A science policy analyst models the spread of misinformation: it starts with 100 false posts and grows by 40% daily. A public alert system, launched on day 0, reduces the daily growth rate by 15 percentage points each day (starting day 1: 40% → 25% → 10% → ...). On which day does the daily increase drop below 1%?

["How a Science Policy Analyst Models and Contains Misinformation Spread: When Does the Daily Growth Drop Below 1%?", "Misinformation spreads rapidly in digital environments, posing serious challenges to public understanding—especially in science communication. A recent science policy analysis models the exponential growth of false posts and evaluates a public alert system designed to slow this spread. This article explores the dynamics of the model: starting with 100 false posts on Day 0, growing initially by 40% per day, then reduced daily by 15 percentage points as an automated alert system activates—from 40% on Day 1, dropping to 25%, 10%, and lower. We determine: On which day does the daily growth rate finally drop below 1%?", "---", "### The Model of Misinformation Spread", "The scenario begins with 100 false posts on Day 0. Without intervention, the number grows by 40% daily:", "[\nF(n) = 100 \ imes (1.4)^n\n]", "But on Day 1, a public alert system launches, beginning a daily reduction of the growth rate by 15 percentage points starting from 40% and decreasing each day:", "- Day 1: 40% growth\n- Day 2: 25% growth\n- Day 3: 10% growth\n- Day 4: —5% (and continues decreasing: 40% − 15k% daily)\n- Eventually, growth approaches negative values, but maximum reduction stops at below 1%.", "This decreasing exponential growth is modeled as:", "[\n\ ext{Growth rate on day } n = 40% - 15% \ imes (n - 1), \quad \ ext{for } n \geq 1\n]", "Let’s define the daily multiplicative growth factor as:", "[\nr(n) = 1 + \frac{40 - 15(n - 1)}{100} = 1 + 0.4 - 0.15(n - 1) = 0.7 + 0.15(2 - (n - 1)) = 0.7 + 0.15(3 - n)\n]", "So,", "[\nr(n) = 0.7 + 0.15(3 - n)\n]", "We want to find the smallest day ( n ) such that:", "[\nr(n) < 0.01 \quad \ ext{(i.e., growth below 1%)}\n]", "Note: Since ( r(n) ) is a decreasing positive quantity that eventually turns negative, we solve:", "[\n0.7 + 0.15(3 - n) < 0.01\n]", "---", "### Solve the Inequality", "[\n0.7 + 0.15(3 - n) < 0.01\n]", "Calculate ( 0.15(3 - n) ):", "[\n0.7 + 0.45 - 0.15n < 0.01\n]", "[\n1.15 - 0.15n < 0.01\n]", "Subtract 1.15:", "[\n-0.15n < 0.01 - 1.15 = -1.14\n]", "Divide by -0.15 (reverse inequality):", "[\nn > \frac{1.14}{0.15} = 7.6\n]", "Thus, ( n > 7.6 ), so the smallest integer ( n ) is ( n = 8 ).", "---", "### Verification by Day", "Let’s verify the growth rates:", "| Day ( n ) | Growth Rate (%) | Factor ( 1 + r(n) ) | Cumulative Increase |\n|-------------|------------------|------------------------|----------------------|\n| 1 | 40% | 1.40 | ×1.40 → 140 |\n| 2 | 25% | 1.25 | ×1.25 → 175 |\n| 3 | 10% | 1.10 | ×1.10 → 192.5 |\n| 4 | —5% | 0.95 | ×0.95 → 183.875 |\n| 5 | —10% | 0.90 | ×0.90 → 165.4875 |\n| 6 | —15% | 0.85 | ×0.85 → 140.6598 |\n| 7 | —20% | 0.80 | ×0.80 → 112.5278 |\n| 8 | —25% | 0.75 | ×0.75 → 84.3963 |\n| 9 | —30% | 0.70 | ×0.70 → 59.07841 |\n| 10 | —35% | 0.65 | ×0.65 → 38.54997 |\n| 11 | —40% | 0.60 | ×0.60 → 23.02998 |\n| 12 | —45% | 0.55 | ×0.55 → 12.71699 |\n| 13 | —50% | 0.50 | ×0.50 → 6.358495 |\n| 14 | —55% | 0.45 | ×0.45 → 2.86122 |\n| 15 | —60% | 0.40 | ×0.40 → 1.144488 |\n| 16 | —65% | 0.35 | ×0.35 → 0.400659 |\n| 17 | —70% | 0.30 | ×0.30 → 0.120198 |", "On Day 17, the growth rate is –70%, so actual spread decreases. But we are tracking when growth rate drops below 1%. At Day 16, growth rate is −65% (<1%), so already below from day 16 onward.", "Wait — our inequality ( n > 7.6 ) suggests day 8, but actual daily application shows the rate crosses below 1% on Day 16, not 8?", "But that contradicts the math? Let’s recheck.", "Ah — critical insight: The model says growth rate drops by 15 percentage points per day starting from 40%, but once it reaches zero and becomes negative, it is still a growth factor <1. The question asks: on which day does the daily increase drop below 1%? That is, when does ( r(n) < 0.01 )?", "But ( r(n) = 0.7 + 0.15(3 - n) )", "Set:", "[\n0.7 + 0.15(3 - n) < 0.01\n]", "[\n0.7 + 0.45 - 0.15n < 0.01\n]", "[\n1.15 - 0.15n < 0.01 \Rightarrow n > \frac{1.14}{0.15} = 7.6\n]", "So starting at ( n = 8 ), ( r(n) < 0.01 ). But on Day 8:", "[\nr(8) = 0.7 + 0.15(3 - 8) = 0.7 - 0.75 = -0.05 \Rightarrow 95% \ ext{ decrease}\n]", "So the rate drops below 1% in absolute value on Day 8, but is already below 1% (in fact, negative). However, the growth factor is ( 1 + r(n) ). When ( r(n) < 0 ), the number shrinks.", "But the question is clear: "When does the daily increase drop below 1%?"", "That is, when ( 1 + r(n) < 1.01 )? No — “increase” implies positive growth. If growth rate is negative, it’s a decrease.", "So “daily increase” drops below 1% means: the growth rate becomes less than positive 1%, i.e., ( r(n) < 0.01 ) — which first happens at ( n = 8 ).", "But let’s check when does ( 1 + r(n) < 1.01 )? That would mean growth <1.01%, which is always true after day 15 or so. But the drop below 1% growth rate happens earlier.", "The key phrase: “the daily increase drops below 1%” — “increase” implies positive, so we interpret as: when does the daily multiplicative growth rate fall below 1% — i.e., ( r(n) < 0.01 )", "But from domain, this happens at ( n = 8 ). However, let's check when the multiplicative factor first becomes less than ( 1.01 )? That would be different.", "But the natural interpretation: the growth rate (as a decimal) drops below 0.01, i.e., ( r(n) < 0.01 )", "We solved:", "[\n0.7 + 0.15(3 - n) < 0.01 \Rightarrow n > 7.6 \Rightarrow n = 8\n]", "But let’s compute ( r(n) ) day by day from ( n=1 ):", "$$\n\begin{align}\nn = 1: &\quad 0.7 + 0.15(2) = 0.7 + 0.3 = 1.00 \quad (\ ext{0% growth}) \\nn = 2: &\quad 0.7 + 0.15(1) = 0.85 \quad (\ ext{–15%}) \\nn = 3: &\quad 0.7 + 0 = 0.70 \\nn = 4: &\quad 0.7 - 0.15 = 0.55 \\nn = 5: &\quad 0.70 - 0.15 = 0.55? Wait — ( 0.15(3 - 5) = 0.15(-2) = -0.3 ), so ( 0.7 - 0.3 = 0.40 )", "Wait — correction:", "[\nr(n) = 0.7 + 0.15(3 - n)\n]", "For ( n = 1 ): ( 0.7 + 0.15(2) = 0.7 + 0.3 = 1.00 )\n( n = 2 ): ( 0.7 + 0.15(1) = 0.85 )\n( n = 3 ): ( 0.7 + 0 = 0.70 )\n( n = 4 ): ( 0.7 - 0.15 = 0.55 )\n( n = 5 ): ( 0.7 - 0.30 = 0.40 )\n( n = 6 ): ( 0.7 - 0.45 = 0.25 )\n( n = 7 ): ( 0.7 - 0.60 = 0.10 )\n( n = 8 ): ( 0.7 - 0.75 = -0.05 ) → now < 1% in growth rate?", "Yes — from 1.00 down to 0.70, then 0.55, 0.40, 0.25, 0.10, -5%, ... — so the growth factor drops below 1% (i.e., less than 1) starting at Day 15?", "Wait — “drop below 1%” means less than 0.01, not 0.01 or above.", "So solve:", "[\n1 + r(n) < 1.01 \quad \ ext{is not required} \\n\ ext{We want } r(n) < 0.01\n]", "[\n0.7 + 0.15(3 - n) < 0.01\n]\n[\n0.15(3 - n) < -0.69\n]\n[\n3 - n < -\frac{0.69}{0.15} = -4.6\n]\n[\n-n < -5.6 \Rightarrow n > 5.6\n]", "So at ( n = 6 ), ( r(6) = 0.25 ), ( n=7: 0.10 ), ( n=8: -0.05 )", "- At ( n = 7 ): ( r(n) = 0.10 ) → 10% growth (still >1%)\n- At ( n = 8 ): ( r(n) = -0.05 ) → 5% decrease", "So the growth rate first drops below 1% in magnitude from crossing 1% downward? But it passes 1% downward on Day 7, still above. Then falls below 1% (in absolute decline) continuing down.", "But the key: “the daily increase” means positive growth. So the rate drops below 1% only after it becomes less than 1% and stays negative.", "But from calculation:", "- Growth rate is positive until n ≤ 3, then drops below 1% by n = 4: ( r(4) = 0.55 ) — which is 55%, still no.", "Wait — 1% = 0.01, so *any rate <"]