Difference between revisions of "Chebyshev polynomials of the first kind"
(Created page with "The Chebyshev polynomials of the first kind are defined recursively by <cmath>T_0(x) = 1,</cmath> <cmath>T_1(x) = x,</cmath> <cmath>T_{n+1}(x) = 2xT_n(x) - T_{n-1}(x),</cmath>...") |
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===First proof=== | ===First proof=== | ||
+ | By the trigonometric definition, <math>T_m(T_n(x)) = \cos(m(\arccos(\cos(n(\arccos(x))))))</math>. | ||
+ | |||
+ | As before, let <math>\arccos x = y</math>. We have <math>\arccos(\cos(ny)) = 2k\pi \pm ny</math> for some integer <math>k</math>. Multiplying by <math>m</math> and distributing gives <math>2mk\pi \pm mny</math>; taking the cosine gives <math>\cos (2mk\pi \pm mny) = \cos mny = \cos ( mn (\arccos x)) = T_{mn}(x)</math>. | ||
+ | |||
+ | For now this proof only applies where the trigonometric definition is defined; that is, for <math>x \in [-1,1]</math>. However, <math>T_{mn}(x)</math> is a degree-<math>mn</math> polynomial, and so is <math>T_m(T_n(x))</math>, so the fact that <math>T_{mn}(x) = T_m(T_n(x))</math> for some <math>mn + 1</math> distinct <math>x \in [-1,1]</math> is sufficient to guarantee that the two polynomials are equal over all real numbers. | ||
===Second proof (Induction)=== | ===Second proof (Induction)=== |
Revision as of 19:03, 28 February 2022
The Chebyshev polynomials of the first kind are defined recursively by
or equivalently by
Contents
Proof of equivalence of the two definitions
In the proof below, will refer to the recursive definition.
For the base case,
for the
base case,
Now for the inductive step, let , so that
. We then assume that
and
, and we wish to prove that
.
From the cosine sum and difference identities we have and
The sum of these equations is
rearranging,
Substituting our assumptions yields
as desired.
Composition identity
For nonnegative integers and
, the identity
holds.
First proof
By the trigonometric definition, .
As before, let . We have
for some integer
. Multiplying by
and distributing gives
; taking the cosine gives
.
For now this proof only applies where the trigonometric definition is defined; that is, for . However,
is a degree-
polynomial, and so is
, so the fact that
for some
distinct
is sufficient to guarantee that the two polynomials are equal over all real numbers.