Chebyshev recurrence: first kind

Chebyshev polynomials of the first kind are a family of polynomials defined by a simple second‑order linear recurrence. For HSC Mathematics Extension 1 students, reasoning with recurrences like this one strengthens the ability to build sequences term‑by‑term — a skill directly drawn from the Sequences and Series topic. Working through this example, you will see how the recurrence generates new polynomial expressions and how a numeric substitution yields a clean final answer.

Problem Statement

$$T_0(x)=1,\qquad T_1(x)=x,\qquad T_{n+1}(x)=2xT_n(x)-T_{n-1}(x).$$

Use the recurrence to find $T_2(x)$ and $T_3(x)$, then evaluate $T_3\left(\frac12\right)$.

---

Hints

Build the polynomials one at a time: first use $T_2=2xT_1-T_0$, then use $T_3=2xT_2-T_1$.

---

Solutions

We start with the given initial values and use the recurrence with $n=1$ to find $T_2(x)$:

$$T_2(x)=2x^2-1.$$

We have now obtained the second-degree Chebyshev polynomial. Next, we set $n=2$ to obtain $T_3(x)$, substituting the expression we just found for $T_2$:

$$T_3(x)=2x(2x^2-1)-x=4x^3-3x.$$

The polynomial simplifies nicely, so we can now substitute $x=\frac12$ directly. Finally, to evaluate $T_3$ at $x=\frac12$, we plug the value into the simplified polynomial:

$$T_3\left(\frac12\right)=4\left(\frac18\right)-3\left(\frac12\right)=\frac12-\frac32=-1.$$

---

Takeaways

• Chebyshev polynomials can be generated recursively from two starting values.
• Work in order; each new polynomial depends on the previous two.
• Substitution is easiest after simplifying the polynomial.

---

Further Readings

HSC Trigonometry, HSC Polys Ext 1, HSC Combinatorics, HSC Polynomials