Typo on page 38?

[artyomboyko]

Good day Sebastian, can you explain in detail how you got in the note -1 in dL/db? Where does the sum sign (sigma) disappear?

[artyomboyko]

Yes. There's a typo there. Found the link.
https://github.com/rasbt/machine-learning-book/tree/main/ERRATA

[rasbt]

Sorry for the typo, but I am glad you already found it!

[artyomboyko]

Can someone explain in detail how it turns out:
$\frac{d}{dw_{j}} \sum_{j} w_{j} x_{j}^{i} = x_{j}^{i}$

How did you get rid of the sigma sign (sum)?

[artyomboyko]

Good day Sebastian. Thanks for your reply.

I don't have enough experience with this kind of thing. Can I write it down as follows:

Since at the moment of computation we are looking for the partial derivative for a particular example from a dataset, we can exclude $\sum_{j}$. Because all other terms of this sum at the moment of calculation will be constants. Therefore, we can write down the following:

$\frac{d}{dw_{j}} \sum_{j} w_{j} x_{j}^{(i)} = \frac{d}{dw_{j}} w_{j} x_{j}^{(i)} = [Using: \quad(f(x) \cdot g(x))' = f'(x) \cdot g(x) + f(x) \cdot g'(x); \quad (x)' = 1; \quad (C)'= 0 ] = $

$$ = \frac{d}{dw_{j}} (w_{j}) \cdot x_{j}^{(i)} + w_{j} \cdot \frac{d}{dw_{j}}(x_{j}^{(i)}) = 1 \cdot x_{j}^{(i)} + w_{j} \cdot 0 = x_{j}^{(i)} $$

Would it be correct in this manner?

[rasbt]

yes this looks correct to me

[artyomboyko]

Thank you so much for your help Sebastian.

[artyomboyko]

Good day Sebastian. How you get -1 on d/db?

$\frac{d}{db} \sum_{j} w_{j} x_{j}^{(i)} = \frac{d}{db} w_{j} x_{j}^{(i)} = [Using: \quad(f(x) \cdot g(x))' = f'(x) \cdot g(x) + f(x) \cdot g'(x); \quad (x)' = 1; \quad (C)'= 0 ] = $

$= \frac{d}{db} (w_{j}) \cdot x_{j}^{(i)} + w_{j} \cdot \frac{d}{db}( x_{j}^{(i)} ) = 0 \cdot x_{j}^{(i)} + w_{j} \cdot 0 = 0 + 0$

Then

$\frac{d}{dw_{j}} (y^{(i)} - \sum_{i} w_j x_{j}^{(i)}+b) = [(x)' = 1, \quad (c)' = 0] = 0 - 0 + 1 = 1$

[rasbt]

I think in the last step it should be 0 - (0+1) instead of 0 - 0+1