How does a tangent line find a root?

1. 1. Solve the tangent instead

   Where $f(x) = x^3 - 4x + 1$ crosses the axis is hard to solve for; where a line crosses is easy. So replace the curve by its tangent at a guess, $x_0 = 3$, where $f(3) = 16$ and $f'(3) = 23$, and solve the line instead.

   Question: The tangent drops $16$ to reach the axis, at a slope of $23$. How far left of $3$ does it get there?

   Slope is rise over run, so the run is rise over slope: $16/23 \approx 0.70$, and the next guess is $x_1 \approx 3 - 0.70 = 2.30$. That is the whole of Newton's method: each step goes back $f(x_n)/f'(x_n)$ from $x_n$.

   $$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$$$x_{n+1}$ $=x_n-\frac{f(x_n)}{f^{\prime }(x_n)}$

   The figure: The curve y = x³ − 4x + 1, which crosses the x-axis three times, near −2.11, 0.25, 1.86. Its tangent at x₀ = 3.00, where f(x₀) = 16.000 and the slope is 23.000, meets the axis at x₁ = 2.3043.

2. 2. Which way to step

   The formula subtracts the ratio, and its signs do the steering. Try a start where the curve falls: at $x_0 = 0$, $f(0) = 1$ and $f'(0) = -4$.

   Question: Where does the next estimate, $x_1$, land?

   $x_1 = 0 - \frac{1}{-4} = 0.25$: the tangent falls to the right, so it meets the axis on the right, $0.004$ short of the root. Adding the ratio instead gives $-0.25$, in amber: the same distance the wrong way, where the tangent is $2$ above the axis, twice as far from it as at the start.

   The figure: The curve y = x³ − 4x + 1, which crosses the x-axis three times, near −2.11, 0.25, 1.86. Its tangent at x₀ = 0.00, where f(x₀) = 1.000 and the slope is −4.000, meets the axis at x₁ = 0.2500. In amber, x₀ + f(x₀)/f′(x₀) = −0.2500, a step the same length the wrong way, where the tangent is 2.000 from the axis.

3. 3. Again, and again

   Each estimate is a new place to draw a tangent. Start again from $x_0 = 3$ and take three steps: watch the estimates, and their distance to the root, under the figure.

   The figure: The curve y = x³ − 4x + 1, which crosses the x-axis three times, near −2.11, 0.25, 1.86. Newton’s method from x₀ = 3.00, after 3 steps, is at 1.869470.

4. 4. Correct digits double

   After three steps $x_3 \approx 1.8695$, $0.0087$ from the root at $1.8608$. Close to the root the curve is nearly straight, so its tangent nearly lands on the root.

   Question: How far from the root will $x_4$ be?

   About $0.00006$, close to $0.0087^2 \approx 0.00008$: too small to see even zoomed in this far. The tangent misses by the curve's bending over the gap, which grows like the gap squared. So the error is roughly squared each step, and the correct decimals double: two at $x_3$, four at $x_4$, eight at $x_5$.

   $$x_{n+1} - r \approx \frac{f''(r)}{2f'(r)}\,(x_n - r)^2$$$x_{n+1}-r$ $\approx \frac{f^{\prime \prime }(r)}{2f^{\prime }(r)}(x_n-r{)}^2$

   The figure: The curve y = x³ − 4x + 1, which crosses the x-axis three times, near −2.11, 0.25, 1.86. Newton’s method from x₀ = 3.00, after 4 steps, is at 1.860871.

5. 5. A flat tangent throws it far

   Start at $x_0 = 1.1$, near the bottom of the dip, between two roots and nearer the one at $1.86$. There $f(1.1) = -2.069$ and $f'(1.1) = -0.37$: the tangent is nearly flat.

   Question: Which root will Newton’s method find?

   The furthest. A nearly flat tangent meets the axis far away: $x_1 = 1.1 - \frac{-2.069}{-0.37} \approx -4.49$, past both nearer roots, and from there the steps close in on $-2.11$. Newton's method follows tangents, not distances, so work out the steps before naming the root. Drag $x_0$ to try other starts.

   The figure: The curve y = x³ − 4x + 1, which crosses the x-axis three times, near −2.11, 0.25, 1.86. Newton’s method from x₀ = 1.10, after 6 steps, is at −2.114915. The root nearest the start, 1.8608, is ringed in amber.

The film, as said, 1:44

1. 0:00

   To find where $f(x) = x^3 - 4x + 1$ crosses the axis, replace the curve by its tangent at a guess, $x_0 = 3$.

2. 0:06

   The tangent drops $16$ at a slope of $23$, so it runs $16/23 \approx 0.70$ to the axis: $x_1 \approx 2.30$.

3. 0:14

   That is the whole method: each step goes back $f(x_n)/f'(x_n)$ from $x_n$.

4. 0:18

   The signs steer. From $x_0 = 0$, $f(0) = 1$ and $f'(0) = -4$: the tangent falls to the right.

5. 0:23

   Subtract the ratio: $x_1 = 0 - \frac{1}{-4} = 0.25$, just $0.004$ short of the root.

6. 0:28

   Adding it instead gives $-0.25$, in amber: the same distance the wrong way, where the tangent is twice as high.

7. 0:35

   Back to $x_0 = 3$. Each estimate is a new place to draw a tangent.

8. 0:39

   Three steps, and $x_3 \approx 1.8695$, $0.0087$ from the root at $1.8608$.

9. 0:47

   Zoom in on the root. Here the curve is nearly straight, so its tangent nearly lands on the root.

10. 0:57

    The fourth step misses by about $0.00006$, close to $0.0087^2$: too small to see even here.

11. 1:04

    The error is roughly squared each step, so the correct decimals double: two at $x_3$, four at $x_4$, eight at $x_5$.

12. 1:10

    Now start at $x_0 = 1.1$, by the bottom of the dip. The nearest root is at $1.86$.

13. 1:16

    There $f'(1.1) = -0.37$: the tangent is nearly flat, and meets the axis far away, at $-4.49$.

14. 1:23

    From there the steps close in on $-2.11$, past both nearer roots.

15. 1:31

    Assuming the nearest root, ringed in amber, is the slip: Newton’s method follows tangents, not distances.

16. 1:37

    Solve the tangent, step, repeat. Near a root each step squares the error; a flat tangent throws the next guess far away.