Fourier Series: Deriving Fejer's Kernel

PROBLEM

Derive Fejer's Kernel (we will use Dirichlet's Kernel), see the other post. 

SOLUTION

Note that $D_n(x)$ is the Dirichlet Kernel.

\begin{eqnarray*}
K_n(x)&=&(1/n)(D_0(x)+D_1(x)+\dots+D_{n-1}(x))\\
&=& (1/n)\sum_0^{j=n-1}\sum_{k=-j}^j e^{ikx}\\
&=& (1/n)\sum_{k=-n-1}^{n-1}(1-|k|)e^{ikx}\\
&=&  \sum_{k=-n-1}^{n-1}(1-|k|/n)e^{ikx}\\
&=& (1/n) \left(\displaystyle\frac{\sin(n+1/2)x}{\sin(x/2)}\right)^2
\end{eqnarray*}

As before the above holds for $0<|x|\leq \pi$. If $x=0$, then $K_n(x)=n$.

Fourier Series: Deriving the Dirichlet Kernel

From my Wavelets course (study guide for exam)
PROBLEM
Derive the formula for the Dirichlet kernel.

SOLUTION
\begin{eqnarray*}
D_n(t)&=& \sum^n_{-n} e^{ikt}=e^{-int}\sum_0^{2n} e^{ijt}, j:=k+n\\
&=& e^{-int}\left(\displaystyle\frac{e^{i(2n+1)t-1}}{e^{it}-1}\right)\\

&=& e^{-int}\left(\displaystyle\frac{e^{i(2n+1)t/2}}{e^{it/2}}\right)\cdot \left(\displaystyle\frac{e^{i(2n+1)t/2}-e^{i(2n+1)t/2}}{e^{it/2}-e^{-it/2}}\right)\\
&=& \left(\displaystyle\frac{\sin(n+1/2)t}{\sin(t/2)}\right)

\end{eqnarray*}

Note that the above is for $0<|t| \leq \pi$. If $t=0$, $D_n(t)=2n+1$.

Complex Analysis: Constant function if f'(z)=0

PROBLEM
Let $G\subset\mathbb{C}$ be a region (open and connected) and let $f:G\rightarrow\mathbb{C}$ be a holomorphic function such that $f'(z)=0$ for all $z\in G$. Prove that $f$ is constant on $G$.

PROOF
Let $G,f$ be as above. Let $z_0\in G$ and define $S:=\{z\in G: f(z)=f(z_0)$. By continuity of $f$ we have that $S$ is closed. Let $a\in S$. Then there exists $\epsilon>0$ such that $B(a,\epsilon)\subset G$. Let $z$ be in this ball and define $g(t):=f(tz+(1-t)a)$ for $0\leq t \leq 1$. Using the chain rule we have that $g'(t)=0$ for $0

Complex Analysis: Nowhere complex differentiable

PROBLEM

Show that $f(z)=Re(z)$ is nowhere complex differentiable.

PROOF

Let $h=\Delta x+i\Delta y$. Then $$\lim_{y\rightarrow 0}\frac{Re(z+h)-Re(z)}{h}=\lim_{y\rightarrow 0}\frac{Re(h)}{h}$

which is 1 if $h=\Delta x$ and 0 if $h=i\Delta y$. 

Ring Theory: A subring of the complex numbers

Problem taken from Jacobson's Basic Algebra, 2.1.4

PROBLEM
Let $I$ be the set of complex numbers of the form $m+n\sqrt{-3}$ where either $m, n$ is an integer or both $m$ and $n$ are halves of odd integers. Show that $I$ is a subring of $\mathbb{C}$.

PROOF
An arbitrary element of $I$ has the form $m/2+n/2\sqrt{-3}$ for some integers $m$ and $n$. First note that $1=2/2+0/2\sqrt{-3}\in I$. Next note that $$m_1/2+n_1/2\sqrt{-3}-m_2/2-n_2/2\sqrt{-3}=(m_1-m_2)/2+(n_1-n_2)\sqrt{-3}\in I$$
so $I$ is closed under addition and subtraction. Finally  $$(m_1/2+(n_1/2)\sqrt{-3})(m_2/2-(n_2/2)\sqrt{-3})=((m_1m_2-3n_1n_2)/4+((m_1n_2-n_1m_2)/4)\sqrt{-3}\in I$$

Notice that $m_1m_2-3n_1n_2$ and $m_1n_2-n_1m_2$ are always even integers, $I$ Is closed under multiplication and hence $I$ is a subring of the complex numbers.

Group Theory: Order of subgroups in Abelian groups

PROBLEM

Let $G$ be an abelian group. For any subgroup $H$ of $G$, $K$ also a subgroup of $G$, $|HK|$ divides $|G|$.

PROOF

Any subgroup of an abelian group is normal, so $H$ and $K$ are both normal subgroups of $G$ which implies $HK$ is a subgroup of $G$. The result follows from Lagrange's theorem which says that the order of any subgroup must divide the order of the group.

Galois Theory: A problem involving irreducible polynomials in a field of nonprime characteristic

PROBLEM
Let $p$ be a prime not equal to the characteristic of $F$. Show that if $a\in F$ then $x^p-a$ is either irreducible in $F[x]$ or has a root in $F$.

PROOF
Let $f=x^p-a$ be as above and let $r$ be a root of $f$ in $E_f$, the extension with roots of $f$. The roots of $f$ are $r\xi^i$  where $\xi$ is a primitive root of unity. Note that $f'(x)=px^{p-1}\neq 0$ since the characteristic of $F$ is assumed not to be $p$. So $(f,f')=1$ which implies $f$ is seperable and $E_f=F(r,\xi)$ is Galois.  We have two cases now: either $r\in F(\xi)$ or $r\not\in F(\xi)$.

 If $r\not\int F(\xi)$ then $|Gal(E_f\backslash F(\xi))|=[E_f:  F(\xi)]>1$. So there exists $\alpha\neq 1\in Gal(E_f\backslash F(\xi))$, $E_f=F(\xi,r)$. So by definition of $\alpha$ there must be another root $r\xi^i$ for $i\neq 0$. Then $\alpha^k(v)=r\xi^{ki}$ implies $\alpha$ has order $p$. So $p=|\alpha| \leq [E_f:F(\xi)]\leq deg(m_r(x))\leq deg(f)=p$. So $f(x)$ is the minimum polynomial and is hence irreducible.

Now consider the case where $r\in F(\xi)$. Let $G$ be $Gal(F(\xi)\backslash F)$ and define $H$ to be a subgroup of $U(\mathbb{Z}_p)$. Note that $H$ is cyclic. Let $G=<\alpha>$ $\alpha_k(r)=r\xi^\ell$ so $\alpha_k(r)\alpha_k(\xi^i)=r\xi^\ell\xi^{ki}=r\xi^{ki+\ell}$. Let $ki+\ell=i$ so $i(1-k)=\ell$. Then $k=1$, $i=1$ implies that $f(x)$ has a root in $F$, else $i=\ell(1+k)^{-\ell}$ so $\alpha_k(r\xi^i)=r\xi^i$ for $\xi^i$ fixed by $\alpha_k$. But $G=<\alpha_k>$, $r\xi^i\in Inv(G)=F$. So $f(x)$ has a root in $F$. 

Ring Theory: A PID identity

PROBLEM
Let D be a principal ideal domain and $h=\gcd(f,g)$. Prove there exists $p$ and $q$ in $D$ such that $pf+gq=h$.

PROOF
We denote the ideal generated by $x$ as $(x)$. Since $D$ is a PID, we have that $(h)=(gcd(f,g))=(f)+(g)$ where $(f), (g)$ were all generated by one element. So by definition of principle idea, there exists $p,q\in D$ such that $pf=(f)$ and $qg=(g)$ so $h=pf+qg$ (up to units of $h$). 

Metric Spaces: a fixed point problem

Problem taken from my first semester analysis course homework

PROBLEM
Let $(X,d)$ be a compact metric space and $f:X\rightarrow X$ be a mapping such that $d(f(x),f(y)) < d(x,y)$ for all $x\neq y$. Show that there exists a unique $x_0\in X$ such that $f(x_0)=x_0$.

PROOF
For all $x,y\in X, x\neq y$ we have that $d(f(x),f(y)) < d(x,y)$. Thus $f:X \rightarrow X$ is continuous. Note that $g(x):=d(x,f(x))$ is also continuous. Recall that continuous functions functions on compact sets have a minimum on $X$. Thus let $d(x_0, f(x_0))=\inf\{d(x,f(x)):x\in X\}$. Assume by way of contradiction that $x_0\neq f(x_0)$. Then $d(f(x_0), f(f(x_0))) < d(x_0,f(x_0))$ which contradicts the fact that $g$ has a minimum at $x_0$. To see uniqueness, assume there exists $y\neq x_0$ where $f(y)=y$. But then $d(x_0, y)=d(f(x_0),f(y)) < d(x_0,y)$ which is a contradiction.