Showing that n+11Tn+1’(x)=sinθsin((n+1)θ) can be done using the differentiation of inverse trigonometric functions as follows.
Un(x)===n+11[cos((n+1)cos−1x)]’=n+1n+11−x2−1[−sin((n+1)cos−1x)]1−x2sin((n+1)cos−1x)sinθsin((n+1)θ)
The second kind Chebyshev polynomial is very useful not only in numerical analysis but also in applied mathematics as a whole, boasting interesting properties along with the first kind Chebyshev polynomial.
Meanwhile, the second kind Chebyshev polynomial can also be defined in reverse using U0(x)=1, U1(x)=2x, and the recursive formula [0]. This is also true for the first kind Chebyshev polynomial, and the reason for naming the first and second kinds is considered to be due to T1(x)=1⋅x and U1(x)=2⋅x.
Proof
[0]
By differentiating both sides of the recursive formula Tn+1(x)=2xTn(x)−Tn−1(X) of the first kind Chebyshev polynomial
Tn+1′(x)=2Tn(x)+2xTn′(x)−Tn−1′(x)
since Tn+1′(x)=(n+1)Un(x),
(n+1)Un(x)=2Tn(x)+2xnUn−1(x)−(n−1)Un−2(x)
and combining as n,
n[Un(x)−2xUn−1(x)+Un−2(x)]=2Tn(x)+Un−2(x)−Un(x)
n[Un(x)−2xUn−1(x)+Un−2(x)]=0
Dividing both sides by n and rearranging yields
Un+1(x)=2xUn(x)−Un−1(x)
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[1]
Since dx=−sinθdθ=−1−x2dθ and sinθ=1−x2,
⟨Un,Um⟩====∫−11Un(x)Um(x)1−x2dx−∫π0sin2θsin((n+1)θ)sin((m+1)θ)sin2θdθ∫0πsin((n+1)θ)sin((m+1)θ)dθ{π/20,n=m,n=m
thus {U0,U1,U2,⋯} forms an orthogonal set.
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[2]
It is self-evident by definition.
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[3]
Case 1. n=0,1
U0(−x)=U1(−x)=1=U0(x)2(−x)=−2x=−U1(x)
Case 2. n≥2 is even
Since the degree of all terms that are not 0 in Un(x) is even, Un(−x)=Un(x)
Case 3. n≥2 is odd
Since the degree of all terms that are not 0 in Un(x) is odd, Un(−x)=−Un(x)
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Implementation
Below is the code for the Chebyshev polynomial written in R.
Since it returns the polynomial itself, it can be directly used for calculations. n is the degree, and by setting kind and turning the print option to true, it will print the coefficients.
The printed coefficients are from the constant term to the higher-order terms in sequence, and since the second kind Chebyshev polynomial is U3(x)=8x3−4x, it is correct. The value of the function is also accurately calculated as U3(3)=8⋅33−4⋅3=216−12=204.
Chebyshev<-function(n,kind=1,print=F){
p<-NAif((round(n)-n)!=0| n<0){stop("Wrong Degree!!")}#degree must be nonnegative integerif(!kind%in%(1:2)){stop("Wrong Kind!!")}#kind must be 1 or 2if(n==0){if(print){print(1)}
p<-function(x){return(1)}return(p)}if(n==1){if(print){print(c(0,kind))}
p<-function(x){return(kind*x)}return(p)}
coef0<-c(1)
coef1<-c(0,kind)for(i in1:(n-1)){
coef2<-(c(0,2*coef1)-c(coef0,0,0))
coef0<-coef1
coef1<-coef2
}if(print){print(coef2)}
p<-function(x){return(sum(coef2*x^(0:n)))}return(p)}
p<-Chebyshev(1,2); p(2)
p<-Chebyshev(3,2,T); p(3)