Python: Computing $e$Prime factorization of a numberPrime factorization for integers with up to 9-digit long...
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Python: Computing $e$
Prime factorization of a numberPrime factorization for integers with up to 9-digit long prime factorsNumber of divisors of a factorialSum of prime factors of binomial coefficients over 9000!${}$Generalized Project Euler 2: A sledgehammer to crack a nutRepeatedly multiplying digits until a single digit is obtainedChecking if the whole number is prime, and if all its digits are tooReformatting Latin vocabulary sets from The BridgePrime pair set finder (Project Euler problem 60)(Codewars): Find the Unknown Digit
$begingroup$
I was trying to solve a problem which stated:
Calculate the first 10 digit prime found in consecutive digits of e.
I was able to solve the problem but I did it by using some 10k digits of e available online. So I tried to write a program which calculates digits of e. The problem is that it simply gives the incorrect answer.
The code and the formula I used are:
$$e = sumlimits_{n=0}^{infty}frac{1}{n!} = frac{1}{1} + frac{1}{1} + frac{1}{1 cdot 2} + frac{1}{1cdot 2 cdot 3} + cdots$$
import math
e=0
x=int(input()) #larger this number, more will be the digits of e
for i in range(x):
e+=(1/(math.factorial(i)))
print(e)
When the user inputs 10, the digits returned are 2.7182815255731922 which is not correct.
Can someone explain why my code does not produce the correct result?
python python-3.x
edited 14 hours ago
mkrieger1
1,3631823
asked yesterday
Aaryan DewanAaryan Dewan
1082
New contributor
Aaryan Dewan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I was trying to solve a problem which stated:
Calculate the first 10 digit prime found in consecutive digits of e.
I was able to solve the problem but I did it by using some 10k digits of e available online. So I tried to write a program which calculates digits of e. The problem is that it simply gives the incorrect answer.
The code and the formula I used are:
$$e = sumlimits_{n=0}^{infty}frac{1}{n!} = frac{1}{1} + frac{1}{1} + frac{1}{1 cdot 2} + frac{1}{1cdot 2 cdot 3} + cdots$$
import math
e=0
x=int(input()) #larger this number, more will be the digits of e
for i in range(x):
e+=(1/(math.factorial(i)))
print(e)
When the user inputs 10, the digits returned are 2.7182815255731922 which is not correct.
Can someone explain why my code does not produce the correct result?
python python-3.x
edited 14 hours ago
mkrieger1
1,3631823
asked yesterday
Aaryan DewanAaryan Dewan
1082
New contributor
Aaryan Dewan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
If you want a really good aproximation of e, you could always use the beautiful(1+9^-(4^6*7))^3^2^85which yields 18457734525360901453873570 digits of e (if you calculate with enough precision), and is a pan-digital formula to boot.
$endgroup$
– Oscar Smith
21 hours ago
1
$begingroup$
This is being discussed on meta, twice.
$endgroup$
– Peilonrayz
13 hours ago
add a comment |
$begingroup$
I was trying to solve a problem which stated:
Calculate the first 10 digit prime found in consecutive digits of e.
I was able to solve the problem but I did it by using some 10k digits of e available online. So I tried to write a program which calculates digits of e. The problem is that it simply gives the incorrect answer.
The code and the formula I used are:
$$e = sumlimits_{n=0}^{infty}frac{1}{n!} = frac{1}{1} + frac{1}{1} + frac{1}{1 cdot 2} + frac{1}{1cdot 2 cdot 3} + cdots$$
import math
e=0
x=int(input()) #larger this number, more will be the digits of e
for i in range(x):
e+=(1/(math.factorial(i)))
print(e)
When the user inputs 10, the digits returned are 2.7182815255731922 which is not correct.
Can someone explain why my code does not produce the correct result?
python python-3.x
edited 14 hours ago
mkrieger1
1,3631823
asked yesterday
Aaryan DewanAaryan Dewan
1082
New contributor
Aaryan Dewan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I was trying to solve a problem which stated:
Calculate the first 10 digit prime found in consecutive digits of e.
I was able to solve the problem but I did it by using some 10k digits of e available online. So I tried to write a program which calculates digits of e. The problem is that it simply gives the incorrect answer.
The code and the formula I used are:
$$e = sumlimits_{n=0}^{infty}frac{1}{n!} = frac{1}{1} + frac{1}{1} + frac{1}{1 cdot 2} + frac{1}{1cdot 2 cdot 3} + cdots$$
import math
e=0
x=int(input()) #larger this number, more will be the digits of e
for i in range(x):
e+=(1/(math.factorial(i)))
print(e)
When the user inputs 10, the digits returned are 2.7182815255731922 which is not correct.
Can someone explain why my code does not produce the correct result?
python python-3.x
python python-3.x
edited 14 hours ago
mkrieger1
1,3631823
asked yesterday
Aaryan DewanAaryan Dewan
1082
New contributor
Aaryan Dewan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 14 hours ago
mkrieger1
1,3631823
asked yesterday
Aaryan DewanAaryan Dewan
1082
New contributor
Aaryan Dewan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 14 hours ago
mkrieger1
1,3631823
edited 14 hours ago
mkrieger1
1,3631823
edited 14 hours ago
mkrieger1
1,3631823
1,3631823
asked yesterday
Aaryan DewanAaryan Dewan
1082
New contributor
Aaryan Dewan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
Aaryan DewanAaryan Dewan
1082
asked yesterday
Aaryan DewanAaryan Dewan
1082
1082
New contributor
Aaryan Dewan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Aaryan Dewan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Aaryan Dewan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
If you want a really good aproximation of e, you could always use the beautiful(1+9^-(4^6*7))^3^2^85which yields 18457734525360901453873570 digits of e (if you calculate with enough precision), and is a pan-digital formula to boot.
$endgroup$
– Oscar Smith
21 hours ago
1
$begingroup$
This is being discussed on meta, twice.
$endgroup$
– Peilonrayz
13 hours ago
add a comment |
$begingroup$
If you want a really good aproximation of e, you could always use the beautiful(1+9^-(4^6*7))^3^2^85which yields 18457734525360901453873570 digits of e (if you calculate with enough precision), and is a pan-digital formula to boot.
$endgroup$
– Oscar Smith
21 hours ago
1
$begingroup$
This is being discussed on meta, twice.
$endgroup$
– Peilonrayz
13 hours ago
$begingroup$
If you want a really good aproximation of e, you could always use the beautiful
(1+9^-(4^6*7))^3^2^85 which yields 18457734525360901453873570 digits of e (if you calculate with enough precision), and is a pan-digital formula to boot.$endgroup$
– Oscar Smith
21 hours ago
$begingroup$
If you want a really good aproximation of e, you could always use the beautiful
(1+9^-(4^6*7))^3^2^85 which yields 18457734525360901453873570 digits of e (if you calculate with enough precision), and is a pan-digital formula to boot.$endgroup$
– Oscar Smith
21 hours ago
1
1
$begingroup$
This is being discussed on meta, twice.
$endgroup$
– Peilonrayz
13 hours ago
$begingroup$
This is being discussed on meta, twice.
$endgroup$
– Peilonrayz
13 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
First of all, don't panic. 10 is just not enough (indeed, it only gives you 5 decimal places. Try 20, and obtain
2.71828182846
which is much closer.
Now, Python uses a native floating point, which may only give you that many digits of precision (say, 30). To get more, you need to work with another representations; fractions.Fraction looks like a good candidate.
Finally, calls to math.factorial waste too much computing power. It is better to compute factorials as you go, e.g.
denom = 1
for i in range(1, x):
e += 1 / denom
denom *= i
answered yesterday
vnpvnp
40.4k232102
$endgroup$
add a comment |
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1 Answer
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$begingroup$
First of all, don't panic. 10 is just not enough (indeed, it only gives you 5 decimal places. Try 20, and obtain
2.71828182846
which is much closer.
Now, Python uses a native floating point, which may only give you that many digits of precision (say, 30). To get more, you need to work with another representations; fractions.Fraction looks like a good candidate.
Finally, calls to math.factorial waste too much computing power. It is better to compute factorials as you go, e.g.
denom = 1
for i in range(1, x):
e += 1 / denom
denom *= i
answered yesterday
vnpvnp
40.4k232102
$endgroup$
add a comment |
$begingroup$
First of all, don't panic. 10 is just not enough (indeed, it only gives you 5 decimal places. Try 20, and obtain
2.71828182846
which is much closer.
Now, Python uses a native floating point, which may only give you that many digits of precision (say, 30). To get more, you need to work with another representations; fractions.Fraction looks like a good candidate.
Finally, calls to math.factorial waste too much computing power. It is better to compute factorials as you go, e.g.
denom = 1
for i in range(1, x):
e += 1 / denom
denom *= i
answered yesterday
vnpvnp
40.4k232102
$endgroup$
add a comment |
$begingroup$
First of all, don't panic. 10 is just not enough (indeed, it only gives you 5 decimal places. Try 20, and obtain
2.71828182846
which is much closer.
Now, Python uses a native floating point, which may only give you that many digits of precision (say, 30). To get more, you need to work with another representations; fractions.Fraction looks like a good candidate.
Finally, calls to math.factorial waste too much computing power. It is better to compute factorials as you go, e.g.
denom = 1
for i in range(1, x):
e += 1 / denom
denom *= i
answered yesterday
vnpvnp
40.4k232102
$endgroup$
First of all, don't panic. 10 is just not enough (indeed, it only gives you 5 decimal places. Try 20, and obtain
2.71828182846
which is much closer.
Now, Python uses a native floating point, which may only give you that many digits of precision (say, 30). To get more, you need to work with another representations; fractions.Fraction looks like a good candidate.
Finally, calls to math.factorial waste too much computing power. It is better to compute factorials as you go, e.g.
denom = 1
for i in range(1, x):
e += 1 / denom
denom *= i
answered yesterday
vnpvnp
40.4k232102
edited yesterday
answered yesterday
vnpvnp
40.4k232102
answered yesterday
vnpvnp
40.4k232102
answered yesterday
vnpvnp
40.4k232102
40.4k232102
add a comment |
add a comment |
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$begingroup$
If you want a really good aproximation of e, you could always use the beautiful
(1+9^-(4^6*7))^3^2^85which yields 18457734525360901453873570 digits of e (if you calculate with enough precision), and is a pan-digital formula to boot.$endgroup$
– Oscar Smith
21 hours ago
1
$begingroup$
This is being discussed on meta, twice.
$endgroup$
– Peilonrayz
13 hours ago