| Version 5 |
Version 4 |
| We can always express $\cos(kt)$ as a polynomial of $\cos(t)$: |
We can always express $\cos(kt)$ as a polynomial of $\cos(t)$: |
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| Examples: |
Examples: |
| \begin{eqnarray*} |
\begin{eqnarray*} |
| \cos(1t)&=&\cos(t)\\ |
\cos(1t)&=&\cos(t)\\ |
| \cos(2t)&=&2(\cos(t))^2-1\\ |
\cos(2t)&=&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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| This fact can be proved using the formula for cosine of angle-sum. |
This fact can be proved using the formula for cosine of angle-sum. |
| If we write $x=\cos t$ we obtain the \emph{Chebyshev polynomials of first kind}, that is |
If we write $x=\cos t$ we obtain the \emph{Chebyshev polynomials of first kind}, that is |
| $$T_n(x)=\cos(nt)$$ |
$$T_n(x)=\cos(nt)$$ |
| where $x=\cos t$. |
where $x=\cos t$. |
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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 hold the recurrence relation: |
These polynomials hold 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 $n=1, 2, \ldots$ |
