| Version 7 |
Version 6 |
| The \emph{Chebyshev polynomials of first kind} are defined by the simple |
The \emph{Chebyshev polynomials of first kind} are defined by the simple |
| formula $$T_n(x)=\cos(nt),$$ where $x=\cos t$. |
formula $$T_n(x)=\cos(nt),$$ where $x=\cos t$. |
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| It is an example of a Trigonometric Polynomial. |
It is an example of a Trigonometric Polynomial. |
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| This can be seen to be a polynomial by expressing $\cos(kt)$ as a polynomial of $\cos(t)$, by using the formula for cosine of angle-sum: |
This can be seen to be a polynomial by expressing $\cos(kt)$ as a polynomial of $\cos(t)$, by using the formula for cosine of angle-sum: |
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| Examples: |
Examples: |
| \begin{eqnarray*} |
\begin{eqnarray*} |
| \cos(1t)&=&\cos(t)\\ |
\cos(1t)&=&\cos(t)\\ |
| \cos(2t)&=&\cos(t)\cos(t) - \sin(t)\sin(t) = 2(\cos(t))^2-1\\ |
\cos(2t)&=&\cos(t)\cos(t) - \sin(t)\sin(t) = 2(\cos(t))^2-1\\ |
| \cos(3t)&=&4(\cos(t))^3-3\cos(t)\\ |
\cos(3t)&=&4(\cos(t))^3-3\cos(t)\\ |
| &\vdots& |
&\vdots& |
| \end{eqnarray*} |
\end{eqnarray*} |
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| So we have |
So we have |
| \begin{eqnarray*} |
\begin{eqnarray*} |
| T_0(x)&=&1\\ |
T_0(x)&=&1\\ |
| T_1(x)&=&x\\ |
T_1(x)&=&x\\ |
| T_2(x)&=&2x^2-1\\ |
T_2(x)&=&2x^2-1\\ |
| T_3(x)&=&4x^3-3x\\ |
T_3(x)&=&4x^3-3x\\ |
| &\vdots& |
&\vdots& |
| \end{eqnarray*} |
\end{eqnarray*} |
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| These polynomials obey the recurrence relation: |
These polynomials obey the recurrence relation: |
| $$T_{n+1}(x) = 2xT_n(x) - T_{n-1}(x)$$ |
$$T_{n+1}(x) = 2xT_n(x) - T_{n-1}(x)$$ |
| for $n=1, 2, \ldots$ |
for $n=1, 2, \ldots$ |
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| Related are the \emph{Chebyshev polynomials of the second kind} that are |
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| defined as $$U_{n-1}(\cos t) = \frac{\sin(n t)}{\sin (t)},$$ which |
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| can similarly be seen to be polynomials through either a similar process as the |
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| above or by the relation $U_{n-1}(t) = n T_n'(t)$. |
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| The first few are: |
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| \begin{eqnarray*} |
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| U_0(x)&=&1\\ |
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| U_1(x)&=&2x\\ |
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| U_2(x)&=&4x^2-1\\ |
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| U_3(x)&=&8x^3-4x\\ |
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| &\vdots& |
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| \end{eqnarray*} |
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| The same recurrence relation also holds for $U$: |
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| $$U_{n+1}(x) = 2xU_n(x) - U_{n-1}(x)$$ |
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| for $n=1, 2, \ldots$. |
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