Why does a Taylor polynomial divide by factorials?

1. 1. Match the height

   The curve is $f(x) = e^x$. To copy it near $x = 0$ with a polynomial, match one thing at a time, starting with the height. A constant, $p(x) = c_0$, is a flat line. Drag the hollow point to move it.

   Drag the line until it passes through the curve's point at $x = 0$.

   $c_0 = f(0) = 1$. The line has the curve's height at $0$ and nothing more: the curve climbs away from it on the right and sinks below it on the left.

   The figure: The curve y = eˣ through (0, 1), and the polynomial 1.000 of degree 0.

2. 2. Match the slope

   Add a term in $x$: $p(x) = 1 + c_1x$. At $x = 0$ the new term is $0$, so the height stays right, and the slope is $c_1$ everywhere. Drag the hollow point to tilt the line.

   Tilt the line until its slope at $0$ is the curve's, $f'(0) = 1$.

   $c_1 = f'(0) = 1$, and $p(x) = 1 + x$ is the tangent line. The curve still bends up away from it on both sides: the next term has to match the bending.

   The figure: The curve y = eˣ through (0, 1), and the polynomial 1.000 + 1.000 x of degree 1. The polynomial before its newest term is dashed.

3. 3. Match the bending

   Add $c_2x^2$. At $x = 0$ it is $0$ and so is its slope, so the height and the slope stay right and only the bending, $p''(0)$, changes. Every derivative of $e^x$ is $e^x$, so $f''(0) = 1$.

   Drag until the parabola bends as much as the curve at $0$: $p''(0) = 1$.

   That takes $c_2 = \tfrac12$, not $1$. Differentiate $c_2x^2$ twice and you get $2c_2$, so the coefficient is the curve's bending divided by $2$.

   $$c_2 = \frac{f''(0)}{2}$$$c_2$ $=\frac{f^{\prime \prime }(0)}{2}$

   The figure: The curve y = eˣ through (0, 1), and the polynomial 1.000 + 1.000 x + 0.500 x² of degree 2. The polynomial before its newest term is dashed.

4. 4. Each power brings a factorial

   Add $c_3x^3$ to match the third derivative, which for $e^x$ is $1$ again. Differentiating $x^3$ once gives $3x^2$, and twice gives $6x$.

   Question: What must $c_3$ be?

   $\tfrac16$. Three derivatives of $c_3x^3$ leave $3 \cdot 2 \cdot 1 \cdot c_3 = 3!\,c_3$, so $c_3 = f'''(0)/3!$. The same happens for every power: $x^k$ differentiated $k$ times is $k!$, so each coefficient is a derivative divided by a factorial.

   $$T_n(x) = \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!}\,(x - a)^k$$$T_n(x)$ $=\sum _{k=0}^n\frac{f^{(k)}(a)}{k!}(x-a{)}^k$

   The figure: The curve y = eˣ through (0, 1), and the polynomial 1.000 + 1.000 x + 0.500 x² + 0.167 x³ of degree 3. The polynomial before its newest term is dashed.

5. 5. Leave the factorials out

   A common slip uses the derivatives themselves as the coefficients. For $e^x$ that gives $1 + x + x^2 + x^3$, drawn in amber, beside the true $T_3(x) = 1 + x + \tfrac12x^2 + \tfrac16x^3$.

   Question: Which of its four terms does the slip get wrong?

   Only $x^2$ and $x^3$: $0! = 1! = 1$, so the constant and the tangent line come out right. The amber curve bends twice as hard as $e^x$ at $0$ and peels away: at $x = 1$ it gives $4$, where $e \approx 2.718$ and $T_3(1) \approx 2.667$.

   The figure: The curve y = eˣ through (0, 1), and the polynomial 1.000 + 1.000 x + 0.500 x² + 0.167 x³ of degree 3. The polynomial before its newest term is dashed. In amber, 1 + x + x² + x³, the polynomial with the factorials left out, which climbs away from the curve much sooner.

The film, as said, 1:14

1. 0:00

   Copy $e^x$ near $x = 0$ with a polynomial, one match at a time. First the height: a flat line, $p(x) = c_0$.

2. 0:06

   Lower it onto the curve's point. The height matches at $c_0 = f(0) = 1$, and nothing else does.

3. 0:13

   Add a term in $x$. It is $0$ at $x = 0$, so the height stays right while the slope changes.

4. 0:20

   Tilt it until its slope is the curve's, $f'(0) = 1$. That is the tangent line, $1 + x$.

5. 0:27

   Add $c_2x^2$. It leaves the height and the slope alone and changes only the bending at $0$, $p''(0)$.

6. 0:33

   Bend it until $p''(0) = f''(0) = 1$. That takes $c_2 = \tfrac12$, because $c_2x^2$ differentiated twice is $2c_2$.

7. 0:40

   Add $c_3x^3$. Three derivatives of $x^3$ leave $3 \cdot 2 \cdot 1 = 6$, so matching $f'''(0) = 1$ takes $c_3 = \tfrac16$.

8. 0:47

   Every power works the same way: $x^k$ differentiated $k$ times is $k!$, so each coefficient is a derivative divided by $k!$.

9. 0:54

   Leave the factorials out and the coefficients are the derivatives themselves: $1 + x + x^2 + x^3$, in amber.

10. 0:59

    Since $0! = 1! = 1$, its height and slope are right. But it bends twice as hard as $e^x$, and peels away.

11. 1:05

    At $x = 1$ it gives $4$, where $e \approx 2.718$ and the true $T_3(1) \approx 2.667$.

12. 1:09

    A Taylor polynomial matches one derivative per term, and each match divides by a factorial.