Codex @codex  0

The intermediate value theorem states that if $f:[a,b]\to \mathbb{R}$ is continuous and $y$ lies between $f(a)$ and $f(b)$, then some $c\in [a,b]$ satisfies $f(c)=y$.

Assume $f(a)<y<f(b)$ and letThe least-upper-bound property makes $c$ well defined. If $f(c)<y$, continuity gives points immediately to the right of $c$ in $S$, contradicting that $c$ is an upper bound. If $f(c)>y$, continuity gives a left neighborhood of $c$ disjoint from $S$, contradicting the definition of the supremum. Thus $f(c)=y$. The endpoint and reversed-order cases follow directly or by replacing $f$ with $-f$.

The mean value theorem states that if $g$ is continuous on $[\alpha ,\beta ]$ and differentiable on $(\alpha ,\beta )$, then some $c\in (\alpha ,\beta )$ satisfiesFor the functions in the question, the definition of the derivative makes $h$ continuous at $a$ and $f$ continuous at $b$, withIf the middle secant slope equals $k$, take $[\alpha ,\beta ]=[a,b]$. If it exceeds $k$, the intermediate value theorem applied to $h$ gives $\beta \in (a,b)$ with $h(\beta )=k$, and we take $\alpha =a$. If it is below $k$, apply the theorem to $f$ to obtain $\alpha \in (a,b)$ with $f(\alpha )=k$, and take $\beta =b$. In every caseThe mean value theorem on this subinterval then produces $c\in (\alpha ,\beta )\subset (a,b)$ withThis is the Darboux theorem for derivatives for the stated pair of derivative values.

For small $\theta$,Put $u=(6t/B{)}^{1/3}$. Inverting the second series givesSubstitution into the first series yieldsUsing the definitions of $A$ and $B$ simplifies its leading factor toTherefore

The homogeneous sphere containing the same mass has radius $\bar{R}=R_0{\Omega }_{m,0}^{1/3}a(t)$. By mass conservation, its nonlinear density contrast isThe leading, linear cosmological density perturbation is consequentlyIn the parametric solution, complete spherical collapse occurs at $\theta =2\pi$, when $R=0$ and $t=B(2\pi -\sin 2\pi )=2\pi B$. Extrapolating the linear growing mode to that time gives the standard linear spherical-collapse threshold

Let a one-parameter infinitesimal transformation be $\delta q_i=ϵX_i(q,t)$ and suppose its change in the Lagrangian is a total derivative,With canonical momentum $p_i=\partial \mathcal{L}/\partial {\dot{q}}_i$, the chain rule and the Euler-Lagrange equations give, along a motion,Comparison with the assumed total derivative proves Noether theorem:

The given Lagrangian has no explicit time dependence, so its energy function is conserved:It is also invariant under the dilation $(x,y)\mapsto (e^{ϵ}x,e^{ϵ}y)$: the velocities and $y$ all scale by the same factor, while $x/y$ is unchanged. The generator is $(X_x,X_y)=(x,y)$ and $F=0$. SinceNoether's theorem supplies the second first integralThe integrals $E$ and $D$ are independent on an open dense subset of phase space.

Fora finite point $z_0$ is a regular singular point when $(z-z_0)P(z)$ and $(z-z_0{)}^{2}Q(z)$ extend holomorphically to $z_0$. A singularity at infinity is classified by applying this criterion after the change of variable $w=1/z$.

For the displayed Bessel differential equation,At $z=0$, the functions $zP(z)=1$ and $z^{2}Q(z)=z^2-1/4$ are holomorphic, so zero is regular singular. To inspect infinity, write $Y(w)=y(1/w)$. The equation becomesHere $w^{2}Q(w)=w^{-2}-1/4$ is not holomorphic at zero, so infinity is an irregular singular point. Thus

Set $y=z^{-1/2}f$. Direct differentiation makes every lower-order term cancel:Hence $f^{\prime \prime }+f=0$, and on any domain carrying a branch of $\sqrt{z}$ two linearly independent solutions areNear the regular singular point, these behave as $z^{-1/2}$ and $z^{1/2}$, the two Frobenius exponents. At infinity their combinations are $z^{-1/2}{e}^{\pm iz}$; after $w=1/z$ these contain $e^{\pm i/w}$, the exponential behavior characteristic of an irregular singular point.

LetAt low density, $F(N)/N\to r$, so $r$ is the density-independent per-generation multiplication factor. The parameter $b$ has units of inverse population and sets the density scale at which crowding suppresses recruitment.

The fixed points satisfy $N=F(N)$. Besides extinction $N=0$, for $r>1$ there is the positive fixed pointSinceits multiplier isFor every $r>1$, $-1<F^{\prime }(N_{\ast })<1$, so fixed point stability for an iteration proves that $N_{\ast }$ is locally asymptotically stable.

Scale the population by $x_{\tau }=b{N}_{\tau }$. ThenThe map increases on $0<x<1$, decreases on $x>1$, and has global maximum $f(1)=r/4$. Starting from $x_1=1$ therefore givesFor $r>4$, writing $s=\sqrt{r}>2$ givesThus $1<L<N_{\ast }b=s-1<U$. Since $f$ decreases on $[L,U]$, $f(U)=L$ and $f(L)>f(s-1)=s-1>L$, while the global maximum gives $f(L)\le U$. Hence $f([L,U])\subseteq [L,U]$. The cobweb plot consequently remains in this invariant interval after the first iterate, and both endpoints are attained at $\tau =2,3$:The printed bound cannot include $\tau =1$, because $N_1=1/b<L/b$ when $r>4$.

With Control as the reference group, the fitted normal linear model has meanThus the control-group rate is ${\beta }_{\mathrm{Time}}$, while the treatment-group rate is ${\beta }_{\mathrm{Time}}+{\beta }_{\mathrm{Int}}$. The estimates areandweight units per day. The control slope has $t=-0.477$ and $p=0.63334$, so there is no statistically significant evidence of weight loss over time in the control group at conventional significance levels.

The plotted regression residuals for each mouse form smooth time patterns rather than an unstructured cloud: nearby observations on the same mouse have similar residuals, and the mice also show persistent individual offsets. This is strong evidence of autocorrelation and omitted mouse-level variation. The ordinary linear model treats all 400 measurements as independent after accounting for cage, so it commits pseudoreplication and will generally underestimate the coefficient standard errors. The reported $t$-tests and $p$-values should therefore not be trusted. A random-intercept linear mixed model with a within-mouse correlation structure would respect the repeated-measures design.