In [here], mathematical constant 
is introduced. Below, a simple equation is presented to compute its value.
, when 
.
Therefore, 
is a infinity loopless irrational number.
1 2 3 4 5 6 7 8 9 10 11 12 13 | #!/usr/bin/env python from math import e s = 1 t = 1 maxn = 21 for x in range(1, maxn): t *= x s += 1.0 / t print "Iteration %d=%.10f, err=%.20f" % (x, s, abs(s - e)) print e |
#!/usr/bin/env python
from math import e
s = 1
t = 1
maxn = 21
for x in range(1, maxn):
t *= x
s += 1.0 / t
print "Iteration %d=%.10f, err=%.20f" % (x, s, abs(s - e))
print eThe equation is efficient in computing its value when iterations reach 17, the error is approximately 0.00000000000000044409
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 | Iteration 1=2.0000000000, err=0.71828182845904509080 Iteration 2=2.5000000000, err=0.21828182845904509080 Iteration 3=2.6666666667, err=0.05161516179237857216 Iteration 4=2.7083333333, err=0.00994849512571205352 Iteration 5=2.7166666667, err=0.00161516179237874979 Iteration 6=2.7180555556, err=0.00022627290348964380 Iteration 7=2.7182539683, err=0.00002786020507672404 Iteration 8=2.7182787698, err=0.00000305861777505356 Iteration 9=2.7182815256, err=0.00000030288585284310 Iteration 10=2.7182818011, err=0.00000002731266057765 Iteration 11=2.7182818262, err=0.00000000226055218988 Iteration 12=2.7182818283, err=0.00000000017287637988 Iteration 13=2.7182818284, err=0.00000000001228572799 Iteration 14=2.7182818285, err=0.00000000000081490370 Iteration 15=2.7182818285, err=0.00000000000005018208 Iteration 16=2.7182818285, err=0.00000000000000222045 Iteration 17=2.7182818285, err=0.00000000000000044409 Iteration 18=2.7182818285, err=0.00000000000000044409 Iteration 19=2.7182818285, err=0.00000000000000044409 Iteration 20=2.7182818285, err=0.00000000000000044409 2.71828182846 |
Iteration 1=2.0000000000, err=0.71828182845904509080 Iteration 2=2.5000000000, err=0.21828182845904509080 Iteration 3=2.6666666667, err=0.05161516179237857216 Iteration 4=2.7083333333, err=0.00994849512571205352 Iteration 5=2.7166666667, err=0.00161516179237874979 Iteration 6=2.7180555556, err=0.00022627290348964380 Iteration 7=2.7182539683, err=0.00002786020507672404 Iteration 8=2.7182787698, err=0.00000305861777505356 Iteration 9=2.7182815256, err=0.00000030288585284310 Iteration 10=2.7182818011, err=0.00000002731266057765 Iteration 11=2.7182818262, err=0.00000000226055218988 Iteration 12=2.7182818283, err=0.00000000017287637988 Iteration 13=2.7182818284, err=0.00000000001228572799 Iteration 14=2.7182818285, err=0.00000000000081490370 Iteration 15=2.7182818285, err=0.00000000000005018208 Iteration 16=2.7182818285, err=0.00000000000000222045 Iteration 17=2.7182818285, err=0.00000000000000044409 Iteration 18=2.7182818285, err=0.00000000000000044409 Iteration 19=2.7182818285, err=0.00000000000000044409 Iteration 20=2.7182818285, err=0.00000000000000044409 2.71828182846
As you can see, the value will not be improved. This is mainly limited by the floating point precision by programing languages, e.g. Python, in this case.
See also: Simple and Efficient C Program to Compute the Mathematic Constant E (Euler’s number)
–EOF (The Ultimate Computing & Technology Blog) —
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