# solving equations with yac in the Ryacas library

How can I solve equation for a variable z? I tried this

library(Ryacas)
yac("Solve((0.07-0.07exp(-z)-0.06508080z) == 0, z)")

I got the wrong answer. The correct answer z=0.148

A obvious answer is z = 0, since then you would have 0.07 - 0.07 \cdot e^0 - 0.0650808 \cdot 0 = 0.07 - 0.07 - 0 = 0.

If you want to solve such equations in R you could rely on the base function uniroot, which takes the intervall to search in for a root as well as the function as arguments and returns the x values for which your function is equal to zero, e.g. f(x) = 0:

func <- function(x) 0.07 - 0.07 * exp(-x) - 0.0650808 * x
# find the obvious answer by including 0 in the interval
uniroot(func, lower = 0, upper = 1)
#> $root #>  0 #> #>$f.root
#>  0
#>
#> $iter #>  0 #> #>$init.it
#>  NA
#>
#> $estim.prec #>  0 # find the second root by adjusting the interval uniroot(func, lower = 0.01, upper = 1) #>$root
#>  0.1475151
#>
#> $f.root #>  1.406651e-07 #> #>$iter
#>  9
#>
#> $init.it #>  NA #> #>$estim.prec
#>  6.103516e-05


Created on 2022-09-01 by the reprex package (v2.0.1)

As an additional hint: You have to explicitly use operators like "*" to tell R what it has to do. E.g. in your above code 0.07exp(-z) is not separated by "*", so R tries to find e.g. a function called 0.07exp which you try to pass an argument, -z, into it (and can obviously not find a suitable function because it wasn't defined).

I hope this clarifies some of your issues regarding R and function solving in R.

Kind regards

1 Like

yacas' newton solver finds your value

yac("Newton((0.07-0.07*Exp(-z)-0.06508080*z),z,0.1,0.0001)")

 "0.147545136027390784"


Its nice to throw equations at wolframAlpha to verify etc.
solve - Wolfram|Alpha (wolframalpha.com)

1 Like

Thank you for your help.

Thank you for your help.

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