1
$\begingroup$

I’m working on a section of a course covering Taylor expansions, and have found that, although there is great notation for simplifying the formula for the coefficients of a general infinite binomial series (i.e. not limited to positive, integer, powers), the computation of the coefficients for any actual series is rather long and messy and (most importantly) unenlightning (i.e. the computations won’t help students gain a better understanding of the expansion).

On Math.SE, I found this answer to a question about the expansion of finite binomial expansions, which is handy, and it made me wonder if anyone had some tips / ideas of how to present the computation of

  1. The general term, and
  2. The first 4-5 nonzero terms

of these expansions, quickly and efficiently? For clarity, I’m speaking of expansions of functions of the form $(a+x)^k = a^k\big(1+\frac{x}{a}\big)^k$, for $a>0$ and $k \in \mathrm{I\!R}$.

Even if an alternative method emphasized some sort of mathematical reasoning more than the brute-force computation, that would be preferable. E.g. solving the coefficients of $f(x) = \frac{1}{(1+x)^2}$ by differentiating the series expansion of $F(x) = \frac{1}{1+x}$; the later method underscores one of the advantages of working with series, and involves related skills, where as Taylor’s formula for the coefficients is just basic derivatives and pattern recognition...


[Edit: addition of specific example, for clarification]

Any type of binomial expansion that will have infinitely many terms, i.e. $(a+x)^k$ where $k$ is not a positive integer.

E.g. 1: Find the Maclaurin series for $f(x) = \sqrt{2-x}$.

E.g. 2: Find the Taylor series for $f(x) = \sqrt[3]{x}$ about $x = 8$.

E.g. 3: Find the Maclaurin series for $f(x) = (1-x)^{3 /4}$.

Currently, I have the following method when the goal is to provide both a general term, as well as the first 3-4 nonzero terms. I’ll use the first example to illustrate:

1. Compute the first few derivatives $\mathbf{f^{(i)}(x)}$ to determine the general pattern for $\mathbf{f^{(n)}(x)}$ On my own, usually the first 3 suffice to identify the general pattern; when illustrating for students, I’ll sometimes include the 4th and 5th derivatives.

N.B. : if the desired series can be obtained by integration or differentiation of a simpler series, I will prioritize that method; see the example in the last paragraph, before the edit, for an example

$$ \begin{array}{ll} f(x) = \sqrt{2-x} \\ \hline % f’(x) = \tfrac12(2-x)^{-1 /2} (-1) & = \quad -\tfrac12(2-x)^{-1 /2} \\ \hline % f’’(x) = (-\tfrac12) (\tfrac 12)(2-x)^{-3/2} (-1)^2 & =\quad -(\tfrac12) (\tfrac 12)(2-x)^{-3/2} \\ \hline % f’’’(x) = (-\tfrac32)(-\tfrac12) (\tfrac 12)(2-x)^{-5/2} (-1)^3 & =\quad -(\tfrac32)(\tfrac12) (\tfrac 12)(2-x)^{-5/2} \\ \hline \end{array} $$

From this we see that, for $n \geq 3$:

$$f^{(n)}(x) = -(\tfrac12)(\tfrac12)(\tfrac32)(\tfrac52)\cdots(\tfrac{2n-3}{2})(2-x)^{-(2n-1)/2},$$

which can be simplified as:

$$ f^{(n)}(x) = - \frac{1\cdot 3 \cdot 5 \; \cdots \; (2n-3)}{2^n}(2-x)^{-(2n-1)/2} $$

2. Evaluate $\mathbf{f^{(n)}(x)}$ at the centre to find the general expression for $\mathbf{f^{(n)}(c )}$

Plugging $c = 0$ into the expression for the $n^{th}$ derivative yields:

$$ f^{(n)}(0) = - \frac{1\cdot 3 \cdot 5 \; \cdots \; (2n-3)}{2^n}(2)^{-(2n-1)/2}, $$

where we can combine the exponents on the powers of 2 to obtain:

$$ f^{(n)}(0) = - \frac{1\cdot 3 \cdot 5 \; \cdots \; (2n-3)}{2^{(4n-1)/2}}, $$

3. Plug the result for $\mathbf{f^{(n)}(c)}$ into Taylor’s formula to obtain the general term.

Recalling that the coefficients of the Taylor series expansion are giving by $a_n = \frac{1}{n!} f^{(n)}(c)$, we have: $$ f(x) = \sqrt{2-x} % = 2 - \frac{1}{2\sqrt{2}}x - \frac{1}{16\sqrt{2}}x^2 - \sum_{n=3}^{\infty} \frac{1\cdot 3 \cdot 5 \; \cdots \; (2n-3)}{n! \; 2^{(4n-1)/2}} x^n, $$

What I dislike about the above is that there are so many simple pitfalls that are only overcome either by extensive practice (in which nothing is learnt other than how to find the expansion of these specific series, which will inevitably fade with time and has little-to-no transfer value), or by writing out many intermediate steps, which makes solving them even longer.

E.g. a common error is to forget to include the factorial terms when writing out the series expansion. An extra line, in which every element of the general term is individually accounted for (e.g. writing $a_3$ as $- \frac{1}{3!} \cdot \frac{1\cdot 3}{2^{11/2}}$ and then computing the final value), avoids these errors, but is quite long to write out for multiple terms (using up time which could be better spent on actually understanding series expansions).

Does anyone know of a more efficient way to get the general term of the sum and/or the first 4-5 nonzero terms of the series? The general formulas that use combinatorial notation are fine for summarizing the results, but don’t seem very practical for generating specific series.

$\endgroup$
  • 1
    $\begingroup$ You mean other than the usual generalization? Maybe you could give an example of $a,k$ that you are looking for, that might help people think of an appropriate answer. Good luck! $\endgroup$ – kcrisman Jun 26 at 2:25
  • $\begingroup$ I personally like deriving the general formula in-class then applying it. The idea that we prove a theorem then apply it is an important idea. There is no need to work out the Taylor coefficients in specific cases when we can work it out in general then make appropriate substitutions. Of course, opinions vary. See: youtube.com/… for my doubtless improvable presentation. $\endgroup$ – James S. Cook Jun 27 at 1:21
  • $\begingroup$ @JamesS.Cook Thanks for sharing. Covering the general derivation goes without saying - I’m with you on that on. Don’t have sufficient WiFi access to check the video just now, but am looking forward to it. I have come up with a reasonable solution in the meantime, which I’ll post, if materially different from your presentation. Appreciate your taking the time to share! Cheers, $\endgroup$ – Rax Adaam Jun 27 at 20:29
2
$\begingroup$
  1. I think the bug can actually be a feature. Kids need work on the "muscles" of computation. It's not like this is the only setting where doing long calculations is needed (can be the norm in physics and engineering problems.) Look at all the questions here about the frustration of dealing with kids that can't perform algebra.

  2. It can be good to look at something new in a slowed down and more mechanical manner, to build familiarity with it. Or even to look at something in differing amounts of assistance. Calculating by hand, calculating by scientific calculator, use of spreadsheets, use of CAS. Graphing by hand (yes on actual paper, with a pencil), graphing with Excel (shows you the data points), graphing with Desmos.

  3. Of course, there is a happy medium. Don't expect kids to calculate stuff out to 20 terms or the like, as Gauss or Euler might. But also don't be totally put off by making them do some grunt algebra.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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