The gradient vanishes exactly when $x^2/4-1=0$ and $y^2/4-1=0$, giving the four points $(\pm 2,\pm 2)$. The Hessian matrix is $\mathrm{diag}(x/2,y/2)$. It is positive definite at $(2,2)$, negative definite at $(-2,-2)$, and indefinite at the two mixed-sign points. Thus $(2,2)$ is a strict local minimum, $(-2,-2)$ a strict local maximum, and $(2,-2),(-2,2)$ are saddle points by the second-derivative test. Their respective function values are $-8/3$, $8/3$, and zero at both saddle points.

Apply Newton's method to $F(t)=1-3\sin t/2-{\sin }^{3}t/3$. The derivative in part (a) givesChoose a starting point near a desired root, such as $t_0=0.6$ for the root in $(0,\pi /2)$; do not divide when $\cos t_k=0$. On $(0,\pi /2)$, $F$ is strictly decreasing and changes sign between its endpoints, so the intermediate value theorem gives exactly one root there. Its derivative is nonzero, giving the usual local convergence of Newton's method. The periodic function has other roots; the iteration does not impose global uniqueness or convergence from every starting value.

The coordinate sum is $2\sin t$, while the sum of their cubes is $6\sin t-4{\sin }^{3}t$. Along the parametrized curve, the composite function is therefore $-3\sin t/2-{\sin }^{3}t/3$. Differentiate, or apply the multivariable chain rule, to obtain

The function is even, so every sine coefficient is zero. Its mean coefficient isFor $n\ge 1$, compute $a_n=(2/\pi ){\int }_0^{\pi }\sin x\cos (nx)\mathrm{d}x$. The product-to-sum identity makes this integral zero for odd $n$ and gives $-4/(\pi (n^2-1))$ for even $n$. Thus the missing Fourier coefficients are $a_n=0$ for odd $n$ and $b_n=0$ for every $n$, with $a_0=4/\pi$. Equivalently,The Fourier series converges to this continuous periodic function; there is no jump value to average.

Under the product order, every point of the open unit disc has both coordinates below one, so $(1,1)$ is an upper bound. Any upper bound $(u,v)$ must have $u\ge 1$ and $v\ge 1$, since the two coordinate projections of the open unit disc are $(-1,1)$. Therefore $(1,1){\le }_P(u,v)$, proving ${\sup }_{{\le }_P}S=(1,1)$.

Under the lexicographic order, every point $(1,v)$, with arbitrary real $v$, is an upper bound, because every point of $S$ has first coordinate strictly below one. Conversely, an upper bound cannot have first coordinate $u<1$: a point $(x,0)\in S$ with $x>u$ would exceed it. If an upper bound has $u>1$, the upper bound $(1,0)$ is strictly smaller. If it is $(1,v)$, the upper bound $(1,v-1)$ is strictly smaller. Hence there is no least upper bound under the lexicographic order. This illustrates the supremum of an open disc under product and lexicographic orders.

For these real-valued functions, $f(n)=O(g(n))$ means that there exist constants $C>0$ and $N\in \mathbb{N}$ such thatThe absolute values matter because neither function is assumed nonnegative. The constants are independent of $n$; finitely many earlier values do not affect Big O notation.

Absorb the finitely many exceptional initial terms in the Big O notation assumption into a constant $C>0$, so that $|a_n|\le Cn$ for every $n\ge 1$. This is possible because each initial term is finite and each index is positive. We prove the stronger bound $|{}^k{b}_n|\le Cn$ for every $k,n\ge 1$.

The case $k=1$ is the chosen bound. If it holds at $k-1$, the triangle inequality givesMathematical induction completes the proof. Thus ${}^k{b}_n=O(n)$ for every fixed $k$; in fact the same constant works for all the iterated Cesàro means. This is the linear-growth bound for iterated Cesàro means.

First prove $a^p\equiv a(\mathrm{mod}p)$ for every nonnegative integer $a$ by mathematical induction. The case $a=0$ is immediate. If it holds at $a$, the binomial theorem and the stated divisibility of the middle binomial coefficients giveThis establishes the induction. For an arbitrary integer $a$, take a nonnegative representative $r$ of its residue class; then $a^p\equiv r^p\equiv r\equiv a(\mathrm{mod}p)$. When $a\not\equiv 0(\mathrm{mod}p)$, part (b) supplies a multiplicative inverse modulo $p$. Multiplying the congruence by that inverse proves Fermat's little theorem:

Choose the representative $1\le a\le p-1$. Since $p$ is prime and does not divide $a$, their greatest common divisor is one. The extended Euclidean algorithm gives the Bezout identityfor some integers $x,y$. Reducing modulo $p$ gives $ax\equiv 1(\mathrm{mod}p)$, so the residue class of $x$ is a multiplicative inverse of $a$. The argument uses the nonzero condition to exclude the only residue with greatest common divisor $p$.

Use modular exponentiation by repeated squaring:Since $17=16+1$, multiply the indicated powers to obtain $27^{17}\equiv 6(\mathrm{mod}7)$. Only the binary expansion of the exponent determines which squared powers are multiplied.