2
$\begingroup$

I know that Maclaurin series are a special case of Taylor series where we set $a=0$, and it is useful for exponential functions (for example). But my question is: Are Taylor series (evaluate at some other point different to zero) really useful in practice?

$\endgroup$
6
  • 3
    $\begingroup$ There are many reasons, ranging from solving limits, creating linear approximations and approximating functions in general, studying asymptotic growth (generally through Laurent series), and so forth. This question is very broad as it stands... What do you have in mind? Try to be a bit more specific $\endgroup$ Sep 18 '16 at 18:08
  • $\begingroup$ my question is principal for approximations. Why do we care to approximate a function at some point different to zero. I mean why evaluate for example f(8) = cos x with a = 5 (or other number different to zero) with taylor formula wikimedia.org/api/rest_v1/media/math/render/svg/… $\endgroup$ Sep 18 '16 at 18:21
  • 1
    $\begingroup$ The simplest answer to that is because you might want to approximate a function elsewhere than around zero. A McLaurin series is usually fine if you want to estimate the value of a function at $x=0.5$, but is often terrible at $x=500.5$. If you want to estimate the function there a Taylor series about $x=500$ might be very useful! $\endgroup$ Sep 18 '16 at 18:45
  • 1
    $\begingroup$ You might then ask why we don't just get the series around $x=500.5$ instead of $x=500$. The key to this is ease of calculation. Imagine you are messing with a square root function and trying to approximate $\sqrt{36.458}$. You'll get a pretty nice estimation if you say it's about $\sqrt{36}=6$. Notice that we are estimating a function by a nearby point not equal to zero.... A Taylor series just makes this more rigorous, algorithmic, and accurate! $\endgroup$ Sep 18 '16 at 18:49
  • $\begingroup$ thank you @BrevanEllefsen that's a good explication $\endgroup$ Sep 18 '16 at 19:20
1
$\begingroup$

An important reason is that the function might not be defined at $0$. Consider the $log$ function.

See this previous questions: taylor series of ln(1+x)?.

As mentioned in the comments, you might need to approximate the function in a region not near $0$. The Maclaurin series might converge very slowly or not at all. You could look for nearer points st which it was easy to calculate the Taylor coefficients.

$\endgroup$
0
$\begingroup$

Taylor series can be used to approximate function at some point (not always $0$). For example, this is a powerful tool for calculating limits.

$\endgroup$
1
  • $\begingroup$ This is more of a comment than an answer. The OP already states the first line in his post as well. Nevertheless, very true. This becomes especially potent when studying asymptotic growth and behavior around a point $\endgroup$ Sep 18 '16 at 18:10

Not the answer you're looking for? Browse other questions tagged or ask your own question.