Skip to main content

calculus and analysis - How to take derivative of parameterized coordinate?


Suppose I have a vector in $\mathbb{R}^n$ but $n$ is not known in advance. I want to be able to write functions which operate on the components of that vector, and then I'd like to be able to take derivatives with respect to the components. As an example, consider the relation: $$\frac{\partial}{\partial x_j} \sum_i x_i$$


Under the assumption that the $x_i$ are independent, I want a call to Simplify[] to return 1. Similarly, calling Simplify[] on $\frac{\partial x_i}{\partial x_j}$ should give KroneckerDelta[i,j]. It's not clear how I should represent generic coordinates like this. I've seen this, but I'm not sure it provides an answer. As the linked post suggests, I could do this for a fixed $n$, but that's not situation I'm working on, especially since I want to see the generic form for any $n$.


For reference, it seems that sympy let's you do something close to this.


from sympy.tensor import IndexedBase, Idx
x = IndexBase('x')
i, j = map(Idx, ['i', 'j'])
x[i]

x[i].diff

#

x[i].diff(x[j])
# ValueError: Can't differentiate wrt the variable: x[j], 1

But despite being able to represent the variables abstractly, I can't seem to differentiate them.


Here is another example. Suppose you wanted to calculate the derivative of the entropy with respect to one of the components of the input distribution (again, assuming all the variables are independent). The final form is the same no matter what $n$-simplex the distribution lives on, so you'd like to be able to do this for any dimension.


$$ \frac{\partial H}{\partial p_j} = - \frac{\partial}{\partial p_j} \sum_i p_i \log p_i = - (\log p_j + 1)$$


Problems like this come up in optimization, when you need to provide the gradient and Hessian to numerical algorithms.


Update: Here are two other related posts:



how to differentiate formally?


How to customize derivative behavior via upvalues?



Answer



I think this can be hacked more or less case by case with UpValues, I think this is one of the most flexible aspects of Mathematica.


For instance if you just want partial derivatives to interact with sums you can just define your sum function MySum (or you can maybe unprotect Sum, not sure if this is possible) and define UpValues


MySum /: D[MySum[s_, i_], x[j_]] := D[s /. i -> j, x[j]]

This gives the desired results for


D[MySum[x[i], i], x[j]]


(1)


-D[MySum[x[i] Log[x[i]], i], x[j]]

(-1 - Log[x[j]])


The x here can also be modified to a more general pattern


Comments

Popular posts from this blog

plotting - How to draw lines between specified dots on ListPlot?

I would like to create a plot where I have unconnected dots and some connected. So far, I have figured out how to draw the dots. My code is the following: ListPlot[{{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4,13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full] I have thought using ListLinePlot command, but I don't know how to specify to the command to draw only selected lines between the dots. Do have any suggestions/hints on how to do that? Thank you. Answer One possibility would be to use Epilog with Line : ListPlot[ {{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4, 13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full, Epilog -> { Line[ ...

equation solving - Invert and fit implicitly defined curve

I need to fit an implicitly defined curve. I thought I could get some data out of Solve , and then using FindFit . Therefore, I would like to find the relation the parametric curve defined by $F(x,y)=0$: Solve[-(1/2) + 1/2 (0.41202 BesselK[0, 0.1 Sqrt[x^2 + y^2]] + (0.101483 x BesselK[1, 0.1 Sqrt[x^2 + y^2]])/Sqrt[x^2 + y^2]) == 0, y] But I can't get an output: Solve was unable to solve the system with inexact coefficients or the system obtained by direct rationalization of inexact numbers present in the system. Since many of the methods used by Solve require exact input, providing Solve with an exact version of the system may help. >> Edit: In particular, I would like to fit the data coming from the curve with the expression of another curve, and not with a function $f(x)$. In particular, since this clearly looks like a cardioid , I would like it to fit to something like it. What other strategies could I try?

dynamic - How can I make a clickable ArrayPlot that returns input?

I would like to create a dynamic ArrayPlot so that the rectangles, when clicked, provide the input. Can I use ArrayPlot for this? Or is there something else I should have to use? Answer ArrayPlot is much more than just a simple array like Grid : it represents a ranged 2D dataset, and its visualization can be finetuned by options like DataReversed and DataRange . These features make it quite complicated to reproduce the same layout and order with Grid . Here I offer AnnotatedArrayPlot which comes in handy when your dataset is more than just a flat 2D array. The dynamic interface allows highlighting individual cells and possibly interacting with them. AnnotatedArrayPlot works the same way as ArrayPlot and accepts the same options plus Enabled , HighlightCoordinates , HighlightStyle and HighlightElementFunction . data = {{Missing["HasSomeMoreData"], GrayLevel[ 1], {RGBColor[0, 1, 1], RGBColor[0, 0, 1], GrayLevel[1]}, RGBColor[0, 1, 0]}, {GrayLevel[0], GrayLevel...