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It appears that the tuple syntax works like this: (variable index, order of derivative) Where something like: base = poly(x*y**2 + x, x, y) deriv_mysterious5 = base.diff((0,1)) print('deriv_mysteri...
Answer
#3: Post edited
by
jrh
·
2020-09-30T21:30:55Z (about 4 years ago)
- It appears that the tuple syntax works like this:
- `(variable index, order of derivative)`
- Where something like:
- ```
- base = poly(x*y**2 + x, x, y)
- deriv_mysterious5 = base.diff((0,1))
- print('deriv_mysterious5 is', deriv_mysterious5)
- ```
- Means: "Take first derivative of `base` with respect to symbol[0]", in this case symbol 0 is x.
- - I was not able to find information on how sympy determines which symbol is symbol 0.
- I think I would recommend the alternative syntax instead:
- ```
- base.diff(<symbol>)
- ```
- e.g.,
- ```
- base.diff(x)
- ```
- Though, interestingly, it seems that for higher order derivatives, `poly.diff` has some behavior that seems to differ from `diff(expr,...)` as seen on the [Calculus](https://docs.sympy.org/latest/tutorial/calculus.html#derivatives) page.
- This syntax works:
- ```
- base.diff(y, y)
- ```
- I was not able to get this syntax working:
- ```
- base.diff(y, 2)
- ```
- But a slightly modified version of this does work,
- ```
base.diff((y, 3))- ```
- It appears that the tuple syntax works like this:
- `(variable index, order of derivative)`
- Where something like:
- ```
- base = poly(x*y**2 + x, x, y)
- deriv_mysterious5 = base.diff((0,1))
- print('deriv_mysterious5 is', deriv_mysterious5)
- ```
- Means: "Take first derivative of `base` with respect to symbol[0]", in this case symbol 0 is x.
- - I was not able to find information on how sympy determines which symbol is symbol 0.
- I think I would recommend the alternative syntax instead:
- ```
- base.diff(<symbol>)
- ```
- e.g.,
- ```
- base.diff(x)
- ```
- Though, interestingly, it seems that for higher order derivatives, `poly.diff` has some behavior that seems to differ from `diff(expr,...)` as seen on the [Calculus](https://docs.sympy.org/latest/tutorial/calculus.html#derivatives) page.
- This syntax works:
- ```
- base.diff(y, y)
- ```
- I was not able to get this syntax working:
- ```
- base.diff(y, 2)
- ```
- But a slightly modified version of this does work,
- ```
- base.diff((y, 2))
- ```
#2: Post edited
by
jrh
·
2020-09-30T21:29:04Z (about 4 years ago)
- It appears that the tuple syntax works like this:
- `(variable index, order of derivative)`
- Where something like:
- ```
- base = poly(x*y**2 + x, x, y)
- deriv_mysterious5 = base.diff((0,1))
- print('deriv_mysterious5 is', deriv_mysterious5)
- ```
- Means: "Take first derivative of `base` with respect to symbol[0]", in this case symbol 0 is x.
- - I was not able to find information on how sympy determines which symbol is symbol 0.
- I think I would recommend the alternative syntax instead:
- ```
- base.diff(<symbol>)
- ```
- e.g.,
- ```
- base.diff(x)
- ```
Though, interestingly, it seems that for higher order derivatives, only this syntax is supported:- ```
base.diff(x, x, x)- ```
- I was not able to get this syntax working:
- ```
base.diff(x, 3)```
- It appears that the tuple syntax works like this:
- `(variable index, order of derivative)`
- Where something like:
- ```
- base = poly(x*y**2 + x, x, y)
- deriv_mysterious5 = base.diff((0,1))
- print('deriv_mysterious5 is', deriv_mysterious5)
- ```
- Means: "Take first derivative of `base` with respect to symbol[0]", in this case symbol 0 is x.
- - I was not able to find information on how sympy determines which symbol is symbol 0.
- I think I would recommend the alternative syntax instead:
- ```
- base.diff(<symbol>)
- ```
- e.g.,
- ```
- base.diff(x)
- ```
- Though, interestingly, it seems that for higher order derivatives, `poly.diff` has some behavior that seems to differ from `diff(expr,...)` as seen on the [Calculus](https://docs.sympy.org/latest/tutorial/calculus.html#derivatives) page.
- This syntax works:
- ```
- base.diff(y, y)
- ```
- I was not able to get this syntax working:
- ```
- base.diff(y, 2)
- ```
- But a slightly modified version of this does work,
- ```
- base.diff((y, 3))
- ```
#1: Initial revision
by
jrh
·
2020-09-30T15:50:48Z (about 4 years ago)
It appears that the tuple syntax works like this:
`(variable index, order of derivative)`
Where something like:
```
base = poly(x*y**2 + x, x, y)
deriv_mysterious5 = base.diff((0,1))
print('deriv_mysterious5 is', deriv_mysterious5)
```
Means: "Take first derivative of `base` with respect to symbol[0]", in this case symbol 0 is x.
- I was not able to find information on how sympy determines which symbol is symbol 0.
I think I would recommend the alternative syntax instead:
```
base.diff(<symbol>)
```
e.g.,
```
base.diff(x)
```
Though, interestingly, it seems that for higher order derivatives, only this syntax is supported:
```
base.diff(x, x, x)
```
I was not able to get this syntax working:
```
base.diff(x, 3)
```