The Mathematical Ninja looked offended. For once, it wasn’t a student that was the guilty party, it was me.

“You’re considering a series on the WHAT?!”

“The binomial expansion,” I said, brightly.

“Drivel!” said the Mathematical Ninja “Drivel, piffle and poppycock! Newtonian claptrap, not worth the space made for it in the formula book!”

This was strong stuff, even for the Mathematical Ninja.

He has more respect for me than for the average student, but not much more. This translates into piercing stares rather than physical violence “Euler series,” he said, as if such a thing were obvious.

“Euler series?” I asked “Not sure I’ve heard of those!”

### The Mathematical Ninja digresses into Stigler’s Law of Eponymy.

“You may know them as Taylor series, but Brooke Taylor didn’t discover them, Euler did.”

“Ah! Why are they called Taylor series?”

“Taylor did discover a kind of series - but you probably call them Maclaurin series. Yet another example of Stigler’s Law.”

“Stigler’s Law?” This conversation was rapidly falling down a rabbit hole.

“No scientific discovery is named for its discoverer,” said the Ninja “An observation made by Merton, of course.”

I rolled my eyes. “So, Tay… Euler series.” “Yes! Much simpler, and you don’t need to use a different method for C2, C4 and FP2 - the same technique works everywhere!” “Take it away, sensei! How about a simple one to show it works - let’s say $(1+x)^3$”

### The Mathematical Ninja’s Guide to Euler Series (which you may call Taylor Series or, here, Maclaurin Series)

His eyebrows said “Molest me not with this pocket calculator stuff,” but he humoured me. “Step 1,” said the Mathematical Ninja, “is to write down $f(x)$ and its derivatives.” “The function $f(x) = (1+x)^3$,” I said. “First derivative is $f^{\prime}(x) = 3(1+x)^2$ and the second derivative is $f^{\prime\prime}(x) = 6(1+x)$.” “Keep going.” I sighed “$f^{\prime\prime\prime}(x) = 6$ and $f^{(4)}$ onwards are all 0 ((Convention: up to three derivatives, use dashes; beyond that, use the number in a bracket)). “Step 2,” said the Ninja, “is to work all of those out at $x=0$.” Luckily, I’d written them down. “$f(0) = 1$, $f^{\prime}(0) = 3$, $f^{\prime\prime}(0) = 6$, $f^{\prime\prime\prime}(0) =6$ All of the others are $0$.” “Yep! And for step 3, you just apply the formula ((It’s in your book, under FP2. You can still use it.)): $$f(x) \\simeq f(0) + \\frac{x}{1!} f^{\\prime}(0) +\\frac{x^2}{2!} f^{\\prime\\prime}(0)+...$$ and so on.” “So I’ve got $1 + \frac x1 \times 3 + \frac {x^2}{2}\times 6 + \frac{x^3}{6}\times 6$ and nothing more which simplifies to $f(x) \simeq 1 + 3x + 3x^2 + x^3$” “Only in this case it’s not $\simeq$, it’s $=$, because you’ve not truncated anything - it’s an exact answer.” “How about a nastier one? $(4-3x)^\frac{1}{2}$”

“Exactly the same principle,” said the Ninja. “So, let me write this in a table, like a student would in an exam:

$n$

$f^{(n)}(x)$

$f^{(n)}(0)$

$\frac{x^n}{n!}$

$0$

$(4-3x)^{\frac{1}{2}}$

$2$

$1$

$-\frac{3}{2}(4-3x)^{-\frac{1}{2}}$

$-\frac{3}{4}$

$x$

$2$

$-\frac{9}{4}(4-3x)^{-\frac{3}{2}}$

$-\frac{9}{32}$

$\frac{x^2}{2}$

$3$

$-\frac{81}{8}(4-3x)^{-\frac{5}{2}}$

$-\frac{81}{512}$

$\frac{x^3}{6}$

” “Good layout,” said the Ninja “And the final answer is $f(x) = 2 - \frac 34 x - \frac{9}{64}x^2 - \frac{27}{1024}x^3$!” “$\simeq$,” said the Ninja “You could take that further if you wanted - but it’s correct so far as it goes and just what you’d get from the binomial formula.” I nodded “Yes, sensei.”

[Updated July 8th to correct error. Thanks to [twit handle=’LargeCardinal’] for the correction.]