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

Question

2700 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-04T12:51:50

#T_0(1.5)# or briefly,

#T_0=1#.

#T_1=1.5#

#T_2=2(1.5)(1.5)T_1-T_0=4.5-1=3.5#,

using #T_n = 2xT_(n-1)-T_(n-2), n>=2#.

#T_3=3(3.5)-1.5=9#

#T_4=3(9)-3.5=23.5#

#T_5=3(23.5)-9=61.5#

#T_6=3(61.5)-23.5=161#

#T_7=3(161)-61.5=421.5#

From wiki Chebyshev Polynomials Table,.

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