How to define and use a linear function

Question

I have the following equation $T = -800n + 12000$. I know that this is a linear function and that the two points $(0, 12000)$ and $(15,0)$ lie on the line.

When I have 2 other values of n between 0 and 15, how is it possible to determine their corresponding T-values with Mathematica.

Answer 1

t[n_] := -800 n + 12000

t[3]
t[7]

returns

9600
6400

This is how you define functions. Note the underscore after the parameter name ("n_"), and that you don't use the underscore in the parameter's references in the function definition.

Note: I changed your T to t. All Mathematica's symbols start with an uppercase letter, and it's a good habit to start your own symbols with a lowercase letter to avoid conflicts.

Answer 2

Although an answer has been accepted already, let me give another simple approach. You don't need to define a function just to evaluate the right hand side of your equation for different n-values. Mathematica can replace the n with numeric values simply be using the /. operator. Say you want to know the t-values for n=3 and n=7, you can write

-800 n - 12000 /. n -> {3, 7}
(*
  Out[4]= {-14400, -17600} 
*)

These rules and replacements are a big thing in Mathematica and are worth a deeper look, because they work on any kind of expression. You should check Replace, ReplaceAll, ReplaceRepeated, ReplaceList and Rule (and RuleDelayed).

Answer 3

The most straightforward approach for defining a function was illustrated by stevenvh.

halirutan's approach is also useful for substituting values into a function.

A pure function represents a third option. It becomes increasingly useful in intermediate to advanced programming when you need a function "on the fly" (i.e. you probably will use it only once and you want the definition right there in the code where it's being used so that you and the kernel don't need to search for what t[n] stands for).

-800 # + 12000 &

---

For a single input:

-800 # + 12000 &[3]

9600

---

Use Map for multiple inputs:

-800 # + 12000 &/@{3,7}

{9600, 6400}

Answer 4

For the record... no function, whether defined by SetDelayed (:=) or Set (=) expliclty or by constructing a pure function (using # and &) is needed. To wit:

   -800 {3, 7} + 12000
{9600,6400}