todai-math

Papers (28)
2025
problems1-3 3 todai-engineering-math 6
2024
problem1 1 problem2 1 problem3 1 todai-engineering-math 6
2023
problem1 1 problem2 1 problem3 1 todai-engineering-math 6
2022
problem1 1 problem2 1 problem3 1 todai-engineering-math__paper1 2 todai-engineering-math__paper2 2 todai-engineering-math__paper3 7
2021
todai-engineering-math__paper1 6 todai-engineering-math__paper2 3 todai-engineering-math__paper3 3
2020
todai-engineering-math 6
2019
todai-engineering-math 6
2018
ist 3 todai-engineering-math 6
2017
ist 3 todai-engineering-math 6
2016
todai-engineering-math 6 translated 3
2015
ist 3
2022 todai-engineering-math__paper3

7 maths questions

QI.1 Complex numbers 2 Contour Integration and Residue Calculus View
In Questions I.1 and I.2, $z$ denotes a complex number, $i$ the imaginary unit, and $|z|$ the absolute value of $z$.
Calculate the following integral, where $C$ is the closed path on the complex plane as shown in Figure 3.1.
$$I_1 = \oint_C \frac{z}{(z-i)(z-1)} \mathrm{d}z$$
(The contour $C$ is a closed path on the complex plane as depicted in Figure 3.1.)
QI.2 Complex numbers 2 Contour Integration and Residue Calculus View
In Questions I.1 and I.2, $z$ denotes a complex number, $i$ the imaginary unit, and $|z|$ the absolute value of $z$.
Consider the definite integral $I_2$ expressed as
$$I_2 = \int_0^{2\pi} \frac{d\theta}{10 + 8\cos\theta}$$
2.1. Find a complex function $G(z)$ when $I_2$ is rewritten as an integral of a complex function as $$I_2 = \oint_{|z|=1} G(z) \mathrm{d}z$$ Note that the integration path is a unit circle centered at the origin on the complex plane oriented counterclockwise. Show the derivation process with your answer.
2.2. Find all singularities of $G(z)$.
2.3. Using the residue theorem, obtain $I_2$.
QII.1 Binomial Distribution Compute Exact Binomial Probability View
Consider a situation where products are produced sequentially. The events producing defective products are independent and identically distributed, and a defective product is produced with a probability of $\phi$ $(0 \leq \phi \leq 1)$. We consider the changes of the probability distributions before and after observing production results. In the following questions, $N (\geq 1)$ denotes the number of products observed.
By defining $v_i = 1$ if the $i$-th product is a defective product, and $v_i = 0$ if it is not defective, we get a series $\boldsymbol{v} = (v_1, \cdots, v_N)$, where the values can be 0 or 1. Let $N_d(\boldsymbol{v})$ be the number of observations with value of 1 in $\boldsymbol{v}$, obtain the occurrence probability of $\boldsymbol{v}$.
QII.2 Continuous Probability Distributions and Random Variables Distribution of Transformed or Combined Random Variables View
Consider a situation where products are produced sequentially. The events producing defective products are independent and identically distributed, and a defective product is produced with a probability of $\phi$ $(0 \leq \phi \leq 1)$. Suppose that the probability of producing a defective product follows the Beta distribution
$$\operatorname{Beta}_{\mathrm{a},\mathrm{b}}(x) = \frac{1}{\mathrm{B}(\mathrm{a},\mathrm{b})} x^{\mathrm{a}-1}(1-x)^{\mathrm{b}-1} \quad (0 \leq x \leq 1),$$
for real numbers $\mathrm{a}(>1)$ and $\mathrm{b}(>1)$. Note that the Beta function $\mathrm{B}(\mathrm{a},\mathrm{b})$ is defined as
$$\mathrm{B}(\mathrm{a},\mathrm{b}) = \int_0^1 t^{\mathrm{a}-1}(1-t)^{\mathrm{b}-1} \mathrm{d}t$$
In the Bayesian estimation, the parameter $\theta$ (in this case, $\phi$) that determines the probability is treated as the random variable and we assume that its distribution is described by $\pi(\theta)$. We calculate $\pi(\theta \mid A)$ by
$$\pi(\theta \mid A) = \frac{\pi(\theta) P(A \mid \theta)}{P(A)}$$
where $\pi(\theta \mid A)$ is the posterior probability, $P(A \mid \theta)$ is the conditional occurrence probability that the event $A$ is observed under $\theta$, and $\pi(\theta)$ is the prior probability.
We assume that $\phi$, the probability of producing a defective product, follows the prior probability $\operatorname{Beta}_{\mathrm{a},\mathrm{b}}(\phi)$. Let $Q(\boldsymbol{v} \mid \phi)$ be the conditional occurrence probability of $\boldsymbol{v}$ under $\phi$ and $Q_{\mathrm{a},\mathrm{b}}(\boldsymbol{v})$ be the occurrence probability of $\boldsymbol{v}$. Obtain the posterior probability after $\boldsymbol{v}$ occurs.
QII.3 Continuous Probability Distributions and Random Variables Distribution of Transformed or Combined Random Variables View
Consider a situation where products are produced sequentially. The events producing defective products are independent and identically distributed, and a defective product is produced with a probability of $\phi$ $(0 \leq \phi \leq 1)$. Suppose that the probability of producing a defective product follows the Beta distribution
$$\operatorname{Beta}_{\mathrm{a},\mathrm{b}}(x) = \frac{1}{\mathrm{B}(\mathrm{a},\mathrm{b})} x^{\mathrm{a}-1}(1-x)^{\mathrm{b}-1} \quad (0 \leq x \leq 1),$$
for real numbers $\mathrm{a}(>1)$ and $\mathrm{b}(>1)$. Note that the Beta function $\mathrm{B}(\mathrm{a},\mathrm{b})$ is defined as
$$\mathrm{B}(\mathrm{a},\mathrm{b}) = \int_0^1 t^{\mathrm{a}-1}(1-t)^{\mathrm{b}-1} \mathrm{d}t$$
By defining $v_i = 1$ if the $i$-th product is a defective product, and $v_i = 0$ if it is not defective, we get a series $\boldsymbol{v} = (v_1, \cdots, v_N)$, where the values can be 0 or 1. Let $N_d(\boldsymbol{v})$ be the number of observations with value of 1 in $\boldsymbol{v}$.
Suppose that $Q(\boldsymbol{v} \mid \phi)$ in Question II.2 is the occurrence probability obtained in Question II.1 and let $a = 2,\ b = 50$, obtain $Q_{2,50}(\boldsymbol{v})$.
QII.4 Continuous Probability Distributions and Random Variables Distribution of Transformed or Combined Random Variables View
Consider a situation where products are produced sequentially. The events producing defective products are independent and identically distributed, and a defective product is produced with a probability of $\phi$ $(0 \leq \phi \leq 1)$. Suppose that the probability of producing a defective product follows the Beta distribution
$$\operatorname{Beta}_{\mathrm{a},\mathrm{b}}(x) = \frac{1}{\mathrm{B}(\mathrm{a},\mathrm{b})} x^{\mathrm{a}-1}(1-x)^{\mathrm{b}-1} \quad (0 \leq x \leq 1),$$
for real numbers $\mathrm{a}(>1)$ and $\mathrm{b}(>1)$. Note that the Beta function $\mathrm{B}(\mathrm{a},\mathrm{b})$ is defined as
$$\mathrm{B}(\mathrm{a},\mathrm{b}) = \int_0^1 t^{\mathrm{a}-1}(1-t)^{\mathrm{b}-1} \mathrm{d}t$$
In Question II.3, with $a=2,\ b=50$, show that the posterior probability becomes the Beta distribution $\operatorname{Beta}_{\mathrm{a}^{\prime},\mathrm{b}^{\prime}}(\phi)$, and obtain $\mathrm{a}^{\prime}$ and $\mathrm{b}^{\prime}$.
QII.5 Continuous Probability Distributions and Random Variables Distribution of Transformed or Combined Random Variables View
Consider a situation where products are produced sequentially. The events producing defective products are independent and identically distributed, and a defective product is produced with a probability of $\phi$ $(0 \leq \phi \leq 1)$. Suppose that the probability of producing a defective product follows the Beta distribution
$$\operatorname{Beta}_{\mathrm{a},\mathrm{b}}(x) = \frac{1}{\mathrm{B}(\mathrm{a},\mathrm{b})} x^{\mathrm{a}-1}(1-x)^{\mathrm{b}-1} \quad (0 \leq x \leq 1),$$
for real numbers $\mathrm{a}(>1)$ and $\mathrm{b}(>1)$. Note that the Beta function $\mathrm{B}(\mathrm{a},\mathrm{b})$ is defined as
$$\mathrm{B}(\mathrm{a},\mathrm{b}) = \int_0^1 t^{\mathrm{a}-1}(1-t)^{\mathrm{b}-1} \mathrm{d}t$$
In Question II.4, where the posterior probability is the Beta distribution $\operatorname{Beta}_{\mathrm{a}^{\prime},\mathrm{b}^{\prime}}(\phi)$ with $a=2,\ b=50$, obtain $\phi$ that gives the maximum likelihood estimate (that maximizes the posterior probability).