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gung - Reinstate Monica
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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.

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?

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.

formatted; edited for English
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gung - Reinstate Monica
  • 147.5k
  • 89
  • 406
  • 717

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

enter image description here

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

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

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

  

miuwhere $\mu$ is the centroid of a group a $a$ between 0-1 enter image description here$[0, 1]$.

If someone can\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 second and the third photos , it whould be great. Thanks in advancestarred equations?

doing this semester a course about machine learning,and while reading a tutorial i came across this question:  enter image description here

X~U[0,d] N->inf delta->0 I need to show that for k=2,the k-means algorithm convergent to the minimum of the squere error. In the solution they wrote the following:

enter image description here

 

miu is the centroid of a group a between 0-1 enter image description here

If someone can please explain me what they did in the second and the third photos , it whould be great. Thanks in advance

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?

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