Regular simpsons rule in numerical methods
0
$begingroup$
In the following code I have implemented simpsons rule. It is working correctly for my first function but for my second function I am getting an error.
So,
Is it possible to do this code without using np.linspace? and if not how do I fix it so my second function works?
def simps(f,a,b,N):
dx = (b-a)/N
x = np.linspace(a,b,N+1)
y = f(x)
S = dx/3 * np.sum(y[0:-1:2] + 4*y[1::2] + y[2::2])
return S
print(simps(lambda x:x**2, 5, 10, 100))
print(simps(lambda x: sin(x),0,pi/2,100))
which gives the output:
291.6666666666667
---------------------------------------------------------------------------
TypeError Traceback (most recent call last)
<ipython-input-143-40325ae966e9> in <module>()
1 print(simps(lambda x:x**2, 5, 10, 100))
----> 2 print(simps(lambda x: sin(x),0,pi/2,100))
<ipython-input-142-860ce9822b06> in simps(f, a, b, N)
3 dx = (b-a)/N
4 x = np.linspace(a,b,N+1)
----> 5 y = f(x)
6 S = dx/3 * np.sum(y[0:-1:2] + 4*y[1::2] + y[2::2])
7 return S
<ipython-input-143-40325ae966e9> in <lambda>(x)
1 print(simps(lambda x:x**2, 5, 10, 100))
---->2 print(simps(lambda x: sin(x),0,pi/2,100))
TypeError: only size-1 arrays can be converted to Python scalars
numerical-methods
asked Dec 4 '18 at 2:04
fr14fr14
38318
$endgroup$
add a comment |
0
$begingroup$
In the following code I have implemented simpsons rule. It is working correctly for my first function but for my second function I am getting an error.
So,
Is it possible to do this code without using np.linspace? and if not how do I fix it so my second function works?
def simps(f,a,b,N):
dx = (b-a)/N
x = np.linspace(a,b,N+1)
y = f(x)
S = dx/3 * np.sum(y[0:-1:2] + 4*y[1::2] + y[2::2])
return S
print(simps(lambda x:x**2, 5, 10, 100))
print(simps(lambda x: sin(x),0,pi/2,100))
which gives the output:
291.6666666666667
---------------------------------------------------------------------------
TypeError Traceback (most recent call last)
<ipython-input-143-40325ae966e9> in <module>()
1 print(simps(lambda x:x**2, 5, 10, 100))
----> 2 print(simps(lambda x: sin(x),0,pi/2,100))
<ipython-input-142-860ce9822b06> in simps(f, a, b, N)
3 dx = (b-a)/N
4 x = np.linspace(a,b,N+1)
----> 5 y = f(x)
6 S = dx/3 * np.sum(y[0:-1:2] + 4*y[1::2] + y[2::2])
7 return S
<ipython-input-143-40325ae966e9> in <lambda>(x)
1 print(simps(lambda x:x**2, 5, 10, 100))
---->2 print(simps(lambda x: sin(x),0,pi/2,100))
TypeError: only size-1 arrays can be converted to Python scalars
numerical-methods
asked Dec 4 '18 at 2:04
fr14fr14
38318
$endgroup$
add a comment |
0
0
0
$begingroup$
In the following code I have implemented simpsons rule. It is working correctly for my first function but for my second function I am getting an error.
So,
Is it possible to do this code without using np.linspace? and if not how do I fix it so my second function works?
def simps(f,a,b,N):
dx = (b-a)/N
x = np.linspace(a,b,N+1)
y = f(x)
S = dx/3 * np.sum(y[0:-1:2] + 4*y[1::2] + y[2::2])
return S
print(simps(lambda x:x**2, 5, 10, 100))
print(simps(lambda x: sin(x),0,pi/2,100))
which gives the output:
291.6666666666667
---------------------------------------------------------------------------
TypeError Traceback (most recent call last)
<ipython-input-143-40325ae966e9> in <module>()
1 print(simps(lambda x:x**2, 5, 10, 100))
----> 2 print(simps(lambda x: sin(x),0,pi/2,100))
<ipython-input-142-860ce9822b06> in simps(f, a, b, N)
3 dx = (b-a)/N
4 x = np.linspace(a,b,N+1)
----> 5 y = f(x)
6 S = dx/3 * np.sum(y[0:-1:2] + 4*y[1::2] + y[2::2])
7 return S
<ipython-input-143-40325ae966e9> in <lambda>(x)
1 print(simps(lambda x:x**2, 5, 10, 100))
---->2 print(simps(lambda x: sin(x),0,pi/2,100))
TypeError: only size-1 arrays can be converted to Python scalars
numerical-methods
asked Dec 4 '18 at 2:04
fr14fr14
38318
$endgroup$
In the following code I have implemented simpsons rule. It is working correctly for my first function but for my second function I am getting an error.
So,
Is it possible to do this code without using np.linspace? and if not how do I fix it so my second function works?
def simps(f,a,b,N):
dx = (b-a)/N
x = np.linspace(a,b,N+1)
y = f(x)
S = dx/3 * np.sum(y[0:-1:2] + 4*y[1::2] + y[2::2])
return S
print(simps(lambda x:x**2, 5, 10, 100))
print(simps(lambda x: sin(x),0,pi/2,100))
which gives the output:
291.6666666666667
---------------------------------------------------------------------------
TypeError Traceback (most recent call last)
<ipython-input-143-40325ae966e9> in <module>()
1 print(simps(lambda x:x**2, 5, 10, 100))
----> 2 print(simps(lambda x: sin(x),0,pi/2,100))
<ipython-input-142-860ce9822b06> in simps(f, a, b, N)
3 dx = (b-a)/N
4 x = np.linspace(a,b,N+1)
----> 5 y = f(x)
6 S = dx/3 * np.sum(y[0:-1:2] + 4*y[1::2] + y[2::2])
7 return S
<ipython-input-143-40325ae966e9> in <lambda>(x)
1 print(simps(lambda x:x**2, 5, 10, 100))
---->2 print(simps(lambda x: sin(x),0,pi/2,100))
TypeError: only size-1 arrays can be converted to Python scalars
numerical-methods
numerical-methods
asked Dec 4 '18 at 2:04
fr14fr14
38318
asked Dec 4 '18 at 2:04
fr14fr14
38318
asked Dec 4 '18 at 2:04
fr14fr14
38318
asked Dec 4 '18 at 2:04
fr14fr14
38318
asked Dec 4 '18 at 2:04
fr14fr14
38318
38318
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
1
$begingroup$
linspace is a very useful function to generate uniformly separated points. Remember to use function from numpy to efficiently map operations on arrays
import numpy as np
def simps(f,a,b,N):
dx = (b-a)/N
x = np.linspace(a,b,N+1)
y = f(x)
S = dx/3 * np.sum(y[0:-1:2] + 4*y[1::2] + y[2::2])
return S
print(simps(lambda x:x**2, 5, 10, 100))
print(simps(lambda x: np.sin(x),0,np.pi/2,100))
With result
291.6666666666667
1.000000000338236
answered Dec 4 '18 at 2:07
caveraccaverac
14.4k31130
$endgroup$
$begingroup$
you have all the answers to my problems haha I appreciate it!
$endgroup$
– fr14
Dec 4 '18 at 2:09
$begingroup$
@fr14 Glad I could help
$endgroup$
– caverac
Dec 4 '18 at 2:11
add a comment |
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1 Answer
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1 Answer
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1
$begingroup$
linspace is a very useful function to generate uniformly separated points. Remember to use function from numpy to efficiently map operations on arrays
import numpy as np
def simps(f,a,b,N):
dx = (b-a)/N
x = np.linspace(a,b,N+1)
y = f(x)
S = dx/3 * np.sum(y[0:-1:2] + 4*y[1::2] + y[2::2])
return S
print(simps(lambda x:x**2, 5, 10, 100))
print(simps(lambda x: np.sin(x),0,np.pi/2,100))
With result
291.6666666666667
1.000000000338236
answered Dec 4 '18 at 2:07
caveraccaverac
14.4k31130
$endgroup$
$begingroup$
you have all the answers to my problems haha I appreciate it!
$endgroup$
– fr14
Dec 4 '18 at 2:09
$begingroup$
@fr14 Glad I could help
$endgroup$
– caverac
Dec 4 '18 at 2:11
add a comment |
1
$begingroup$
linspace is a very useful function to generate uniformly separated points. Remember to use function from numpy to efficiently map operations on arrays
import numpy as np
def simps(f,a,b,N):
dx = (b-a)/N
x = np.linspace(a,b,N+1)
y = f(x)
S = dx/3 * np.sum(y[0:-1:2] + 4*y[1::2] + y[2::2])
return S
print(simps(lambda x:x**2, 5, 10, 100))
print(simps(lambda x: np.sin(x),0,np.pi/2,100))
With result
291.6666666666667
1.000000000338236
answered Dec 4 '18 at 2:07
caveraccaverac
14.4k31130
$endgroup$
$begingroup$
you have all the answers to my problems haha I appreciate it!
$endgroup$
– fr14
Dec 4 '18 at 2:09
$begingroup$
@fr14 Glad I could help
$endgroup$
– caverac
Dec 4 '18 at 2:11
add a comment |
1
1
1
$begingroup$
linspace is a very useful function to generate uniformly separated points. Remember to use function from numpy to efficiently map operations on arrays
import numpy as np
def simps(f,a,b,N):
dx = (b-a)/N
x = np.linspace(a,b,N+1)
y = f(x)
S = dx/3 * np.sum(y[0:-1:2] + 4*y[1::2] + y[2::2])
return S
print(simps(lambda x:x**2, 5, 10, 100))
print(simps(lambda x: np.sin(x),0,np.pi/2,100))
With result
291.6666666666667
1.000000000338236
answered Dec 4 '18 at 2:07
caveraccaverac
14.4k31130
$endgroup$
linspace is a very useful function to generate uniformly separated points. Remember to use function from numpy to efficiently map operations on arrays
import numpy as np
def simps(f,a,b,N):
dx = (b-a)/N
x = np.linspace(a,b,N+1)
y = f(x)
S = dx/3 * np.sum(y[0:-1:2] + 4*y[1::2] + y[2::2])
return S
print(simps(lambda x:x**2, 5, 10, 100))
print(simps(lambda x: np.sin(x),0,np.pi/2,100))
With result
291.6666666666667
1.000000000338236
answered Dec 4 '18 at 2:07
caveraccaverac
14.4k31130
answered Dec 4 '18 at 2:07
caveraccaverac
14.4k31130
answered Dec 4 '18 at 2:07
caveraccaverac
14.4k31130
answered Dec 4 '18 at 2:07
caveraccaverac
14.4k31130
14.4k31130
$begingroup$
you have all the answers to my problems haha I appreciate it!
$endgroup$
– fr14
Dec 4 '18 at 2:09
$begingroup$
@fr14 Glad I could help
$endgroup$
– caverac
Dec 4 '18 at 2:11
add a comment |
$begingroup$
you have all the answers to my problems haha I appreciate it!
$endgroup$
– fr14
Dec 4 '18 at 2:09
$begingroup$
@fr14 Glad I could help
$endgroup$
– caverac
Dec 4 '18 at 2:11
$begingroup$
you have all the answers to my problems haha I appreciate it!
$endgroup$
– fr14
Dec 4 '18 at 2:09
$begingroup$
you have all the answers to my problems haha I appreciate it!
$endgroup$
– fr14
Dec 4 '18 at 2:09
$begingroup$
@fr14 Glad I could help
$endgroup$
– caverac
Dec 4 '18 at 2:11
$begingroup$
@fr14 Glad I could help
$endgroup$
– caverac
Dec 4 '18 at 2:11
add a comment |
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