactly did your source say about it? This could help people address your particular issues; you should edit the information into the question. $\endgroup$
    – David K
    Nov 11, 2019 at 6:21
  • $\begingroup$ Polynomials are the simplest types of functions, so if you approximate a function with a polynomial function then whatever calculation you want to do might be greatly simplified. $\endgroup$
    – littleO
    Nov 11, 2019 at 6:38

2 Answers 2


How exactly do you plan to compute $\sin(\cos(x))$? How can you guarantee that this computation is accurate? With the Taylor partial expansion, once you give me $ \sin(1)$ and $\cos(1)$ (whose values themselves are not immediately clear) at least its a polynomial, and I know how to use polynomials, even by hand.

For the Taylor approximation, we only need two evaluations total, of well known functions, $\sin(1)$ and $\cos(1)$. The direct evaluation requires (as you said) lots of inputs, and I can't tell you off-hand how Wolfram computes $\sin\cos x$...

Admittedly, in a world with Wolfram|Alpha at everyone's fingertips, maybe its harder to appreciate what can be done with pen and paper...


There are many applications of the taylor series. The main reason they are often used, is that generally speaking polynomials are easier to deal with analytically than your average nonlinear function. Also you get relatively good control over the truncation error which is also vital for many applications in calculus.

Whether or not calculating a Taylor series make sense for your problem heavily depends on the task at hand. If you just want to plot the function it will most likely not be the best option to calculate a high order Taylor approximation.

  • $\begingroup$ Can you give some example(s) of when a Taylor series should be used? $\endgroup$ Nov 11, 2019 at 6:20
  • $\begingroup$ For example the proof of the convergence of a finite difference scheme usually is done via Taylor series. $\endgroup$
    – maxmilgram
    Nov 11, 2019 at 6:26
  • 1
    $\begingroup$ Taylor series approximations are often used in physics applications, when you find yourself with an equation along the lines of y = e^(-x/a), where a is very large and x is approx 0. Instead of working with e^(-x/a), a first order taylor expansion is used, y = 1 - x/a is a reasonable approximation, but more importantly we have approximated a nonlinear function with a linear function. $\endgroup$ Nov 11, 2019 at 7:34

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .