The mathematical constant e

[Mikam]

Hi
The mathematical constant e, also called Euler's constant, is the base of the Exp and Log where e=2.718.
i.e 2.718 e 4.607 = 100.27
Is their any way using GCB to execute this function.
Thanks in advance

[Anobium]

Yes, I think so. If you can calculate e in integer maths then we can.

Look on the web for some example of where this has been done in the past, for other languages, then, the port to GCB should be relatively simple.
If you have v0.94 (any of the release candidates) or v0.95 or greater looks at your \GreatCowBasic\Include folder . You will see maths.h and trig[n]places.h these are the current release of how we use integer maths to provide factorised results.

[Mikam]

Although the first part is straightforward i.e 4 where 2.718 multiplies itself 4 times=54.598, i am not able to derive the 0.607 to into the equation for GCB. to give the final value of 100.27.
Sorry if i being a bit thick on this one.

[Anobium]

Attach the code you have please. Add lots of comments so we cam help.

[Mikam]

~~~~
;Chip Settings

chip 16F887,8

;Defines (Constants)

define LCD_IO 8

define LCD_RW PORTD.6

define LCD_Enable PORTD.7

define LCD_RS PORTD.5

define LCD_DATA_PORT PORTB

;Variables
Dim OA, OB, OC, i, j As long
Dim answer As long

'#include <maths.h>
'Dim p_base As byte
'ver 17dec15.a
'GCGB v0.94.28 2015-11-23 (When compling)
'
main:
'j is derived from x-y curve, this value can be from 0 to 5
'values would be multiplied to remove the decimal point

'The difficulty for me is to use the remaining constant e=0.6
'The answer should be around 100000
i=2.781 'Constant e
j=4.6

OA=(27812781)/1000
OB=(OA2781)/1000
OC=(OB*2781)/1000
locate 0,0
print OA
locate 1,0
print OB
locate 2,0
print OC

wait 1 S
goto main

'*
the display shows
7733
21505
59805
​

[Anobium]

You will have to remind me. How do we resolve the remaining 0.6? What is the mathematical intent?

[Bert]

(from Stackoverflow.com) Express the exponent as fraction and separate both parts (if the numbers get too big, you can use continued fraction expansion to get a good approximation):

4.6 = 46/10 = 23/5

10^4.6 = 10^(23/5) = (10^23)^(1/5)

The integer parts shouldn't be a problem (see http://en.wikipedia.org/wiki/Exponentiation_by_squaring). Then you can calculate the root by one of the algorithms described here: http://en.wikipedia.org/wiki/Nth_root_algorithm

[Anobium]

Excellent. gcb supports sqroot so this is a practical approach.

[Mikam]

Thanks for your guidance. I will give it a go.