level 5
米味儿🎏
楼主
n = 2
F = Integrate[(D[Subscript[u, n] = \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(k = 1\), \(n\)]\(
\*SuperscriptBox[\((1 - t)\), \(3\)]
\*SuperscriptBox[\(t\), \(3\)]
\*SuperscriptBox[\(t\), \(k - 1\)]
\*SubscriptBox[\(a\), \(k\)]\)\), {t, 2}])^2, {t, 0, 1}] +
c (Integrate[(Subscript[u, n] = \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(k = 1\), \(n\)]\(
\*SuperscriptBox[\((1 - t)\), \(3\)]
\*SuperscriptBox[\(t\), \(3\)]
\*SuperscriptBox[\(t\), \(k - 1\)]
\*SubscriptBox[\(a\), \(k\)]\)\))^2, {t, 0, 1}] - 1)
Solve[Do[Print[D[F, Subscript[a, i]]], {i, n}],
D[F, c], {Do[Print[Subscript[a, i]], {i, n}], c}]
2019年01月19日 09点01分
1
F = Integrate[(D[Subscript[u, n] = \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(k = 1\), \(n\)]\(
\*SuperscriptBox[\((1 - t)\), \(3\)]
\*SuperscriptBox[\(t\), \(3\)]
\*SuperscriptBox[\(t\), \(k - 1\)]
\*SubscriptBox[\(a\), \(k\)]\)\), {t, 2}])^2, {t, 0, 1}] +
c (Integrate[(Subscript[u, n] = \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(k = 1\), \(n\)]\(
\*SuperscriptBox[\((1 - t)\), \(3\)]
\*SuperscriptBox[\(t\), \(3\)]
\*SuperscriptBox[\(t\), \(k - 1\)]
\*SubscriptBox[\(a\), \(k\)]\)\))^2, {t, 0, 1}] - 1)
Solve[Do[Print[D[F, Subscript[a, i]]], {i, n}],
D[F, c], {Do[Print[Subscript[a, i]], {i, n}], c}]
