Written by Colin+ in core 2, core 4, ninja maths.

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.

"What would you use instead?" I asked.

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!"

"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 $(1+x)^3$"

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 01.

"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 formula2:

$$f(x) \simeq f(0) + \frac{x}{1!} f^{\prime}(0) +\frac{x^2}{2!} f^{\prime\prime}(0)+... $$ 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 @LargeCardinal for the correction.]

## twentythree

RT @icecolbeveridge: [FCM] Why the binomial expansion is stupid and you shouldn’t use it: http://t.co/Cer5QXhmci

## twentythree

@icecolbeveridge that’s great, and yes, much simpler IMO

## twentythree

that’s great, and yes, much simpler

## LargeCardinal

@icecolbeveridge Brilliant! Sending to all my maths students and teacher friends! đź™‚

## LargeCardinal

@icecolbeveridge Ps-minor correction; formula in 8th para from end should end “…+3x^2 + x^3” (unless I have Monday fatigue…)

## MathbloggingAll

Why the binomial expansion is stupid and you shouldnâ€™t use it http://t.co/8MoPBxkt5x

## Nathan Briggs

NICE!

## Nathan Briggs

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