主题:【第七届原创】一个极限的求证过程

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Daichaozheng
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改进后的程序如下:



Dim x As Double, y(301) As Double, sh As Double, ch As Double, n As Integer, fw As Double, fw1 As Double

Private Sub Command1_Click()

fw = Val(Text1.Text)

fw1 = fw / 150

n = 1

For i = 1 To 301

If i = 150 Then GoTo 10

x = -fw + i * fw1

sh = (Exp(x) - Exp(-x)) / 2

ch = (Exp(x) + Exp(-x)) / 2

y(i) = (2 * x ^ 3 * sh ^ 2 + 2 * x * sh ^ 2 + x * ch ^ 2 * sh ^ 2 + 3 * ch * sh ^ 3 - 6 * x ^ 3 * ch ^ 2 - x * sh ^ 4) / x ^ 5 / sh ^ 4

10 Next i

If y(149) > y(151) Then y(0) = y(149) Else y(0) = y(151)

If 2 * y(149) - y(148) < 2 * y(151) - y(152) Then y(150) = 2 * y(149) - y(148) Else y(150) = 2 * y(151) - y(152)

y(150) = (y(150) + y(0)) / 2

a = 32 / 315

Text2.Text = Str$(a)

Text3.Text = Str$(y(150))

Picture1.Cls

Picture1.Scale (0, 0)-(100, 5000)

For i = 1 To 5000

a = 256 - i / 10

If a < 0 Then a = 0

Picture1.Line (0, i)-(100, i), RGB(0, 0, a)

Next i

Picture1.Scale (-1.1 * Val(Text1.Text), 0.13)-(1.1 * Val(Text1.Text), -0.02)

Picture1.Line (-Val(Text1.Text), 0)-(Val(Text1.Text), 0), RGB(256, 256, 256)

Picture1.Line (0, 0)-(0, 0.12), RGB(256, 256, 256)

For i = 0 To 0.115 Step 0.01

Picture1.Line (-0.08, i)-(0.08, i), RGB(256, 256, 256)

Next i

For i = 0 To 0.115 Step 0.001

Picture1.Line (-0.03, i)-(0.03, i), RGB(256, 256, 256)

Next i

For i = 0 To 0.115 Step 0.01

Picture1.CurrentX = 0.02

Picture1.CurrentY = i + 0.0035

Picture1.Print i

Next i

For i = 1 To 300

X1 = -fw + i * fw1

X2 = X1 + fw1

Y1 = y(i)

Y2 = y(i + 1)

Picture1.Line (X1, Y1)-(X2, Y2), RGB(128, 256, 256)

Next i

SavePicture Picture1.Image, "c:\tu.jpeg"

End Sub

Private Sub end_Click()

End

End Sub

Private Sub Form_Activate()

Picture1.Scale (0, 0)-(100, 5000)

For i = 1 To 5000

a = 256 - i / 10

If a < 0 Then a = 0

Picture1.Line (0, i)-(100, i), RGB(0, 0, a)

Next i

End Sub

放大图形的上部如下:

zyl3367898
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