I am interested on finding translation and multiplicative identities for Chebyshev polynomials in the way they exist for Bernoulli ones:
$$T_n (x+y) = ?$$
$$T_n (ax) = ?$$
Thanks
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Inverse fact this is an implícit definición as is $$T_n (cos(x)) $$ – 24th_moonshine Jul 10 '16 at 21:12
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I ask for an explícit one. Thanks – 24th_moonshine Jul 10 '16 at 21:13
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Your quest has a "natural" limitation, explaining why the first formula $T_n(a+b)$ you will get is with very few interest. The intuitive reason; the Chebyshev polynomials natural definition is $T_n (cos(x))=cos(nx)$. It means that their "natural" definition domain is [-1,+1]. Taking $T_n(a+b)$, you will have cases where either $a+b<-1$or $a+b>1$ – Jean Marie Jul 10 '16 at 21:32
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Sometimes in little pot there is best jam – 24th_moonshine Jul 11 '16 at 05:38
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One thing is how you define them, and another if they are easy to be extended to broader domains. It is quite similar as what happens on gamma and zeta functions, defined in half complex plane but extended easily afterwards to whole complex domain. – 24th_moonshine Jul 11 '16 at 05:49