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京都大学 情報学研究科 通信情報システム専攻 2024年2月実施 専門基礎A [A-1]

Author

祭音Myyura

Description

(1)

Find the limit

limx0sinhxx\lim_{x \to 0} \frac{\sinh x}{x}

(2)

Evaluate the integral

D={(x,y)0x+y1,0xy1}D = \{(x,y) \mid 0 \leq x+y \leq 1, 0 \leq x-y \leq 1\}
I=Dx2 dxdyI = \iint_D x^2 \ dxdy

(3)

Given the matrix:

A=(ab0d)A = \begin{pmatrix} a & b \\ 0 & d \end{pmatrix}

(i) Find the eigenvalues and corresponding eigenvectors of AA, assuming it has distinct eigenvalues.

(ii) Using the eigenvalues and eigenvectors from (i), find AnA^n, where nn is a positive integer.

(iii) If the eigenvalues are the same, find AnA^n for a positive integer nn.

Kai

(1)

We know the Maclaurin series for sinhx\sinh x is:

sinhx=x+x36+O(x5)\sinh x = x + \frac{x^3}{6} + O(x^5)

Thus,

sinhxx=1+x26+O(x4)\frac{\sinh x}{x} = 1 + \frac{x^2}{6} + O(x^4)

As x0x \to 0, the limit is:

limx0sinhxx=1\lim_{x \to 0} \frac{\sinh x}{x} = 1

(2)

We express xx and yy in terms of uu and vv: x=u+v2x=\frac{u+v}{2}, y=uv2y=\frac{u-v}{2}. The integrand x2x^{2} becomes (u+v2)2(\frac{u+v}{2})^{2}. The Jacobian of the transformation is calculated as J=12|J|=\frac{1}{2}.

The integral in the uvuv-coordinate system is given by:

Dx2dA=0101(u+v)2412dudv=0101(u+v)28dudv\iint _{D}x^{2}\,dA=\int _{0}^{1}\int _{0}^{1}\frac{(u+v)^{2}}{4}\cdot \frac{1}{2}\,du\,dv=\int _{0}^{1}\int _{0}^{1}\frac{(u+v)^{2}}{8}\,du\,dv

We integrate with respect to uu first, then vv:

0118[(u+v)33]01dv=01124[(1+v)3v3]dv=011+3v+3v224dv\int _{0}^{1}\frac{1}{8}\left[\frac{(u+v)^{3}}{3}\right]_{0}^{1}\,dv=\int _{0}^{1}\frac{1}{24}[(1+v)^{3}-v^{3}]\,dv=\int _{0}^{1}\frac{1+3v+3v^{2}}{24}\,dv

Evaluating the final integral:

124[v+3v22+v3]01=124(1+32+1)=124(72)=748\frac{1}{24}\left[v+\frac{3v^{2}}{2}+v^{3}\right]_{0}^{1}=\frac{1}{24}\left(1+\frac{3}{2}+1\right)=\frac{1}{24}\left(\frac{7}{2}\right)=\frac{7}{48}

(3)

(i)

The characteristic equation is:

det(AλI)=0\text{det}(A - \lambda I) = 0
(aλb0dλ)\begin{pmatrix} a - \lambda & b \\ 0 & d - \lambda \end{pmatrix}

The determinant is:

(aλ)(dλ)=0(a - \lambda)(d - \lambda) = 0

Thus, the eigenvalues are λ1=a\lambda_1 = a and λ2=d\lambda_2 = d.

The corresponding eigenvectors are:

  • For λ1=a\lambda_1 = a, the eigenvector is (10)\begin{pmatrix} 1 \\ 0 \end{pmatrix}.
  • For λ2=d\lambda_2 = d, the eigenvector is (bda1)\begin{pmatrix} \frac{b}{d-a} \\ 1 \end{pmatrix}.

(ii)

If AA has distinct eigenvalues, it is diagonalizable. We can write A=PDP1A=PDP^{-1}, where DD is the diagonal matrix of eigenvalues and PP is the matrix whose columns are the corresponding eigenvectors:

P=(1bda01),D=(a00d)P=\left(\begin{matrix}1&\frac{b}{d-a}\\ 0&1\end{matrix}\right),\quad D=\left(\begin{matrix}a&0\\ 0&d\end{matrix}\right)

The inverse of PP is P1=(1bda01)P^{-1}=\left(\begin{matrix}1&-\frac{b}{d-a}\\ 0&1\end{matrix}\right).Then An=PDnP1A^{n}=PD^{n}P^{-1}:

An=(1bda01)(an00dn)(1bda01)A^{n}=\left(\begin{matrix}1&\frac{b}{d-a}\\ 0&1\end{matrix}\right)\left(\begin{matrix}a^{n}&0\\ 0&d^{n}\end{matrix}\right)\left(\begin{matrix}1&-\frac{b}{d-a}\\ 0&1\end{matrix}\right)
An=(anbdnda0dn)(1bda01)=(anbanda+bdnda0dn)A^{n}=\left(\begin{matrix}a^{n}&\frac{bd^{n}}{d-a}\\ 0&d^{n}\end{matrix}\right)\left(\begin{matrix}1&-\frac{b}{d-a}\\ 0&1\end{matrix}\right)=\left(\begin{matrix}a^{n}&-\frac{ba^{n}}{d-a}+\frac{bd^{n}}{d-a}\\ 0&d^{n}\end{matrix}\right)
An=(anbdnanda0dn)A^{n}=\left(\begin{matrix}a^{n}&b\frac{d^{n}-a^{n}}{d-a}\\ 0&d^{n}\end{matrix}\right)

(iii)

If a=da = d, then AA is not diagonalizable, and we use the Jordan form to find AnA^n:

An=(ab0a)n=(annban10an)A^n = \begin{pmatrix} a & b \\ 0 & a \end{pmatrix}^n = \begin{pmatrix} a^n & nba^{n-1} \\ 0 & a^n \end{pmatrix}