Question

The Chebyshev polynomials $T_n(x)$ can be written as $$T_n(x)=\cos (n{\cos }^{-1}x)$$. (a) Verify that these functions do satisfy the Chebyshev equation. (b) Use de Moivre's theorem to show that an alternative expression is $$T_n(x)=\sum _{r\mathrm{even}}^n(-1{)}^{r/2}\frac{n!}{(n-r)!r!}{x}^{n-r}{(1-x^2)}^{r/2}$$.

Short answer

Verify derivatives satisfy Chebyshev equation, then expand $T_n(x)$ using de Moivre's theorem and binomial expansion to get the alternative form.

Step-by-step solution

Understanding Chebyshev polynomials

Given the definition of the Chebyshev polynomials as $T_n(x)=\cos (n{\cos }^{-1}x)$, first, recall the Chebyshev equation, which is $(1-x^2)y^{″}-x{y}^{\prime }+n^{2}y=0$.

Finding Derivatives of $T_n(x)$

Find the first and second derivatives of $T_n(x)$. Use the chain rule to derive: $$y^{\prime }=-n\sin (n{\cos }^{-1}x)\cdot \frac{1}{\sqrt{1-x^2}}$$ and $$y^{″}=-n(-n{\sin }^2(n{\cos }^{-1}x)\cdot \frac{1}{1-x^2}+\cos (n{\cos }^{-1}x)\cdot \frac{1}{1-x^2}).$$

Substitute Derivatives into the Chebyshev Equation

Substitute $y$, $y^{\prime }$, and $y^{″}$ back into the Chebyshev equation: $$(1-x^2)y^{″}-x{y}^{\prime }+n^{2}y=0.$$ Simplify each term to verify the equation holds.

Apply de Moivre's Theorem

Recall de Moivre's theorem which states that $\cos (n\theta )=\mathrm{\Re }((\cos \theta +i\sin \theta {)}^n)$. Use this to express $T_n(x)=\cos (n{\cos }^{-1}x)$ in terms of powers of $x$ and trigonometric identities.

Expand Using Binomial Theorem

Expand $({\cos }^{-1}x+i\sin ({\cos }^{-1}x){)}^n$ as a binomial series: $$T_n(x)=\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})(i\sin ({\cos }^{-1}x){)}^k(\cos ({\cos }^{-1}x){)}^{n-k}.$$

Separate Real and Imaginary Parts

Extract the real part of the expansion by focusing on the even powers of $i$. This results in terms containing only $x$. Simplify each component to form the full expression.

Final Expression in Alternative Form

Combine terms according to whether the power of $x$ is even. The final expression is: $$T_n(x)=\sum _{r\text{ even}}^n(-1{)}^{r/2}\frac{n!}{(n-r)!r!}{x}^{n-r}{(1-x^2)}^{r/2}.$$

Key concepts

Chebyshev equation

The Chebyshev equation is important in understanding the behavior of Chebyshev polynomials. The equation is given as $$(1-x^2)y^{″}-x{y}^{\prime }+n^{2}y=0$$ Here, $y$ is a function of $x$, $y^{\prime }$ is the first derivative of $y$, and $y^{″}$ is the second derivative of $y$. Chebyshev polynomials, $T_n(x)$, satisfy this differential equation. To verify that $T_n(x)=\cos (n{\cos }^{-1}x)$ satisfies this equation, you first find the first and second derivatives of $T_n(x)$. Then, substitute these derivatives into the equation. Simplify the terms to confirm that the equation holds true. This step-by-step differentiation and substitution help establish the relationship between Chebyshev polynomials and their governing differential equation. Understanding how to manipulate and work with this equation is key to both verifying the polynomial's properties and solving related differential equations.

de Moivre's Theorem

de Moivre's theorem links complex numbers and trigonometry, and it's vital for expanding Chebyshev polynomials in their alternative form. This theorem states: $$\cos (n\theta )=\mathrm{\Re }((\cos \theta +i\sin \theta {)}^n)$$ Here, $\mathrm{\Re }$ refers to the real part of the complex number. Using de Moivre's theorem, the Chebyshev polynomials can be rewritten in terms of powers of $x$. In the context of Chebyshev polynomials, we apply the theorem by starting with the given definition $T_n(x)=\cos (n{\cos }^{-1}x)$. We then express $\cos (n{\cos }^{-1}x)$ using de Moivre's theorem, which allows involving imaginary numbers. Though the expressions involve $i$, our interest lies in the real part, facilitating the transformation into a polynomial form.

Binomial Theorem

The binomial theorem is another critical tool in expressing Chebyshev polynomials in an alternative form. The binomial theorem states: $$(a+b{)}^n=\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})a^{n-k}{b}^k$$ We use this theorem to expand $$(\cos ({\cos }^{-1}x)+i\sin ({\cos }^{-1}x){)}^n$$ By treating $a\equiv \cos ({\cos }^{-1}x)$ and $b\equiv i\sin ({\cos }^{-1}x)$. This helps reframe our polynomial into a series involving powers of $x$. The resulting series contains both real and imaginary parts. We focus on extracting the real part to correlate directly with the Chebyshev polynomial. Breaking down this expansion into simpler sub-steps can solidify comprehension. Each term in the series combines different powers of $x$, refined in later steps.

Trigonometric Identities

Trigonometric identities simplify complex polynomial expressions, such as those found in Chebyshev polynomials. Some key identities include: - $si{n}^2(x)+co{s}^2(x)=1$ - $cos(2x)=co{s}^2(x)-si{n}^2(x)$ These identities help break down and simplify polynomials. For instance, when expressing Chebyshev polynomials in terms of $x$, trigonometric identities allow transforming and simplifying trigonometric terms like $\sin ({\cos }^{-1}x)$ and $\cos ({\cos }^{-1}x)$ directly into polynomial terms. Using known identities aids in combining or reducing terms, making complex expressions more manageable. For Chebyshev polynomial, converting trigonometric parts into powers of $x$ leads to accessible, polynomial forms aiding in broader applications.