# Physical applications of higher terms of Taylor series

Depressingly many of the physical "applications" of Taylor series that I can find in textbooks and online are actually just applications of linear approximation, since they only take the constant and linear term of a Taylor series. For instance, this how you get the Newtonian limit of the kinetic energy of special relativity, the $1/r^3$ behavior of an electric dipole far away, and the standard approximation to the period of a pendulum. Since the whole point of a Taylor series is that we can go on from the linear approximation to get quadratic and higher approximations, I find this unsatisfying.

So what are some good physical applications of the higher order terms in a Taylor expansion? (I'm looking for something more than just "we can get a better approximation than the linear one by including more terms" — ideally, a situation where the higher terms give some new qualitative insight.)

• I am not sure what you want. If you regard Taylor series as approximation, then approximations are the only application you can get. If you want physical applications of higher derivatives, then this exists of course, but is not really about Taylor expansion per se. May 5, 2014 at 7:03
• There are all sorts of uses for the fact that a function is linear to first order with a certain slope, above and beyond compting numerical values of the function. I'm looking for something similar for the higher terms. May 5, 2014 at 13:59
• For a suggestion, see my answer to the Math StackExchange question What are power series used for? (a reference request). Specifically, I suspect this is one of those things you'll probably have to try my "personal books and library books research collected into a folder" idea for. Earlier in my learning and teaching of math I encountered many situations like you're in, but worse as it was before the internet and I was often in very isolated rural areas, so I wound up making my own lists. May 5, 2014 at 14:40
• Seeing the title of this question, I feel compelled to share the following (possibly apocryphal) anecdote: math.stackexchange.com/a/28899/37122 May 5, 2014 at 17:37
• @DaveLRenfro I am in the process of making my list. I already looked at all the calculus books on my shelf and didn't find anything. Unfortunately none of the questions at your link is what I'm looking for, either. May 5, 2014 at 22:19

Here's one that I just thought of, by modifying a problem in the ODEs section of my textbook.

Question: In the presence of air resistance, does a thrown ball take longer to go up or to come down?

We assume that the force of air resistance is proportional to velocity, with constant of proportionality $p$. Thus, we have

$$F = m a = - p v - m g.$$

Since $a=v'$, this is a differential equation for $v$. For simplicity, let's divide out by $m$ and write $q=p/m$, so it becomes

$$v' = - q v - g.$$

The differential equation is separable, so we can solve it to get

$$v = \left(v_0 + \frac{g}{q}\right) e^{-q t} - \frac{g}{q}.$$

Integrating again, we get the height as a function of time:

$$y = \left(\frac{v_0}{q} + \frac{g}{q^2}\right)(1-e^{-q t}) - \frac{g t}{q}.$$

We can solve $v = 0$ to find the time $t_{\mathrm{max}}$ at which the ball reaches its maximum height, but it's not so easy to solve $y=0$ to find the time at which it reaches the ground. The book answers the question by using a clever argument to conclude that $y(2t_{\mathrm{max}})>0$. But we can also approximate $y$ by a power series in the small parameter $q$.

Let's substitute the second-degree Taylor polynomial for $e^{-q t}$, namely

$$1 - q t + \frac{1}{2} q^2 t^2,$$

into $v$. We get

$$v \approx \left(v_0 + \frac{g}{q}\right) \left(1 - q t + \frac{1}{2} q^2 t^2\right) - \frac{g}{q}$$

Multiplying this out, we get

$$v\approx v_0 + \frac{g}{q} - (v_0 q + g) t + \frac{1}{2} (v_0 q^2 + g q) t^2 - \frac{g}{q}.$$

Canceling the $\frac{g}{q}$s and discarding the term involving $q^2$ (since $q$ is small), we obtain

$$v\approx v_0 + (v_0 q + g) t + \frac{1}{2} g q t^2 .$$

to first order in $q$. Note that all the $q$s in the denominators canceled out, and this is a linear approximation to $v$ as a function of $q$. In principle, we could have derived it by simply taking the derivative of $v$ with respect to $q$ at $q=0$. However, it's not at all obvious from the formula for $v$ that it's even differentiable at $q=0$! (It's not even necessarily obvious that it has a limit as $q\to 0$, but of course the physical considerations imply that it must.) The power series expansion is much nicer.

We can also integrate $v$ to get the height

$$y \approx v_0 t + \frac{1}{2} (v_0 q + g) t^2 + \frac{1}{6} g q t^3.$$

Now we can answer the original question. Putting $v=0$ and solving for $t$ with the quadratic formula, we get

$$t_{\mathrm{max}} \approx \frac{1}{q} + \frac{v_0}{g} \pm \frac{1}{q}\left(1+\frac{v_0^2 q^2}{g^2}\right)^{1/2}.$$

Using the linear Taylor polynomial for the binomial series, we get

$$t_{\mathrm{max}} \approx \frac{1}{q} + \frac{v_0}{g} \pm \frac{1}{q}\left(1+\frac{v_0^2 q^2}{2g^2}\right).$$

We must use the minus sign to cancel out the impossible $1/q$s, yielding

$$t_{\mathrm{max}} \approx \frac{v_0}{g} - \frac{v_0^2 q}{2 g}.$$

Of course, the first term, $\frac{v_0}{g}$, is the obvious value in the no-air-resistance case $v = v_0 - g t$.

We can also set $y=0$ and solve for $t$ to find the time when the ball lands. The solution $t=0$ (when it was thrown) factors out and we can use the quadratic formula again:

$$t_{\mathrm{lands}} \approx \frac{3}{2q} + \frac{3v_0}{2g} \pm \frac{3}{2q}\left(1 - \frac{2qv_0}{3g} + \frac{v_0^2 q^2}{g^2}\right)^{1/2}.$$

Using the second-degree Taylor polynomial for the binomial series this time (since there is a power of $q$ rather than just $q^2$ inside the square root) but discarding the resulting $q^4$ term, we get

$$t_{\mathrm{lands}} \approx \frac{3}{2q} + \frac{3v_0}{2g} \pm \frac{3}{2q}\left(1 - \frac{qv_0}{3g} + \frac{v_0^2 q^2}{2g^2} - \frac{q^2 v_0^2}{18 g^2}\right).$$

Again we must take the minus sign to cancel the impossible $\frac{3}{2q}$, and we get

$$t_{\mathrm{lands}} \approx \frac{2 v_0}{g} - \frac{2 v_0^2 q}{3g^2}.$$

Observe that $t_{\mathrm{lands}} > 2 t_{\mathrm{max}}$.

• That's impressive but 10 equations too many! After your second equation, you can solve to get $v=(g/q)(-1+\exp(qt_{\max}-qt))$. So for $u>0$, $t=t_{\max}\pm u$, $v=(g/q)(-1+\exp(\mp qu))\simeq (g/q)(-1+1\mp qu +q^2u^2)$, and $|v|\simeq(g/q)\;|qu\mp q^2u^2|$. The speeds are therefore greater for $t=t_{\max}-u$ than for $t=t_{\max}+u$, which means that it goes up more quickly than it comes down.
– user173
May 6, 2014 at 3:28
• @MattF. thanks! I do like the approach that actually solves for $t_{\mathrm{lands}}$ (approximately) as being more direct, even if there is some sneakier argument that is shorter. May 6, 2014 at 15:05
• Unsure how sneaky at all this could be called or whether I'm missing something, but doesn't the mundane solution simply note that gravity and air friction both oppose the ascent, thus shortening it compared to the descent where they compete? That's what I would have pointed out as a student. On the other hand, I do enjoy being able to draw a good quantitative conclusion too. Jun 16, 2021 at 23:35

First f all, check out this great paper. It has some interesting examples, especially in Appendix A.

A typical example is relativistic mechanics: the usual first order approximation yields the unsatisfactory Newtonian mechanics (see for example here), which is in many applications not enough to explain what is happening.

ADDED: Since OP asked for an example, calculations in special relativity usually boil down to the use of the binomial series, hence they are mathematically suitable for a calculus class. Whether people understand the underlying physics is a different question.

For example, Section 15.5 in these great notes explains how in the relativistic harmonic oscillator the period depends on the amplitude (Section 15.5.1, page 311), Formula (15.132).

• That does look like a really nice paper. Can you point to a specific example of a calculation in relativistic mechanics that uses a higher-order Taylor expansion (and, hopefully, that a calculus student could understand)? May 5, 2014 at 23:45
• @MikeShulman the relativistic energy is a function of velocity via the so-called gamma function; $\gamma = 1/\sqrt{1-v^2/c^2}$. The binomial series provides $\gamma = 1+ \frac{1}{2}\frac{v^2}{c^2}+ \cdots$. We sometimes hear $E=mc^2$, but, the real story is $E = \gamma mc^2$ where $m$ is the rest mass (it is a constant). So, the second order Taylor term in in the relativistic energy and the constant term are interesting; $E = mc^2 + \frac{m}{2}v^2+ \cdots$ we find the relativistic energy reveals the previously hidden rest energy $mc^2$ and the classical KE term. Higher order terms... Jul 25, 2017 at 4:35
• @MikeShulman terms are physically relevant, we must keep all of them to give an entire account of the relativistic energy. Only when $v<<c$ do we have $v/c << 1$ and hence the classical KE term and the hidden rest energy give an apt description of physics. Of course, in as much as the rest energy is constant, we can analyze physical motion in terms of the classical KE $\frac{m}{2}v^2$ alone. Jul 25, 2017 at 4:37

Much of quantum mechanics is only known in a perturbative framework. Essentially, the basic objects of interest are series. Terms are calculated by Feynman diagrams of increasingly complex diagrams for higher order corrections. So, the series description is pragmatically fundamental. I suppose in principle, non-pertubative solutions exist, however, if no one can hack the math then for a physical purposes (literally in my answer) the series generated by Feynman diagrams represents the best view we have of quantum field theory.

Also, low order applications are interesting. We use $F=mg$ to approximate $F = \frac{GmM}{(R_E+h)^2}$ at $h=0$. In fact, $F=mg$ is just the constant term in the Taylor series for gravity. I submit to you this is still interesting, and, in view of the final I just graded, still sufficiently challenging for many undergraduates.

Applications where you use all the terms in a series are perhaps nicely found in math itself. For example, the alternating harmonic series convergence to $\ln(2)$ as seen from the expansion of $\ln(1+x)$. This is a nice example which uses all the terms in the series. Many examples of this nature are found in every textbook, they make formidable homework problems when taken out of context.

• Identifying the constant term is interesting, but specifically not what I asked for. I also have plenty of "applications" inside of math, but I wanted physical ones. Quantum mechanics is promising, but (say) getting Feynman diagrams out of path integrals requires a good deal more math than my calc 2 students have; can you give me a specific example? May 6, 2014 at 15:09
• @MikeShulman specific Feynman diagram calculation accessible to calculus II students... I don't have one handy. On the other hand, perhaps the energy eigenstates of the hydrogen atom might work for your purposes. I'll think about it, may I'll add something later. May 6, 2014 at 17:07

I don't know if this is the sort of thing you're thinking about -- it seems to me like a too-obvious example, but it seems to fit your question pretty well. Suppose you have a particle moving along some path $r(t)$, and suppose we expand $r(t)$ in a Taylor series as $r(t) = a_0 + a_1 t + (1/2) a_2 t^2 + (1/6) a_3 t^3 + ....$ Then $a_0$ tells us the initial location of the particle, $a_1$ its initial velocity, $a_2$ its initial acceleration, $a_3$ its initial jerk, $a_4$ the initial "jounce" or "snap" (beyond that the derivatives don't have standard names).

A second example, one that maybe isn't "physical" enough but also might give insight into the higher-order terms: Compound interest. If a bank account balance is growing according to $B(t)=A e^{rt}$, and we expand the RHS as a Taylor series, the constant term tells you the initial balance, and the linear term tells you how the balance would grow if we were only considering simple interest; the higher-order terms take into account the effects of compounding. (You can adapt this example to deal with any kind of exponential growth or decay.)

• The first example doesn't seem to really say anything about Taylor series; it's just a statement about higher derivatives which they've seen in calc 1. I like the second example better, but "the higher terms take compounding into effect" is a bit vague --- one might just as well say that "the difference between $A e^{rt}$ and its linear approximation" is what takes compounding into effect. What mileage do you get out of knowing that you can approximate the compounding by a polynomial? May 6, 2014 at 15:07
• In the first example, I think of the linear approximation (1st two terms) as "Where the particle would be if it were not accelerating"; the quadratic approx (1st 3 terms) as "Where the particle would be if it had constant acceleration"; etc. Each additional term is a correction that takes into account a change in the previous term. May 6, 2014 at 15:13
• @MichaelE2 In calc 1 the students learned that the first derivative of position is velocity and the second derivative is acceleration. If I write down the Taylor series of the position function and say "see, you can tell by taking the derivative of the series that the first derivative is velocity and the second derivative is acceleration" then I think they're going to feel "so what, we knew that already; what did we gain by writing it as a power series?" But mweiss's second comment makes clearer what he meant, and that makes sense. May 7, 2014 at 18:13

While this isn't a 'mainstream' topic, I feel like homotopy analysis methods are worth a glance here. The idea is to approximate some non-linear equation by a linear one, take a continuous connection between them, solve the linear problem, and 'slide' the solution over to the non-linear case. If $L\{u(t)\}=0$ is the linear operator and $N\{u(t)\}$ is the non-linear operator, we introduce a new parameter $q$ and consider the equation

$$(1-q)L\{u(t;q)\}+qN\{u(t;q)\}=0$$

where the solution now depends on $q$. $q=0$ is the linear case and $q=1$ is the full non-linear equation you're trying to solve. What we do is expand the dependence on $q$ as a Taylor series $u(x;q)=\sum u_n(x)q^n$. This leads to a recursion relation for $u_n$ in terms of $u_{n-1}$ that can be solved iteratively. Fiddling around with parameters to ensure that the radius of convergence includes $q=1$ then gives you the non-linear solution as an infinite series. While not physics per se, it has major applications when dealing with nonlinear waves and similar things. The wikipedia page says it better than I can. It might not be appropriate to go through it in detail (it could be done with only an ODE course, but it would probably take as much time to understand as a major topic of the curriculum), but it's a good example of where arbitrarily high terms in the Taylor series become important. It's not just a formality; the first or second order approximations are rarely of any meaningful accuracy.

Another major example (though I again apologize for it not being purely physics) would be in numerical approximations, where Taylor series are bread and butter. Being able to expand Taylor series is fundamental in deriving many methods for numerically solving ODEs, and is crucial for being able to estimate bounds and convergence orders. If you only take linear approximations you'll have a horrible time of everything.