91 - Using the ‘SymPy’ object directly

library(caracas)

Using SymPy directly

First we get the SymPy object:

sympy <- get_sympy()

sympy$diff("2*a*x", "x")
#> 2*a
sympy$solve("x**2 - 1", "x")
#> [[1]]
#> -1
#> 
#> [[2]]
#> 1

Elaborate example

How can we minimise the amount of material used to produce a cylindric tin can that contains 1 litre. The cylinder has diameter \(d\) and height \(h\). The question is therefore: What is \(d\) and \(h\)?

We introduce the variables d (diameter) and h (height):

d <- sympy$symbols('d')
h <- sympy$symbols('h')

The problem is a constrained optimisation problem, and we solve it by a Lagrange multiplier, and therefore we introduce lam (the Lagrange multiplier):

lam <- sympy$symbols('lam')

We now set up the problem:

area_str <- "Pi/2 * d**2 + Pi * h * d"
vol_str <- "Pi/4 * d**2 * h"
lap_str <- paste0("(", area_str, ") - lam*((", vol_str, ") - 1)")
lap <- sympy$parsing$sympy_parser$parse_expr(
  lap_str,
  local_dict = list('d' = d, 'h' = h, 'lam' = lam))

We can now find the gradient:

grad <- sympy$derive_by_array(lap, list(d, h, lam))
grad
#> [-Pi*d*h*lam/2 + Pi*d + Pi*h, -Pi*d**2*lam/4 + Pi*d, -Pi*d**2*h/4 + 1]

And find the critical points:

sol <- sympy$solve(grad, list(d, h, lam), dict = TRUE)
sol
#> [[1]]
#> [[1]]$d
#> 2**(2/3)/Pi**(1/3)
#> 
#> [[1]]$h
#> 2**(2/3)/Pi**(1/3)
#> 
#> [[1]]$lam
#> 2*2**(1/3)*Pi**(1/3)
#> 
#> 
#> [[2]]
#> [[2]]$d
#> 2**(2/3)*(-1 - sqrt(3)*I)/(2*Pi**(1/3))
#> 
#> [[2]]$h
#> 2**(2/3)*(-1 - sqrt(3)*I)/(2*Pi**(1/3))
#> 
#> [[2]]$lam
#> -2**(1/3)*Pi**(1/3) + 2**(1/3)*sqrt(3)*I*Pi**(1/3)
#> 
#> 
#> [[3]]
#> [[3]]$d
#> 2**(2/3)*(-1 + sqrt(3)*I)/(2*Pi**(1/3))
#> 
#> [[3]]$h
#> 2**(2/3)*(-1 + sqrt(3)*I)/(2*Pi**(1/3))
#> 
#> [[3]]$lam
#> -2**(1/3)*Pi**(1/3) - 2**(1/3)*sqrt(3)*I*Pi**(1/3)

We take the one with the real solution:

sol[[1]]
#> $d
#> 2**(2/3)/Pi**(1/3)
#> 
#> $h
#> 2**(2/3)/Pi**(1/3)
#> 
#> $lam
#> 2*2**(1/3)*Pi**(1/3)

We now have a short helper function to help getting appropriate R expressions (such a function will be included in later versions of this package):

to_r <- function(x) {
  x <- as.character(x)
  x <- gsub("Pi", "pi", x, fixed = TRUE)
  x <- gsub("**", "^", x, fixed = TRUE)
  x <- parse(text = x)
  return(x)
}

sol_d <- to_r(sol[[1]]$d)
sol_d
#> expression(2^(2/3)/pi^(1/3))
eval(sol_d)
#> [1] 1.083852
sol_h <- to_r(sol[[1]]$h)
sol_h
#> expression(2^(2/3)/pi^(1/3))
eval(sol_h)
#> [1] 1.083852

(It is left as an exercise to the reader to show that the critical point indeed is a minimum.)

Simple example with assumptions

x <- sympy$symbols('x')
x$assumptions0
#> $commutative
#> [1] TRUE
x <- sympy$symbols('x', positive = TRUE)
x$assumptions0
#> $positive
#> [1] TRUE
#> 
#> $extended_nonpositive
#> [1] FALSE
#> 
#> $infinite
#> [1] FALSE
#> 
#> $extended_positive
#> [1] TRUE
#> 
#> $real
#> [1] TRUE
#> 
#> $commutative
#> [1] TRUE
#> 
#> $finite
#> [1] TRUE
#> 
#> $imaginary
#> [1] FALSE
#> 
#> $extended_nonzero
#> [1] TRUE
#> 
#> $negative
#> [1] FALSE
#> 
#> $nonpositive
#> [1] FALSE
#> 
#> $extended_negative
#> [1] FALSE
#> 
#> $nonzero
#> [1] TRUE
#> 
#> $hermitian
#> [1] TRUE
#> 
#> $zero
#> [1] FALSE
#> 
#> $extended_nonnegative
#> [1] TRUE
#> 
#> $nonnegative
#> [1] TRUE
#> 
#> $complex
#> [1] TRUE
#> 
#> $extended_real
#> [1] TRUE
eq <- sympy$parsing$sympy_parser$parse_expr("x**2 - 1",
                                            local_dict = list('x' = x))
sympy$solve(eq, x, dict = TRUE)
#> [[1]]
#> [[1]]$x
#> 1

Another example with assumptions

x <- sympy$symbols('x', positive = TRUE)
eq <- sympy$parsing$sympy_parser$parse_expr("x**3/3 - x",
                                            local_dict = list('x' = x))
eq
#> x**3/3 - x
grad <- sympy$derive_by_array(eq, x)
grad
#> x**2 - 1
sympy$solve(grad, x, dict = TRUE)
#> [[1]]
#> [[1]]$x
#> 1