Why is there a difference between starting with a “0” and “3” for approx
I was attempting an approximated value of pi through the formula of
pi= 3 + (4/(2*3*4)) - (4/(4*5*6)) + (4/(6*7*8)) - … (and so on). However, my code (shown below) had 2 separate answers (3.1415926535900383 and 3.141592653590042) when:
- approx variable started with "0" and "3" respectively
- n=10000
Does anyone know why?
def approximate_pi(n):
approx=0
deno=2
if n == 1:
return 3
for x in range(n-1):
if x%2:
approx -= 4/((deno)*(deno+1)*(deno+2))
else:
approx += 4/((deno)*(deno+1)*(deno+2))
deno+=2
return approx+3
and
def approximate_pi(n):
approx=3
deno=2
if n == 1:
return 3
for x in range(n-1):
if x%2:
approx -= 4/((deno)*(deno+1)*(deno+2))
else:
approx += 4/((deno)*(deno+1)*(deno+2))
deno+=2
return approx
python python-3.x
edited Nov 12 '18 at 8:28
jww
52.7k39221483
asked Nov 12 '18 at 4:48
Zi Yang Chow
1
add a comment |
I was attempting an approximated value of pi through the formula of
pi= 3 + (4/(2*3*4)) - (4/(4*5*6)) + (4/(6*7*8)) - … (and so on). However, my code (shown below) had 2 separate answers (3.1415926535900383 and 3.141592653590042) when:
- approx variable started with "0" and "3" respectively
- n=10000
Does anyone know why?
def approximate_pi(n):
approx=0
deno=2
if n == 1:
return 3
for x in range(n-1):
if x%2:
approx -= 4/((deno)*(deno+1)*(deno+2))
else:
approx += 4/((deno)*(deno+1)*(deno+2))
deno+=2
return approx+3
and
def approximate_pi(n):
approx=3
deno=2
if n == 1:
return 3
for x in range(n-1):
if x%2:
approx -= 4/((deno)*(deno+1)*(deno+2))
else:
approx += 4/((deno)*(deno+1)*(deno+2))
deno+=2
return approx
python python-3.x
edited Nov 12 '18 at 8:28
jww
52.7k39221483
asked Nov 12 '18 at 4:48
Zi Yang Chow
1
@tripleee - I thought this had some close votes due to duplicates. Did something happen to them?
– jww
Nov 12 '18 at 9:33
add a comment |
I was attempting an approximated value of pi through the formula of
pi= 3 + (4/(2*3*4)) - (4/(4*5*6)) + (4/(6*7*8)) - … (and so on). However, my code (shown below) had 2 separate answers (3.1415926535900383 and 3.141592653590042) when:
- approx variable started with "0" and "3" respectively
- n=10000
Does anyone know why?
def approximate_pi(n):
approx=0
deno=2
if n == 1:
return 3
for x in range(n-1):
if x%2:
approx -= 4/((deno)*(deno+1)*(deno+2))
else:
approx += 4/((deno)*(deno+1)*(deno+2))
deno+=2
return approx+3
and
def approximate_pi(n):
approx=3
deno=2
if n == 1:
return 3
for x in range(n-1):
if x%2:
approx -= 4/((deno)*(deno+1)*(deno+2))
else:
approx += 4/((deno)*(deno+1)*(deno+2))
deno+=2
return approx
python python-3.x
edited Nov 12 '18 at 8:28
jww
52.7k39221483
asked Nov 12 '18 at 4:48
Zi Yang Chow
1
I was attempting an approximated value of pi through the formula of
pi= 3 + (4/(2*3*4)) - (4/(4*5*6)) + (4/(6*7*8)) - … (and so on). However, my code (shown below) had 2 separate answers (3.1415926535900383 and 3.141592653590042) when:
- approx variable started with "0" and "3" respectively
- n=10000
Does anyone know why?
def approximate_pi(n):
approx=0
deno=2
if n == 1:
return 3
for x in range(n-1):
if x%2:
approx -= 4/((deno)*(deno+1)*(deno+2))
else:
approx += 4/((deno)*(deno+1)*(deno+2))
deno+=2
return approx+3
and
def approximate_pi(n):
approx=3
deno=2
if n == 1:
return 3
for x in range(n-1):
if x%2:
approx -= 4/((deno)*(deno+1)*(deno+2))
else:
approx += 4/((deno)*(deno+1)*(deno+2))
deno+=2
return approx
python python-3.x
python python-3.x
edited Nov 12 '18 at 8:28
jww
52.7k39221483
asked Nov 12 '18 at 4:48
Zi Yang Chow
1
edited Nov 12 '18 at 8:28
jww
52.7k39221483
asked Nov 12 '18 at 4:48
Zi Yang Chow
1
edited Nov 12 '18 at 8:28
jww
52.7k39221483
edited Nov 12 '18 at 8:28
jww
52.7k39221483
edited Nov 12 '18 at 8:28
jww
52.7k39221483
52.7k39221483
asked Nov 12 '18 at 4:48
Zi Yang Chow
1
asked Nov 12 '18 at 4:48
Zi Yang Chow
1
asked Nov 12 '18 at 4:48
Zi Yang Chow
1
1
@tripleee - I thought this had some close votes due to duplicates. Did something happen to them?
– jww
Nov 12 '18 at 9:33
add a comment |
@tripleee - I thought this had some close votes due to duplicates. Did something happen to them?
– jww
Nov 12 '18 at 9:33
@tripleee - I thought this had some close votes due to duplicates. Did something happen to them?
– jww
Nov 12 '18 at 9:33
@tripleee - I thought this had some close votes due to duplicates. Did something happen to them?
– jww
Nov 12 '18 at 9:33
add a comment |
3 Answers
3
active
oldest
votes
I think is because you can't have exact float numbers in pc. More info you can get here: Why can't decimal numbers be represented exactly in binary?
answered Nov 12 '18 at 5:16
ChaosPredictor
1,91311624
add a comment |
It is because of the way float is in python. If you do not have the digit before the decimal it gives 3 extra precision digits(from my trial). This changes the answer because when you start with 0, you get a different calculation altogether.
answered Nov 12 '18 at 5:51
Knl_Kolhe
113
add a comment |
An approximation algorithm approximates. Neither number is the true value of π. Your two versions start at different starting points, so why are you surprised that they give you slightly different approximations? What matters is that the longer you run them, the further they will both converge to the true value.
Note that this is not an artifact of the finite-precision representation of floats. While floating point rounding will affect your results, you would see differences even with unlimited precision arithmetic.
answered Nov 12 '18 at 9:29
alexis
33.5k954114
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
I think is because you can't have exact float numbers in pc. More info you can get here: Why can't decimal numbers be represented exactly in binary?
answered Nov 12 '18 at 5:16
ChaosPredictor
1,91311624
add a comment |
I think is because you can't have exact float numbers in pc. More info you can get here: Why can't decimal numbers be represented exactly in binary?
answered Nov 12 '18 at 5:16
ChaosPredictor
1,91311624
add a comment |
I think is because you can't have exact float numbers in pc. More info you can get here: Why can't decimal numbers be represented exactly in binary?
answered Nov 12 '18 at 5:16
ChaosPredictor
1,91311624
I think is because you can't have exact float numbers in pc. More info you can get here: Why can't decimal numbers be represented exactly in binary?
answered Nov 12 '18 at 5:16
ChaosPredictor
1,91311624
answered Nov 12 '18 at 5:16
ChaosPredictor
1,91311624
answered Nov 12 '18 at 5:16
ChaosPredictor
1,91311624
answered Nov 12 '18 at 5:16
ChaosPredictor
1,91311624
1,91311624
add a comment |
add a comment |
It is because of the way float is in python. If you do not have the digit before the decimal it gives 3 extra precision digits(from my trial). This changes the answer because when you start with 0, you get a different calculation altogether.
answered Nov 12 '18 at 5:51
Knl_Kolhe
113
add a comment |
It is because of the way float is in python. If you do not have the digit before the decimal it gives 3 extra precision digits(from my trial). This changes the answer because when you start with 0, you get a different calculation altogether.
answered Nov 12 '18 at 5:51
Knl_Kolhe
113
add a comment |
It is because of the way float is in python. If you do not have the digit before the decimal it gives 3 extra precision digits(from my trial). This changes the answer because when you start with 0, you get a different calculation altogether.
answered Nov 12 '18 at 5:51
Knl_Kolhe
113
It is because of the way float is in python. If you do not have the digit before the decimal it gives 3 extra precision digits(from my trial). This changes the answer because when you start with 0, you get a different calculation altogether.
answered Nov 12 '18 at 5:51
Knl_Kolhe
113
answered Nov 12 '18 at 5:51
Knl_Kolhe
113
answered Nov 12 '18 at 5:51
Knl_Kolhe
113
answered Nov 12 '18 at 5:51
Knl_Kolhe
113
113
add a comment |
add a comment |
An approximation algorithm approximates. Neither number is the true value of π. Your two versions start at different starting points, so why are you surprised that they give you slightly different approximations? What matters is that the longer you run them, the further they will both converge to the true value.
Note that this is not an artifact of the finite-precision representation of floats. While floating point rounding will affect your results, you would see differences even with unlimited precision arithmetic.
answered Nov 12 '18 at 9:29
alexis
33.5k954114
add a comment |
An approximation algorithm approximates. Neither number is the true value of π. Your two versions start at different starting points, so why are you surprised that they give you slightly different approximations? What matters is that the longer you run them, the further they will both converge to the true value.
Note that this is not an artifact of the finite-precision representation of floats. While floating point rounding will affect your results, you would see differences even with unlimited precision arithmetic.
answered Nov 12 '18 at 9:29
alexis
33.5k954114
add a comment |
An approximation algorithm approximates. Neither number is the true value of π. Your two versions start at different starting points, so why are you surprised that they give you slightly different approximations? What matters is that the longer you run them, the further they will both converge to the true value.
Note that this is not an artifact of the finite-precision representation of floats. While floating point rounding will affect your results, you would see differences even with unlimited precision arithmetic.
answered Nov 12 '18 at 9:29
alexis
33.5k954114
An approximation algorithm approximates. Neither number is the true value of π. Your two versions start at different starting points, so why are you surprised that they give you slightly different approximations? What matters is that the longer you run them, the further they will both converge to the true value.
Note that this is not an artifact of the finite-precision representation of floats. While floating point rounding will affect your results, you would see differences even with unlimited precision arithmetic.
answered Nov 12 '18 at 9:29
alexis
33.5k954114
answered Nov 12 '18 at 9:29
alexis
33.5k954114
answered Nov 12 '18 at 9:29
alexis
33.5k954114
answered Nov 12 '18 at 9:29
alexis
33.5k954114
33.5k954114
add a comment |
add a comment |
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@tripleee - I thought this had some close votes due to duplicates. Did something happen to them?
– jww
Nov 12 '18 at 9:33