Question

作为一个例子，我想尝试找到

的偏导数

f(x) = \sum_{i=1}^n x_i^2

千里马中的

（预期的输出将是\frac{\partial f}{\partial x_k} = 2x_k）我试过跟随，但似乎索引变量没有像我预期的那样处理，有人能解释我做错了吗？

如果您将n和k替换为实际数字，则相同的命令有效，但不是以这种形式：

f(x) := 1/2 * sum( x[i]^2, i, 1, n);
print(diff(f(x),x[k]));

Try it online!

Answer 1

Maxima can't handle derivative with respect to a indexed variable by default. I wrote a couple of small packages to handle these problems. Perhaps this is useful to you.

See: https://pastebin.com/MmYJRqCq (sum_kron_delta, summation of Kronecker delta) and: https://pastebin.com/UGCPgvAn (diff_sum, derivative of summation wrt indexed variable)

Here's an example applied to your problem. I'll assume you have downloaded the code above to your computer.

(%i1) load ("sum_kron_delta.mac");
(%o1)                         sum_kron_delta.mac
(%i2) load ("diff_sum.mac");
(%o2)                            diff_sum.mac
(%i3) 'diff ('sum (x[i]^2, i, 1, n), x[j]);
                             n
                            ====
                            \
(%o3)                     2  >    x  kron_delta(i, j)
                            /      i
                            ====
                            i = 1

Note that you have to write 'diff('sum(... that is, with the quote mark ' to indicate that diff and sum are nouns (formal expressions) instead of verbs (functions which are called). This is necessary in the implementation of diff_sum and sum_kron_delta because they work with simplification rules. (It's a long story, which I can explain if there's interest.)

I see we got the kron_delta summation, but we need to cause the simplification rules to be applied. We could also write expand(%, 0, 0) here instead of ''%.

(%i4) ''%;
(%o4) 2 (if (1 <= j) and (j <= n) and %elementp(j, integers) then x  else 0)
                                                                   j

At this point we have the final result, which we can simplify further with additional data.

(%i5) assume (j >= 1, j <= n);
(%o5)                          [j >= 1, n >= j]
(%i6) ''%o4;
(%o6)            2 (if %elementp(j, integers) then x  else 0)
                                                    j
(%i7) declare (j, integer);
(%o7)                                done

(%i8) ''%o6;
(%o8)                                2 x
                                        j

If this seems fruitful to you, I'll be happy to go into details.

使用索引变量的衍生工具

1 个答案: