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I realize that the Maclaurin Series is a special form of the Taylor Series where the series is centered at $x=0$, but I have to wonder what's special about it such that it deserves its own special designation? On that point, how would you know (or care) which point to choose as the center of a Taylor Series?

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    $\begingroup$ Certainly $0$ is nice. So nice that we would rather keep $0$ and change the function. For example, we expand $\ln(1+x)$ about $x=0$ instead of expanding $\ln x$ about $x=1$. $\endgroup$ – André Nicolas Jul 15 '13 at 4:17
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Expanding on the comment above, the idea is that we really like the expression

$$ \sum_{k=0}^\infty a_k z^k, $$ simply because it is easy to manipulate and involves less writing than a series with powers of $(z-a)$. So a lot of the time we like to shift our function so that the "point of interest" is simply $0$ (mathematicians try to be efficient, I suppose).

Typically we expand in a Taylor series (or more generally, a Laurent Series) about the point $z=a$ to investigate the behavior of $f$ near $a$. Is $f$ well behaved, or does it blow up? Can it be approximated using polynomials? If so, how good is this approximation and how far away from $a$ will it hold? This third question is the basis of many classical numerical analysis algorithms, including numerical differentiation and integration, as well as solution methods for ODEs.

The analysis of these methods relies heavily on Taylor series - for example, say we're at $x=a$ and want to approximate the value of the function $f$ at $a+h$, a little distance away. The Taylor series about $x=a$ reads:

$$ f(x)=f(a)+f^\prime(a)(x-a)+\frac{f^{\prime\prime}(a)}{2}(x-a)^2+O((x-a)^3) $$ where the "big-O-$(x-a)^3$" means a quantity that grows as a constant multiple of $(x-a)^3$. If we evaluate this Taylor approximation at $x=a+h$, we arrive at the nice, simple expression

$$ f(a+h)=f(a)+hf^\prime(a)+\frac{h^2}{2}f^{\prime\prime}(a)+O(h^3) $$ This says that if we know the value of the function and its first and second derivatives at $x=a$, we can approximate the value of $f$ at $a+h$ to an accuracy of $h$-cubed. So, for instance, if $h=0.1$, our approximation will only be off by a constant multiple of $0.001$. (This constant, incidentally, will depend on how bad the third derivative is near $a$).

Of course, I'm only using this "numerical" idea as an example of why we might expand the Taylor series at a location other than 0 - the idea has plenty of other uses.

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I think that Taylor series expansions around zero are not so special so as to deserve their own name, and in fact when I teach this material I do not use the term "Maclaurin series" (except to warn students in passing that others may use the term). In this part of calculus students already plenty of things to memorize, like many hard-to-keep-distinct convergence tests.

From a hard-nosed perspective there cannot be any truly distinguished expansion point for a Taylor series. The fact that in many of the simplest standard examples of elementary functions $0$ is an especially nice expansion point is an artifact of the fact that the coordinate system has been chosen so as to make $0$ a distinguished point: think e.g. about $\sin x, \cos x, e^x$. As soon as we start changing from one coordinate system to another we will certainly have to expand around nonzero points. This comes up for instance in the theory of analytic continuation in the complex variable case.

In practice, you want to expand around a point $c$ such that you are interested in the behavior of the function near $c$. The Taylor series is not guaranteed to converge at any point other than $x = c$; if it does converge, it is not guaranteed to be equal to the function. The way you show that the Taylor series $T(x) = T_{f,c}(x) = f(x)$ is to consider the remainder and apply various estimates on the derivatives of $f$. These derivatives typically grow quickly as you move too far away from the expansion point.

For example, if you are trying to compute $(26.5)^{\frac{1}{3}}$ using Taylor series, then $c= 27$ is a good expansion point: then $x =26.5$ is close enough to $c$ so that the convergence of the series is rapid, and since $27$ is a perfect cube, the Taylor series coefficients will take an especially simple form.

As another example, you might think about the Taylor series of $f(x) = \frac{1}{1+x^2}$ at various central points $c$. The radius of convergence of the Taylor series at $c$ is $\sqrt{c^2+1}$ for reasons that can only really be understood by thinking about the complex variable case. (Thus in a precise quantitative sene, $c = 0$ is the the worst expansion point in this case!) In a later course, the choice of expansion point is related to the various Laurent series expansions of a meromorphic function: one certainly cannot get away with always expanding around $0$!

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