Chiu Chang Suann Jing/Examples/Example 10

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Example of Problem from Chiu Chang Suann Jing

Of $2$ water weeds, one grows $3$ feet and one grows $1$ foot on the first day.
The growth of the first becomes every day half of that of the preceding day
while the other grows twice as much as the previous day.
In how many days will the two grow to equal heights?


Solution

$2 \frac 6 {13}$ days, at which time they will grow to $4 \frac {11} {13}$ feet.


Proof

The growth of the $2$ plants is governed by a geometric sequence.

Let the height of the two plants after the first day be $a_1$ and $a_2$ respectively.

Let the common ratio of the growth rates of the two plants be $r_1$ and $r_2$ respectively.

Let $d$ be the number of days after which they reach the same height.

We have:

\(\ds \frac {a_1 \paren {r_1^d - 1} } {r_1 - 1}\) \(=\) \(\ds \frac {a_2 \paren {r_2^d - 1} } {r_2 - 1}\) Sum of Geometric Sequence: Corollary 1
\(\ds \leadsto \ \ \) \(\ds \frac {3 \paren {1 - \frac 1 {2^d} } } {1 - \frac 1 2}\) \(=\) \(\ds \frac {2^d - 1} {2 - 1}\) plugging in the numbers
\(\ds \leadsto \ \ \) \(\ds 6 \paren {1 - \frac 1 {2^d} }\) \(=\) \(\ds 2^d - 1\) simplification
\(\ds \leadsto \ \ \) \(\ds 2^{2 d} - 7 \times 2^d + 6\) \(=\) \(\ds 0\) rearrangement, and multiplying by $2^d$
\(\ds \leadsto \ \ \) \(\ds \paren {2^d - 1} \paren {2^d - 6}\) \(=\) \(\ds 0\) factorising

Thus we have that $2^d = 1$ or $2^d = 6$.

Hence either $d = 0$, which does not work, or $d = \log_2 6$.

This gives $d \approx 2.585$, and a common height of $5$ feet.


There is believed to be a mistake here, possibly a typo.
In particular: The mechanism arrived at in Chiu Chang Suann Jing must be different from that used here. I suspect that a linear growth rate is assumed throughout the day rather than the continuous exponential one.
You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by reviewing it, and either correcting it or adding some explanatory material as to why you believe it is actually correct after all.
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Sources

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