Parking space for various notes about the sinc function and slow motion step by step proof of value at $x=0$
Value at $x=0$ Proof
Extreme Value Theorem
If $f$ maps to an ordered set such as $\mathbb{R}$ or $\mathbb{Z}$ and is continuous on closed interval $[a,b]$, then
$$
\boxed{
\begin{align}
&\exists c \in [a,b] : \forall x \in [a,b], f(c) \geq f(x) \\
&\exists d \in [a,b] : \forall x \in [a,b], f(d) \leq f(x) \\
\end{align}
}
$$
Rolle’s Theorem
For a function $f$ that is continuous on interval $[a,b]$ and differentiable on $(a,b)$ and points $a$ and $b$ satisfy $f(a) = f(b)$, then only three possible cases exist of which $f$ must satisfy at least one:
$$
\begin{align}
&\text{Case 1: } f \text{ is constant on } [a,b] \\
&\implies \\
&\exists k \in \mathbb{R} : \forall x \in [a,b], f(x) = k \\
&\implies \\
&\forall x \in (a,b), f'(x) = 0 \\
\\
&\text{Case 2: } f \text{ has a maximum on } (a,b) \\
&\implies \\
&\exists c \in (a,b) : f(c) = \max(f) \text{ on } [a,b] &&\text{ Extreme Value Theorem} \\
&\implies \\
&f'(c) = 0 \\
\\
&\text{Case 3: } f \text{ has a minimum on } (a,b) \\
&\implies \\
&\exists c \in (a,b) : f(c) = \min(f) \text{ on } [a,b] &&\text{ Extreme Value Theorem} \\
&\implies \\
&f'(c) = 0 \\
\end{align}
$$
The result of which is the conclusion that
$$
\begin{align}
&f \text{ is continuous on } [a,b] \land \\
&f \text{ is differentiable on } (a,b) \land \\
&f(a) = f(b) \\
&\implies \\
\end{align}
$$
$$
\boxed{
\exists c \in (a,b) : f'(c) = 0 \\
}
$$
Mean Value Theorem
The MVT starts with a very similar premise as Rolle’s theorem, but widens the scope and applicability.
For a function $f$ that is continuous on interval $[a,b]$ and differentiable on $(a,b)$, we can create an auxiliary function used for this proof defined as
$$
\begin{align}
g(x) &= f(x) – \left( \frac{f(b) – f(a)}{b – a} \right) (x – a) \\
\end{align}
$$
Because $g$ is a linear combination of $f$ and $x$, we also know that $g$ is continuous on interval $[a,b]$ and differentiable on $(a,b)$. (TBD how this is proven).
Note also that this helper function achieves all the conditions for Rolle’s Theorem, namely
$$
\begin{align}
g(a) &= f(a) \\
\\
g(b) &= f(b) – \left( \frac{f(b) – f(a)}{b – a} \right) (b – a) \\
g(b) &= f(b) – [f(b) – f(a)] \\
g(b) &= f(a) \\
\\
&\implies g(a) = g(b) \\
\end{align}
$$
From Rolle’s Theorem, $\exists c \in (a,b) : g'(c) = 0$. From differentiating our helper function $g$ we also can find that
$$
\begin{align}
g'(x) &= f'(x) – \left( \frac{f(b) – f(a)}{b – a} \right) \\
g'(c) &= f'(c) – \left( \frac{f(b) – f(a)}{b – a} \right) \\
0 &= f'(c) – \left( \frac{f(b) – f(a)}{b – a} \right) \\
f'(c) &= \frac{f(b) – f(a)}{b – a} \\
\end{align}
$$
In conclusion,
$$
\begin{align}
&f \text{ is continuous on } [a,b] \land \\
&f \text{ is differentiable on } (a,b) \\
&\implies \\
\end{align}
$$
$$
\boxed{
\exists c \in (a,b) : f'(c) = \frac{f(b) – f(a)}{b – a} \\
}
$$
Cauchy Mean Value Theorem
Cauchy Mean Value Theorem further generalizes the learnings of Rolle’s Theorem and the MVT for use with two functions.
With two functions $f$ and $g$ that are both continuous on interval $[a,b]$ and differentiable on $(a,b)$ and $\forall x \in [a,b], g'(x) \neq 0$, we again use an auxiliary function
$$
\begin{align}
h(x) &= f(x) – \lambda g(x) &&\lambda \in \mathbb{R} \\
\end{align}
$$
At interval endpoints we have
$$
\begin{align}
h(a) &= f(a) – \lambda g(a) \\
h(b) &= f(b) – \lambda g(b) \\
\end{align}
$$
Again, $h$ is a linear combination of continuous differentiable functions, so it is expected to also be continuous and differentiable. We achieve the final criteria of Rolle’s Theorem by using a specific $\lambda$ such that
$$
\begin{align}
h(a) &= h(b) \\
f(a) – \lambda g(a) &= f(b) – \lambda g(b) \\
f(b) – f(a) &= \lambda [g(b) – g(a)] \\
\lambda &= \frac{f(b) – f(a)}{g(b) – g(a)} \\
\end{align}
$$
Using this specific $\lambda$ to more explicitly define our auxiliary function we have
$$
\begin{align}
h(x) &= f(x) – \frac{f(b) – f(a)}{g(b) – g(a)} g(x) \\
\end{align}
$$
Now we can now say from Rolle’s Theorem that
$$
\begin{align}
\exists c \in (a,b) : h'(c) = 0 &\implies \\
h'(x) &= f'(x) – \frac{f(b) – f(a)}{g(b) – g(a)} g'(x) \\
h'(c) &= f'(c) – \frac{f(b) – f(a)}{g(b) – g(a)} g'(c) \\
0 &= f'(c) – \frac{f(b) – f(a)}{g(b) – g(a)} g'(c) \\
f'(c) &= \frac{f(b) – f(a)}{g(b) – g(a)} g'(c) \\
\frac{f'(c)}{g'(c)} &= \frac{f(b) – f(a)}{g(b) – g(a)} \\
\end{align}
$$
In conclusion,
$$
\begin{align}
&f, g \text{ are continuous on } [a,b] \land \\
&f, g \text{ are differentiable on } (a,b) \land \\
&\forall x \in [a,b], g'(x) \neq 0 \\
&\implies \\
\end{align}
$$
$$
\boxed{
\exists c \in (a,b) : \frac{f'(c)}{g'(c)} = \frac{f(b) – f(a)}{g(b) – g(a)} \\
}
$$
L’Hôpital’s Rule
L’Hôpital’s Rule is used in cases where we want to find a limit when the functions in both the denominator and the numerator approach 0. More explicitly, if we want to find
$$
\begin{align}
\lim_{x \to c} \frac{f(x)}{g(x)} \\
\\
f(c) = 0 \\
g(c) = 0 \\
\end{align}
$$
In this case, we do not use the generic interval $[a,b]$ and instead inspect the interval used in the limit, $[x,c]$. Otherwise, we use a similar premise as above where functions $f$ and $g$ are both continuous on interval $[x,c]$ and differentiable on $(x,c)$ and $g'(c) \neq 0$.
Now we define a new variable $\xi$ between the input variable $x$ and the limit point $c$.
$$
\begin{align}
\xi \in (x, c) \\
\end{align}
$$
From CMVT
$$
\begin{align}
\exists \xi \in (x,c) : \frac{f'(\xi)}{g'(\xi)} = \frac{f(c) – f(x)}{g(c) – g(x)} \\
\end{align}
$$
Since this approach is used in cases where $f(c) = g(c) = 0$, this is simplified to
$$
\begin{align}
\exists \xi \in (x,c) : \frac{f'(\xi)}{g'(\xi)} = \frac{- f(x)}{- g(x)} &&f(c) = g(c) = 0 \\
\exists \xi \in (x,c) : \frac{f'(\xi)}{g'(\xi)} = \frac{f(x)}{g(x)} &&f(c) = g(c) = 0 \\
\end{align}
$$
As we don’t know what $\xi$ is, this finding is not yet that helpful. However, note that as $x \to c$, we also have $\xi \to c$ i.e. because the limit point is the same in both cases, it does not matter which input variable is used, so we may rewrite this as
$$
\begin{align}
\lim_{x \to c} \frac{f'(\xi)}{g'(\xi)} = \lim_{x \to c} \frac{f(x)}{g(x)} &&f(c) = g(c) = 0 \\
\lim_{x \to c} \frac{f'(x)}{g'(x)} = \lim_{x \to c} \frac{f(x)}{g(x)} &&f(c) = g(c) = 0 \\
\end{align}
$$
So in summary
$$
\begin{align}
&f, g \text{ are continuous around limit point } c \land \\
&f, g \text{ are differentiable around limit point } c \land \\
&\forall x \text{ around limit point c}, g'(x) \neq 0 \land \\
&f(c) = g(c) = 0 \\
&\implies \\
\end{align}
$$
$$
\boxed{
\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}
}
$$
Sinc Function
The $\text{sinc}$ function is defined as
$$
\begin{align}
\text{sinc}(x) &= \frac{\sin(x)}{x} \\
\end{align}
$$
Note that in the case where $x=0$, the $\text{sinc}$ function cannot be directly determined due to the denominator approaching 0, so instead we use L’Hôpital’s Rule to determine $\text{sinc}(0)$
$$
\begin{align}
\lim_{x \to c} \frac{f(x)}{g(x)} &= \lim_{x \to c} \frac{f'(x)}{g'(x)} \\
\\
f(x) &= \sin(x) \\
f'(x) &= \cos(x) \\
g(x) &= x \\
g'(x) &= 1 \\
\\
\lim_{x \to c} \frac{\sin(x)}{x} &= \lim_{x \to c} \frac{\cos(x)}{1} \\
\lim_{x \to c} \frac{\sin(x)}{x} &= \cos(c) \\
\lim_{x \to 0} \frac{\sin(x)}{x} &= \cos(0) \\
\lim_{x \to 0} \frac{\sin(x)}{x} &= 1 \\
\end{align}
$$
Frequency Response Approximations
When the sinc function describes a signal amplitude spectrum, the power in the spectrum may also be estimated as a constant 1 at “low values” and then $1/x$ for “higher values”. These two functions then converge at $1 = 1/x$ or $x = 1$.
$$
\begin{array}{ll}
P_{sinc}(x) \approx \cases{
\begin{array}{ll}
1 &|x| < 1 \\
\frac{1}{x} &|x| \geq 1 \\
\end{array}
} \\
P_{sinc,dB}(x) \approx \cases{
\begin{array}{ll}
20 log_{10}(1) &|x| < 1 \\
20 log_{10}(\frac{1}{x}) &|x| \geq 1 \\
\end{array}
} \\
P_{sinc,dB}(x) \approx \cases{
\begin{array}{ll}
0 &|x| < 1 \\
-20 log_{10}(x) &|x| \geq 1 \\
\end{array}
} \\
\end{array}
$$
On a magnitude plot, this looks like a flat line until $x = 1$, then a downward slope of 20 dB/decade. Similarly for a $sinc^2$ plot
$$
\begin{array}{ll}
P_{sinc^2}(x) \approx \cases{
\begin{array}{ll}
1 &|x| < 1 \\
\frac{1}{x^2} &|x| \geq 1 \\
\end{array}
} \\
P_{sinc,dB}(x) \approx \cases{
\begin{array}{ll}
0 &|x| < 1 \\
-40 log_{10}(x) &|x| \geq 1 \\
\end{array}
} \\
\end{array}
$$
This plot would have the same inflection point, but a steeper slope (40 dB/decade)