- #1
zetafunction
- 391
- 0
given the 'normalized' Chebyshev and Legendre Polynomials
[tex] \frac{L_{2n}(x)}{L_{2n}(0)} [/tex] and [tex] \frac{T_{2n}(x)}{T_{2n}(0)} [/tex]
for n even and BIG 2n--->oo
then it would be true that (in this limit) [tex] \frac{L_{2n}(x)}{L_{2n}(0)}=\frac{sin(x)}{2x} [/tex] and [tex] \frac{T_{2n}(x)}{T_{2n}(0)}=J_{0}(2x) [/tex]
here 'T' is used for Chebyshev polynomials and 'L' is used for Legendre ones , J0 is the zeroth order Bessel function.
Curiously enough the sign of the polynomials and the Taylor series representation for the functions follow both a sequence
[tex] 1+ \sum_{n\ge 1}a(2n)(-1)^{n}x^{2n} [/tex] with ALL the a(2n) being positive numbers.
[tex] \frac{L_{2n}(x)}{L_{2n}(0)} [/tex] and [tex] \frac{T_{2n}(x)}{T_{2n}(0)} [/tex]
for n even and BIG 2n--->oo
then it would be true that (in this limit) [tex] \frac{L_{2n}(x)}{L_{2n}(0)}=\frac{sin(x)}{2x} [/tex] and [tex] \frac{T_{2n}(x)}{T_{2n}(0)}=J_{0}(2x) [/tex]
here 'T' is used for Chebyshev polynomials and 'L' is used for Legendre ones , J0 is the zeroth order Bessel function.
Curiously enough the sign of the polynomials and the Taylor series representation for the functions follow both a sequence
[tex] 1+ \sum_{n\ge 1}a(2n)(-1)^{n}x^{2n} [/tex] with ALL the a(2n) being positive numbers.