CMI 2024 | Data Science | Part B | Question: 8

Question

Let $\alpha$ be a fixed real number and let $f$ be a function from $\mathbb{R}^{2}$ to $\mathbb{R}^{2}$ defined as
\[
f(x, y)=(x \cos \alpha-y \sin \alpha, x \sin \alpha+y \cos \alpha)
\]

Write down an expression for $f^{10}(x, y)$, where $f^{10}$ denotes the function obtained by composing $f$ with itself $10$ times.

Answer 1

By doing for F(F(x))=>
xcos(2a) - ysin(2a), xsin(2a) + ycos(2a)...

Hence, concluding..
We get..
xcos(10a) - ysin(10a), xsin(10a) + ycos(10a)...
.

Answer 2

\[
f(x,y) = (x \cos \alpha - y \sin \alpha, \; x \sin \alpha + y \cos \alpha)
\]
Composing \(f\) with itself \(n\) times gives a rotation by \(n\alpha\).  
So,
\[
f^{10}(x,y) = (x \cos (10\alpha) - y \sin (10\alpha), \; x \sin (10\alpha) + y \cos (10\alpha))
\]

FULL METHOD :

\[
R(\alpha) =
\begin{bmatrix}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{bmatrix}
\]
So, applying the rotation to a vector \(\begin{bmatrix} x \\ y \end{bmatrix}\):
\[
f(x,y) =
R(\alpha) \begin{bmatrix} x \\ y \end{bmatrix}
=
\begin{bmatrix}
x \cos \alpha - y \sin \alpha \\
x \sin \alpha + y \cos \alpha
\end{bmatrix}
\]
Composing \(f\) with itself \(n\) times gives a rotation by \(n\alpha\), i.e.
\[
f^{n}(x,y) =
R(n\alpha) \begin{bmatrix} x \\ y \end{bmatrix}
\]
Therefore, for \(n=10\):
\[
f^{10}(x,y) =
\begin{bmatrix}
x \cos (10\alpha) - y \sin (10\alpha) \\
x \sin (10\alpha) + y \cos (10\alpha)
\end{bmatrix}
\]