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I am doing a course about machine learning this semester, and while reading a tutorial I came across this question:

enter image description here

\begin{align} X &\sim U[0,d] \\ N &\rightarrow \inf \\ \delta &\rightarrow 0 \end{align}

I need to show that for k=2, the k-means algorithm convergent to the minimum of the square error. In the solution they wrote the following:

$$x^{(0)} = \frac{\mu^{(0)}_1+\mu^{(0)}_2}{2} = \alpha d \tag{*}$$

where $\mu$ is the centroid of a group $a$ between $[0, 1]$.

\begin{align} \mu_1^{(1)} &= \frac{1}{2} \alpha d \tag{*} \\ \mu_2^{(1)} &= \alpha d + \frac{}{2} = \frac{1+\alpha}{2}d \tag{*} \\ x^{(1)} &= \frac{\mu_1^{(1)}+\mu_2^{(1)}}{2} = \frac 1 2 \alpha d + \frac 1 4 d \tag{*} \\ \\ \mu_1^{(n)} &= \frac 1 2 x^{(n-1)} \tag{*} \\ \mu_1^{(n)} &= \frac{x^{(n-1)}+d} 2 \tag{*} \end{align}

Can please explain me what they did in the starred equations?

The photo at the top is the question itself. As for the first starred equation, I thought that what they wrote means that given $\mu$ 1 and 2, $X$ will be approximately the average between them. $(n)$ means iteration number $n$. As for the lower set of starred equations, I don't understand how they calculated the centroids for every iteration from the definition of the centroid.

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  • $\begingroup$ OK. the first photo is the question itself. As for the second photo, I thought that what they wrote means that given miu 1 and 2, X will be approximately the average between them. (n) means iteration number n. as for the thirs photo, i dont understand how from the definition of the centroid they calculated the centroids for every iteration. i hope its clear now $\endgroup$
    – Eduard
    Commented Apr 4, 2018 at 15:02
  • $\begingroup$ I edited your post to use the $\LaTeX$ formatting our site makes possible. Please ensure it still says what you want it to. $\endgroup$ Commented Apr 4, 2018 at 15:09
  • $\begingroup$ @gung delta meand the square , aka the distance between 2 points. $\endgroup$
    – Eduard
    Commented Apr 4, 2018 at 15:10

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