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Substitute $x^3$ into the Taylor series for the exponential function:The general nonzero term is $x^{3n}/n!$; the power series coefficients of powers not divisible by three are zero. The radius of convergence is infinite.

The power series for $e^{y^3}$ has uniform convergence on every compact interval, so termwise integration from zero to $x$ is valid. ThusThis power series converges for every real $x$, including negative $x$ with the usual orientation of a definite integral.

The Maclaurin series of the hyperbolic cosine and cosine areBoth are even, so write their quotient as $1+a{x}^2+b{x}^4+O(x^6)$. Multiplication by the denominator gives coefficients $a-1/2$ and $b-a/2+1/24$. Matching the numerator yields $a=1$ and $b=1/2$. ThusThe denominator is nonzero near the expansion point, so division of the power series is valid there.

Let $S_N={\sum }_{n=1}^N(-1{)}^{n+1}/n$. Integrate the finite geometric series identity ${\sum }_{k=0}^{N-1}(-t{)}^k=[1-(-t{)}^N]/(1+t)$ over $0\le t\le 1$:The remainder has absolute value at most ${\int }_0^1{t}^{N}dt=1/(N+1)\to 0$. Hence the partial sums converge, withThis proves both convergence and the value of the alternating harmonic series without substituting an endpoint into an unproved infinite power series. Its convergence is conditional because the corresponding harmonic series of absolute values diverges.

Put $u=\ln (1+x)=x-x^2/2+x^3/3+O(x^4)$. Since $u=O(x)$, the outer Taylor series requires $u^2=x^2-x^3+O(x^4)$ and $u^3=x^3+O(x^4)$. ThereforeBoth logarithms are defined on a neighbourhood of zero, so this composition of power series is valid there.

Set $\ell =\ln 2$. Applying the exponential series and the Maclaurin series of the sine givesMultiply the two power series, retaining every contribution through degree four:Since $0<\ln 2<1$, the coefficient of $x^4$ is nonzero, so these are indeed the first two nonzero terms. The remainder is in Big O notation.

Use the exponential series to obtain the Maclaurin series of the hyperbolic sine and hyperbolic cosine:Dividing these power series gives the hyperbolic tangent expansionThe displayed remainder uses Big O notation as $x\to 0$.

For $n>1$, the Taylor series of the natural logarithm converges absolutely at $1/n$:Multiply this power series by $n+1/2$ and subtract $1$. The constant term cancels the subtracted $1$, and the coefficient of $1/n$ is $-1/2+1/2=0$. For $k\ge 2$, the coefficient of $n^{-k}$ isHenceIn particular $u_n=1/(12n^2)-1/(12n^3)+O(n^{-4})$, in Big O notation.

For $k\ge 2$, writeUse positivity from part (d) and bound the absolute terms of the power series in part (c) by a geometric series. For $n\ge 2$,For $n=1$, $u_1=\frac{3}{2}\ln 2-1<\frac{1}{2}<1$, since $\ln 2={\int }_1^{2}dt/t<1$. Thus a bound valid for every positive integer $n$ is

Classification and existence. For $y^{\prime \prime }+p(x)y^{\prime }+q(x)y=0$, an ordinary point $x_0$ is one where $p,q$ are analytic. There are two independent analytic power series solutions, with arbitrary initial values $y(x_0),y^{\prime }(x_0)$. A regular singular point is a singular point where $(x-x_0)p(x)$ and $(x-x_0{)}^{2}q(x)$ are analytic. The Frobenius method then constructs solutions beginning with $(x-x_0{)}^{\sigma }$, where $\sigma$ solves the indicial equation. At least one Frobenius solution exists; two such solutions exist without logarithms when the two exponents differ by a noninteger. Repeated exponents or an integer difference can require a logarithmic second solution.

For this equation,Both are analytic at zero, while $p,q$ themselves have poles there. Thus zero is a regular singular point.

Indicial equation and recurrence. Multiply by $4x$ and insert $y=x^{\sigma }{\sum }_{n\ge 0}{a}_n{x}^n$, $a_0\ne 0$:The lowest power $x^{\sigma -1}$ has coefficient $[4\sigma (\sigma -1)+2\sigma ]a_0$, givingThe coefficient of $x^{n+\sigma }$ isFor either exponent the common factor is nonzero for $n\ge 0$, so the recurrence relation reduces toNormalizing each leading coefficient to one, the two Frobenius solutions areMore explicitly, $y_2=\sqrt{x}{\sum }_{n\ge 0}{2}^{n}n!x^n/(2n+1)!$. The distinct leading powers establish linear independence, so $y=C_1{y}_1+C_2{y}_2$ on the positive real side of the singular point. The displayed exponential function is the requested closed-form solution; substitution gives $y_1^{\prime }=y_1/2,y_1^{\prime \prime }=y_1/4$, which makes the differential expression zero.

Convergence. For either exponent the ratio of successive terms in the coefficient power series isThus both coefficient series have $\boxed{R=\infty }$ by the ratio test. This does not remove the $\sqrt{x}$ branch point of $y_2$: its analytic coefficient factor is entire, while the full Frobenius solution requires a square-root branch away from zero in the complex plane. The differential equation has no other finite singular points to limit continuation of these coefficient series.

Divide the Chebyshev differential equation by $1-x^2$. Its coefficients $-x/(1-x^2)$ and $m^2/(1-x^2)$ are analytic at zero, so $\boxed{x=0\text{ is an ordinary point}.}$ The finite singular points of a differential equation are $x=\pm 1$.

Insert $y={\sum }_{n\ge 0}{a}_n{x}^{n+\sigma }$, $a_0\ne 0$. The coefficient of the lowest power $x^{\sigma -2}$ is $\sigma (\sigma -1)a_0$, hence $\sigma =0$ or $1$. The next power gives $\sigma (\sigma +1)a_1=0$, and the remaining powers yieldFor $\sigma =0$, $a_1$ is free andChoosing $a_1=0$ selects the even power series, with $y(0)=a_0,y^{\prime }(0)=0$. For $\sigma =1$, the next-power equation forces $a_1=0$ andThis selects the odd power series, with $y(0)=0,y^{\prime }(0)=a_0$. Their different initial values prove linear independence. Although the Frobenius method produces two possible leading exponents, zero is an ordinary point, and a general $\sigma =0$ series already includes both initial constants.

Let $M=|m|$ for integer $m$. In the even or odd power series whose parity matches $M$, the numerator in the recurrence relation vanishes at the coefficient of $x^M$. All coefficients of lower powers in that parity are nonzero, and all higher ones vanish. Therefore a polynomial solution exists, of polynomial degree $M$, containingFor nonnegative $m$ these are the Chebyshev polynomials $T_m$. The normalized low-degree cases follow directly from the recurrence relation:Enforcing $T_m(1)=1$ givesTo see that normalization at one is always possible, set $x=\cos \theta$ in the Chebyshev differential equation. For $Y(\theta )=y(\cos \theta )$, the differential expression $(1-x^2)y^{\prime \prime }-x{y}^{\prime }$ becomes $Y^{\prime \prime }$. Thus $Y=\cos (M\theta )$ gives $T_M(x)=\cos (M\arccos x)$ on $[-1,1]$ and has value one at $x=1$. It is a polynomial because the cosine addition formula gives $T_{M+1}=2x{T}_M-T_{M-1}$. Negative integer $m$ produces the same solution as $M=|m|$.

For the normalized second-order linear differential equation $y^{\prime \prime }+P(z)y^{\prime }+Q(z)y=0$, an ordinary point $z_0$ has analytic $P,Q$ near $z_0$. At a regular singular point, which is not ordinary, $(z-z_0)P(z)$ and $(z-z_0{)}^{2}Q(z)$ are analytic. Ordinary points admit two independent convergent power series solutions. Regular singular points instead call for the Frobenius method, which permits leading powers and, in some cases, logarithms.

Here zero is an ordinary point. Substitute $y={\sum }_{n\ge 0}{a}_n{z}^n$ into the Hermite differential equation and equate coefficients:The even and odd initial coefficients are independent, givingThese are the power series solutions of the Hermite differential equation, with the present parameter equal to twice the parameter in that article. The ratio of successive terms within each nonterminating parity is asymptotically $2|z{|}^2/n$, so both series have infinite radius of convergence.

For a nonzero polynomial of degree $m$, its highest coefficient in $-2z{y}^{\prime }+\lambda y$ forces $\lambda =2m$. Conversely at that value the matching parity terminates. HenceThe polynomial solutions are scalar multiples of the Hermite polynomials. Two examples are $\boxed{\lambda =0:y=1;\lambda =2:y=z}$; direct substitution verifies both. The other independent solution generally remains nonpolynomial.

An ordinary point of the normalized second-order equation is a point where both coefficient functions $p(x)$ and $q(x)$ are analytic in a neighborhood. Ordinary continuity is enough for a local existence theorem, but analyticity is the condition used for a convergent local power series solution.

The exponential power series gives $e^{-1/z}={\sum }_{m=0}^{\infty }(-1{)}^m/(m!z^m)$ for $z\ne 0$. Hence all singular terms areInfinitely many negative powers occur, so zero is an essential singularity.

Use separation of variables, $\Phi =F(r)H(\theta )$, in the Laplace equation in polar coordinates. Dividing by $FH/r^2$ givesSingle-valued $2\pi$-periodic angular factors have integer $n\ge 0$. For $n\ge 1$, they are $\cos n\theta ,\sin n\theta$, and the radial equation $r^2{F}^{\prime \prime }+r{F}^{\prime }-n^{2}F=0$ has solutions $r^n,r^{-n}$. The zero angular mode gives $F=a_0+b_0\log r$. These separated products give the requested solutions; arbitrary sums may use independent coefficients for each radial-angular combination.

Boundedness at the origin removes the logarithm and all negative radial powers. The boundary function is even, so its sine Fourier coefficients vanish. With the convention $f(\theta )=a_0+{\sum }_{n\ge 1}{a}_n\cos n\theta$, its Fourier coefficients areThus the bounded interior harmonic extension isFor $r<R$ the series and its derivatives converge locally uniformly, so it is harmonic. It is a rotated and scaled instance of the harmonic measure of a semicircle in a disk. For example, summing the power series for the complex arctangent, or differentiating the following expression, gives the equivalent real formThis also proves boundedness directly.

At the two jumps, the boundary interpretation matters. At every continuity point the radial limit equals the prescribed value. At $\theta =\pm \pi /2$, the radial limit is $C/2$, the midpoint of the two one-sided values. Consequently for $C\ne 0$ there is no harmonic solution continuous on the closed disk with exactly the printed boundary values. The standard bounded Dirichlet problem imposes the trace away from the jumps, or almost everywhere; assigning zero separately at those two boundary points does not alter the interior solution.

For uniqueness let $w=\tilde{\Phi }-\Phi$. Formally the divergence theorem applied to $w\nabla w$ givesFor continuous zero boundary data, this immediately forces $w=0$. With the actual discontinuous data, one should justify the two exceptional points rather than assume finite Dirichlet energy there. Let $|w|\le M$ and choose a cutoff ${\eta }_{ϵ}$ which is zero within ${ϵ}^2$ of those points, one beyond $ϵ$, and linear in logarithmic distance between. Its gradient satisfies $\int |\nabla {\eta }_{ϵ}{|}^2=O(1/|\log ϵ|)$. Applying the divergence theorem to ${\eta }_{ϵ}^{2}w\nabla w$ givesThe resulting energy tends to zero. Hence $\nabla w=0$ on every compact subset and $w$ is constant; its zero boundary trace on the open arcs makes it zero. This is the logarithmic cutoff proof of harmonic uniqueness with finitely many boundary exceptions.

For the exterior harmonic extension of disk boundary data, boundedness instead excludes $r^n$ and the logarithm. The same boundary coefficients giveThis is ${\Phi }_{\mathrm{in}}(R^2/r,\theta )$, using inversion in a circle. Its limit at infinity is $C/2$. To see why that value is forced independently of this construction, average any bounded exterior harmonic function over a circle. The mean satisfies $(r{\overline{\Phi }}^{\prime }{)}^{\prime }=0$, hence is $A+B\log r$. Boundedness forces $B=0$, and the boundary mean fixes $A=C/2$. A constant limit at infinity must therefore be $C/2$.

Finally suppose two bounded exterior solutions have this limit. Their difference $w$ tends to zero. Under planar inversion in a circle, it is bounded and harmonic near the origin, so the isolated point there is removable. Its local expansion gives $w=O(r^{-1})$, ${\partial }_{r}w=O(r^{-2})$ in the original exterior. Apply the preceding divergence theorem energy argument on $R<r<L$, with logarithmic cutoffs at the two boundary jumps. After removing those cutoffs, the inner boundary term vanishes; the outer one is ${\int }_{r=L}w{\partial }_{r}wds=O(L^{-2})$. Letting $L\to \infty$ forces $\nabla w=0$, and the zero limit gives $w=0$. Thus the bounded exterior solution is unique as well.

The Cauchy-Riemann equations are $u_x=v_y$ and $u_y=-v_x$. A holomorphic function is smooth, so equality of mixed partial derivatives givesThus its real and imaginary parts are harmonic functions.

An entire function is holomorphic at every finite point of the complex plane. The globally convergent power series for the exponential implies that $e^z$ and $e^{-z}$ are entire. Their difference therefore givesat every complex $z$, proving that the hyperbolic sine is entire. For powers, the base case $f_1(z)=z$ has derivative one everywhere. If $z^n$ is entire with derivative $n{z}^{n-1}$, the product rule gives $(z^{n+1}{)}^{\prime }=z^n+zn{z}^{n-1}=(n+1)z^n$ everywhere. Induction proves that every positive integer power is entire.

A function has a pole of order $n$ at $z_0$ when it is holomorphic in a punctured neighbourhood and $g(z)=(z-z_0{)}^{n}f(z)$ extends holomorphically to $z_0$ with $g(z_0)\ne 0$. An isolated essential singularity is neither a removable singularity nor a pole; equivalently its Laurent series has infinitely many nonzero negative-power terms. In particularhas an essential singularity at zero.

For $f_N$, the Laurent series about infinity isTo classify the point correctly, use $w=1/z$. Then $f_N(1/w)=w^{-N}{e}^w$, and $e^w$ is holomorphic with value one at zero. ThereforeThe pole at infinity is compatible with an essential singularity at the finite point zero. Negative powers of $z$ are regular powers of the local coordinate $w$ near infinity, so they do not make that point essential.

The convergent power series defining the matrix exponential gives $e^A=U{e}^D{U}^{†}$, with $e^D=\mathrm{diag}(e^{{\lambda }_1},\dots ,e^{{\lambda }_n})$. Hence the matrix exponential determinant identity isThe diagonalization proof uses the given Hermitian hypothesis; the identity itself holds for every square real or complex matrix.

Insert the power series and equate the coefficient of $x^n$. In the Legendre differential equation this givesHence the recurrence relation isThe free coefficients $a_0,a_1$ generate the independent even and odd series, respectively. They converge at least for $|x|<1$, up to the nearest singular points of the equation.

For a nonnegative integer $\ell$, the chain of the same parity as $\ell$ reaches $n=\ell$, where the numerator of the recurrence relation vanishes. The next coefficient and every later coefficient in that chain are zero, producing a polynomial of degree $\ell$. Normalizing it at $x=1$ gives the Legendre polynomial $P_{\ell }$.

The opposite-parity chain never reaches $n=\ell$. Its other possible zero, $n=-\ell -1$, is negative and cannot occur, so a nonzero initial coefficient produces a nonterminating power series. It is linearly independent of $P_{\ell }$ and can be scaled to the Legendre function of the second kind $Q_{\ell }$.

The degree statement requires $\ell \ge 0$, as the examples in part (c) indicate. For a negative integer $\ell$, the same equation has the nonnegative degree $-\ell -1$, because $\ell (\ell +1)=(-\ell -1)(-\ell )$. A polynomial cannot literally have negative degree.