A more challenging function

In the previous section, you may have wondered ``why bother with an iterative solution when I can easily see that $f(x)$ $=$ $x^2$ is a solution?'' However, what if you were modelling a problem where things were a bit messier? Suppose you knew that for some constant $K$ you had:

$$f'(x) = -f(x) - K(f(x) - \exp(-x))$$ (1)

$$f(0) = 2.$$ (2)

Once again you have an unknown function, $f(x)$, defined in terms of its derivative.¹However, our current definition is more complicated, since the right-hand side is an expression in both $x$ and our unknown function $f(x)$. You might have studied differential equations (which Equation 1 is) and be able to use a well-studied bag of tricks (integrals, integrating factors, etc.)²to work out with paper and pencil that

$$f(x) = \exp(-(K+1)x) + \exp(-x)$$ (3)

However, if what you need is an approximation, you can proceed as in the previous section:

1. Decide on a step size that you'll increment $x$ by (we used 0.1 in the last section, but your mileage may vary), and call it $h$.
2. Denote your approximate function as $f_x$ to distinguish it from the exact function $f(x)$. Also denote the approximate derivative as $f'_x$, since it won't always be identical to $f'(x)$.
3. Since you are given the exact value $f(0)$ $=$ 2, start things out on the right foot by setting $f_0$ $=$ 2, and the initial value of $x$ $=$ 0.
4. $f_h$ is $f'_0\times h$ $+$ $f_0$, and $f'_0$ is calculated by by substituting $f_0$ for $f(0)$ in Equation 1.
5. $f_{2h}$ is $f'_h\times h$ $+$ $f_h$, substituting $f_h$ for $f(h)$ in Equation 1.
6. And so on...

Since you've been given the exact expression for $f(x)$, you can always compare your approximation $f_x$ to $f(x)$.