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For real constants $a,c$, apply Picard iteration to the initial value problem $y^{\prime }=ay/(1-x)$, $y(0)=c$, starting with $y_0(x)=c$. The recurrence isSince ${\int }_0^x{\log }^k(1-t)/(1-t)dt=-{\log }^{k+1}(1-x)/(k+1)$, mathematical induction gives the displayed partial sums. The exponential series converges at every finite argument, so $y_n(x)\to c(1-x{)}^{-a}$ for every real $x<1$, uniformly on each compact interval in that range. This is not a Taylor series in $x$, so its convergence is not restricted to $|x|<1$. The differential equation's coefficient is singular at $x=1$ even when a special solution has a removable singularity there.

For a real angle $\alpha$ and a unit vector $\mathbf{n}$, the rotation about $\mathbf{n}$ sendsThe vector triple product identity gives the equivalent expression $\mathbf{x}+\sin \alpha \mathbf{n}\times \mathbf{x}+(1-\cos \alpha )\mathbf{n}\times (\mathbf{n}\times \mathbf{x})$. To see why it is a rotation, decompose $\mathbf{x}$ into its parallel and perpendicular components relative to $\mathbf{n}$. The parallel component is fixed, while the perpendicular component becomes $\cos \alpha {\mathbf{x}}_{\perp }+\sin \alpha \mathbf{n}\times {\mathbf{x}}_{\perp }$, the usual plane rotation. It is also the matrix exponential of the linear operator $H\mathbf{x}=\alpha \mathbf{n}\times \mathbf{x}$: the identity $H^3=-{\alpha }^{2}H$ reduces its exponential series to these three terms.

For $a>0$, this integral is an entire function of $x$: the exponential series converges uniformly on bounded subsets after multiplication by the integrable bound $e^{|x|}{u}^{a-1}$. ThusFor positive real $x$, a change of variables gives $F_a(x)=x^{-a}\gamma (a,x)$, where here $\gamma (a,x)={\int }_0^x{e}^{-t}{t}^{a-1}dt$ denotes the lower incomplete gamma function. The singular-looking factor $x^{-a}$ is removed by the integral over $[0,1]$. Applying integration by parts to the derivative of $u^a{e}^{-xu}$ gives $x{F}_a^{\prime }+a{F}_a=e^{-x}$; differentiating and eliminating $e^{-x}$ proves $x{F}_a^{\prime \prime }+(a+1+x)F_a^{\prime }+a{F}_a=0$.

The two identities areTo prove them, expand $e^{\pm x\log x}$ in its exponential series. Since $|x\log x|\le 1/e$ on $(0,1)$, the Weierstrass M-test justifies termwise integration. Substituting $x=e^{-t}$ gives ${\int }_0^1{x}^k(\log x{)}^{k}dx=(-1{)}^{k}k!/(k+1{)}^{k+1}$ through the gamma function. Inserting these moments into the series and shifting $n=k+1$ proves both formulas. Each resulting numerical series has absolute convergence.

Multiply the two exponential series in $e^{-(x-a{)}^2}=e^{-a^2}{e}^{2ax}{e}^{-x^2}$ to obtain the displayed Taylor expansion. Its first three nonzero terms depend on the real parameter $a$. For $a\ne 0,\pm 1/\sqrt{2}$, they end at degree two. At $a=0$, odd coefficients vanish and the first three terms are $1-x^2+x^4/2$. At $a=\pm 1/\sqrt{2}$, the quadratic coefficient vanishes, leaving $e^{-1/2}(1+2ax-4a{x}^3/3)$ as the first three nonzero terms. Parameter-dependent vanishing coefficients must be checked when a question counts nonzero terms rather than prescribing an expansion order.