How to compute an integral of a given function without SymPy?
I'm having difficulties creating a simple for loop code to integrate a given function without using SymPy. I'm thinking using some sort of Riemann approximation but I don't know how to do this exactly. The code I have so far is:
def xsquared(x):
n = 2
return x**n
def integral(fun, xmin, xmax):
total = 0
for a in range(xmin, xmax):
x = a
total += fun(x*1.235)
return total
print(integral(xsquared, 0, 4))
The output gives 21.3 but how do I do this without inputting a number referring to the "fun(x*1.235)" part?
Any help would be appreciated.
python python-3.x math
edited Nov 13 '18 at 5:37
ChaosPredictor
1,91311624
asked Nov 12 '18 at 23:56
complexsets
102
add a comment |
I'm having difficulties creating a simple for loop code to integrate a given function without using SymPy. I'm thinking using some sort of Riemann approximation but I don't know how to do this exactly. The code I have so far is:
def xsquared(x):
n = 2
return x**n
def integral(fun, xmin, xmax):
total = 0
for a in range(xmin, xmax):
x = a
total += fun(x*1.235)
return total
print(integral(xsquared, 0, 4))
The output gives 21.3 but how do I do this without inputting a number referring to the "fun(x*1.235)" part?
Any help would be appreciated.
python python-3.x math
edited Nov 13 '18 at 5:37
ChaosPredictor
1,91311624
asked Nov 12 '18 at 23:56
complexsets
102
add a comment |
I'm having difficulties creating a simple for loop code to integrate a given function without using SymPy. I'm thinking using some sort of Riemann approximation but I don't know how to do this exactly. The code I have so far is:
def xsquared(x):
n = 2
return x**n
def integral(fun, xmin, xmax):
total = 0
for a in range(xmin, xmax):
x = a
total += fun(x*1.235)
return total
print(integral(xsquared, 0, 4))
The output gives 21.3 but how do I do this without inputting a number referring to the "fun(x*1.235)" part?
Any help would be appreciated.
python python-3.x math
edited Nov 13 '18 at 5:37
ChaosPredictor
1,91311624
asked Nov 12 '18 at 23:56
complexsets
102
I'm having difficulties creating a simple for loop code to integrate a given function without using SymPy. I'm thinking using some sort of Riemann approximation but I don't know how to do this exactly. The code I have so far is:
def xsquared(x):
n = 2
return x**n
def integral(fun, xmin, xmax):
total = 0
for a in range(xmin, xmax):
x = a
total += fun(x*1.235)
return total
print(integral(xsquared, 0, 4))
The output gives 21.3 but how do I do this without inputting a number referring to the "fun(x*1.235)" part?
Any help would be appreciated.
python python-3.x math
python python-3.x math
edited Nov 13 '18 at 5:37
ChaosPredictor
1,91311624
asked Nov 12 '18 at 23:56
complexsets
102
edited Nov 13 '18 at 5:37
ChaosPredictor
1,91311624
asked Nov 12 '18 at 23:56
complexsets
102
edited Nov 13 '18 at 5:37
ChaosPredictor
1,91311624
edited Nov 13 '18 at 5:37
ChaosPredictor
1,91311624
edited Nov 13 '18 at 5:37
ChaosPredictor
1,91311624
1,91311624
asked Nov 12 '18 at 23:56
complexsets
102
asked Nov 12 '18 at 23:56
complexsets
102
asked Nov 12 '18 at 23:56
complexsets
102
102
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
To compute integral by Riemann means that you are computing the limit of Riemann's sum by making partitions innner (Wikipedia).
For that, your function could have a new parameter interval: dx or maybe you could just guess the interval by splitting the full range into N equal sized intervals. Here is an example asking dx as argument.
Then, your function should be:
def riemann(fun, xmin, xmax, dx):
total = 0
a = xmin
while a < xmax:
total += fun(a + dx/2)*dx
a += dx
return total
Example Outputs
print(riemann(xsquared, 0, 4, 0.1))
> 21.330000000000013
print(riemann(xsquared, 0, 4, 0.25))
> 21.3125
print(riemann(xsquared, 0, 4, 0.5))
> 21.25
Analythic resolution gives: 64/3 ~ 21.33333
You're aproximating then the integral by computing the area of the rectangle having:
height: the function value at interval middle pointfun(a + dx/2)
width: the interval length (dx)
Note: if xmax < xmin, you should verify that dx < 0.
answered Nov 13 '18 at 0:11
Cheche
828218
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
To compute integral by Riemann means that you are computing the limit of Riemann's sum by making partitions innner (Wikipedia).
For that, your function could have a new parameter interval: dx or maybe you could just guess the interval by splitting the full range into N equal sized intervals. Here is an example asking dx as argument.
Then, your function should be:
def riemann(fun, xmin, xmax, dx):
total = 0
a = xmin
while a < xmax:
total += fun(a + dx/2)*dx
a += dx
return total
Example Outputs
print(riemann(xsquared, 0, 4, 0.1))
> 21.330000000000013
print(riemann(xsquared, 0, 4, 0.25))
> 21.3125
print(riemann(xsquared, 0, 4, 0.5))
> 21.25
Analythic resolution gives: 64/3 ~ 21.33333
You're aproximating then the integral by computing the area of the rectangle having:
height: the function value at interval middle pointfun(a + dx/2)
width: the interval length (dx)
Note: if xmax < xmin, you should verify that dx < 0.
answered Nov 13 '18 at 0:11
Cheche
828218
add a comment |
To compute integral by Riemann means that you are computing the limit of Riemann's sum by making partitions innner (Wikipedia).
For that, your function could have a new parameter interval: dx or maybe you could just guess the interval by splitting the full range into N equal sized intervals. Here is an example asking dx as argument.
Then, your function should be:
def riemann(fun, xmin, xmax, dx):
total = 0
a = xmin
while a < xmax:
total += fun(a + dx/2)*dx
a += dx
return total
Example Outputs
print(riemann(xsquared, 0, 4, 0.1))
> 21.330000000000013
print(riemann(xsquared, 0, 4, 0.25))
> 21.3125
print(riemann(xsquared, 0, 4, 0.5))
> 21.25
Analythic resolution gives: 64/3 ~ 21.33333
You're aproximating then the integral by computing the area of the rectangle having:
height: the function value at interval middle pointfun(a + dx/2)
width: the interval length (dx)
Note: if xmax < xmin, you should verify that dx < 0.
answered Nov 13 '18 at 0:11
Cheche
828218
add a comment |
To compute integral by Riemann means that you are computing the limit of Riemann's sum by making partitions innner (Wikipedia).
For that, your function could have a new parameter interval: dx or maybe you could just guess the interval by splitting the full range into N equal sized intervals. Here is an example asking dx as argument.
Then, your function should be:
def riemann(fun, xmin, xmax, dx):
total = 0
a = xmin
while a < xmax:
total += fun(a + dx/2)*dx
a += dx
return total
Example Outputs
print(riemann(xsquared, 0, 4, 0.1))
> 21.330000000000013
print(riemann(xsquared, 0, 4, 0.25))
> 21.3125
print(riemann(xsquared, 0, 4, 0.5))
> 21.25
Analythic resolution gives: 64/3 ~ 21.33333
You're aproximating then the integral by computing the area of the rectangle having:
height: the function value at interval middle pointfun(a + dx/2)
width: the interval length (dx)
Note: if xmax < xmin, you should verify that dx < 0.
answered Nov 13 '18 at 0:11
Cheche
828218
To compute integral by Riemann means that you are computing the limit of Riemann's sum by making partitions innner (Wikipedia).
For that, your function could have a new parameter interval: dx or maybe you could just guess the interval by splitting the full range into N equal sized intervals. Here is an example asking dx as argument.
Then, your function should be:
def riemann(fun, xmin, xmax, dx):
total = 0
a = xmin
while a < xmax:
total += fun(a + dx/2)*dx
a += dx
return total
Example Outputs
print(riemann(xsquared, 0, 4, 0.1))
> 21.330000000000013
print(riemann(xsquared, 0, 4, 0.25))
> 21.3125
print(riemann(xsquared, 0, 4, 0.5))
> 21.25
Analythic resolution gives: 64/3 ~ 21.33333
You're aproximating then the integral by computing the area of the rectangle having:
height: the function value at interval middle pointfun(a + dx/2)
width: the interval length (dx)
Note: if xmax < xmin, you should verify that dx < 0.
answered Nov 13 '18 at 0:11
Cheche
828218
edited Nov 13 '18 at 0:23
answered Nov 13 '18 at 0:11
Cheche
828218
answered Nov 13 '18 at 0:11
Cheche
828218
answered Nov 13 '18 at 0:11
Cheche
828218
828218
add a comment |
add a comment |
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