When finding the Maclaurin series representation for sin(x)/x, I decided to multiply the Maclaurin series for each individual function first.

The Maclaurin series for sin(x) is: $\sum_{n=0 }^{\infty}\frac{(-1)^{n}x^{2n+1}}{(2n+1)!}$

The Maclaurin series for 1/x is: $\sum_{n=0 }^{\infty}(1-x)^{n}$

So wouldn't the Maclaurin series representation for the both of these would be multiplying their power series together in order to obtain:

$\sum_{n=0 }^{\infty}(1-x)^{n}\frac{(-1)^{n}x^{2n+1}}{(2n+1)!}$

However, I've seen solutions that just divide the Maclaruin series for sin(x) by x; but why are you allowed to do that? Doesn't "x" also change with each terms (it has its own Maclaurin series)

  • 5
    $\begingroup$ No, instead just take the series for sin(x) and divide each term by x $\endgroup$
    – Henry Lee
    Mar 22 '20 at 22:13
  • $\begingroup$ But I'm asking why that works exactly because doesn't "x" also change with each term in the series so you'd need the series representation of it too right? For example, if you had x/sin(x), wouldn't you do the maclaurin series for x and then divide each term in that series by sin(x) $\endgroup$
    – MT0820
    Mar 22 '20 at 22:29
  • $\begingroup$ @MT0820 does $x$ even have a Maclaurin series? $\endgroup$
    – an4s
    Mar 22 '20 at 22:30
  • $\begingroup$ Im sorry I meant 1/(x*sinx) $\endgroup$
    – MT0820
    Mar 22 '20 at 22:31
  • $\begingroup$ Is your question about $\sin(x)/x$ or $1/(x \sin x)$? Those are two completely different things. (Also: @an4s: Yes, $x$ has a Maclaurin series: The Maclaurin series of $x$ is just $x$ itself.) $\endgroup$ Mar 22 '20 at 22:35

There are several misconceptions in your post. I think the main thing here is to remember that the $\sum$ symbol means: $$\sum_{n=0}^\infty a_n = a_0 + a_1 + a_2 + \cdots$$ If this is kept in mind, then a lot of what I'm about to say will be clearer.

First, you cannot multiply $\sum a_n$ with $\sum b_n$ to get $\sum a_n b_n$. In other words: $$\left( \sum_{n=1}^\infty a_n \right)\left( \sum_{n=1}^\infty b_n \right) \neq \sum_{n=1}^\infty a_n b_n$$ meaning that $$(a_1 + a_2 + a_3 + \cdots) (b_1 + b_2 + b_3 + \cdots) \neq (a_1b_1 + a_2b_2 + a_3b_3 + \cdots)$$

Second: Remember that a Taylor series centered at $c$ has the form $$\sum_{n=0}^\infty a_n(x-c)^n = a_0 + a_1(x-c) + a_2(x-c)^2 + \cdots$$ and a Maclaurin series is just a Taylor series with $c = 0$: $$\sum_{n=0}^\infty a_n x^n = a_0 + a_1x + a_2x^2 + \cdots$$ Here, the $a_n$'s don't depend on $x$, meaning they shouldn't have any $x$'s in them.

What does this mean for you? It means (first of all) that the series you wrote down $$\sum_{n=0}^\infty (1-x)^n \frac{(-1)^nx^{2n}}{(2n+1)!}$$ is not a Taylor series (so, in particular, it's not a Maclaurin series). Also, your series for $1/x$ is a Taylor series centered at $x = 1$, but it is not a Maclaurin series.

Finally, if your Maclaurin series is $$\sum_{n=0}^\infty a_nx^n = a_0 + a_1x + a_2x^2 + a_3x^3 + \cdots$$ then dividing by $x$ gives \begin{align*} \frac{1}{x}\sum_{n=0}^\infty a_nx^n & = \frac{1}{x}\left(a_0 + a_1x + a_2x^2 + a_3x^3 + \cdots\right) \\ & = \frac{a_0}{x} + a_1 + a_2x + a_3x^2 + \cdots \end{align*} This last series is not a Maclaurin series because of the $a_0/x$ term --- unless $a_0 = 0$, in which case the $a_0/x$ term goes away.

  • $\begingroup$ Wouldn't multiplying these series just be the Cauchy Product? $\endgroup$
    – MT0820
    Mar 22 '20 at 22:38
  • $\begingroup$ Yes, but the series you wrote for $\sin(x)/x$ was not the Cauchy product of your series for $\sin(x)$ and that for $1/x$. You just wrote $\sum a_n b_n$. $\endgroup$ Mar 22 '20 at 22:39

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