The Price of a Nudge

Scenario

Amber's test kiln has one knob this afternoon: temperature offset $t$, in a unit where $t=0$ is the scheduled cone and $t>0$ is hotter. The energy the studio pays — fuel plus the risk of a bloated glaze — is

$$e(t) = t^3 - 3t + 0.5.$$

They are sitting at $t = 0.5$. The question is not "what is a derivative?" The question is: if we nudge the knob a tiny bit hotter, how much extra energy do we pay, per unit nudge?

That price is the whole of gradient descent in one dimension. Lesson 2 will call the same price a slope and draw it as an arrow that points uphill. Lesson 3 will walk the other way by a step $\eta$ times that price. Autodiff exists to compute this price when there are $10^{11}$ knobs. If the 1D price is fuzzy, every later picture is theater.

A few answers look almost right and are wrong:

• $-0.875$ — you reported $e(0.5)$. That is the height of the hill, not the slope. You can stand high with a flat slope, or low with a steep one.
• $-2$ — you formed a coarse chord from $0$ to $1$: $\frac{e(1)-e(0)}{1} = \frac{(1-3+0.5)-0.5}{1} = -2$. That is a whole cone away. It is a cousin of the price, not the price at $0.5$.
• $0.75$ — you took the slope as $3t^2$ and dropped the linear term $-3t$. At $t=0.5$, $3(0.25) = 0.75$. But the linear term is doing four times as much work as the cubic term here.

Keep every term. The exact price of a nudge at $0.5$ is $-2.25$.

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Height is not slope

Let us evaluate the kiln's baseline state:

$$e(0.5) = (0.5)^3 - 3(0.5) + 0.5 = 0.125 - 1.5 + 0.5 = -0.875.$$

The kiln is currently below its nominal scheduled energy of $0.5$ (at $t=0$). That number is a height. It tells you where you stand on the hill, but it does not tell you which way the ground tilts under your boots. You can stand high on a wide plateau where nudging changes nothing, or you can stand at sea level on the edge of a sheer cliff.

Now define the price as a limit, so that "tiny nudge" is a mathematical contract rather than a loose vibe:

$$e'(t) = \lim_{h \to 0} \frac{e(t+h) - e(t)}{h}.$$

The numerator is extra energy. The denominator is the nudge. The ratio is extra energy per unit nudge.

Deriving the slope: expand the cubic

Do not simply recite the power rule as a magic spell. Look at what happens to the cubic term under a nudge $h$:

$$(t+h)^3 = t^3 + 3t^2 h + 3t h^2 + h^3.$$

Now assemble the nudged energy $e(t+h)$:

$$e(t+h) = \left(t^3 + 3t^2 h + 3t h^2 + h^3\right) - 3(t+h) + 0.5.$$

Subtract the baseline energy $e(t) = t^3 - 3t + 0.5$:

$$e(t+h) - e(t) = 3t^2 h + 3t h^2 + h^3 - 3h.$$

Divide by the nudge $h$:

$$\frac{e(t+h) - e(t)}{h} = 3t^2 - 3 + 3t h + h^2.$$

As the nudge $h$ shrinks to zero, the terms $3th$ and $h^2$ vanish completely, leaving the exact instantaneous derivative:

$$e'(t) = 3t^2 - 3.$$

At Amber's operating point $t = 0.5$:

$$e'(0.5) = 3(0.5)^2 - 3 = 3(0.25) - 3 = 0.75 - 3 = -2.25.$$

The negative sign has an immediate physical meaning: a hotter nudge lowers energy. Uphill is to the left (colder). The knob at $0.5$ wants to turn hotter if you want to walk downhill.

At the stationary points where the price is zero:

$$e'(t) = 0 \implies 3t^2 - 3 = 0 \implies t^2 = 1 \implies t = \pm 1.$$

The second derivative is $e''(t) = 6t$. At $t = 1$, $e''(1) = 6 > 0$, so $t = 1$ is a local minimum ($e(1) = -1.5$). At $t = -1$, $e''(-1) = -6 < 0$, so $t = -1$ is a local maximum ($e(-1) = 2.5$).

[!IMPORTANT] Zero price means a stationary point, not automatically a bottom. Later in Lesson 4 and Lesson 13 we will encounter two-well landscapes and saddle points where the slope is zero yet you are standing on a ridge.

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A chord is a cousin, not a twin

Suppose you do not have an analytical formula for $e(t)$. You only have a black-box kiln that you can query at two temperatures. You choose a nudge width $h = 0.5$ and compute the two-sided difference quotient:

$$\frac{e(t+h) - e(t-h)}{2h}.$$

Two-sided differences are vastly superior to one-sided differences. Why? Because the Taylor expansions on both sides cancel the leading error term:

$$e(t + h) = e(t) + e'(t)h + \frac{1}{2}e''(t)h^2 + \frac{1}{6}e'''(t)h^3 + \dots$$ $$e(t - h) = e(t) - e'(t)h + \frac{1}{2}e''(t)h^2 - \frac{1}{6}e'''(t)h^3 + \dots$$

Subtract the two equations: the baseline $e(t)$ cancels, and crucially, the quadratic curvature term $\frac{1}{2}e''(t)h^2$ cancels as well:

$$e(t+h) - e(t-h) = 2e'(t)h + \frac{1}{3}e'''(t)h^3.$$

Divide by $2h$:

$$\frac{e(t+h) - e(t-h)}{2h} = e'(t) + \frac{1}{6}e'''(t)h^2.$$

For Amber's cubic, the third derivative is constant: $e'''(t) = 6$. Therefore $\frac{1}{6}(6)h^2 = h^2$, giving an exact identity:

$$\frac{e(t+h) - e(t-h)}{2h} = e'(t) + h^2.$$

The error is not "approximately" $h^2$; for any cubic with leading coefficient 1, it is identically $h^2$!

In contrast, a one-sided chord retains the $\frac{1}{2}e''(t)h$ term. For Amber's kiln at $t=0.5$, $e''(0.5) = 6(0.5) = 3$, so its error is $1.5h + h^2$.

The Home Numbers Table

Here is the exact comparison of prices and chords around $t = 0.5$:

Quantity	Formula	Numerical Value	Error vs $e'(0.5) = -2.25$
Hill Height	$e(0.5)$	$-0.875$	N/A (different physical unit)
Instantaneous Price	$e'(0.5)$	$-2.25$	$0.00$ (exact)
Two-sided chord ($h = 0.5$)	$\frac{e(1.0) - e(0.0)}{1.0}$	$-2.00$	$+0.25 = h^2$
Two-sided chord ($h = 0.1$)	$\frac{e(0.6) - e(0.4)}{0.2}$	$-2.24$	$+0.01 = h^2$
Two-sided chord ($h = 0.01$)	$\frac{e(0.51) - e(0.49)}{0.02}$	$-2.2499$	$+0.0001 = h^2$
One-sided chord ($h = 0.5$)	$\frac{e(1.0) - e(0.5)}{0.5}$	$-1.25$	$+1.00 = 1.5h + h^2$
One-sided chord ($h = 0.1$)	$\frac{e(0.6) - e(0.5)}{0.1}$	$-2.09$	$+0.16 = 1.5h + h^2$

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One timid downhill hop

How do we use this price to train the knob? Gradient descent updates the parameter by subtracting the slope scaled by a small step size $\eta$:

$$t \leftarrow t - \eta\, e'(t).$$

Amber, being cautious with the studio kiln, picks $\eta = 0.05$ from $t = 0.5$:

$$t' = 0.5 - 0.05(-2.25) = 0.5 + 0.1125 = 0.6125.$$

Notice how the two minus signs interact:

1. The formula has a minus sign because we want to go downhill (opposite the gradient).
2. The slope $e'(0.5)$ is $-2.25$ (negative).
3. The product $- \eta e'(t)$ is positive ($+0.1125$), moving the knob rightward to $t' = 0.6125$.

Evaluating the new energy:

$$e(0.6125) = (0.6125)^3 - 3(0.6125) + 0.5 \approx 0.2298 - 1.8375 + 0.5 = -1.1077.$$

The energy fell from $-0.875$ to $-1.1077$. The hop succeeded.

The Wrong-Sign Disaster

What happens if you accidentally add the slope ($t \leftarrow t + \eta e'(t)$)?

$$t_{\text{wrong}} = 0.5 + 0.05(-2.25) = 0.5 - 0.1125 = 0.3875.$$

Evaluating energy:

$$e(0.3875) = (0.3875)^3 - 3(0.3875) + 0.5 \approx -0.6043.$$

Energy rose by $+0.2707$ toward the local maximum at $t = -1$. You walked uphill. In a production neural network, dropping that minus sign causes the loss to explode immediately.

[!NOTE] How far to hop is Lesson 3's business. In Lesson 3 we will study the learning rate $\eta$ in depth, and in Lesson 5 we will meet the Lipschitz constant $L$ that sets its ceiling. Here, we keep $\eta = 0.05$ fixed solely to verify the direction of descent.

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Interactive Workbench: Chord vs Tangent

Explore the relationship between the chord span $h$, chord symmetry, and descent hops in the interactive laboratory below:

OPT-P2-NUDGE

Now examine the physical geometry on the 3D concept rig:

OPT-P2-ANATOMY

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Try it: Transfer example

Work this problem on scratch paper before revealing the answer:

Consider the quadratic energy:

$$g(t) = t^2 - 4t \quad \text{at} \quad t = 1.$$

1. Find the exact derivative $g'(1)$.
2. Compute the two-sided chord with $h = 0.5$.
3. Compute the one-sided chord with $h = 0.5$.
4. Why is the two-sided chord exact here?

Reveal solution

1. Exact derivative: $$g'(t) = 2t - 4 \implies g'(1) = 2(1) - 4 = -2.$$
2. Two-sided chord ($h = 0.5$): Evaluate at $1 + 0.5 = 1.5$ and $1 - 0.5 = 0.5$: $$g(1.5) = (1.5)^2 - 4(1.5) = 2.25 - 6 = -3.75.$$ $$g(0.5) = (0.5)^2 - 4(0.5) = 0.25 - 2 = -1.75.$$ $$\frac{g(1.5) - g(0.5)}{2(0.5)} = \frac{-3.75 - (-1.75)}{1.0} = \frac{-2.0}{1.0} = -2.$$ The two-sided chord gives $-2$ exactly!
3. One-sided chord ($h = 0.5$): $$g(1) = 1^2 - 4(1) = -3.$$ $$\frac{g(1.5) - g(1)}{0.5} = \frac{-3.75 - (-3)}{0.5} = \frac{-0.75}{0.5} = -1.5.$$
4. Why is the two-sided chord exact? Recall our Taylor error term: $\frac{1}{6}g'''(t)h^2$. For any quadratic function, $g''(t) = 2$ and $g'''(t) = 0$. The third derivative vanishes everywhere! Because there is no cubic or higher-order curvature, the Taylor series terminates, leaving zero chord error.

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Runnable code verification

Run this in Python to confirm the calculations and observe machine precision limits:

e  = lambda t: t**3 - 3*t + 0.5
de = lambda t: 3*t**2 - 3

print(e(0.5), de(0.5))                       # -0.875 -2.25

# Two-sided chord error shrinks as h**2
for h in (0.5, 0.1, 0.01):
    two = (e(0.5+h) - e(0.5-h)) / (2*h)
    print(f"h={h:<4} chord={two:.6f} error={round(two - de(0.5), 6)} (= h**2)")

# Downhill hop
print("hop landing:", 0.5 - 0.05 * de(0.5))  # 0.6125

# The floor of finite differences: catastrophic cancellation
h_micro = 1e-13
two_micro = (e(0.5 + h_micro) - e(0.5 - h_micro)) / (2 * h_micro)
print(f"h=1e-13: error = {abs(two_micro - de(0.5)):.6e}")  # ~4.4e-4 (worse than h=0.01!)

The Floor of Numerical Differentiation

Why does $h = 10^{-13}$ yield an error around $4 \times 10^{-4}$, which is hundreds of times worse than $h = 0.01$?

At $h = 10^{-13}$, $e(0.5 + h)$ and $e(0.5 - h)$ agree in their first 13 decimal digits. When you subtract two numbers that agree to 13 digits in standard 64-bit IEEE floating-point arithmetic (which has roughly 15 to 17 digits of mantissa), the leading digits cancel completely. The remaining difference is dominated by round-off noise. Dividing that noise by $2h = 2 \times 10^{-13}$ magnifies the error enormously.

Tiny is a scale, not a personality trait. Finite differences check analytic derivatives at scales like $h = 10^{-5}$; pushing $h$ to machine epsilon causes catastrophic cancellation.

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Failure modes this chapter names

• Reporting height as slope: Confusing where you stand ($-0.875$) with the slope of the ground ($-2.25$).
• Dropping a term of $e'(t)$: Taking only $3t^2$ ($0.75$) and dropping $-3$.
• Treating a coarse chord as the derivative: Assuming $h=0.5$ gives the exact slope ($-2.00$ instead of $-2.25$).
• Stepping $+ \eta e'$ and calling it descent: Walking uphill toward the local maximum.
• Calling a one-sided chord two-sided: Forgetting that one-sided chords carry an $O(h)$ error that is much larger than $h^2$.

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Why autodiff is this price, not a second calculus

When the knob is a weight $w_j$ in a neural network, $e$ is the training loss, and $\frac{\partial e}{\partial w_j}$ is "how much extra loss we pay if we nudge $w_j$."

Reverse-mode automatic differentiation (backprop) computes every such 1D price in a single backward sweep. It does not invent a brand-new form of calculus. It is the chain rule, scheduled efficiently (which we will formalize in Chapter P5).

The desk problem asks for a two-sided chord with a whole-number $h$ so you can do it by hand. Expect it to differ from the exact slope by $h^2$. The desk problem also does a hop like this on a different cubic. Work that problem on paper and type the integers into the desk quiz.

Next: P3 The Uphill List — what happens when the hill has more than one knob.