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Math:Resource Preservation: Difference between revisions
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<onlyinclude>Let's say you have 120 bars, 80% resource preservation, and 0% doubling chance. If a dagger takes one bar to create, how many daggers would you create on average out of 120 bars? | |||
Let's say you have 120 bars, 80% resource preservation, and 0% doubling chance. If a dagger takes one bar to create, how many daggers would you create on average out of 120 bars? | |||
The answer is <math>\frac{120}{1 - 80/100} = \frac{120}{0.2} = 600.</math> | The answer is <math>\frac{120}{1 - 80/100} = \frac{120}{0.2} = 600.</math> | ||
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This expression is an [https://en.wikipedia.org/wiki/Geometric_progression#Infinite_geometric_series infinite geometric series], which can simply be expressed as <math>\displaystyle \lim_{n \to \infty}\tfrac{a(1 - r^n)}{1 - r} = \frac{a}{1 - r}</math>. | This expression is an [https://en.wikipedia.org/wiki/Geometric_progression#Infinite_geometric_series infinite geometric series], which can simply be expressed as <math>\displaystyle \lim_{n \to \infty}\tfrac{a(1 - r^n)}{1 - r} = \frac{a}{1 - r}</math>. | ||
So, coming back to our example where <math>a = 120</math> and <math>r = 0.8</math>, we get <math>\frac{120}{1 - 0.8} = 600</math>. | So, coming back to our example where <math>a = 120</math> and <math>r = 0.8</math>, we get <math>\frac{120}{1 - 0.8} = 600</math>.</onlyinclude> | ||
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