如何最简洁的解3个一元一次方程

weichanghe2000 · 发表于 2010-1-6 09:57:39

本帖最后由 weichanghe2000 于 2010-1-9 18:26 编辑

大家好，下面这段代码是用来产生3个一元一次方程，但是我运行这段代码后，确不能解得 Subscript[U, 1]，Subscript[U, 2]，Subscript[U, 3]. 必须另外再写一个Solve 才能解得。这样如果方程为多元多次，将变得很麻烦。
因此，怎样才能够用简洁的方法，解得Subscript[U, 1]，Subscript[U, 2]，Subscript[U, 3]呢？谢谢。

下面是代码，复制后直接得到 Notebook公式，方便大家查看

Notebook[{
Cell[BoxData[{
RowBox[{
RowBox[{
RowBox[{"{",
RowBox[{
SubscriptBox["\[Sigma]f", "1"], ",",
SubscriptBox["\[Sigma]f", "2"], ",",
SubscriptBox["\[Sigma]f", "3"], ",",
SubscriptBox["\[Sigma]f", "4"]}], "}"}], "=",
RowBox[{"{",
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RowBox[{
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RowBox[{"{",
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SubscriptBox["\[Theta]", "2"], ",",
SubscriptBox["\[Theta]", "3"], ",",
SubscriptBox["\[Theta]", "4"]}], "}"}], "=",
RowBox[{"{",
RowBox[{"1.97", ",", "0.53", ",", "0.26", ",", "0.34"}], "}"}]}],
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RowBox[{
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SubscriptBox["U", "0"], "=", "0"}], ";",
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SubscriptBox["U", "4"], "=", "120"}], ";"}], "\[IndentingNewLine]",
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SubscriptBox["L", "1"], "-",
SubscriptBox["L", "0"]}], "=", "1"}], ",",
RowBox[{
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SubscriptBox["L", "2"], "-",
SubscriptBox["L", "1"]}], "=", "3.5"}], ",",
RowBox[{
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SubscriptBox["L", "3"], "-",
SubscriptBox["L", "2"]}], "=", "3"}], ",",
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WindowMargins->{{-21, Automatic}, {Automatic, 103}},
AutoIndent->True,
LineIndent->3,
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StyleDefinitions->"Default.nb"
]

复制代码

ljwxhlzp · 发表于 2010-1-6 16:57:16

{Subscript[\[Sigma]f, 1], Subscript[\[Sigma]f, 2],
Subscript[\[Sigma]f, 3], Subscript[\[Sigma]f, 4]} = {5.60, 15.82,
53.07, 98.09};
{Subscript[\[Theta], 1], Subscript[\[Theta], 2], Subscript[\[Theta],
3], Subscript[\[Theta], 4]} = {1.97, 0.53, 0.26, 0.34};
L = {1, 3.5, 3, 2.5};
{Subscript[U, 0], Subscript[U, 4]} = {0, 120};
Solve[Subscript[\[Theta], #]*((Subscript[U, #] -
Subscript[\[Sigma]f, #]) Cosh[
Subscript[\[Theta], #]*L[[#]]] + (Subscript[\[Sigma]f, #] -
Subscript[U, # - 1]))/ Sinh[Subscript[\[Theta], #]*L[[#]]] ==
Subscript[\[Theta], # +
1]*((Subscript[U, # + 1] -
Subscript[\[Sigma]f, # +
1]) + (Subscript[\[Sigma]f, # + 1] - Subscript[U, #]) Cosh[
Subscript[\[Theta], # + 1]*L[[# + 1]]] )/
Sinh[Subscript[\[Theta], # + 1]*L[[# + 1]]] & /@ {1, 2,
3}, {Subscript[U, 1], Subscript[U, 2], Subscript[U, 3]}]

复制代码

weichanghe2000 · 发表于 2010-1-6 17:09:24

2# ，ljwxhlzp ，连For循环都没用，就直接解出Subscript[U, 1]，Subscript[U, 2]，Subscript[U, 3]来了。厉害啊。谢谢。

weichanghe2000 · 发表于 2010-1-8 16:24:39

本帖最后由 weichanghe2000 于 2010-1-8 16:28 编辑

一组 U0 和U4，可以用 2# ，ljwxhlzp 的方法，
但是当 U0={0,0,0,0,0,0,0,0,0,0}，对应 U4={100,110，1120，130，140，150，160，170，180，190} 共10组数据时，
怎样一次性解得 U1 U2 U3 呢？难道要手动执行 10次 2# ，ljwxhlzp 的程序吗？ljwxhlzp  你能否将你的程序更改下，适合于多组  U0 和U4 数据？等待指教。谢谢。

这时方程和系数仍然为下面的：

Cell[BoxData[{
RowBox[{
  RowBox[{
RowBox[{"{",
RowBox[{
   SubscriptBox["\[Sigma]f", "1"], ",",
   SubscriptBox["\[Sigma]f", "2"], ",",
   SubscriptBox["\[Sigma]f", "3"], ",",
   SubscriptBox["\[Sigma]f", "4"]}], "}"}], "=",
RowBox[{"{",
RowBox[{"5.60", ",", "15.82", ",", "53.07", ",", "98.09"}],
"}"}]}], ";"}], "\n",
RowBox[{
  RowBox[{
RowBox[{"{",
RowBox[{
   SubscriptBox["\[Theta]", "1"], ",",
   SubscriptBox["\[Theta]", "2"], ",",
   SubscriptBox["\[Theta]", "3"], ",",
   SubscriptBox["\[Theta]", "4"]}], "}"}], "=",
RowBox[{"{",
RowBox[{"1.97", ",", "0.53", ",", "0.26", ",", "0.34"}], "}"}]}],
  ";"}], "\n",
RowBox[{
  RowBox[{"L", "=",
RowBox[{"{",
RowBox[{"1", ",", "3.5", ",", "3", ",", "2.5"}], "}"}]}],
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RowBox[{
  RowBox[{
RowBox[{"{",
RowBox[{
   SubscriptBox["U", "0"], ",",
   SubscriptBox["U", "4"]}], "}"}], "=",
RowBox[{"{",
RowBox[{"0", ",", "120"}], "}"}]}], ";"}], "\n",
RowBox[{"Solve", "[",
  RowBox[{
RowBox[{
RowBox[{"(",
   RowBox[{
   RowBox[{
   FractionBox[
      RowBox[{
      SubscriptBox["\[Theta]", "#1"], " ",
      RowBox[{"(",
      RowBox[{
         RowBox[{
         RowBox[{"(",
         RowBox[{
            SubscriptBox["U", "#1"], "-",
            SubscriptBox["\[Sigma]f", "#1"]}], ")"}], " ",
         RowBox[{"Cosh", "[",
         RowBox[{
            SubscriptBox["\[Theta]", "#1"], " ",
            RowBox[{
            "L", "\[LeftDoubleBracket]", "#1",
            "\[RightDoubleBracket]"}]}], "]"}]}], "+",
         RowBox[{"(",
         RowBox[{
         SubscriptBox["\[Sigma]f", "#1"], "-",
         SubscriptBox["U",
            RowBox[{"#1", "-", "1"}]]}], ")"}]}], ")"}]}],
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      RowBox[{
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      RowBox[{
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         SubscriptBox["U",
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         SubscriptBox["\[Sigma]f",
            RowBox[{"#1", "+", "1"}]]}], ")"}], "+",
         RowBox[{
         RowBox[{"(",
         RowBox[{
            SubscriptBox["\[Sigma]f",
            RowBox[{"#1", "+", "1"}]], "-",
            SubscriptBox["U", "#1"]}], ")"}], " ",
         RowBox[{"Cosh", "[",
         RowBox[{
            SubscriptBox["\[Theta]",
            RowBox[{"#1", "+", "1"}]], " ",
            RowBox[{"L", "\[LeftDoubleBracket]",
            RowBox[{"#1", "+", "1"}], "\[RightDoubleBracket]"}]}],
         "]"}]}]}], ")"}]}],
      RowBox[{"Sinh", "[",
      RowBox[{
      SubscriptBox["\[Theta]",
         RowBox[{"#1", "+", "1"}]], " ",
      RowBox[{"L", "\[LeftDoubleBracket]",
         RowBox[{"#1", "+", "1"}], "\[RightDoubleBracket]"}]}],
      "]"}]]}], "&"}], ")"}], "/@",
RowBox[{"{",
   RowBox[{"1", ",", "2", ",", "3"}], "}"}]}], ",",
RowBox[{"{",
RowBox[{
   SubscriptBox["U", "1"], ",",
   SubscriptBox["U", "2"], ",",
   SubscriptBox["U", "3"]}], "}"}]}], "]"}]}], "Input"]

ljwxhlzp · 发表于 2010-1-9 17:18:02

{Subscript[\[Sigma]f, 1], Subscript[\[Sigma]f, 2],
Subscript[\[Sigma]f, 3], Subscript[\[Sigma]f, 4]} = {5.60, 15.82,
53.07, 98.09};
{Subscript[\[Theta], 1], Subscript[\[Theta], 2], Subscript[\[Theta],
3], Subscript[\[Theta], 4]} = {1.97, 0.53, 0.26, 0.34};
L = {1, 3.5, 3, 2.5};
U0 = {0, 0, 0, 0, 0, 0, 0, 0, 0, 0};
U4 = {100, 110, 120, 130, 140, 150, 160, 170, 180, 190};
(Reap@Do[
Subscript[U, 0] = U0[[a]];
Subscript[U, 4] = U4[[a]];
Sow@Solve[
Subscript[\[Theta], #]*((Subscript[U, #] -
Subscript[\[Sigma]f, #]) Cosh[
Subscript[\[Theta], #]*
L[[#]]] + (Subscript[\[Sigma]f, #] - Subscript[
U, # - 1]))/ Sinh[Subscript[\[Theta], #]*L[[#]]] ==
Subscript[\[Theta], # +
1]*((Subscript[U, # + 1] -
Subscript[\[Sigma]f, # +
1]) + (Subscript[\[Sigma]f, # + 1] - Subscript[
U, #]) Cosh[
Subscript[\[Theta], # + 1]*L[[# + 1]]] )/
Sinh[Subscript[\[Theta], # + 1]*L[[# + 1]]] & /@ {1, 2,
3}, {Subscript[U, 1], Subscript[U, 2], Subscript[U, 3]}][[
1]], {a, 1, 10}])[[2, 1]]

复制代码

weichanghe2000 · 发表于 2010-1-9 18:00:59

本帖最后由 weichanghe2000 于 2010-1-11 18:05 编辑

5# ，ljwxhlzp ，厉害啊。Reap 和 Sow 函数从来没用过，真的实现了我的效果。值得学习。

ljwxhlzp · 发表于 2010-1-9 19:39:32

这儿的高手很多，在他们面前，我只是个初学者，看看他们的帖子，你自己就会写了。

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