knitr::opts_chunk$set( collapse = TRUE, comment = "#>" )
library(caracas)
inline_code <- function(x) { x } if (!has_sympy()) { # SymPy not available, so the chunks shall not be evaluated knitr::opts_chunk$set(eval = FALSE) inline_code <- function(x) { deparse(substitute(x)) } }
SymPy
directlyFirst we get the SymPy
object:
sympy <- get_sympy()
sympy$diff("2*a*x", "x") sympy$solve("x**2 - 1", "x")
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$derive_by_array(lap, list(d, h, lam)) grad
And find the critical points:
sol <- sympy$solve(grad, list(d, h, lam), dict = TRUE) sol
We take the one with the real solution:
sol[[1]]
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]]$d) sol_d eval(sol_d) sol_h <- to_r(sol[[1]]$h) sol_h eval(sol_h)
(It is left as an exercise to the reader to show that the critical point indeed is a minimum.)
x <- sympy$symbols('x') x$assumptions0 x <- sympy$symbols('x', positive = TRUE) x$assumptions0 eq <- sympy$parsing$sympy_parser$parse_expr("x**2 - 1", local_dict = list('x' = x)) sympy$solve(eq, x, dict = TRUE)
x <- sympy$symbols('x', positive = TRUE) eq <- sympy$parsing$sympy_parser$parse_expr("x**3/3 - x", local_dict = list('x' = x)) eq grad <- sympy$derive_by_array(eq, x) grad sympy$solve(grad, x, dict = TRUE)
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