Why the Maclaurin series gives you Pascal’s Triangle

The Mathematical Ninja, some time ago, pointed out a curiosity about Pascal's Triangle and the Maclaurin1 (or Taylor2 ) series of a product:

$\diffn{n}{(uv)}{x} = uv^{(n)} + n u'v^{(n-1)} + \frac{n(n-1)}{2} u'' v^{(n-2)} + ...$, where $v^{(n)}$ means the $n$th derivative of $v$ - which looks a lot like Pascal's Triangle.

Naturally I thought that was A Pretty Neat Trick, and showed it to a student, who promptly asked: why?

Oh, bugger.

I had no idea why. I did what I usually do when I don't know why something works: I asked Barney. Barney hasn't responded yet, so I presume he doesn't know, either. Neither did the Googles. It was time for the last resort: work it out myself.

I've now figured it out, although it needed some heavy artillery; I hope it provides an insight into how a mathematical brain works.

The Taylor Series

I'm going to use a less fiddly version of the Taylor Series than you usually see in FP whatever. I'm going to assert without proof that for any sensible3 function $u$ or $v$, for $x$ close enough to $x_0$:

$u(x) = u(x_0) + hu'(x_0) + \frac 12 h^2 u''(x_0) + ... + \frac{1}{i!} h^i u^{(i)}(x_0) + ...$, where the superscript $(i)$ means the $r$th derivative, and $h := (x - x_0)$.

I could also write, more neatly:

$v(x) = \sum_{j=0}^{\infty} \frac{h^j}{j!} v^{(j)}(x_0)$. (To me, that's beautifully compact).

The big gnus

I've also got, for $f(x) = u(x)v(x)$, $f(x) = \sum_{k=0}^{\infty} \frac{h^k}{k!} f^{(k)}(x_0)$.4

So, taking a deep breath, I've got:

$\sum_{k=0}^{\infty} \frac{h^k}{k!} f^{(k)}(x_0) = \left(\sum_{i=0}^{\infty} \frac{h^i}{i!} u^{(i)}(x_0)\right)\left(\sum_{j=0}^{\infty} \frac{h^j}{j!} v^{(j)}(x_0)\right)$. Take a moment to marvel. I'm now going to match coefficients.

Cover me; I'm going in.

For any $k$ I happen to pick on the left, I'll need to get all of the $h^k$ terms that come out of the infinite product on the right. Happily, there's a simple way to find them: it's every pair of terms where the powers of $h$ sum to $k$ - for instance, if I have $k=3$, I'll need the $h^0$ term from the $u$ expansion multiplied by the $h^3$ term from $v$, plus the $h^1$ and $h^2$ terms, plus the $h^2$ and $h$ terms, plus the $h^3$ and $h^0$ terms - there are no other ways to make $h^3$ from those sums.

So, what do those terms look like?

On the left, I've got $\frac{h^k}{k!} f^{(k)}(x_0)$. On the right, I've got $\sum_{r=0}^{k} \frac{h^r}{r!}u^{(r)}(x_0) \times \frac{h^{(k-r)}}{(k-r)!} v^{(k-r)}(x_0)$.

That simplifies to $\frac{h^k}{r!(k-r)!} u^{(r)}v^{(k-r)}$.

Dividing both sides by $h^k$ and multiplying by $k!$ gives:

$f^{(k)}(x_0) = \frac{k!}{r!(k-r)!} u^{(r)}(x_0)v^{(k-r)}(x_0)$ - and that factorial gubbins at the beginning is just ${}^{k}C_r$ - which is the Pascal's triangle coefficient we were looking for!

An aside

This makes me wonder about semiderivatives - if Pascal's triangle can be extended to negative and fractional powers, and Pascal's triangle is implicated in the product rule, can you use these rules to figure out (say) $\diffn{\frac {1}{2}}{y}{x}$? I figure it'll need some gamma function magic, but I just figured out that $\diffn{\frac {1}{2}}{x^2}{x} \simeq 1.505x^{\frac{3}{2}}$. What that means, on the other hand, is an open question to me.

Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.

  1. Discovered by Taylor []
  2. Discovered by Euler []
  3. i.e., one where all the derivatives exist []
  4. It's not strictly necessary to use different letters for the count variable, but it's helpful for clarity shortly. []

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