vector derivative of vector summation

Question

Want to find, $$ \frac{d}{dV} \sum_{k=1}^{n} {V_{k}}^{2} $$

Where $V$ is a $n$-dimensional vector.

Is following correct? If not then what is the answer?

Let $$ V = [v_{1}, v_{2}, v_{3}, \dots , v_{n}]$$

$$ \frac{d}{dV} \sum_{k=1}^{n} {V_{k}}^{2} = [2v_{1}, 2v_{2}, 2v_{3}, \dots , 2 v_{n}]$$

correct. your summation index should start from $0$ and the vector is of dimension $n+1$, or first element should be $v_1$ — gunes, Aug 27 '20 at 12:58
Correction in vector index made in the question. Thank you @gunes — Ajey, Aug 27 '20 at 13:26
I assume that by $\frac{d}{dV}$ you mean the gradient operator? — NDewolf, Aug 27 '20 at 14:21
@NDewolf : it is one part of the optimization, so yes it is a gradient operator. — Ajey, Aug 27 '20 at 14:28

score 0 · Accepted Answer · answered Sep 07 '20 at 17:13

0

The derivative is correct as mentioned in the comments.

answered Sep 07 '20 at 17:13

Ajey

1 Answers1