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A cosine written as a polynomial: the recurrence, the equal ripple between $-1$ and $1$, and the theorem that no monic polynomial of the same degree stays smaller.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to derive the Chebyshev recurrence and leading coefficient from the cosine definition, state the equal ripple property, prove that the monic Chebyshev polynomial minimises the largest size among monic polynomials, and say why the word monic cannot be dropped.
You know that an interval and a weight define an orthogonal family, and that every such family has a three-term recurrence and interior roots. The Chebyshev family is the one whose weight makes it a cosine, and being a cosine gives it properties no other family has.
$T_n$ is defined by $T_n(\cos\theta) = \cos n\theta$. A polynomial is monic when its leading coefficient is $1$. The monic Chebyshev polynomial is $\tilde{T}_n = T_n/2^{n-1}$. A family equioscillates when it attains its largest size with alternating signs at a sequence of points; $T_n$ does so at $n + 1$ points of $[-1, 1]$.
Define $T_n$ by $T_n(\cos\theta) = \cos n\theta$. The multiple-angle identity $\cos(n+1)\theta + \cos(n-1)\theta = 2\cos\theta\cos n\theta$ becomes the recurrence $T_{n+1} = 2xT_n - T_{n-1}$, which shows that $T_n$ really is a polynomial and that its leading coefficient doubles each degree: $2^{n-1}$. Three consequences follow directly from the cosine. On $[-1, 1]$, $|T_n| \le 1$. It attains $+1$ and $-1$ alternately at $n + 1$ points — the equal ripple. And its $n$ roots, at $\cos\dfrac{(2k+1)\pi}{2n}$, lie inside and crowd towards the ends. The theorem that matters uses the second of these:
> Among all monic polynomials of degree $n$, $\tilde{T}_n = T_n/2^{n-1}$ has the smallest largest size on $[-1, 1]$, namely $2^{1-n}$.
The proof is three moves: a better monic competitor would make the difference $\tilde{T}_n - q$ alternate in sign at $n + 1$ points, giving it $n$ roots, while cancelling leading terms leaves it of degree at most $n - 1$ — so it is zero, and there is no competitor. The word monic is what makes the statement non-trivial: without it the zero polynomial wins. And monic is exactly the shape of the node product $\prod(x - x_i)$ in the interpolation error, which is why the next lesson is about the roots of $T_n$.
Another way: steps
Another way: picture
Draw $T_6$ on $[-1, 1]$: a wave that touches the ceiling at $+1$ and the floor at $-1$, seven times in all, with the turning points crowded towards the two ends and stretched out in the middle. Every other monic polynomial scaled to degree six has to poke through one of those two lines somewhere — that is the whole theorem, and the picture is why the alternation count is what the proof consumes.
The minimality theorem is quoted without the word monic, which makes it plainly false — the zero polynomial has the smallest largest value of anything. The other half of the mistake is expecting the bounded behaviour to continue outside $[-1, 1]$. It reverses there: beyond the interval $T_n$ grows faster than any other monic polynomial of its degree, which is a second extremal property and the reason the family also appears in acceleration methods for iterations.
$T_0 = 1$, $T_1 = x$.
The two starting values.
$T_2 = 2x^{2} - 1$, $T_3 = 4x^{3} - 3x$, $T_4 = 8x^{4} - 8x^{2} + 1$.
Leading coefficients $2, 4, 8$.
$T_2$ at $x = 0, \pm 1$: $-1, +1, +1$ — three alternation points for degree two.
$n + 1$ of them.
Monic degree $3$: $\tilde{T}_3 = x^{3} - \tfrac34 x$, largest size $\tfrac14$ on $[-1, 1]$.
$2^{1-3} = \tfrac14$.
Compare $x^{3}$, also monic: its largest size is $1$, four times worse.
The obvious choice is the bad one.
$T_4 = 8x^{4} - 8x^{2} + 1$, so the leading coefficient is $8$.
Dividing: $\tilde{T}_4 = x^{4} - x^{2} + \tfrac18$.
Its largest size on $[-1, 1]$ is $\tfrac18 = 2^{1-4}$, and no monic quartic does better.
What is the leading coefficient of the Chebyshev polynomial $T_{3}$?
Answer:
Give the leading coefficient of $T_{5}$, $T_{6}$ and $T_{7}$.
| Degree | Leading coefficient | |
|---|---|---|
| Degree $5$ | 5 | |
| Degree $6$ | 6 | |
| Degree $7$ | 7 |
Match each property of $T_n$ on $[-1, 1]$ to the fact about cosine it comes from.
| $\cos(n+1)\theta + \cos(n-1)\theta = 2\cos\theta\cos n\theta$ | A cosine never leaves $[-1, 1]$ | $\cos n\theta$ is $\pm 1$ at $\theta = k\pi/n$ | $\cos n\theta$ vanishes at $\theta = (2k+1)\pi/2n$ | |
|---|---|---|---|---|
| The three-term recurrence | ||||
| $|T_n(x)| \le 1$ on $[-1, 1]$ | ||||
| $n + 1$ points where $T_n$ is $+1$ or $-1$, alternating | ||||
| $n$ roots, crowded towards the ends |
Select every statement that is true of the Chebyshev polynomial $T_n$ on $[-1, 1]$.
This task has no paper form; do it on a device.
Among **all** monic polynomials of degree $4$, what is the smallest possible value of $\max_{-1 \le x \le 1}|p(x)|$? Give a fraction.
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Put the five steps of the proof that the monic Chebyshev polynomial has the smallest largest size into order.
Number the steps in order (write the number in the box):
You can derive the Chebyshev properties from the cosine and prove the minimality theorem. Say in your own words why the proof counts sign changes rather than computing anything.
8. Your turn: the monic Chebyshev polynomial of degree $4$ and its largest size, step 3