The equation to find the partial derivative of a cost function with respect to a parameter θj is given in the book 'Hands on Machine Learning with scikit-learn, keras and tensorflow ':

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  • m = number of instances in the dataset
  • x = input vector for the prediction
  • y = label for the input vector

I am not able to understand what the last scalar x(i)j means. Could someone please tell me what the variable means.


From the context you have provided, my reading is that $x^{(i)}_j$ is the $j$-th element of the $i$-th input vector $\mathbf{x}^{(i)}$, where there are $i = 1,..., m$ training instances.

Addressing comments.

In response to:

So in the case where $j = 1$, that is, $\theta_j = \theta_1$, then $x^{(i)}_j$ would be the 1st element of $\mathbf{x}^{(i)}$.

Yes you are entirely correct.

And in response to:

E.g : If $\mathbf{x}^{(i)} = [1,2,3,4,5]$ and $\theta_j = \theta_1$ then $x^{(i)}_j$ = 2 (considering 0 to be the first index).

This is just my view, but my personal preference, and also advice, is to make a distinction between formal "mathematics-indexing" which you see in mathematics in print, and implementational "Python-indexing". This is to avoid confusion.

In the case of mathematics-indexing, which is the convention for indexing vectors like $\mathbf{x}^{(i)}$ in print, then in your case $j = 1, ... 5$. In this case, then in your example $x^{(i)}_1 = 1$, i.e. the first element of the vector $\mathbf{x}^{(i)}$.

In the case of (what I am assuming to be) Python-indexing, which is what you will need for implementation, then $j = 0, ..., 4$. For your example, under this convention, then you are correct, $x^{(i)}_1 = 2$. Having now checked the book this is correct, Geron is using $j = 0, ..., n$ to index $(n+1)$ elements of the vector $\mathbf{x}^{(i)}$.

  • $\begingroup$ So if θj = θ1, then x(i)j would be the 1st element of x(i). E.g : If x(i) = [1,2,3,4,5] and θj = θ1 then x(i)j = 2 (considering 0 to be the first index). $\endgroup$ – Arya Man Apr 1 at 20:04
  • $\begingroup$ Thank you, that helps. $\endgroup$ – Arya Man Apr 1 at 20:42
  • $\begingroup$ Having now looked at the book - you are correct on both counts, the author is using using $j= 0, ..., n$ to index the $(n+1)$ elements of $\mathbf{x}^{(i)}$. $\endgroup$ – microhaus Apr 1 at 20:45
  • $\begingroup$ Sorry, but could I ask a small follow-up question? If θj = 5 then should the jth element in θ also be 5? E.g - If on the left hand side θj= θ3 = 5 then should the θ on the right-hand-side of the equation be [θ0, θ1, θ2, 5, θ4, θ5, θ6]? $\endgroup$ – Arya Man Apr 2 at 0:31
  • $\begingroup$ I will use a new response to answer your new question at stats.stackexchange.com/questions/517820/… and address this final comment in the new response $\endgroup$ – microhaus Apr 2 at 11:47

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