91Ó°ÊÓ

The Chebyshev polynomials of the first kind, \(T_n(x)\), are defined by the recursive relations: \(T_0(x) = 1\) \(T_1(x) = x\) \(T_{n+1}(x) = 2xT_n(x) - T_{n-1}(x)\) for \(n \geq 1\)

We want to show that: \(2 T_{m}(x) T_{n}(x) = T_{m+n}(x) + T_{m-n}(x)\) for \(m > n\) To prove (a), let's recall the identity: \(T_m(x) = cos(m \cdot arccos(x))\) Now multiply \(T_{m}(x)\) with \(T_{n}(x)\), then we have: \(2T_{m}(x) T_{n}(x) = 2 \cos(m \cdot \arccos(x))\cos(n \cdot \arccos(x))\) Applying the sum-to-product trigonometric identity, this is equal to: \(T_{m+n}(x) + T_{m-n}(x) = cos((m+n) \cdot \arccos(x)) + cos((m-n) \cdot \arccos(x))\) Thus, Property (a) holds true.

We want to show that: \(T_{m}\left(T_{n}(x)\right)=T_{m n}(x)\) To prove (b), let's use the identity \(T_m(x) = cos(m \cdot arccos(x))\). And remember that \(T_n(x) = cos(n \cdot \arccos(x))\). Thus, \(T_{m}(T_{n}(x)) = T_{m}(cos(n \cdot \arccos(x)))\). Now substitute \(y = cos(n \cdot \arccos(x))\) and we get: \(T_m(y) = cos(m \cdot \arccos(y))\) Then, by the composition of functions, we have: \(T_{m}(T_{n}(x)) = cos(m \cdot \arccos(cos(n \cdot \arccos(x))))\) Using the double angle formula for cosine, we can simplify this expression: \(T_{m}(T_{n}(x)) = cos(m \cdot (n \cdot \arccos(x)))\) As a result, we get: \(T_{m}(T_{n}(x)) = cos(mn \cdot \arccos(x))\) This is equal to: \(T_{mn}(x) = cos(mn \cdot \arccos(x))\) Thus, Property (b) holds true.