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For the first derivative, $f^{\prime }(x)=\frac{1}{2}(1+x{)}^{-1/2}$, which is the stated formula at $r=1$. If the formula holds at $r$, differentiating multiplies its coefficient by $\frac{1}{2}-r=-(2r-1)/2$. The factorial coefficients satisfyThe sign changes, proving by mathematical induction that

Apply the given Taylor formula with integral remainder at $x=1/2$. Its polynomial part is $1+2S_n$, where $S_n$ is the first $n$ terms of the requested series, sinceTo justify passage to the infinite series, for $0\le t\le 1/2$,Therefore the integral remainder obeysThe last inequality follows from $\left(\genfrac{}{}{0ex}{}{2n}{n}\right)\le {\sum }_{j=0}^{2n}\left(\genfrac{}{}{0ex}{}{2n}{j}\right)=4^n$, a direct binomial theorem bound. Hence $\sqrt{3/2}=1+2\lim S_n$, and

For a nonempty set $S\subseteq \mathbb{R}$, a real number $s$ is its supremum if every $x\in S$ satisfies $x\le s$ and every upper bound $u$ of $S$ satisfies $s\le u$. Equivalently, $s$ is an upper bound and, for every $\epsilon >0$, some $x\in S$ has $s-\epsilon <x\le s$. The least upper bound axiom states that every nonempty set of real numbers bounded above has a real supremum.

Suppose $(a_n)$ is nondecreasing and bounded above. Apply the least upper bound axiom to $S=\{a_n:n\ge 1\}$, obtaining $s=\sup S$. For any $\epsilon >0$, $s-\epsilon$ cannot be an upper bound, so there is $N$ with $a_N>s-\epsilon$. For every $n\ge N$, monotonicity givesso $|a_n-s|<\epsilon$. This is precisely the definition of a convergent sequence. If $(a_n)$ is nonincreasing and bounded below, apply the proved case to $(-a_n)$; its limit is the negative of the infimum of the original terms. Thus the bounded monotone sequence theorem follows directly from the axiom:

Write $A={\sum }_i{a}_i^2$, $B={\sum }_i{b}_i^2$, $C={\sum }_i{a}_i{b}_i$. If $B=0$, every $b_i=0$ and the claimed inequality is immediate. Otherwise the sum of squaresholds for every real $t$. Choosing $t=C/B$ gives $0\le A-C^2/B$, and hence the Cauchy-Schwarz inequality in its stronger absolute-value form:In particular $C\le \sqrt{A}\sqrt{B}$, as requested. For unit vectors, $A=B=1$, so this same argument givesThe absolute value is essential for opposite-pointing vectors; it follows from the squared inequality without a sign assumption on their dot product.

Split the vector $x$ into its normal and tangential orthogonal projections:A reflection in the plane negates the first component and preserves the second. Thus $Hx=x-2(n\cdot x)n$, giving the Householder reflectionSince $n\cdot n=1$ and $n\cdot u=n\cdot v=0$,The columns of a matrix representation of a linear map are the coordinates of the images of the ordered basis vectors. Consequently, in the orthonormal basis $(u,v,n)$,Equivalently, with the orthogonal matrix $Q=(uvn)$, the change of basis formula is $N=Q^{T}MQ$.

For the additive function, setting both arguments to zero gives $f(0)=0$, and then $f(-x)=-f(x)$. Continuity at the given point $x_0$ impliesFor any real $x$, $f(x+h)-f(x)=f(h)$, so $f$ is continuous at every point. This proves the continuity part of additive function continuous at one point is linear.

Let $c=f(1)$. Repeated addition and negation give $f(m)=mc$ for every integer $m$. For integers $p$ and positive $q$,so $f(r)=cr$ for every rational number $r$. Given any real $x$, choose rational $r_n\to x$. Continuity and density of the rational numbers giveConversely each real multiple of $x$ satisfies the Cauchy functional equation.

For the multiplicative equation, assume first that $g$ is not identically zero and choose $x_1$ with $g(x_1)\ne 0$. Then $g(x_1)=g(x_1)g(0)$ forces $g(0)=1$. Nextmakes every value nonzero, andmakes every value positive. Thus $h(x)=\log g(x)$ is a continuous real additive function, since $h(x+y)=h(x)+h(y)$. The proved classification gives $h(x)=kx$ for a real constant $k$. This solves the continuous exponential functional equation:Both types satisfy the equation and continuity, and every nonzero solution is everywhere strictly positive.

Use integration by parts on ${\sin }^{n-1}x\sin x$, for $n\ge 2$:Both boundary terms are zero. Therefore the Wallis integrals satisfyStarting with $I_0=\pi /2$ and $I_1=1$, repeated application givesThese formulas also hold at $n=0$ by empty products.

For $0<x<\pi /2$, $0<\sin x<1$, so $0<{\sin }^{n}x<{\sin }^{n-1}x$ when $n\ge 1$. Integrating the strict inequalities proves $0<I_n<I_{n-1}$. In particular $I_{2n+1}/I_{2n}<1$. For $n\ge 1$, the recurrence and $I_{2n-1}>I_{2n}$ giveThusAt $n=0$ the same displayed bounds are $0<2/\pi <1$, also true directly from $I_1<I_0$.

The squeeze theorem now gives $I_{2n+1}/I_{2n}\to 1$. Substituting the evaluated integrals showsand henceThis is twice the finite-product form of the Wallis product; the proof used only the integral recurrence and strict comparison of its integrands.

For a standard collinear Lorentz boost, put $\beta =v/c$ and $\Gamma =(1-{\beta }^2{)}^{-1/2}$. Its equations are $x^{\prime }=\Gamma (x-\beta ct)$ and $c{t}^{\prime }=\Gamma (ct-\beta x)$. Define the rapidity $\varphi =\mathrm{arctanh}\beta$. Since ${\cosh }^2\varphi -{\sinh }^2\varphi =1$ and $\cosh \varphi >0$, $\cosh \varphi =\Gamma$ and $\sinh \varphi =\Gamma \beta$. ThereforeAdding and subtracting, and using $\cosh \varphi \mp \sinh \varphi =e^{\mp \varphi }$, givesA second boost multiplies these factors by $e^{\mp {\varphi }^{\prime }}$, so the combined rapidity is $\varphi +{\varphi }^{\prime }$. The hyperbolic tangent addition formula then gives the relativistic velocity addition lawVelocities are signed along the common axis, so this formula also covers oppositely directed boosts.

For steady inviscid flow at atmospheric pressure, Bernoulli's equation along a freely falling streamline gives $u^2/2+gz=u_0^2/2+gH$. The downward speed is thereforeFor the incompressible jet, mass conservation fixes its volume flux: $\pi R^{2}u=\pi a^2{u}_0$. ConsequentlyFar from impact, the inviscid falling jet and radial impact film is thin and nearly horizontal. Let $v$ be its radial speed and $h$ its thickness. The same volume flux now crosses a cylindrical surface of area $2\pi rh$, giving $v=a^2{u}_0/(2rh)$. At the free surface, atmospheric pressure and Bernoulli's equation give $v^2=u_0^2+2g(H-h)$. Eliminate $v$ to obtainThis description assumes a steady incompressible jet, negligible surface tension, and negligible dissipation between the tube and the thin far-field film. It is a free-surface relation, not an assertion that pressure is atmospheric throughout the film or that the impact region itself is thin.

For every $u\in U$, write $u=(u-\alpha u)+\alpha u$. The first term lies in $\ker \alpha$, since ${\alpha }^2=\alpha$, and the second lies in $\mathrm{im}\alpha$. If $v$ belongs to both spaces, write $v=\alpha w$; then $\alpha v={\alpha }^{2}w=\alpha w=v$, whereas membership in the kernel gives $\alpha v=0$. Thus the intersection is zero andIn a basis assembled from bases of these two spaces, the idempotent linear map has matrix $\mathrm{diag}(0_{d-r},I_r)$, where $d=\dim U$ and $r=\mathrm{rank}\alpha$. ThereforeThis basis consists entirely of eigenvectors, so $\alpha$ is diagonalizable over the reals, including the extreme cases $r=0$ and $r=d$.

For constrained optimization in minimization form, let $x\in X$, $g_i(x)\le 0$ and $h_j(x)=0$, and define the LagrangianThe Lagrangian sufficiency theorem says: if $x_{\ast }$ is feasible, ${\lambda }_i\ge 0$, ${\lambda }_i{g}_i(x_{\ast })=0$ for every $i$, and $x_{\ast }$ minimizes $L(\cdot ,\lambda ,\nu )$ globally over $X$, then $x_{\ast }$ minimizes $f$ over the feasible set. Equality multipliers ${\nu }_j$ may have either sign.

Indeed, for every feasible $x$,The first inequality uses the inequality-multiplier signs and the equality constraints; the last equality uses complementary slackness. This proves global optimality without assuming convexity or differentiability. The maximization version reverses the relevant signs, or follows by replacing $f$ with $-f$.

When $X$ is convex and the Lagrangian is differentiable and convex, the first-order condition ${\nabla }_{x}L(x_{\ast })=0$ implies its global minimum, because $L(x)\ge L(x_{\ast })+\nabla L(x_{\ast })\cdot (x-x_{\ast })$. Thus feasible stationarity plus complementary slackness is sufficient in that convex setting. Mere stationarity in a general nonconvex problem is not the theorem's global-minimum hypothesis.

Use the Pearson chi-squared goodness-of-fit test. Under the null hypothesis, each count has the specified binomial distribution, and no parameter is estimated from the data. The seven mutually exclusive classes give $7-1=6$ degrees of freedom. Their expected counts are all at least five, so the usual chi-squared asymptotic approximation does not require pooling. The computed statistic isSince $9.02<12.59$, do not reject fairness at the 5% level. The approximate $p$-value is $0.172$. This conclusion concerns agreement with the specified independent fair-coin count model; a nonrejection is not proof of fairness.

Uniform convergence on an interval $I$ means that for every $ϵ>0$ there is $N$ such that $|f_n(x)-f(x)|<ϵ$ for every $x\in I$ and every $n\ge N$. Equivalently, the supremum of this difference tends to zero. A uniform limit of continuous functions is continuous: at a point $x_0$, choose one $f_N$ uniformly within $ϵ/3$ of $f$, then use continuity of $f_N$ to make $|f_N(x)-f_N(x_0)|<ϵ/3$. The triangle inequality gives $|f(x)-f(x_0)|<ϵ$. Thus $f$ is integrable on the compact interval, andFor the half-line example, differentiating gives $f_n^{\prime }(x)=n^{-2}{e}^{-x/n}(1-x/n)$. The maximum is attained at $x=n$, where $f_n(n)=1/(en)$. Therefore $f_n\to 0$ uniformly on $[0,\infty )$. Nevertheless, the substitution $x=ny$ givesThe integral does not tend to zero. This is the phenomenon that uniformly vanishing functions can retain a nonzero integral: the finite-length estimate above has no finite analogue for the entire half-line.