質問編集履歴

1

追記

2020/08/13 03:56

投稿

melonattacker
melonattacker

スコア13

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  print("u_t:", u_t) # ここが①のu_tと等しければ正解
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  ```
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+ # 追記
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+ `scipy.optimize.newton`を用いて、解くことができました。
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+ ```Python
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+ import numpy as np
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+ import math
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+ from scipy import integrate, optimize
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+ beta = 0.3
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+ chi = 1.0
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+ productivity = 100.0
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+ interest_rate = 0.05
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+ price_mu = 0.01
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+ utility_mu = 0.03
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+ utility_sigma = 0.3
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+ theta: float = utility_sigma / math.sqrt(2 * utility_mu)
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+ y = lambda u: math.sqrt(1 / (2 * math.pi * theta ** 2)) * math.e ** (- u ** 2 / (2 * theta ** 2))
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+ def N_t(u_t):
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+ iy, err = integrate.quad(y, -np.inf, u_t)
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+ userbase = 1 - iy
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+ return userbase - math.exp(-(u_t) + math.log(chi / (productivity * beta)) - ((1 - beta) / beta) * math.log((1 - beta) / (interest_rate - price_mu)))
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+ u_t = optimize.newton(N_t, 0)
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+ iy, err = integrate.quad(y, -np.inf, u_t)
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+ n_t: float = 1 - iy
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+ print(u_t)
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+ print(n_t)
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+ ```