Parking space for various notes about the sinc function and slow motion step by step proof of value at $x=0$<\/p>\n\n\n\n
If $f$ maps to an ordered set such as $\\mathbb{R}$ or $\\mathbb{Z}$ and is continuous on closed interval $[a,b]$, then<\/p>\n\n\n\n
$$
\\boxed{
\\begin{align}
&\\exists c \\in [a,b] : \\forall x \\in [a,b], f(c) \\geq f(x) \\\\
&\\exists d \\in [a,b] : \\forall x \\in [a,b], f(d) \\leq f(x) \\\\
\\end{align}
}
$$<\/p>\n\n\n\n
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:<\/p>\n\n\n\n
$$
\\begin{align}
&\\text{Case 1: } f \\text{ is constant on } [a,b] \\\\
&\\implies \\\\
&\\exists k \\in \\mathbb{R} : \\forall x \\in [a,b], f(x) = k \\\\
&\\implies \\\\
&\\forall x \\in (a,b), f'(x) = 0 \\\\
\\\\
&\\text{Case 2: } f \\text{ has a maximum on } (a,b) \\\\
&\\implies \\\\
&\\exists c \\in (a,b) : f(c) = \\max(f) \\text{ on } [a,b] &&\\text{ Extreme Value Theorem} \\\\
&\\implies \\\\
&f'(c) = 0 \\\\
\\\\
&\\text{Case 3: } f \\text{ has a minimum on } (a,b) \\\\
&\\implies \\\\
&\\exists c \\in (a,b) : f(c) = \\min(f) \\text{ on } [a,b] &&\\text{ Extreme Value Theorem} \\\\
&\\implies \\\\
&f'(c) = 0 \\\\
\\end{align}
$$<\/p>\n\n\n\n
The result of which is the conclusion that<\/p>\n\n\n\n
$$
\\begin{align}
&f \\text{ is continuous on } [a,b] \\land \\\\
&f \\text{ is differentiable on } (a,b) \\land \\\\
&f(a) = f(b) \\\\
&\\implies \\\\
\\end{align}
$$<\/p>\n\n\n\n
$$
\\boxed{
\\exists c \\in (a,b) : f'(c) = 0 \\\\
}
$$<\/p>\n\n\n\n
The MVT starts with a very similar premise as Rolle’s theorem, but widens the scope and applicability.<\/p>\n\n\n\n
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<\/p>\n\n\n\n
$$
\\begin{align}
g(x) &= f(x) – \\left( \\frac{f(b) – f(a)}{b – a} \\right) (x – a) \\\\
\\end{align}
$$<\/p>\n\n\n\n
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).<\/p>\n\n\n\n
Note also that this helper function achieves all the conditions for Rolle’s Theorem, namely<\/p>\n\n\n\n
$$
\\begin{align}
g(a) &= f(a) \\\\
\\\\
g(b) &= f(b) – \\left( \\frac{f(b) – f(a)}{b – a} \\right) (b – a) \\\\
g(b) &= f(b) – [f(b) – f(a)] \\\\
g(b) &= f(a) \\\\
\\\\
&\\implies g(a) = g(b) \\\\
\\end{align}
$$<\/p>\n\n\n\n
From Rolle’s Theorem, $\\exists c \\in (a,b) : g'(c) = 0$. From differentiating our helper function $g$ we also can find that<\/p>\n\n\n\n
$$
\\begin{align}
g'(x) &= f'(x) – \\left( \\frac{f(b) – f(a)}{b – a} \\right) \\\\
g'(c) &= f'(c) – \\left( \\frac{f(b) – f(a)}{b – a} \\right) \\\\
0 &= f'(c) – \\left( \\frac{f(b) – f(a)}{b – a} \\right) \\\\
f'(c) &= \\frac{f(b) – f(a)}{b – a} \\\\
\\end{align}
$$<\/p>\n\n\n\n
In conclusion,<\/p>\n\n\n\n
$$
\\begin{align}
&f \\text{ is continuous on } [a,b] \\land \\\\
&f \\text{ is differentiable on } (a,b) \\\\
&\\implies \\\\
\\end{align}
$$<\/p>\n\n\n\n
$$
\\boxed{
\\exists c \\in (a,b) : f'(c) = \\frac{f(b) – f(a)}{b – a} \\\\
}
$$<\/p>\n\n\n\n
Cauchy Mean Value Theorem further generalizes the learnings of Rolle’s Theorem and the MVT for use with two functions.<\/p>\n\n\n\n
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<\/p>\n\n\n\n
$$
\\begin{align}
h(x) &= f(x) – \\lambda g(x) &&\\lambda \\in \\mathbb{R} \\\\
\\end{align}
$$<\/p>\n\n\n\n
At interval endpoints we have<\/p>\n\n\n\n
$$
\\begin{align}
h(a) &= f(a) – \\lambda g(a) \\\\
h(b) &= f(b) – \\lambda g(b) \\\\
\\end{align}
$$<\/p>\n\n\n\n
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<\/p>\n\n\n\n
$$
\\begin{align}
h(a) &= h(b) \\\\
f(a) – \\lambda g(a) &= f(b) – \\lambda g(b) \\\\
f(b) – f(a) &= \\lambda [g(b) – g(a)] \\\\
\\lambda &= \\frac{f(b) – f(a)}{g(b) – g(a)} \\\\
\\end{align}
$$<\/p>\n\n\n\n
Using this specific $\\lambda$ to more explicitly define our auxiliary function we have<\/p>\n\n\n\n
$$
\\begin{align}
h(x) &= f(x) – \\frac{f(b) – f(a)}{g(b) – g(a)} g(x) \\\\
\\end{align}
$$<\/p>\n\n\n\n
Now we can now say from Rolle’s Theorem that<\/p>\n\n\n\n
$$
\\begin{align}
\\exists c \\in (a,b) : h'(c) = 0 &\\implies \\\\
h'(x) &= f'(x) – \\frac{f(b) – f(a)}{g(b) – g(a)} g'(x) \\\\
h'(c) &= f'(c) – \\frac{f(b) – f(a)}{g(b) – g(a)} g'(c) \\\\
0 &= f'(c) – \\frac{f(b) – f(a)}{g(b) – g(a)} g'(c) \\\\
f'(c) &= \\frac{f(b) – f(a)}{g(b) – g(a)} g'(c) \\\\
\\frac{f'(c)}{g'(c)} &= \\frac{f(b) – f(a)}{g(b) – g(a)} \\\\
\\end{align}
$$<\/p>\n\n\n\n
In conclusion,<\/p>\n\n\n\n
$$
\\begin{align}
&f, g \\text{ are continuous on } [a,b] \\land \\\\
&f, g \\text{ are differentiable on } (a,b) \\land \\\\
&\\forall x \\in [a,b], g'(x) \\neq 0 \\\\
&\\implies \\\\
\\end{align}
$$<\/p>\n\n\n\n
$$
\\boxed{
\\exists c \\in (a,b) : \\frac{f'(c)}{g'(c)} = \\frac{f(b) – f(a)}{g(b) – g(a)} \\\\
}
$$<\/p>\n\n\n\n
L’H\u00f4pital’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<\/p>\n\n\n\n
$$
\\begin{align}
\\lim_{x \\to c} \\frac{f(x)}{g(x)} \\\\
\\\\
f(c) = 0 \\\\
g(c) = 0 \\\\
\\end{align}
$$<\/p>\n\n\n\n
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 $g'(c) \\neq 0$.<\/p>\n\n\n\n
Now we define a new variable $\\xi$ between the input variable $x$ and the limit point $c$.<\/p>\n\n\n\n
$$
\\begin{align}
\\xi \\in (x, c) \\\\
\\end{align}
$$<\/p>\n\n\n\n
From CMVT<\/p>\n\n\n\n
$$
\\begin{align}
\\exists \\xi \\in (x,c) : \\frac{f'(\\xi)}{g'(\\xi)} = \\frac{f(c) – f(x)}{g(c) – g(x)} \\\\
\\end{align}
$$<\/p>\n\n\n\n
Since this approach is used in cases where $f(c) = g(c) = 0$, this is simplified to<\/p>\n\n\n\n
$$
\\begin{align}
\\exists \\xi \\in (x,c) : \\frac{f'(\\xi)}{g'(\\xi)} = \\frac{- f(x)}{- g(x)} &&f(c) = g(c) = 0 \\\\
\\exists \\xi \\in (x,c) : \\frac{f'(\\xi)}{g'(\\xi)} = \\frac{f(x)}{g(x)} &&f(c) = g(c) = 0 \\\\
\\end{align}
$$<\/p>\n\n\n\n
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<\/p>\n\n\n\n
$$
\\begin{align}
\\lim_{x \\to c} \\frac{f'(\\xi)}{g'(\\xi)} = \\lim_{x \\to c} \\frac{f(x)}{g(x)} &&f(c) = g(c) = 0 \\\\
\\lim_{x \\to c} \\frac{f'(x)}{g'(x)} = \\lim_{x \\to c} \\frac{f(x)}{g(x)} &&f(c) = g(c) = 0 \\\\
\\end{align}
$$<\/p>\n\n\n\n
So in summary<\/p>\n\n\n\n
$$
\\begin{align}
&f, g \\text{ are continuous around limit point } c \\land \\\\
&f, g \\text{ are differentiable around limit point } c \\land \\\\
&\\forall x \\text{ around limit point c}, g'(x) \\neq 0 \\land \\\\
&f(c) = g(c) = 0 \\\\
&\\implies \\\\
\\end{align}
$$<\/p>\n\n\n\n
$$
\\boxed{
\\lim_{x \\to c} \\frac{f(x)}{g(x)} = \\lim_{x \\to c} \\frac{f'(x)}{g'(x)}
}
$$<\/p>\n\n\n\n
The $\\text{sinc}$ function is defined as<\/p>\n\n\n\n
$$
\\begin{align}
\\text{sinc}(x) &= \\frac{\\sin(x)}{x} \\\\
\\end{align}
$$<\/p>\n\n\n\n
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’H\u00f4pital’s Rule to determine $\\text{sinc}(0)$<\/p>\n\n\n\n
$$
\\begin{align}
\\lim_{x \\to c} \\frac{f(x)}{g(x)} &= \\lim_{x \\to c} \\frac{f'(x)}{g'(x)} \\\\
\\\\
f(x) &= \\sin(x) \\\\
f'(x) &= \\cos(x) \\\\
g(x) &= x \\\\
g'(x) &= 1 \\\\
\\\\
\\lim_{x \\to c} \\frac{\\sin(x)}{x} &= \\lim_{x \\to c} \\frac{\\cos(x)}{1} \\\\
\\lim_{x \\to c} \\frac{\\sin(x)}{x} &= \\cos(c) \\\\
\\lim_{x \\to 0} \\frac{\\sin(x)}{x} &= \\cos(0) \\\\
\\lim_{x \\to 0} \\frac{\\sin(x)}{x} &= 1 \\\\
\\end{align}
$$<\/p>\n\n\n\n
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 $x = 1$.<\/p>\n\n\n\n
$$
\\begin{array}{ll}
\tP_{sinc}(x) \\approx \\cases{
\t\\begin{array}{ll}
\t\t1 \t\t\t&|x| < 1 \\\\
\t\t\\frac{1}{x}\t&|x| \\geq 1 \\\\
\t\\end{array}
\t} \\\\
\t
\tP_{sinc,dB}(x) \\approx \\cases{
\t\\begin{array}{ll}
\t\t20 log_{10}(1) \t\t\t\t&|x| < 1 \\\\
\t\t20 log_{10}(\\frac{1}{x})\t&|x| \\geq 1 \\\\
\t\\end{array}
\t} \\\\
\tP_{sinc,dB}(x) \\approx \\cases{
\t\\begin{array}{ll}
\t\t0 \t\t\t\t&|x| < 1 \\\\
\t\t-20 log_{10}(x)\t&|x| \\geq 1 \\\\
\t\\end{array}
\t} \\\\
\\end{array}
$$<\/p>\n\n\n\n
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<\/p>\n\n\n\n
$$
\\begin{array}{ll}
\tP_{sinc^2}(x) \\approx \\cases{
\t\\begin{array}{ll}
\t\t1 \t\t\t&|x| < 1 \\\\
\t\t\\frac{1}{x^2}\t&|x| \\geq 1 \\\\
\t\\end{array}
\t} \\\\
\tP_{sinc,dB}(x) \\approx \\cases{
\t\\begin{array}{ll}
\t\t0 \t\t\t\t&|x| < 1 \\\\
\t\t-40 log_{10}(x)\t&|x| \\geq 1 \\\\
\t\\end{array}
\t} \\\\
\\end{array}
$$<\/p>\n\n\n\n
This plot would have the same inflection point, but a steeper slope (40 dB\/decade)<\/p>\n","protected":false},"excerpt":{"rendered":"
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 […]<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":736,"menu_order":3,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"class_list":["post-971","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/neilfoxman.com\/index.php?rest_route=\/wp\/v2\/pages\/971","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/neilfoxman.com\/index.php?rest_route=\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/neilfoxman.com\/index.php?rest_route=\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/neilfoxman.com\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/neilfoxman.com\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=971"}],"version-history":[{"count":62,"href":"https:\/\/neilfoxman.com\/index.php?rest_route=\/wp\/v2\/pages\/971\/revisions"}],"predecessor-version":[{"id":1128,"href":"https:\/\/neilfoxman.com\/index.php?rest_route=\/wp\/v2\/pages\/971\/revisions\/1128"}],"up":[{"embeddable":true,"href":"https:\/\/neilfoxman.com\/index.php?rest_route=\/wp\/v2\/pages\/736"}],"wp:attachment":[{"href":"https:\/\/neilfoxman.com\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=971"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}