Identify a shifted exponential from evidence

Try

A synthetic cooling investigation records a temperature of 80 °C at time 0 hours, 50 °C at 2 hours, and 35 °C at 4 hours. The room temperature was not recorded. Can these observations identify a cooling model with an unknown limiting temperature?

The raw output ratios are 50/80=0.625 and 35/50=0.7. They differ, so a pure proportional-decay model y=Aq^(t/2) cannot match all three. However the changes across the equal two-hour intervals are −30 and −15 °C. The second change is half the first. This suggests testing a shifted exponential: a constant baseline plus a proportionally changing excess.

Write the family as y(t)=k+Aq^(t/2). Here t is numerical elapsed hours; y and k are temperatures in °C; A is an initial temperature difference in °C; q is a positive dimensionless factor per two hours. For nonconstant behavior require A≠0 and q≠1. The family is an assumption to investigate, not a conclusion established merely by three readings.

Predict

Predict whether subtracting 20 °C would make the remaining amounts proportional. Then predict whether a 20 °C limiting temperature is ever reached at a finite time under this formula.

Time t (hours)	Temperature y (°C)	Excess y−20 (°C)	Change from previous reading (°C)
0	80	60	Not applicable
2	50	30	−30
4	35	15	−15
6	Prediction 27.5	Prediction 7.5	Prediction −7.5

The excess halves every two hours. Thus the candidate is y(t)=20+60(1/2)^(t/2). The temperature itself does not halve: the baseline stays fixed. For every finite t, the exponential factor is positive, so the predicted temperature is strictly greater than 20. On the physical domain t≥0 its range is (20,80]. The lower endpoint is approached, not attained.

Experiment

Plot the three observations before using the candidate curve. On the same scale, plot their excesses above 20. Compare the equal-time changes on both traces: subtracting a constant moves the curve vertically without changing those differences. The proportional ratios appear only after subtracting the appropriate baseline.

Now change only the middle observation from 50 to 49 °C. Before revealing the result, recalculate both changes and predict whether the inferred baseline rises or falls.

Check the changed middle reading

The changes are −31 and −14, giving q=14/31. The inferred baseline is k=(49−(14/31)80)/(1−14/31)=399/17≈23.47 °C; A=80−k=961/17≈56.53 °C. A one-degree change in one reading changes the inferred baseline by about 3.47 degrees. Three-point exact fitting is sensitive to measurement errors and does not independently validate a limiting temperature.

Understand

For equally spaced readings y₀,y₁,y₂ at times 0,h,2h with h>0, write yⱼ=k+Aqʲ. Define d₁=y₁−y₀ and d₂=y₂−y₁. Subtracting cancels k:

d₁=A(q−1), and d₂=Aq(q−1)=qd₁.

If d₁≠0, the factor is q=d₂/d₁. For an allowed nonconstant model, require q>0 and q≠1. Since y₁−q y₀=k(1−q), recover k=(y₁−q y₀)/(1−q), then A=y₀−k. Substituting these values checks all three readings. In the worked case q=1/2, k=(50−40)/(1/2)=20 and A=60.

These conditions are part of the method, not inconvenient details to discard. If both changes are zero, a constant model with A=0 works for many factors, so the factor is not identified. If the first change is zero but the second is not, no allowed model works: the first equation would force A=0. Equal nonzero changes give q=1, but that makes k+Aqʲ constant, contradicting the observations. Opposite-sign changes give q<0, and a nonzero first change followed by zero gives q=0; neither is a positive-base exponential model. Near q=1, the division by 1−q can amplify small input errors. An apparently precise algebraic baseline can therefore be uncertain.

Unequal time intervals invalidate the shortcut d₂/d₁. For example, the worked candidate at times 0,2,6 gives temperatures 80,50,27.5. The changes −30 and −22.5 span two and four hours; their ratio 0.75 is not the two-hour factor. If the baseline is independently known, subtract it first and use the actual gap: for readings at times u and v, the per-hour factor b satisfies b^(v−u)=(y(v)−k)/(y(u)−k), provided the nonzero excesses have a positive ratio. That is a different inference problem from recovering an unknown baseline.

Apply

Freeze the original candidate before inspecting a new measurement at 6 hours: the recorded temperature is 27.7 °C with stated uncertainty ±0.3 °C. The original-scale residual, observed minus predicted, is 27.7−27.5=0.2 °C. The prediction belongs to the interval [27.4,28.0], so it is compatible with this measurement allowance. This one check supports compatibility; it does not prove the model, identify the true room temperature, or establish uncertainty bounds on every future prediction. The fitting readings were treated as exact for the algebraic demonstration; actual uncertain fitting readings would introduce additional prediction uncertainty.

Suppose instead the new reading were 28.2±0.3 °C. Its interval [27.9,28.5] excludes 27.5, so the frozen prediction would fail that check. Changing k after seeing the result makes this observation fitting or development evidence. Record that change and obtain another unused check rather than relabeling the revised fit as an independent success.

Finally express the original model using elapsed minutes m. Since t=m/60, the rule is y=20+60(1/2)^(m/120). The one-hour factor for the excess is √(1/2), while the two-hour factor is 1/2. The baseline does not change with the time unit. Explain aloud why dividing the factor 1/2 by 60 would not perform a unit conversion: successive time steps multiply factors rather than add or divide them linearly.

Continue

Keep the equal spacing, chosen family, recovered baseline, factor interval, physical domain and unused prediction with your model. A nonzero offset can conceal proportional change in the raw outputs. Subtracting a fitted offset reveals a candidate mechanism; it does not establish that mechanism physically.

The next lesson constructs a logarithmic response directly from input/output data, rather than merely using logarithms to solve an exponential target time. Retain original-response residuals when comparing its predictions. For broader fitting practice, return to polynomial fitting and validation: a more flexible fit and an independently supported model remain different outcomes in either family.