Substitute the exponential definitions of hyperbolic cosine and hyperbolic sine. The cross terms cancel:This proves the identity for real arguments, and the same algebra proves it for complex arguments.

Using the principal complex square root, $\sqrt{i}=e^{i\pi /4}$. Its two square roots differ by a sign, givingEquivalently their arguments are $\pi /8$ and $9\pi /8$ modulo $2\pi$. If instead the right-hand square root is allowed to take both values, the union has four elements $e^{i(\pi /8+k\pi /2)}$, $k=0,1,2,3$, satisfying $z^4=i$. Only the two boxed values square to the principal right-hand side.

The values of the complex logarithm at $-1$ are $i(2m+1)\pi$, $m\in \mathbb{Z}$. For any such value and any integer $n$,This is an integer multiple of $\pi$, so its sine is zero. In particular the result holds with the principal value $\log (-1)=i\pi$. The argument does not require treating the multivalued logarithm as a single globally defined function.

Expanding the determinant gives $\det (\lambda I-M)=(\lambda -1)[(\lambda -1{)}^2-25]$. Solving each eigenvalue equation yieldsTheir pairwise inner products are $16-25+9=0$, $12-12=0$ and $12-12=0$. Their squared Euclidean norms are $50,50,25$, respectively, so division by $\sqrt{50},\sqrt{50},5$ gives an orthonormal basis. This also follows from the spectral theorem for real symmetric matrices: symmetry makes eigenvectors for distinct eigenvalues orthogonal.

The characteristic polynomial isThus the eigenvalues and corresponding eigenvectors over $\mathbb{C}$ areAll nonzero scalar multiples of these vectors are also eigenvectors. Distinct eigenvalues make them linearly independent. Put them in the columns ofThe equations $A{v}_{\pm }={\lambda }_{\pm }{v}_{\pm }$ are exactly $AB=BC$, giving the diagonalizationThere is no diagonalization over $\mathbb{R}$, since the matrix has no real eigenvalues.

First take $a>0$ and $|x|<a$, and put $u=\arcsin (x/a)$, $q=\sqrt{a^2-x^2}$. Then $u^{\prime }=1/q$. The chain rule givesConsequentlywhich proves the identity on the open domain of differentiability. The squared inverse sine is unchanged when $a$ is replaced by $-a$, so the identity also holds for negative $a$ with $|x|<|a|$. The value $a=0$ is excluded.

For a series with eventually nonzero terms $a_n$, suppose $r={\lim }_{n\to \infty }|a_{n+1}/a_n|$ exists. The ratio test says that $r<1$ implies absolute convergence, while $r>1$ implies divergence because the terms fail to tend to zero. At $r=1$ the test is inconclusive.

Continuity at $x_0$ requires that $f(x_0)$ be defined and that ${\lim }_{x\to x_0}f(x)=f(x_0)$. In the epsilon-delta definition, for each $\epsilon >0$ there must be a $\delta >0$ such that $|f(x)-f(x_0)|<\epsilon$ whenever $|x-x_0|<\delta$ within the domain. Differentiability requires the finite two-sided limitIn particular, merely having a punctured limit of a function does not imply continuity unless the assigned value agrees with that limit.

For an integrable discrete random variable, its expectation and variance are $\mu ={\sum }_{n\in \mathbb{Z}}n{p}_n$ and ${\sum }_{n\in \mathbb{Z}}(n-\mu {)}^2{p}_n$, respectively. The first formula requires $\sum |n|p_n<\infty$ for a finite ordinary mean; the variance may be infinite. A probability density function gives $P(\alpha <Y<\beta )={\int }_{\alpha }^{\beta }f(y)dy$.

Conditioning on $X$ and using independence givesThe interchange is justified by Tonelli's theorem, since all integrands are nonnegative. Consequently $Z$ has an absolutely continuous distribution, withIts integral is ${\sum }_n{p}_n=1$. This is the convolution of independent random variables, specifically a Gaussian smoothing of an integer-valued distribution for the given normal density. The specified weights are normalized because $2(3/{\pi }^2){\sum }_{n\ge 1}{n}^{-2}=1$.

Set $g(x,y)=y-(x-2{)}^2$. The gradient of $g$ never vanishes, so the Lagrange multiplier equations $\nabla f=\lambda \nabla g$ apply at a constrained stationary point. They are $4x=-2\lambda (x-2)$, $2y=\lambda$, and $y=(x-2{)}^2$. Eliminating $\lambda ,y$ givesWith $t=x-2$, this becomes $t^3+t+2=(t+1)(t^2-t+2)=0$. The quadratic factor is $(t-1/2{)}^2+7/4>0$, so the only real possibility is $t=-1$. ThereforeThe level curve through this point is the ellipse $2x^2+y^2=3$, meeting the coordinate axes at $(\pm \sqrt{3/2},0)$ and $(0,\pm \sqrt{3})$. Its tangent line at $(1,1)$ has slope $-2x/y=-2$, equal to the derivative $2(x-2)$ of the constraint parabola there. The two curves are tangent.

The parabola touches this ellipse from outside; every other point on the constraint lies on a larger contour. An exact check of this geometric argument iswith equality only at $x=1$. Thus the point is the unique constrained global minimum.

Use the real Fourier series convention $f(x)\sim a_0/2+{\sum }_{n\ge 1}[a_n\cos (n\pi x)+b_n\sin (n\pi x)]$ for period two. The Fourier coefficients areThe last two integrals follow by integration by parts. Thus the Fourier series of a half-wave rectified ramp isOn $(-1,1)$, $f(x)=|x|+x$. Its even part is $f_e(x)=|x|$ and its odd part is $f_o(x)=x$, so the cosine and sine terms separate:At a jump, the Fourier series converges to the average of the one-sided limits. Thus the first series for $f$ takes the value $1$ at odd integers, while the Fourier series of a periodic sawtooth for $f_o$ takes the value $0$ there. The even function has a continuous periodic extension.

At $x=0$, the formula for $f_e$ gives $0=1/2-(4/{\pi }^2){\sum }_{r\ge 0}(2r+1{)}^{-2}$. At $x=1/2$, the formula for $f_o$ has only odd-index terms and gives $1/2=(2/\pi ){\sum }_{r\ge 0}(-1{)}^r/(2r+1)$. Hence