質問編集履歴
1
追記
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File without changes
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@@ -40,4 +40,38 @@
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print("n_t:", n_t)
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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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51
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from scipy import integrate, optimize
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52
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53
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beta = 0.3
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54
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chi = 1.0
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55
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productivity = 100.0
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56
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interest_rate = 0.05
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57
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price_mu = 0.01
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utility_mu = 0.03
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59
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utility_sigma = 0.3
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60
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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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```
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