Unit IV · Lecture 26 of 40 · B.Sc. Data Science, 1st Year, Semester II
Polynomials (linear, quadratic, cubic) work well for data that changes by additive amounts. But many real-world phenomena change by multiplicative factors — each step multiplies the previous value by a constant factor. For such data, exponential and power curves are far more appropriate than polynomials.
If you invest ₹100 at 10% annual interest, after 1 year you have ₹110, after 2 years ₹121, after 3 years ₹133.10. The increase is multiplicative (×1.10 each year), not additive (+₹10 each year). A straight line would predict ₹130 after 3 years (wrong), while an exponential curve predicts ₹133.10 (correct). The exponential model captures the multiplicative growth naturally.
1. Population growth: Bacteria, humans, epidemics
2. Compound interest: Investments, debt
3. Radioactive decay: Half-life of isotopes
4. Viral spread: Early-stage pandemic growth
5. Technology adoption: Moore's Law, internet users
6. Cooling/heating: Newton's Law of Cooling
1. Scatter plot: Points follow a J-shaped curve (rapid increase) or a decay curve (rapid decrease then flattening).
2. Constant ratio: The ratio yᵢ₊₁/yᵢ is approximately constant across consecutive observations.
3. Log-linear plot: If plotting log(y) vs x gives a straight line, the data is exponential.
4. Residuals from linear fit: If residuals show a systematic curved pattern (not random), consider exponential.
Form: Exponential with base b
Linearise: log y = log a + x·log b
Let: Y = log y, A = log a, B = log b
Becomes: Y = A + Bx (straight line!)
Use when: Growth/decay rate is proportional to current value
Form: Natural exponential
Linearise: ln y = ln a + bx
Let: Y = ln y, A = ln a
Becomes: Y = A + bx (straight line!)
Use when: Continuous growth/decay (calculus-based models)
Form: Power curve (allometric)
Linearise: log y = log a + b·log x
Let: Y = log y, X = log x, A = log a
Becomes: Y = A + bX (straight line!)
Use when: Relationship between log(x) and log(y) is linear
Many non-linear curves can be transformed into straight lines by applying a suitable transformation to x, y, or both. Once linearised, we fit a straight line using least squares, then transform back to get the original curve parameters.
When we linearise and fit a line to the transformed data, we are minimising the sum of squared errors in the transformed space, not the original space:
For y = abˣ: We minimise Σ(log yᵢ − log ŷᵢ)², not Σ(yᵢ − ŷᵢ)²
This means the fit is optimal in the log scale, which gives more weight to smaller values and less weight to larger values. This is often desirable (it prevents large values from dominating the fit), but it's important to understand the distinction.
Given: n data points (x₁,y₁), (x₂,y₂), …, (xₙ,yₙ)
Step 1: Take logarithms of both sides:
log y = log a + x · log b
Step 2: Define transformed variables:
Y = log y, A = log a, B = log b
The equation becomes: Y = A + Bx
Step 3: Apply least squares to (x, Y) pairs:
B = [nΣxY − Σx·ΣY] / [nΣx² − (Σx)²]
A = Ȳ − B·x̄
Step 4: Transform back:
a = antilog(A) = 10ᴬ (if using log₁₀)
b = antilog(B) = 10ᴮ
Step 5: Write the fitted equation: ŷ = a · bˣ
A bacteria culture grows as follows. Fit a curve of the form y = abˣ.
| x (hours) | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| y (thousands) | 2.5 | 6.0 | 14.0 | 35.0 | 85.0 |
Take log₁₀ of each y value: Y = log y
| x | y | Y = log₁₀(y) | xY | x² |
|---|---|---|---|---|
| 1 | 2.5 | 0.3979 | 0.3979 | 1 |
| 2 | 6.0 | 0.7782 | 1.5564 | 4 |
| 3 | 14.0 | 1.1461 | 3.4383 | 9 |
| 4 | 35.0 | 1.5441 | 6.1764 | 16 |
| 5 | 85.0 | 1.9294 | 9.6470 | 25 |
| 15 | 5.7957 | 21.2160 | 55 |
n = 5, Σx = 15, ΣY = 5.7957, ΣxY = 21.2160, Σx² = 55
x̄ = 3, Ȳ = 1.15914
B = [nΣxY − Σx·ΣY] / [nΣx² − (Σx)²]
= [5(21.2160) − 15(5.7957)] / [5(55) − 225]
= [106.080 − 86.9355] / [275 − 225]
= 19.1445 / 50 = 0.38289
A = Ȳ − B·x̄ = 1.15914 − 0.38289(3) = 1.15914 − 1.14867 = 0.01047
a = 10ᴬ = 10^0.01047 = 1.0244
b = 10ᴮ = 10^0.38289 = 2.4148
The bacteria population multiplies by approximately 2.41× every hour. This represents approximately 141% growth per hour.
| x | y (observed) | ŷ = 1.0244 × 2.4148ˣ | Residual |
|---|---|---|---|
| 1 | 2.5 | 1.0244 × 2.4148 = 2.474 | +0.026 |
| 2 | 6.0 | 1.0244 × 5.831 = 5.973 | +0.027 |
| 3 | 14.0 | 1.0244 × 14.082 = 14.426 | −0.426 |
| 4 | 35.0 | 1.0244 × 34.006 = 34.836 | +0.164 |
| 5 | 85.0 | 1.0244 × 82.120 = 84.124 | +0.876 |
At x = 6 (6 hours):
ŷ = 1.0244 × 2.4148⁶ = 1.0244 × 198.34 = 203.2 thousand
At x = 10 (10 hours):
ŷ = 1.0244 × 2.4148¹⁰ = 1.0244 × 6,973 = 7,143 thousand ≈ 7.1 million
Note: Exponential models predict explosive growth. In reality, growth eventually slows due to resource limits (logistic growth). Exponential models are reliable only for short-term predictions.
Step 1: Take natural logarithm (ln) of both sides:
ln y = ln a + bx
Step 2: Define: Y = ln y, A = ln a
Equation becomes: Y = A + bx
Step 3: Apply least squares to (x, Y) pairs:
b = [nΣxY − Σx·ΣY] / [nΣx² − (Σx)²]
A = Ȳ − b·x̄
Step 4: Transform back:
a = eᴬ (antilog using base e)
Step 5: Fitted equation: ŷ = a · eᵇˣ
y = abˣ and y = aeᵇˣ are equivalent forms:
If y = abˣ, then b = eᵏ where k = ln(b), so y = aeᵏˣ.
If y = aeᵇˣ, then eᵇ = B, so y = aBˣ.
Use y = abˣ when the growth factor per period is known.
Use y = aeᵇˣ when the continuous growth rate is known (common in calculus-based models).
a: The initial value (y when x = 0). ŷ(0) = a.
b (in aeᵇˣ): The continuous growth rate.
• b > 0 → exponential growth
• b < 0 → exponential decay
• b = 0 → constant (no growth/decay)
Doubling time: t₂ = ln(2)/b
Half-life: t₁/₂ = ln(2)/|b| (for decay)
The amount of a radioactive substance remaining over time is given below. Fit y = aeᵇˣ and find the half-life.
| Time (x, hours) | 0 | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|
| Amount (y, grams) | 100 | 72 | 50 | 37 | 26 | 19 |
Y = ln(y) for each data point:
| x | y | Y = ln(y) | xY | x² |
|---|---|---|---|---|
| 0 | 100 | 4.6052 | 0.0000 | 0 |
| 1 | 72 | 4.2767 | 4.2767 | 1 |
| 2 | 50 | 3.9120 | 7.8240 | 4 |
| 3 | 37 | 3.6109 | 10.8327 | 9 |
| 4 | 26 | 3.2581 | 13.0324 | 16 |
| 5 | 19 | 2.9444 | 14.7220 | 25 |
| 15 | 22.6073 | 50.6878 | 55 |
n = 6, Σx = 15, ΣY = 22.6073, ΣxY = 50.6878, Σx² = 55
x̄ = 2.5, Ȳ = 3.76788
b = [6(50.6878) − 15(22.6073)] / [6(55) − 225]
= [304.1268 − 339.1095] / [330 − 225]
= −34.9827 / 105 = −0.33317
A = 3.76788 − (−0.33317)(2.5) = 3.76788 + 0.83293 = 4.60081
a = eᴬ = e^4.60081 = 99.58
Half-life = ln(2) / 0.33317 = 0.6931 / 0.3332 = 2.08 hours
The substance decays at a continuous rate of 33.3% per hour. It takes approximately 2.08 hours for half the substance to decay.
Step 1: Take logarithm of both sides:
log y = log a + b · log x
Step 2: Define: Y = log y, X = log x, A = log a
Equation becomes: Y = A + bX
Step 3: Apply least squares to (X, Y) pairs:
b = [nΣXY − ΣX·ΣY] / [nΣX² − (ΣX)²]
A = Ȳ − b·X̄
Step 4: Transform back:
a = antilog(A) = 10ᴬ
Step 5: Fitted equation: ŷ = a · xᵇ
1. When a log-log plot (log y vs log x) gives a straight line.
2. Allometric growth in biology (organ size vs body size).
3. Scaling laws in physics (period of pendulum vs length: T = kL^0.5).
4. Engineering relationships (drag force vs velocity: F = kv²).
5. Zipf's law in linguistics, Pareto distributions in economics.
Exponential (y = abˣ): The base is constant, the exponent varies with x. Growth rate is proportional to current value.
Power (y = axᵇ): The exponent is constant, the base varies with x. The relationship is a fixed power law.
How to tell them apart:
• Plot log(y) vs x → linear? → Exponential
• Plot log(y) vs log(x) → linear? → Power curve
Fit a power curve y = axᵇ to the following data:
| x | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| y | 2.0 | 5.7 | 10.4 | 16.0 | 22.4 |
X = log₁₀(x), Y = log₁₀(y):
| x | y | X = log x | Y = log y | XY | X² |
|---|---|---|---|---|---|
| 1 | 2.0 | 0.0000 | 0.3010 | 0.0000 | 0.0000 |
| 2 | 5.7 | 0.3010 | 0.7559 | 0.2275 | 0.0906 |
| 3 | 10.4 | 0.4771 | 1.0170 | 0.4852 | 0.2276 |
| 4 | 16.0 | 0.6021 | 1.2041 | 0.7250 | 0.3625 |
| 5 | 22.4 | 0.6990 | 1.3502 | 0.9438 | 0.4886 |
| 2.0792 | 4.6282 | 2.3815 | 1.1693 |
n = 5, ΣX = 2.0792, ΣY = 4.6282, ΣXY = 2.3815, ΣX² = 1.1693
X̄ = 0.41584, Ȳ = 0.92564
b = [5(2.3815) − (2.0792)(4.6282)] / [5(1.1693) − (2.0792)²]
= [11.9075 − 9.6229] / [5.8465 − 4.3231]
= 2.2846 / 1.5234 = 1.4997 ≈ 1.5
A = 0.92564 − 1.5(0.41584) = 0.92564 − 0.62376 = 0.30188
a = 10^0.30188 = 2.004 ≈ 2.0
The exponent b = 1.5 indicates that y grows as the 1.5th power of x. This is a common scaling law — for example, the relationship between body mass and metabolic rate in animals often follows a power law with exponent around 0.75.
Click each curve type to see it fitted to sample data:
Exponential growth
Exponential decay
Power curve
Enter data and choose which exponential model to fit:
| Feature | y = a + bx | y = abˣ | y = aeᵇˣ | y = axᵇ |
|---|---|---|---|---|
| Type | Linear | Exponential | Natural Exp. | Power |
| Transform | None | Y = log y | Y = ln y | Y = log y, X = log x |
| Linearised form | y = a + bx | Y = A + Bx | Y = A + bx | Y = A + bX |
| Parameters | 2 (a, b) | 2 (a, b) | 2 (a, b) | 2 (a, b) |
| Shape | Straight line | J-curve / decay | J-curve / decay | Power law curve |
| Growth type | Additive | Multiplicative | Continuous mult. | Power scaling |
| Best for | Constant rate of change | Constant ratio | Continuous processes | Scaling laws |
| Diagnostic plot | y vs x → line | log y vs x → line | ln y vs x → line | log y vs log x → line |
Q1: To fit y = abˣ, we take logarithms and obtain:
Q2: For the power curve y = axᵇ, the linearised form is:
Q3: A requirement for fitting exponential curves is:
Q4: If log(y) vs x gives a straight line, the appropriate model is:
Q5: In y = aeᵇˣ, if b < 0, the curve represents:
Q6: The half-life of a substance with decay model y = ae⁻⁰·⁵ˣ is:
Q7: The difference between y = abˣ and y = axᵇ is:
Q8: When fitting exponential curves by linearisation, we minimise errors in:
1. Fit y = abˣ to: x = 1, 2, 3, 4, 5 and y = 3, 9, 27, 81, 243. What do you observe?
2. The population of a city (in thousands) over decades is: x = 0, 1, 2, 3, 4 and y = 50, 70, 98, 137, 192. Fit y = aeᵇˣ and predict the population at x = 5.
3. Fit a power curve y = axᵇ to: x = 1, 2, 3, 4, 5 and y = 3, 12, 27, 48, 75. Verify by computing predicted values.
4. For the data in Problem 2, also fit a straight line. Compare the SSE of both models. Which fits better?
5. Explain why we cannot fit an exponential curve if one of the y values is zero or negative.
6. A radioactive substance has a half-life of 5 hours. Write the decay equation y = aeᵇˣ and find the value of b.
7. How would you decide between fitting y = abˣ and y = axᵇ to a given dataset? Describe the diagnostic procedure.
8. BONUS: Show that for y = abˣ, the ratio yᵢ₊₁/yᵢ = b (constant) for equally spaced x values. Use this to estimate b directly from the data in Problem 1 and compare with the least squares estimate.
Lecture 27: Correlation — Karl Pearson's Coefficient of Correlation
Lecture 28: Rank Correlation — Spearman's Coefficient
Lecture 29: Regression Lines & Their Properties
Unit IV · Lecture 26 · BDS113 Introduction to Statistics · B.Sc. Data Science, Semester II