Using Chebyshev nth degree polynomial #T_n (x) = cos ( n ( arc cos ( x ) ) ), |x | < = 1,# how do you prove that #cos ( 10 ( arc cos ( 0.25) ) ) = 0.816894531#, nearly?

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Answer 1 · 2017-01-05T12:50:05

#1673/2048=0.816894531#, nearly

Let #x = cos theta#.

Here, x =1/4 (rational), and so, I expect the answer to be rational (

also when x =sqrt(rational).).

Using the expansion

#cos 10theta = 512 cos^10theta-1280 cos^8theta+1120 cos^6theta#

#-400 cos^4theta+50 cos^2theta-1#,

#T_(10) (x)=cos(10 arc cos (x))#

#= 512 x^10-1280 x^8+1120 x^6-400 x^4+50 x^2-1#.

So, #T_(10)(0.25)=T_(10)(1/4)#

#=512(1/4)^10-1280(1/4)&8+1120(1/4)^6-400(1/4)^4-59(1/4)^2-1#

#=1673/2048#, upon simplification.