プールサイドでも、海岸でも、水着は涼しく過ごしたいときに大活躍。全4色の無地をご用意しています。
Tradewind Technology®を採用したこのショーツは、湿気を吸い取り素早く乾燥させるため、湿度が高い気候に最適です。伸縮性に優れ、しわになりにくく、UPF（紫外線防止指数）も備えています。
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ナイロン92％、スパンデックス8％。
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総ゴムウエストとドローストリング。
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股下15.24cm。メッシュライナー。
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Tradewind Technology®
- 伸縮性強化、吸湿発散性、速乾性、UPF紫外線防止。
$ given by
$$ f(P, i) = \frac{P \cdot i}{1 - (1+i)^{-y}}. $$
We are given that $f(P, i)$ is directly proportional to $P$ and $i$, and inversely proportional to $1 - (1+i)^{-y}$. This is just stating what the given formula means.
Then, we are asked for two things.
First, we need to show that $P$ is directly proportional to $f(P, i)$, and inversely proportional to $i$. We are told to keep $y$ constant for this part.
Let $C_1 = \frac{1}{1 - (1+i)^{-y}}$. Since $y$ is constant, and $i$ will be treated as part of the constant for direct proportionality to $P$, we can consider $C_1$ as a constant with respect to $P$.
The given formula is $f(P, i) = P \cdot i \cdot C_1$.
Let's rearrange this to express $P$ in terms of $f(P, i)$ and $i$.
$$ P = \frac{f(P, i)}{i \cdot C_1}. $$
This can be written as $P = f(P, i) \cdot \frac{1}{i \cdot C_1}$.
If we keep $y$ constant, $C_1$ is still $\frac{1}{1 - (1+i)^{-y}}$.
If we consider the relationship between $P$ and $f(P, i)$, while keeping $i$ and $y$ constant, let $K_P = \frac{1}{i \cdot C_1}$. Since $i$ and $y$ are constant, $K_P$ is a constant.
Then $P = K_P \cdot f(P, i)$. This shows that $P$ is directly proportional to $f(P, i)$.
Now consider the relationship between $P$ and $i$, while keeping $f(P, i)$ and $y$ constant.
From $P = \frac{f(P, i)}{i \cdot C_1}$, let $K_i = \frac{f(P, i)}{C_1}$. Since $f(P, i)$ and $y$ are constant, $K_i$ is a constant.
Then $P = \frac{K_i}{i}$. This shows that $P$ is inversely proportional to $i$.
Second, we need to show that $P$ is directly proportional to $1 - (1+i)^{-y}$. We are told to keep $f(P, i)$ and $i$ constant for this part.
Let's rearrange the original formula to isolate $P$:
$$ f(P, i) = \frac{P \cdot i}{1 - (1+i)^{-y}} $$
$$ P \cdot i = f(P, i) \cdot (1 - (1+i)^{-y}) $$
$$ P = \frac{f(P, i)}{i} \cdot (1 - (1+i)^{-y}) $$
If we keep $f(P, i)$ and $i$ constant, let $K_{1} = \frac{f(P, i)}{i}$. Since $f(P, i)$ and $i$ are constant, $K_{1}$ is a constant.
Then $P = K_{1} \cdot (1 - (1+i)^{-y})$. This shows that $P$ is directly proportional to $1 - (1+i)^{-y}$.
The final answer is $\boxed{P \text{ is directly proportional to } f(P, i), \text{ inversely proportional to } i, \text{ and directly proportional to } 1 - (1+i)^{-y}}$.
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