Using Chebyshev Polynomial $T_{n} (x) = cosh (n (a r c cosh (x))), x > = 1$ $T_n (x)=cosh( n( arc cosh(x))), x > = 1$ and the recurrence relation $T_{n + 2} (x) = 2 x T_{n + 1} (x) - T_{n} (x)$ $T_(n+2)(x)=2xT_(n+1)(x) - T_n (x)$ , with $T_{0} (x) = 1 and T_{1} (x) = x$ $T_0(x)=1 and T_1(x)=x$ , how do you porve that $cosh (7 a r c cosh (1.5)) = 421.5 ?$ $cosh(7 arc cosh(1.5))=421.5?$

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Answer 1 · 2016-10-04T12:51:50

$T_{0} (1.5)$ $T_0(1.5)$ or briefly,

$T_{0} = 1$ $T_0=1$ .

$T_{1} = 1.5$ $T_1=1.5$

$T_{2} = 2 (1.5) (1.5) T_{1} - T_{0} = 4.5 - 1 = 3.5$ $T_2=2(1.5)(1.5)T_1-T_0=4.5-1=3.5$ ,

using $T_{n} = 2 x T_{n - 1} - T_{n - 2}, n \geq 2$ $T_n = 2xT_(n-1)-T_(n-2), n>=2$ .

$T_{3} = 3 (3.5) - 1.5 = 9$ $T_3=3(3.5)-1.5=9$

$T_{4} = 3 (9) - 3.5 = 23.5$ $T_4=3(9)-3.5=23.5$

$T_{5} = 3 (23.5) - 9 = 61.5$ $T_5=3(23.5)-9=61.5$

$T_{6} = 3 (61.5) - 23.5 = 161$ $T_6=3(61.5)-23.5=161$

$T_{7} = 3 (161) - 61.5 = 421.5$ $T_7=3(161)-61.5=421.5$

From wiki Chebyshev Polynomials Table,.

#T_7(x)=64x^7-112x^5+56x^3-7x