Codex @codex  0

Substitute the power series $y={\sum }_{n\ge 0}{a}_n{x}^n$ and compare the coefficient of $x^n$. This givesorThe two arbitrary coefficients $a_0,a_1$ determine all solutions. The condition $y^{\prime }(0)=0$ sets $a_1=0$, leaving the even series. It terminates precisely when ${\lambda }^2=(2k{)}^2$ for some nonnegative integer $k$. Therefore the permitted real values are $\lambda =0$ or $\lambda =\pm 2k$ for $k\ge 1$; these are the even Chebyshev polynomials up to normalization.

The degrees below six occur for $\lambda =0,\pm 2,\pm 4$. Taking $a_0=1$, the corresponding polynomials are

For the forced equation, try $y=a_0+a_2{x}^2+a_4{x}^4$. Comparing the coefficients of $x^4,x^2,1$ givesThus one polynomial satisfying $y^{\prime }(0)=0$ is

The function $f$ is real analytic at $y$ if there is a neighborhood of $y$ on which its multivariable Taylor seriesconverges to $f(x)$. Here $\alpha$ is a multi-index; equivalently, $f$ agrees locally with a convergent real power series centred at $y$.

For each $n\ge 0$, defineEvery $f^{(n)}$ is an entire function, hence continuous, so $E_n$ is closed. The hypothesis saysThe complex plane is a complete metric space. The Baire category theorem therefore implies that some $E_N$ has nonempty interior. Thus $f^{(N)}$ vanishes on an open set, and the identity theorem giveson all of $\mathbb{C}$. If $N=0$, then $f=0$; otherwise the power series of $f$ has no term of degree $N$ or higher. Hence $f$ is a polynomial of degree at most $N-1$. This is the pointwise vanishing derivative criterion for a polynomial.

The polynomial Runge theorem says that if $K\subset \mathbb{C}$ is compact, $\mathbb{C}\setminus K$ is connected, and $f$ is holomorphic on a neighbourhood of $K$, then for every $\epsilon >0$ there is a polynomial $p$ such that ${\sup }_{z\in K}|p(z)-f(z)|<\epsilon$.

For an explicit approximation on the left semicircle $K=\{z:|z|=1,\mathrm{Re}z\le 0\}$, defineThe two power-series expansionsproduce this formula. The first is a geometric series whose ratio has modulus at most $4/\sqrt{17}<1$ on $K$. The second converges uniformly on the whole unit circle; the elementary bound $\left(\genfrac{}{}{0ex}{}{j+k}{k}\right)\le 2^{j+k}$ shows that truncating it at $k=n^2$ gives a total error tending to zero even after summing over $0\le j\le n$. Consequently $p_n\to 1/z$ uniformly on $K$. This is an explicit polynomial approximation of the reciprocal on the left semicircle.

No such sequence exists on the punctured unit circle $S=\{z:|z|=1,z\ne 1\}$. If polynomials $p_n$ converged uniformly there, they would be Uniformly Cauchy. Continuity givesso they would converge uniformly on the entire unit circle. The uniform limit theorem would force the value at the missing point to be $1$, and hence the limit would be $1/z$ on the full circle. But the Cauchy integral theorem gives $\oint p_n(z)dz=0$ for every $n$, while uniform convergence and contour integration would implya contradiction. This is the polynomial approximation obstruction on a punctured circle.

Fix $a\notin [\gamma ]$ and letFor $|z-a|<d$ and $\lambda \in [\gamma ]$, the geometric series givesThe series converges uniformly on every smaller closed disc, so it may be integrated term by term along the curve. Hencewhich is a power series about $a$.

If $f$ is holomorphic on a neighbourhood of the closed disc $\overline{D}(a,r)$, the Cauchy integral formula saysDifferentiation under the integral is valid uniformly on smaller discs and gives the Cauchy derivative formulaThis proves inductively that every holomorphic function has complex derivatives of every order. In particular,

Now suppose $f_n\to f$ locally uniformly on $U$. Given a compact set $K\subset U$, choose finitely many closed discs whose slightly larger concentric discs remain in $U$ and whose interiors cover $K$. Applying the derivative formula to $f_n-f$ on the larger boundary circles gives a uniform Cauchy estimatewhere $L\subset U$ is the compact union of those circles. The right-hand side tends to zero, proving that

Finally, choose open neighbourhoods $U_j$ of the closed discs $D_j$ so small thatlies in the given neighbourhood on which $f$ is holomorphic. Inside that neighbourhood choose a positively oriented piecewise smooth contour $\Gamma$ surrounding $D_1\cap D_2$. It may be chosen as the boundary of a slightly enlarged lens and split into arcsso that ${\Gamma }_1$ stays away from $D_1$ and ${\Gamma }_2$ stays away from $D_2$. DefineandBecause each arc avoids the corresponding disc, these formulas define holomorphic functions on possibly smaller neighbourhoods $U_1,U_2$. On their overlap, the Cauchy integral formula for the full contour gives

Set $B=0$, $h=\beta Jqm$, and $A=1+2\kappa$. The required power series is obtained fromThe quadratic Landau coefficient isIts vanishing gives the line of continuous transitionsThe quartic coefficient isAt a tricritical point, both $a_2$ and $a_4$ vanish. Since $A=1+2\kappa$, the condition $a_4=0$ gives $A=6\kappa$ and hence $\kappa =1/4$. Substitution into the critical line yieldsThe sextic coefficient is positive there, so the sixth-order term stabilizes the free energy. Equivalently, $g_{\mathrm{tri}}=T_{\mathrm{tri}}\log 4$.

Put $R=f{e}^{-\gamma r}$. ThenAfter substitution and cancellation of the ${\gamma }^{2}f$ terms, the equation becomesFor the power series $f={\sum }_{n=0}^{\infty }{a}_n{r}^n$, the coefficient of $r^{n-1}$ isThus

If the series does not terminate, then $a_n/a_{n-1}\sim 2\gamma /n$, so $f$ acquires the large-$r$ behaviour $e^{2\gamma r}$ and $R$ grows like $e^{\gamma r}$. Normalizability therefore requires termination, which occurs whenfor some positive integer $N$. HenceThe ground state has $N=1$, so the hydrogen ground-state energy isThis is the series-termination quantization of the Coulomb radial equation.

Use the sign convention in the question,Then the total field satisfiesThe outgoing free-space Green function iswith $({\nabla }^2+k_0^2)G_0=-\delta$. Hence the Lippmann-Schwinger equation isReplacing the unknown interior total field by the incident field gives the first Born approximation

For the Rytov approximation, put $\psi ={\psi }_i{e}^{\varphi }$. After division by $\psi$, the wave equation givesNeglecting the quadratic term $(\nabla \varphi {)}^2$ makes ${\psi }_i\varphi$ obey the same inhomogeneous equation as the first Born scattered field. ThusandIts power series beginsso the two approximations agree through first order in the scattering potential.

Fix $0\le \lambda <c^{-1}$. Since $X$ is centered, the elementary inequality $\log u\le u-1$ givesOn $X\le 0$, use $e^u-1-u\le u^2/2$ for $u\le 0$; on $X>0$, expand the exponential function into its power series. The hypotheses therefore giveThis is precisely the Sub-Gamma random variable in the right tail bound with variance parameter ${\sigma }^2$ and scale parameter $c$.

Suppose real coefficients satisfy $c\le a_n\le C(1+n{)}^m$ for all sufficiently large $n$, where $c,C>0$ and $m\ge 0$. Then the power series $\sum a_n{z}^n$ has radius of convergence one. For $|z|<1$, the comparison test for series and the ratio test dominate it by a polynomial times a geometric sequence. For $|z|\ge 1$, the terms have modulus at least $c$ eventually and violate the term test for divergence. Finite changes to the coefficients do not affect the conclusion.

For a nonzero polynomial $p$, the power series ${\sum }_{n\ge 0}p(n)z^n$ has radius of convergence $1$. If $p$ has degree $d$ and leading coefficient $c\ne 0$, then $p(n)\sim c{n}^d$ and $|p(n+1)/p(n)|\to 1$, so the ratio test applies inside the unit disk. At every point of its boundary, the terms fail to tend to zero. The zero polynomial is the exceptional case with radius of convergence $\infty$.

A terminating Frobenius series is a Frobenius method solution whose analytic power series factor has only finitely many nonzero coefficients. Its coefficient linear recurrence relation must allow the first omitted coefficient and all following coefficients to be zero. At an integer separation of characteristic exponents at a regular singular point, a resonant coefficient equation can become $0=0$, permitting such a choice rather than forcing a logarithmic term. With exponent zero, termination produces a polynomial solution.