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Chebyshev recurrence: first kind

Authors
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    Name
    Vu Hung
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Problem Statement

The Chebyshev polynomials of the first kind satisfy

T0(x)=1,T1(x)=x,Tn+1(x)=2xTn(x)−Tn−1(x).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 T2(x)T_2(x) and T3(x)T_3(x), then evaluate T3(12)T_3\left(\frac12\right).

Hints

Build the polynomials one at a time: first use T2=2xT1−T0T_2=2xT_1-T_0, then use T3=2xT2−T1T_3=2xT_2-T_1.

Solutions

T2(x)=2x2−1.T_2(x)=2x^2-1.

Then

T3(x)=2x(2x2−1)−x=4x3−3x.T_3(x)=2x(2x^2-1)-x=4x^3-3x.

So

T3(12)=4(18)−3(12)=12−32=−1.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.

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