We will prove that: \(D_n(x)=\sum_{k=-n}^{n} e^{ikx} = 1+2\sum_{k=1}^{n} cos(kx) = \frac{sin((n+\frac{1}{2})x)}{sin(\frac{x}{2})} \forall n \in \mathbb{N}\)

Part (1): Note that, by Euler's Formula: \( e^{ikx}=cos(kx)+isin(kx)\)

\(\Rightarrow\) \(\sum_{k=-n}^{n} e^{ikx} = \sum_{k=-n}^{n} cos(kx) +isin(kx)\)

\( =\sum_{k=-n}^{0} cos(kx) +isin(kx) +\sum_{k=0}^{n} cos(kx) +isin(kx)\)

Using the fact that: \( sin(-nx)=-sin(nx)\) and \(cos(-nx)=cos(nx)\):

\(\sum_{k=-n}^{0} cos(kx) +isin(kx) +\sum_{k=0}^{n} cos(kx) +isin(kx)\) =\( \sum_{k=-n}^{n} cos(kx)\)

Because all \(isin(kx)\) terms in {\(\mathbb{-n,...,0}\)} are negative and all \(isin(kx)\) terms in {\(\mathbb{0,...,n}\)} are positive, hence they cancel.

Since \(cos(0)=1\) and \(cos(-nx)=cos(nx)\):

\(\sum_{k=-n}^{n}cos(kx)=1+2\sum_{k=1}^{n} cos(kx)\)

Then we have: \(\sum_{k=-n}^{n} e^{ikx} = 1+2\sum_{k=1}^{n} cos(kx)\)

Part (2): We will now prove using mathematical induction that \(1+2\sum_{k=1}^{n} cos(kx) = \frac{sin((n+\frac{1}{2})x)}{sin(\frac{x}{2})}\)

We establish a base case as \(n=1\):

\(1+2\sum_{k=1}^{1} cos(kx) = \frac{sin(\frac{3x}{2})}{sin(\frac{x}{2})}\)

\(\Rightarrow\) \(1+2cos(x) = \frac{sin(\frac{3x}{2})}{sin(\frac{x}{2})}\)

Now it is not very apparent that this is actually true. However, Lagrange's trigonometric formula states: \( \sum_{k=1}^{n} cos(kx) =\frac{-1}{2}+ \frac{sin((n+\frac{1}{2})x)}{2sin(\frac{x}{2})}\)

If we add \(\frac{1}{2}\) to both sides of this formula,multiply both sides by two, and let \(n=1\) we arrive at the desired equivalence relation and hence prove the base case. Now we establish the inductive case as \(n=m\):

\(\Rightarrow\) \(1+2\sum_{k=1}^{m} cos(kx) = \frac{sin((m+\frac{1}{2})x)}{sin(\frac{x}{2})}\)

\(\Rightarrow\) \(2cos((m+1)x)+1+2\sum_{k=1}^{m} cos(kx) = 2cos((m+1)x)+\frac{sin((m+\frac{1}{2})x)}{sin(\frac{x}{2})}\) (By hypothesis)

\(\Rightarrow\) \(1+2\sum_{k=1}^{m+1} cos(kx) = \frac{(2cos((m+1)x))(sin(\frac{x}{2}))+sin((m+\frac{1}{2})x)}{sin(\frac{x}{2})}\)

Using the fact that \(cos(u)sin(v)=\frac{1}{2}(sin(u+v)-sin(u-v))\):

\(=\frac{sin((m+\frac{3}{2})x)-sin((m+\frac{1}{2})x)+sin((m+\frac{1}{2})x)}{sin(\frac{x}{2})}\)

\(\Rightarrow\) \(1+2\sum_{k=1}^{m+1} cos(kx)=\frac{sin((m+\frac{3}{2})x)}{sin(\frac{x}{2})}\)

So we may conclude that if the statement is true for some \(n=m\) then it is true for some \(n=m+1\). The proof follows by induction.

Then, compiling all of this, we have: \(D_n(x)=\sum_{k=-n}^{n} e^{ikx} = 1+2\sum_{k=1}^{n} cos(kx) = \frac{sin((n+\frac{1}{2})x)}{sin(\frac{x}{2})}\)

Which was to be proved.

QED

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