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Let's say the task is to find the Taylor series at the origin of the function $$f(x) = \frac{3x}{1-x-2x^2}$$

The formula is $T^n_0 =\sum^n_{k=0} \frac{f^{(k)}(0)}{k!}x^k$. If I follow this formula, I need to at least compute the 4th or 5th derivatives to find the general expression which takes a long time and very likely ends up with a wrong expression.

Peter Mortensen
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ElleryL
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    Use partial fraction decomposition, and $\frac{1}{1-x}=\sum_n x^n$. – OR. Apr 04 '15 at 15:57
  • No don't use partial fraction decomposition unless you need the closed form for the coefficients. See my answer. =) – user21820 Apr 05 '15 at 06:00
  • Hey, partial fractions isn't of much help to you when you try to do something like $\dfrac x{1-x-x^2}$. However — calling the coefficients $a_n$ — it isn't too hard to come up with the recurrence relation $a_n-a_{n-1}-a_{n-2}=0$, and then once you calculate a few terms, you realize that it's the Fibonacci sequence. (In your case, you have $a_0=0$, $a_1=3$, and $a_n-a_{n-1}-2a_{n-2}=0$, I believe. This makes it easier to get the first few coefficients, if that's all you need rather than a general formula.) – Akiva Weinberger Apr 05 '15 at 07:17
  • I would say that you almost never use the definition to compute a Taylor series. Using the definition means computing $f^(n)(0)$ for every $n$, which means in general finding a general formula for $f^(n)(x)$ for $x$ close to $0$ and every $n$. To do this, you need to guess a formula and prove it by induction. Your example shows that computing the first derivative functions to guess the formula is a very annoying task. – Taladris Apr 05 '15 at 07:27
  • @columbus8myhw: That's exactly my answer. – user21820 Apr 05 '15 at 08:16
  • @user21820 Oh, sorry, didn't notice :( – Akiva Weinberger Apr 05 '15 at 08:17

3 Answers3

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HINT: Note that $1-x-2x^2=(1-2x)(1+x)$, so

$$\frac{3x}{1-x-2x^2}=\frac1{1-2x}-\frac1{1+x}\;.$$

Now use a known Taylor series expansion to expand each of the fractions on the right-hand side, and combine the series into a single series.

Added: Let me emphasize Math$1000$’s comment below: once you know a few power series, you should always try to make use of them to get new ones. Here it’s just a matter of adding a couple, but sometimes you may have to work a bit harder: differentiating or integrating one, multiplying it by some power of $x$, or applying some combination of these manipulations.

Brian M. Scott
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    I'd like to add that you really should learn to do stuff like this. Don't just take derivatives mindlessly...that's not the point of the exercise. – Math1000 Apr 04 '15 at 16:00
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There’s another way, not as good as applying what you know about geometric series and others of your acquaintance. But it’s completely effective.

Have you learned how to divide polynomials? It’s taught in high school here in the States. To get a series expansion for a rational function, do a long division in the same spirit, but simply arrange your polynomials in ascending order of degree rather than descending, as you did in high school. Just watch out for careless errors, since they can badly spoil your result.

Lubin
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Solution

$\dfrac{3z}{1-z-2z^2} = 3z \sum_{k=0}^\infty (z+2z^2)^k$ as $z \to 0$ since $z+2z^2 \to 0$. [So there is power series.]

Let $(a_k:k\in\mathbb{N})$ be a sequence such that $\dfrac{3z}{1-z-2z^2} = \sum_{k=0}^\infty a_k z^k$.

Then $3z = (1-z-2z^2) \sum_{k=0}^\infty a_k z^k = a_0 + (a_1-a_0) z + \sum_{k=0}^\infty (a_{k+2}-a_{k+1}-2a_k) z^{k+2}$

Thus $a_0 = 0$ and $(a_1-a_0) = 3$ and $a_{k+2}-a_{k+1}-2a_k = 0$ for all $k \in \mathbb{N}_{\ge 2}$.

Notes

This method works even when the denominator is a large polynomial and partial fraction decomposition would be ridiculous to do by hand. However, if you want a closed-form solution you still need to solve the recurrence, which would give exactly the same answer. If you look carefully this also shows the relationship between partial fraction decomposition (including the case with multiple roots) and recurrence relations.

user21820
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