变量符号数量上乘积的偏导数之和 [英] Sum of partial derivatives of a product over a symbolic number of variables

问题描述

我希望SymPy评估如下表达式:

I would like SymPy to evaluate an expression like the following:

我将如何定义符号和表达式，以便SymPy能够很好地处理它?我想将N保留为一个符号，即不制作x的实际有限列表.我已经尝试了IndexedBase和Sum/Product的各种组合，但是没有使其正常工作.

How would I define the symbols and the expression so that SymPy could handle it nicely? I would like to keep N as just a symbol, i.e. not make an actual finite list of x's. I have tried various combinations of IndexedBase and Sum /Product, but didn't get it working right.

推荐答案

理想情况是:

x = IndexedBase("x")
i, j, N = symbols("i j N")
expr = Sum(Product(exp(-x[j]**2), (j, 1, N)).diff(x[i]), (i, 1, N))

到目前为止，这是未评估的，expr是

So far this is unevaluated, expr is

Sum(Derivative(Product(exp(-x[j]**2), (j, 1, N)), x[i]), (i, 1, N)) 

可以使用方法doit对其进行评估.不幸的是，产品的差异化尚不能奏效:expr.doit()退货

The method doit can be used to evaluate it. Unfortunately the differentiation of a product doesn't quite work yet: expr.doit() returns

N*Derivative(Product(exp(-x[j]**2), (j, 1, N)), x[i])

将乘积重写为求和前的总和有帮助:

Rewriting the product as the sum prior to differentiation helps:

expr = Sum(Product(exp(-x[j]**2), (j, 1, N)).rewrite(Sum).diff(x[i]), (i, 1, N))
expr.doit()

返回

Sum(Piecewise((-2*exp(Sum(log(exp(-x[j]**2)), (j, 1, N)))*x[i], (1 <= i) & (i <= N)), (0, True)), (i, 1, N))

这是区分的正确结果.可悲的是，我们在Piecewise中有一个无关紧要的条件，并且log(exp(...))应该已经简化了. SymPy不会从外部总和的上下文中推断出(1 <= i) & (i <= N)是True，并且还犹豫要简化log(exp，以为x[j]可能很复杂.因此，我采用分段方式求助于外科手术，将其替换为第一部分，并强行扩大原木:

which is the correct result of differentiation. Sadly we have that extraneous condition in Piecewise, and also log(exp(...)) that should have been simplified. SymPy doesn't infer that (1 <= i) & (i <= N) is True from the context of the outer sum, and it also hesitates to simplify log(exp thinking x[j] might be complex. So I resort to surgical procedure with Piecewise, replacing it by the first piece, and to forcefully expanding logs:

e = expr.doit()
p = next(iter(e.atoms(Piecewise)))
e = expand_log(e.xreplace({p: p.args[0][0]}), force=True)

现在e是

Sum(-2*exp(Sum(-x[j]**2, (j, 1, N)))*x[i], (i, 1, N))

很遗憾，无法使exp(Sum(..))再次成为产品.

Couldn't get exp(Sum(..)) to become a Product again, unfortunately.

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