# Proving the Dirichlet Kernel

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

Note by Ethan Robinett
3 years, 9 months ago

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- 3 years, 9 months ago