Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2005 / Analysis

If $f$ is holomorphic on a neighborhood of the closed disc, the differentiated Cauchy integral formula iswith positive orientation. On the circle the integrand has modulus at most $M/r^{k+1}$ and the contour has length $2\pi r$. Thus

Take a circle of radius $R\ge S$ about zero. The Cauchy estimate gives, for each integer $k>n$,Therefore all these derivatives vanish. The Taylor series of an entire function converges to it everywhere, and henceAs usual $n$ is a nonnegative integer; if a real exponent is allowed, the degree is at most its integer part. A negative growth exponent forces $f=0$.

For a bounded entire function, the preceding result with $n=0$ gives Liouville's theorem. If a nonconstant polynomial $p$ had no complex root, $1/p$ would be entire. The leading term gives $|p(z)|\to \infty$ at infinity, so $1/p$ is bounded outside a large disc and is also bounded on the remaining compact disc. Liouville's theorem would make $1/p$, and hence $p$, constant. This proves the fundamental theorem of algebra.

Finally, away from the zeros of $\cos z$, the holomorphic quotient $h=f/\cos z$ has modulus at most $C$. At each zero it therefore has a removable singularity. These isolated points have no finite accumulation, so removing all of them gives a bounded entire function. Liouville's theorem gives a constant $c$ with $|c|\le C$, andConversely every such multiple satisfies the assumed bound.

Integrate $e^{iz}/z$ around the upper half-disc with the upper half-disc of radius $\epsilon$ about zero removed. The real segments run from $-R$ to $-\epsilon$ and $\epsilon$ to $R$; the small semicircle is clockwise. There are no poles inside this indented contour.

The large arc contribution tends to zero: its modulus is at most ${\int }_0^{\pi }{e}^{-R\sin \theta }d\theta$, which tends to zero by dominated convergence. On the small arc $z=\epsilon e^{i\theta }$, $\theta$ decreases from $\pi$ to zero, so its integral tends to ${\int }_{\pi }^{0}id\theta =-i\pi$. The contour theorem consequently givesTaking imaginary parts is legitimate for each truncated integral. The imaginary integrand $\sin x/x$ is continuous at zero after assigning its limiting value one, so the inner indentation can be removed. Therefore

Let $\omega =e^{2\pi i/3}$ and integrate $1/(1+z^3)$ around the positively oriented sector from argument zero to $2\pi /3$, with outer radius $R>1$. The single pole inside is $z_0=e^{i\pi /3}$, with residue $1/(3z_0^2)=1/(3\omega )$. The outward real radial segment contributes $I_R={\int }_0^{R}dx/(1+x^3)$; the return radial segment contributes $-\omega I_R$ because $(\omega x{)}^3=x^3$.

On the circular arc, $|1+z^3|\ge R^3-1$, so its integral has modulus at most $(2\pi R/3)/(R^3-1)\to 0$. Thus the residue theorem yieldsSince $\omega (1-\omega )=\omega -{\omega }^2=i\sqrt{3}$,The ordinary improper integral converges at both endpoints, so no principal-value interpretation is needed here.

For $y^{\prime \prime }+p(x)y^{\prime }+q(x)y=0$, a singular point at zero is regular when $xp(x)$ and $x^{2}q(x)$ extend analytically to zero. Hereso these products are analytic at zero. The equation remains singular there because $q$ has a pole.

Substitute $y={\sum }_{n\ge 0}{a}_n{x}^{n+\sigma }$. The lowest power gives the indicial equation $\sigma (\sigma -1+\lambda )=0$. Comparing the next powers givesFor $0<\lambda <1$ the denominators do not vanish. With $\sigma =0$, $a_1=-a_0/\lambda$ and $a_2=0$, soWith $\sigma =1-\lambda$, the recurrence becomesConsequently $y_2=a_0{x}^{1-\lambda }(1-x{)}^{\lambda }$ on $0<x<1$, with the binomial series for its last factor. The nonzero coefficient ratios tend to one, giving radius one for that power-series factor. The factor $x^{1-\lambda }$ specifies the Frobenius branch; the different leading powers prove independence.

When $\lambda =1$, both indicial roots become zero and the analytic Frobenius solution in both cases is a multiple of $1-x$. To obtain the missing solution use reduction of order. Away from zero and one, $p=1/x$, henceThus the real general solution on any interval avoiding the singular points isIts Wronskian with $1-x$ is $1/x\ne 0$, proving independence. For complex solutions choose a logarithm branch on a suitable simply connected region.