For the suggested substitution, put $y=xv(x)$. Then the product rule gives $y^{\prime }=v+x{v}^{\prime }$, while the right side of the ordinary differential equation becomes $v+v^{-3}$. Cancelling $v$ leaves the separable differential equationIntegration yields $v^4/4=\log x+C$. The initial condition gives $v(1)=1$, so $v^4=1+4\log x$. The positive root is forced by the initial value and continuity:This is the maximal connected interval containing $1$ on which the real solution exists. At the left endpoint $y=0$, where the original ordinary differential equation is undefined; extending the fourth-root formula further left would also cease to be real.

A critical point requires $f_x=f_y=0$. Since $y^4+y^2+1>0$ for every real $y$, the first partial derivative forces $x=0$. The second then reduces to $3y^2-3=0$, so the only critical points are $(0,1)$ and $(0,-1)$.

At these points the Hessian matrix is diagonal:The first is a positive-definite matrix, giving a strict local minimum; the second is an indefinite matrix, giving a saddle point. Their values are $f(0,1)=-1$ and $f(0,-1)=3$. HenceThe minimum is local: along $x=0$, $f(0,y)=y^3-3y+1$ is unbounded below as $y\to -\infty$.

At the origin the needed partial derivatives areThe Taylor series through total degree two is thereforewhich givesHere total degree is what matters: $x^2{y}^2$ is quartic, not quadratic. The exact omitted polynomial is $y^3+x^2{y}^2+x^2{y}^4$, which verifies the stated Big O notation remainder directly.

The multivariable chain rule follows the change of both coordinates along the parametric curve:Substituting the partial derivatives from part (a), with $x=1+s^2$ and $y=e^s$, yieldsThe first term comes from changing $x$, and the remaining terms from changing $y$.

Take each partial derivative while holding the other coordinate fixed. Differentiating the polynomial term by term givesSince the function is a polynomial, both orders of the mixed partial derivative agree everywhere.

In a symmetric relation, each of the $n$ diagonal pairs $(a,a)$ can be chosen independently. For each of the $\left(\genfrac{}{}{0ex}{}{n}{2}\right)$ unordered pairs of distinct elements, either both ordered pairs $(a,b),(b,a)$ occur or neither occurs. Every such choice is free and produces a distinct symmetric relation. HenceFor the empty set this correctly gives one binary relation, the empty one.

A binary relation $R$ on $A$ is a subset of $A\times A$. Its properties are quantified over elements of $A$:

Thus an equivalence relation is a binary relation that is simultaneously reflexive, symmetric and transitive. The equivalence class of $a$ is the set of all elements related to it:Antisymmetry permits diagonal pairs and does not mean that a binary relation fails to be a symmetric relation.

For finite sets, the inclusion-exclusion principle statesAn element present in $r$ of the sets is counted ${\sum }_{k=1}^r(-1{)}^{k+1}(\genfrac{}{}{0ex}{}{r}{k})=1$ times, which explains the alternating corrections.

There are $\lfloor 999/d\rfloor$ positive integers strictly below $1000$ divisible by $d$. For the divisors $2,5,7$, the single counts are $499,199,142$. Their pairwise intersections correspond to divisibility by $10,14,35$, giving $99,71,28$. The triple intersection has $\lfloor 999/70\rfloor =14$ elements. Therefore

The base case is $F_5=5$, so $5\mid F_5$. To relate successive blocks of five, use the Fibonacci addition formula:This identity can also be obtained simply by applying the Fibonacci recurrence five times. If the mathematical induction hypothesis is $5\mid F_{5n}$, thenis a sum of two multiples of five. HenceThe generalisation is Fibonacci multiple-index divisibility:Indeed, $F_{mn+m}=F_m{F}_{mn+1}+F_{m-1}{F}_{mn}$, so precisely the same mathematical induction works with $5$ replaced by $m$.

Mathematical induction has two obligations. Establish the base case $S(0)$ if the natural numbers start at zero, or $S(1)$ if they start at one. Then prove the induction step: for an arbitrary natural $k$ in the relevant range, assume $S(k)$ and deduce $S(k+1)$.

The base case starts the chain, and the step propagates truth to every subsequent natural number:

Repeated squaring computes $a,a^2,a^4,a^8,\dots$ by successively squaring. The binary representation of the exponent specifies which of these values to multiply. For modular exponentiation, reduce after every multiplication; only a number of steps proportional to the logarithm of the exponent is needed.

Here $13\equiv 2(\mathrm{mod}11)$ and $17=16+1$. Successive squares giveTherefore

Use modular arithmetic to keep all intermediate numbers small:Thus $34\cdot 45\cdot 55+89\equiv 2\cdot 5\cdot 7+1=71\equiv 7\equiv -1(\mathrm{mod}8)$. Cubing gives

In modular arithmetic, calculations may be performed after replacing each input by its remainder. For a positive modulus $m$, write $[a{]}_m$ for the remainder in $\{0,\dots ,m-1\}$. ThenEquivalently, if $a\equiv a^{\prime }(\mathrm{mod}m)$ and $b\equiv b^{\prime }(\mathrm{mod}m)$, addition and multiplication give $a+b\equiv a^{\prime }+b^{\prime }$ and $ab\equiv a^{\prime }{b}^{\prime }$, and a fixed nonnegative integer exponent gives $a^k\equiv (a^{\prime }{)}^k$. One cannot in general reduce the exponent modulo $m$ by this rule.