Question

1. be The Chebyshev polynomials of the second type So, S1, S2, ... follow the Sj+1(t) = 2t S;(t) - Sj-1(t), j = 1,2,... with So(t) = 1 and Si(t) = 2t. Prove for Sj(t), the leading term is 23+), i.e., S;(t) = 23+3 + L.D.T

Answer 1

. polynomials So, so, S2... be the chebyshen follow the Sjri(t) = 2t S; (+) - s574) 9=1,2, .. with solt) = 1 S(+) =24 for Prove 2° Sylt), the heading term is lie S.(t) = 2 + + lower Degece term (h.D. to? (h. D.T.) Induction : soil proof By * n=1 Sy(+) = 2t + otº S24+) = 2+ť + L.O.T valid for not (i=1) * n=2 o Sy(t) = 245/4+) - Solt) Syty(+) = 2+(2+) - 1 S2(t) = 4+2.1 solt)=1 S (+) = 2t

-S of S2 (4) = 24 - 2 $24) = 2ť + LD.I valid for n=2 Strong Induction * Let given statement is true / valid for nej- PC3-1) Sj,44) = Su-2)(t) = 2 tsjąlt) - Sjişte S; -, (t) = 23! " + L.O.T. -O here hower Degre is ajal we have to prove that given statement toe for n=3 By help of o e [ 5; (t) = 2! + + L..D.T.] Now is By given statement S;lt) = 27. Sj., (+) - Sinets .: j = (j-1) + → sy 6) = 24) (234.+1.0.7) = $1,64) S;(t) = 28-14. £4+ + 2.t.(L.DT) +- $j.24) . Degree in Silt) = 24 + (L.DT)" - 33-2, 1. u Degree ai

o By siz has maximum Degace 3-2 [By Induction] Sp(t)=2 + + (Polynomial of Degree i) . 1 + (polynomial of Degree < 5-2) S; (t) = 2* + L.D.T. so given statement is true for all i. HENCE PYOVED STRORG INDUCTION 1 Base Case :- Prove the statement for first step 2. Induction Hypothesis - Assume- kleok - the steps you are currently steping on pina), naj - STRONG s. The steps that you have steped on before including the current one (in our case nasel) (pin) , nein) Inductive step . 3 Prove next step THANK YOU!

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