- #1
Muh. Fauzi M.
- 17
- 1
This is something Chebyshev polynomial problems. I need to show that:
##\sum_{r=0}^{n}T_{2r}(x)=\frac{1}{2}\big ( 1+\frac{U_{2n+1}(x)}{\sqrt{1-x^2}}\big )##
by using two type of solution :
##T_n(x)=\cos(n \cos^{-1}x)## and ##U_n(x)=\sin(n \cos^{-1}x)## with ##x=\cos\theta##,
I have form the complex superposition:
##T_n(x)+iU_n(x)=(x+i\sqrt{1-x^2})^n##
and expand it by binomial theorem to get :
##T_n(x)=x^n-\dbinom{n}{2}x^{n-2}(1-x^2)+\dbinom{n}{4}x^{n-4}(1-x^2)^2-...##
and
##U_n(x)=\sqrt{1-x^2}\big[ \dbinom{n}{1}x^{n-1}-\dbinom{n}{3}(1-x^2)+... \big]##
I try to change ##T_n(x)## to ##T_{2r}(x)## and ##U_n(x)## to ##U_{2n+1}(x)##, but still stuck and can't solve the problem.
Any one can help solve this?
##\sum_{r=0}^{n}T_{2r}(x)=\frac{1}{2}\big ( 1+\frac{U_{2n+1}(x)}{\sqrt{1-x^2}}\big )##
by using two type of solution :
##T_n(x)=\cos(n \cos^{-1}x)## and ##U_n(x)=\sin(n \cos^{-1}x)## with ##x=\cos\theta##,
I have form the complex superposition:
##T_n(x)+iU_n(x)=(x+i\sqrt{1-x^2})^n##
and expand it by binomial theorem to get :
##T_n(x)=x^n-\dbinom{n}{2}x^{n-2}(1-x^2)+\dbinom{n}{4}x^{n-4}(1-x^2)^2-...##
and
##U_n(x)=\sqrt{1-x^2}\big[ \dbinom{n}{1}x^{n-1}-\dbinom{n}{3}(1-x^2)+... \big]##
I try to change ##T_n(x)## to ##T_{2r}(x)## and ##U_n(x)## to ##U_{2n+1}(x)##, but still stuck and can't solve the problem.
Any one can help solve this?