Question

how to calculate Exponential function by using vba

Asked by: edirol

In inputbox input x and n
exponential function : e^x=1+ x/1! + x^2/2! +.....x^n/n!       ; n!=1*2*3...*n
inside the function we can use t(n)=x^n/n!  and for recurrence formula t(n)=t(n-1)*x/n
The partial sum to the n-th clause is expressed with s(n) which is ; s(n)=s(n-1)+t(n)
then e=1+s(n)
the calculation will repeat until i=n and |t(n)|<0.001

the problem is when i run the program the answer is no same with the exponential function above.


Sub e2()
    Dim n As Double
    Dim x As Double
    Dim i As Double
    Dim tn As Double
    Dim sn As Double
    Dim t1 As Double
    Dim e As Double
   
    x = CInt(InputBox("x?"))
    n = CInt(InputBox("n?"))
   
    Do
        For i = 1 To n
            t1 = x / (n - 1)
            tn = (x / (n - 1)) * (x / n)
            sn = tn + t1
            e = 1 + sn
           
            If x = 0 Then Exit For
       
        Next i
    Loop Until tn < 0.001 Or -tn < 0.001
   
    MsgBox "e^" & CStr(x) & "=" & CStr(e)
       
End Sub

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Asked On
2008-11-17 at 21:28:32ID23913275
Tags

vba programming

Topics

Algorithms

,

Microsoft Excel Spreadsheet Software

,

Math & Science

Participating Experts
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Points
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Comments
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Answers

 

by: WaterStreetPosted on 2008-11-17 at 21:35:42ID: 22982247

instead of    

x = CInt(InputBox("x?"))
   n = CInt(InputBox("n?"))


try
   x = CDbl(InputBox("x?"))
   n = CDbl(InputBox("n?"))



 

by: edirolPosted on 2008-11-17 at 21:45:37ID: 22982281

thanks WaterStreet

i`d tried CDbl but the answer is still wrong.
for example x=2,n=3 the answer should be e^2=6.3333
i think there are something wrong with the formulas for t1,tn

 

by: zorvekPosted on 2008-11-17 at 22:19:17ID: 22982415

Sub e2()

   Dim n As Double
   Dim x As Double
   Dim e As Double
   Dim Laste As Double
   
   x = CInt(InputBox("x?"))
   
   e = 1
   x = 3
   n = 1
   Do
      Laste = e
      e = e + x ^ n / Factorial(n)
      n = n + 1
   Loop While Abs(e - Laste) > 0.0000001
   
   MsgBox "e^" & x & "=" & e
       
End Sub

Public Function Factorial(ByRef Value As Double) As Double

   If Value = 1 Then
      Factorial = 1
      Exit Function
   Else
      Factorial = Value * Factorial(Value - 1)
   End If

End Function

Kevin

 

by: deightonPosted on 2008-11-18 at 02:46:15ID: 22983373

here's a solution that avoids using the ^ operator, there's not much point in doing that, cos you could hard code e as a constant and do e^n in code.


Anyway, your keeping of a running total for the value seemed to be in error.

       Dim n As Double
       Dim x As Double
       Dim i As Double
       Dim tn As Double
       Dim sn As Double
       Dim t1 As Double
       Dim e As Double

     x = CInt(InputBox("x?"))
   n = CInt(InputBox("n?"))


       e = 0

       tn = 1
       e = e + tn

       'now e is the first term in the taylor expansion (call this term 0)

       Do

           'add on term 1, 2, 3

           i = i + 1
           tn = tn * x / i       'this gets the next term from the last

           e = e + tn

           'If x = 0 Then Exit For -    not needed, gives tn=0 when x = 0, so it ends below

           'the end ing condition wasn't true
       Loop Until tn < 0.001 And tn >= 0 Or -tn < 0.001 And tn <= 0

      msgbox ("e^" & CStr(x) & "=" & CStr(e))

 

by: edirolPosted on 2008-11-18 at 02:46:58ID: 22983380

thanks Kevin.
i`d tried your solution and it is working!!!however i have to make the program by using the
functions in the question.

Actually i found that i  misunderstood the question.This is the question again ;
Exponential function : e^x=1+ x/1! + x^2/2! +.....x^n/n!       ; n!=1*2*3...*n
Inside the function we can use t(n)=x^n/n!  and for recurrence formula t(n)=t(n-1)*x/n
The partial sum to the n-th clause is expressed with s(n) which is ; s(n)=s(n-1)+t(n)
Then e=1+s(n)
The calculation will repeat until n=100 and |t(n)|<0.001

Sub e21()
    Dim x As Double
    Dim n As Double
    Dim i As Double
    Dim t1 As Double
    Dim tn As Double
    Dim sn As Double
    Dim e As Double
   
     
    x = CDbl(InputBox("x?"))
   
    Do
        For n = 1 To 100
            t1 = x / (n - 1)
            tn = (x / (n - 1)) * (x / n)
            sn = tn + t1
            e = 1 + sn
           
            If x = 0 Then Exit For
       
        Next n
    Loop Until tn < 0.001 Or -tn < 0.001
   
    MsgBox "e^" & CStr(x) & "=" & CStr(e)
End Sub

I`d  tried to run the program but it comes into debug.
For example if x=3
     e^3=20.0855....

 

by: deightonPosted on 2008-11-18 at 02:59:36ID: 22983414

for one thing, you've got looping Until tn < 0.001 Or -tn < 0.001

, but if say tn=3, then -tn < .001 and the loop ends



 

by: deightonPosted on 2008-11-18 at 03:23:09ID: 22983540

also if you look at

  For n = 1 To 100
           t1 = x / (n - 1)




then where n=1, you have an immediate division by zero in the second line

 

by: edirolPosted on 2008-11-18 at 03:55:05ID: 22983681

already tried to change the functions.but still wrong

 

by: deightonPosted on 2008-11-18 at 04:12:20ID: 22983753

your calculation of the terms in the series is not right.  The first term has to be calculated before you go into the loop, but that's always 1.

t0 = 1
t1 = t0 * x / 1
t2 = t1 * x / 2
t3 = t2 * x / 3

t(n) = t(n-1) * x / n


but you don't want to store all those values, you want to calculate the next term from the previous and throw away the old one - so lets call the nth term variable 'tn'

first consider this

---------------------------------------------------------------------
dim tn as double

x=2
tn = 1
for n = 1 to 10
  tn = tn * x/n
 msgbox("term " + cstr(n) + " = " + cstr(tn))
next
---------------------------------------------------------------------

so looking at the code above, I calculate each of the first 10  term in the series.  But I want to add them all up, including t0, so then I need running_total
---------------------------------------------------------------------
x=2
tn = 1
running_total = tn

for n = 1 to 100
   tn = tn * x/n
 running_total = running_total + tn
 
next
---------------------------------------------------------------------
'at the end of this, the value is in running total

do you see what this bit does
  tn = tn * x/n


it takes the last term and works out the latest term from it.


 

by: edirolPosted on 2008-11-18 at 05:02:48ID: 22984059

Thanks deighton ;
 I`d  changed  my program by referring your functions and it works.
But how can i make that program to calculate until abs(tn)<0.0001?
I did `Step Into ` and in `Locals Window`  when Abs(tn)<0.0001 it should
stop running,but it still run until n=100.

Sub ee()

    Dim x As Double
    Dim n As Double
    Dim tn As Double
    Dim running_total As Double

    x = CDbl(InputBox("x"))
    tn = 1
    running_total = tn

    Do
        For n = 1 To 100
            tn = tn * x / n
            running_total = running_total + tn
        Next n
    Loop Until Abs(tn) < 0.00001

    MsgBox CStr(running_total)
End Sub

 

by: deightonPosted on 2008-11-18 at 05:28:00ID: 22984250

if you look at that, you have a for loop 'nested' in the Do loop, so you will always do 100 iterations, then you will break out of the second loop only if abs(tn) < .00001,

if you get rid of the do loop and have

 For n = 1 To 100
           tn = tn * x / n
           running_total = running_total + tn
           if Abs(tn) < 0.00001 then exit for
       Next n


you will iterate for a maximum of 100 times, but leaving the loop if tn < .00001

is one way of doing that

you could also use a do loop with a counter and the loop until n>100 or abs(tn)<.000001


 

by: edirolPosted on 2008-11-18 at 06:16:48ID: 22984619


At last !!!Thanks deighton.
Anyway can u explain why does tn = 1& sn = tn

 

by: deightonPosted on 2008-11-18 at 07:12:58ID: 22985261

tn = 1

this is setting tn to the first term in the series, which is always 1

tn = tn * x / n

this calculates the next value in the series each time

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