Second-kind Chebyshev recurrence

Problem Statement

The second-kind Chebyshev polynomials satisfy

$$U_0(x)=1,\qquad U_1(x)=2x,\qquad U_{n+1}(x)=2xU_n(x)-U_{n-1}(x).$$

Use this recurrence to find $U_3(x)$ and $U_4(x)$.

---

Hints

First find $U_2(x)$, then continue to $U_3(x)$ and $U_4(x)$.

---

Solutions

$$U_2(x)=2x(2x)-1=4x^2-1.$$ $$U_3(x)=2x(4x^2-1)-2x=8x^3-4x.$$ $$U_4(x)=2x(8x^3-4x)-(4x^2-1)=16x^4-12x^2+1.$$

---

Takeaways

• The first-kind and second-kind Chebyshev polynomials use the same recurrence form.
• The starting values determine which family is produced.
• Organize recurrence work line by line to avoid algebra slips.

---

Further Readings

• HSC Vectors: https://vumaths.com/booklets/hsc-vectors/
• HSC Proofs: https://vumaths.com/booklets/hsc-proofs/
• HSC Last Resorts: https://vumaths.com/booklets/hsc-last-resorts/
• HSC Mechanics: https://vumaths.com/booklets/hsc-mechanics/

---

Connect with me

• Instagram: https://www.instagram.com/vuhung16/
• Substack: https://vuhung16.substack.com/
• GitHub: https://github.com/vuhung16au/