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Imagine I have the following function

$$ \vec{f}(\vec{x}) = x \vec{x}, x = | \vec{x} |, \vec{x} \in R^3 $$

That is, the function is essentially a quadratic function, but contains a vector direction as well. Intuitively from single variable calculus I would expect the gradient $ \nabla \vec{f} = (\partial \vec{f}/ \partial x_1,\partial \vec{f}/ \partial x_2,\partial \vec{f}/ \partial x_3) $ to be proportional to $2x$, however I also would expect it to be a 3x3 matrix.

My most naive attempt would be to do

$$ \vec{f} = x_1^2 \vec{e}_1 + x_2^2 \vec{e}_2 + x_3^2 \vec{e}_3 $$

and say that

$$ \nabla \vec{f} = 2 x_1 \vec{e}_1 + 2 x_2 \vec{e}_2 + 2 x_3 \vec{e}_3 $$

But it would mean that every gradient w.r.t. a vector would always be a diagonal matrix, which seems wrong to me. What I really want to create is the Jacobian $ \partial \vec{f}_i / \partial x_j $ but I think I get a little bit confused about what I do with the base vectors $ \vec{e_i} $ during the partial derivative.

pingul
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2 Answers2

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You sound like you just want the Jacobian but are trying to force it into a singular gradient vector, and I think the main part you are getting confused on is how to differentiate that initial function f. Don't bother thinking of them as vectors for a second and just go through them normally for multivariable, treating all variables which you are not differentiating in regard to as just constants:

$\vec{f} = [f_1, f_2, f_3]^T$

$f_1(\vec{x}) = (x_1^2 + x_2^2 + x_3^2)^{1/2} * x_1$

$\frac{\partial f_1(\vec{x})}{\partial x_1} = (x_1^2 + x_2^2 + x_3^2)^{1/2} +\frac{x_1}{2(x_1^2 + x_2^2 + x_3^2)^{1/2}} * 2*x_1 = (x_1^2 + x_2^2 + x_3^2)^{1/2} +\frac{x_1^2}{(x_1^2 + x_2^2 + x_3^2)^{1/2}}$ by the chain rule.

By analogy, you can get $\frac{\partial f_2(\vec{x})}{\partial x_2}$ and $\frac{\partial f_3(\vec{x})}{\partial x_3}$. For the others, let's look at $\frac{\partial f_1(\vec{x})}{\partial x_2}$ for an example:

$\frac{\partial f_1(\vec{x})}{\partial x_2} = \frac{x_1*x_2}{(x_1^2 + x_2^2 + x_3^2)^{1/2}}$

To summarize, if $i = j$:

$\frac{\partial f_i(\vec{x})}{\partial x_j} = x +\frac{x_j^2}{x}$

Else:

$\frac{\partial f_i(\vec{x})}{\partial x_j} = \frac{x_i*x_j}{x}$

You can then plug this into a Jacobian form

cmitch
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    Thank you @cmitch this is very useful. I think the issue stemmed from being a little bit lazy and not properly expanding $x$ -- that would have made the cross derivatives more apparent! From plugging it in Jacobian form I get that $ \nabla \vec{f} = x I + \frac{1}{x} \vec{x}^t \vec{x} $ which is closer to something I would have expected. Thanks for the clear explanation, much appreciated! – pingul Jul 16 '20 at 00:34
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Recall the formula for gradient of scalar times vector: $$\nabla(a\vec{v}) = \vec{v}\otimes\nabla a + a\nabla\vec{v}.$$ In our case we have $f(\vec{x})= |\vec{x}|\vec{x}$ so $$\nabla{|\vec{x}|} = \nabla\sqrt{x_1^2+x_2^2+x_3^2}= \frac{\vec{x}}{|\vec{x}|}, \quad \nabla \vec{x} = \nabla(x_1,x_2,x_3) = I$$ where $I$ is the identity matrix. Therefore $$\nabla f(\vec{x}) = \vec{x}\otimes \frac{\vec{x}}{|\vec{x}|} + |\vec{x}|I = \left[\frac{x_ix_j}{|\vec{x}| } + |\vec{x}|\delta_{ij}\right]_{1\le i,j \le 3}.$$

mechanodroid
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  • This is very useful -- thanks! My memory for formulas have certainly gone down since uni, so showing the steps to get to the solution is very useful. Just as a nit: I'm imagining the first equation is supposed to $ \vec{x} $ instead of $ \vec{v} $? Many thanks! – pingul Jul 27 '20 at 09:28
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    @pingul Rather it is supposed to be $\vec{v}$ and not $\vec{x}$, thanks. Glad to be helpful. – mechanodroid Jul 27 '20 at 10:32