Here $J_4=(X^2+1)$ is a principal ideal. The evaluation homomorphism $p\mapsto p(i)$ from $\mathbb{R}[X]$ onto $\mathbb{C}$ has this kernel of a ring homomorphism: division by $X^2+1$ leaves a remainder $a+bX$, and $a+bi=0$ with real $a,b$ forces both to vanish. Surjectivity follows because every complex number is $a+bi$. ConsequentlyEquivalently, $X^2+1$ is an irreducible polynomial over the real numbers.

The conditions set the constant and linear coefficients to zero, so $\boxed{J_3=(X^2)}$. This is a principal ideal, but it is not maximal: there is a strict chain $(X^2)⊊(X)⊊\mathbb{R}[X]$. In the quotient ring, the nonzero class of $X$ has square zero, so the quotient is not a field.

It is not an ideal. The constant polynomial one belongs to $J_2$, but its product with $X$ does not: the derivative of $X$ at zero is one. Thus $J_2$ fails closure under multiplication by arbitrary elements of $\mathbb{R}[X]$, a required ideal property. Closure under addition alone would not suffice.

The evaluation homomorphism $\mathbb{R}[X]\to \mathbb{R}$, $p\mapsto p(0)$, is surjective and has kernel of a ring homomorphism $J_1=(X)$. Hence $J_1$ is an ideal, and the first isomorphism theorem for rings gives $\boxed{\mathbb{R}[X]/J_1\cong \mathbb{R}}$. Since this quotient is a field, $\boxed{J_1\text{ is maximal}}$.

Yes. In ${\mathbb{F}}_7$, the minimal polynomial splits asThe three factors are distinct, so the minimal polynomial criterion for diagonalizability applies. For example the eigenvectors $v_{\lambda }=(1,{\lambda }^2,\lambda {)}^{\mathsf{T}}$ for $\lambda =1,2,4$ give an eigenbasis over this finite field. The contrast with characteristic three is that the shift polynomial now splits without repeated roots.

No. In characteristic three, $X^3-1=(X-1{)}^3$. This is still the minimal polynomial, because the three cyclic basis vectors remain independent. Its repeated factor violates the minimal polynomial criterion for diagonalizability. Equivalently, the only possible eigenvalue is one; a diagonalizable matrix with that sole eigenvalue would be the identity matrix, whereas this matrix is not the identity.

No. The same cyclic-vector argument gives minimal polynomial $X^3-1=(X-1)(X^2+X+1)$ over the finite field ${\mathbb{F}}_2$. The quadratic has no root in ${\mathbb{F}}_2$: its value is one at both zero and one. Thus the minimal polynomial does not split into linear factors over the specified field, violating the minimal polynomial criterion for diagonalizability. It becomes diagonalizable over a suitable larger finite field, but that is a different question.

Yes. The matrix is a cyclic shift operator on three coordinates. Since $e_1,A{e}_1,A^2{e}_1$ are the three standard basis vectors, no polynomial of degree below three annihilates $A$. Also $A^3=I$, so its minimal polynomial is $X^3-1$. Over the complex numbers this has the distinct roots $1,\omega ,{\omega }^2$, where $\omega =e^{2\pi i/3}$. The minimal polynomial criterion for diagonalizability therefore proves diagonalizability.

Explicitly, $v_{\lambda }=(1,{\lambda }^2,\lambda {)}^{\mathsf{T}}$ satisfies $A{v}_{\lambda }=\lambda v_{\lambda }$ when ${\lambda }^3=1$. These three eigenvectors form an eigenbasis. This is a special case of diagonalization of a cyclic shift over a field.

Use the Cayley transformFor $\mathrm{Im}z>0$, its denominator is nonzero, and direct subtraction gives $1-|w{|}^2=4\mathrm{Im}z/|z+i{|}^2>0$. Thus the image lies in the open unit disk. Conversely, for $|w|<1$ the inverse is $z=i(1+w)/(1-w)$, with $\mathrm{Im}z=(1-|w{|}^2)/|1-w{|}^2>0$. Hence it maps the upper half-plane bijectively onto the disk.

A Möbius transformation has the form $M(z)=(az+b)/(cz+d)$ with $ad-bc\ne 0$, extended to the Riemann sphere by its limiting values at its pole and at infinity. If $c\ne 0$, it sends $-d/c$ to infinity and infinity to $a/c$; if $c=0$, infinity maps to infinity.

Represent $M$ by the invertible matrix $\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$, with nonzero scalar multiples representing the same map. Substitution shows that composition corresponds to matrix multiplication. The product determinant is the product of two nonzero determinants, so the composition is again a Möbius transformation. Solving $w=M(z)$ giveswith the same extended-value convention. This inverse proves that $M$ is a bijection of the extended complex plane.

If a holomorphic function $g$ on the punctured complex plane satisfied $e^{g(z)}=z$, the chain rule would give $e^{g(z)}{g}^{\prime }(z)=1$, hence $g^{\prime }(z)=1/z$. This contradicts the previous part. Therefore there is no global holomorphic logarithm on the punctured plane. A branch of the complex logarithm requires a suitable smaller domain.

Take the unit circle $\gamma (t)=e^{it}$, $0\le t\le 2\pi$, in the punctured complex plane. Direct parameter integration givesA global holomorphic antiderivative would make this integral zero by the previous part. Thus there is no such antiderivative on the punctured plane. Local primitives can exist without combining into one single-valued function on this domain.

The chain rule gives $dF(\gamma (t))/dt=F^{\prime }(\gamma (t)){\gamma }^{\prime }(t)=f(\gamma (t)){\gamma }^{\prime }(t)$. The fundamental theorem of calculus therefore yieldsbecause the closed curve has equal endpoints. The existence of a single-valued holomorphic antiderivative on the whole neighborhood is the decisive hypothesis.

For a smooth curve $\gamma :[a,b]\to \mathbb{C}$ and continuous $f$ on its image, the contour integral is defined byThe derivative carries the direction and speed of traversal; the complex-valued real-parameter integral is defined by integrating its real and imaginary parts.

First solve for $v=y^{\prime }$. Multiplication by $e^x$ and use of $v(1)=0$ giveIntegrating from zero to $x$, and using $y(0)=0$, yields $y(x)=-{\int }_0^x{e}^{-s}{\int }_s^1{e}^{t}f(t)dtds$. Reversing the order of integration givesThus $G(x,t)$ equals $G_1$ for $t<x$ and $G_2$ for $x<t$. It is continuous at $x=t$, vanishes at $x=0$, and has $G_x(1,t)=0$. Its derivative jumps from $-1$ to zero across $x=t$, so the jump is $+1$, matching the coefficient of $y^{\prime \prime }$. Both pieces satisfy $G_{xx}+G_x=0$ away from the jump. These facts also verify the Green representation directly. The homogeneous solutions are $A+B{e}^{-x}$, and the two boundary conditions force $A=B=0$, proving uniqueness.

For $f(t)=e^{-t}$, the formula givesHence $y(x)=-x{e}^{-x}$. Indeed $y^{\prime }=(x-1)e^{-x}$ and $y^{\prime \prime }=(2-x)e^{-x}$, which verify the equation and both boundary conditions.

The first iterate isOn the prescribed box, $F(x,y)=1+x^2+y^2$ is continuous, $|F|\le 102$, and $|F(x,y)-F(x,v)|\le 20|y-v|$. The initial value is distance nine from the nearer horizontal edge. Thus on $|x|\le h=9/102=3/34$, the Picard map keeps functions within the box: its values differ from one by at most $102h=9$.

Successive differences satisfy, by induction,The Weierstrass M-test proves uniform convergence on this local interval. The uniform limit is continuous, and the Lipschitz estimate allows passage to the limit inside the integral, giving the displayed integral equation. Differentiation gives $y^{\prime }=1+x^2+y^2$, $y(0)=1$. The same local Lipschitz bound proves local uniqueness.

Direct differentiation gives $u^{\prime }={\sec }^2(x+\pi /4)=1+u^2$, $u(0)=1$, and $u\to +\infty$ as $x\uparrow \pi /4$. On the common positive existence interval put $w=y-u$. It satisfies $w^{\prime }=(y+u)w+x^2$, $w(0)=0$, so the integrating-factor formula givesThus $y\ge u$. Let $x_{\ast }>0$ be the maximal forward existence endpoint. The solution is strictly increasing because $y^{\prime }\ge 1$. If $x_{\ast }$ were finite and $y$ remained bounded, its finite limit and the smooth right side would allow local continuation, contradicting maximality. Comparison with $u$ prevents existence through $\pi /4$. ConsequentlyThere is no contradiction with Picard's theorem, which gives local existence and uniqueness. Its box estimate gives only a short guaranteed interval, and the Lipschitz constant for $y^2$ is not globally bounded in $y$.

After division by $z(z+2)$, the coefficients areTheir only finite singularities are $z=0$ and $z=-2$. At either point $z_0$, the products $(z-z_0)p$ and $(z-z_0{)}^{2}q$ are analytic, so both points are regular singular points. All other finite points are ordinary.

The lowest power after substituting the Frobenius series gives $\sigma (2\sigma -1)=0$. The coefficient of $z^{n+\sigma }$ then givesFor both exponents the denominator is nonzero for every $n\ge 0$, so each arbitrary nonzero $a_0$ determines a series. Unless it terminates, its coefficient ratio tends to $-1/2$, proving convergence for $|z|<2$. For the half-integer exponent choose a square-root branch. Their leading powers $1$ and $z^{1/2}$ show that the two local solutions are linearly independent.

If $\alpha =m$ is a positive integer, choose $\sigma =0$. The recurrence numerator is nonzero for $0\le n<m$ and vanishes at $n=m$. Thus $a_m\ne 0$, $a_{m+1}=0$, and all later coefficients vanish: a polynomial solution of exact degree $m$ exists. It is proportional to the Chebyshev polynomial $T_m(z+1)$, since the substitution $t=z+1$ transforms the equation into $(t^2-1)y^{\prime \prime }+t{y}^{\prime }-m^{2}y=0$.