# How do you find the first three iterates for the function f(x)=-2x+5 given the initial value x_0=2?

Dec 16, 2017

${x}_{1} = 1$, ${x}_{2} = 3$ and ${x}_{3} = - 1$

Bonus: ${x}_{n} = \frac{1}{3} {\left(- 2\right)}^{n} + \frac{5}{3}$

#### Explanation:

To get the next iterate, apply the function $f \left(x\right)$ to the previous one:

${x}_{1} = f \left({x}_{0}\right) = f \left(2\right) = - 2 \left(\textcolor{b l u e}{2}\right) + 5 = 1$

${x}_{2} = f \left({x}_{1}\right) = f \left(1\right) = - 2 \left(\textcolor{b l u e}{1}\right) + 5 = 3$

${x}_{3} = f \left({x}_{2}\right) = f \left(3\right) = - 2 \left(\textcolor{b l u e}{3}\right) + 5 = - 1$

Bonus - General formula for a term

To find the general formula for a term of this sequence of iterates, first note that for large $x$ we have $f \frac{x}{x} \approx - 2$. So we should expect a formula like ${x}_{n} = a {\left(- 2\right)}^{n} + b$.

Let's try:

$1 = {x}_{1} = - 2 a + b$

$3 = {x}_{2} = 4 a + b$

Adding twice the first equation to the second, we find:

$3 b = 5$

So $b = \frac{5}{3}$

Subtracting the first equation from the second we get:

$6 a = 2$

So $a = \frac{1}{3}$

So let's try:

${x}_{n} = \frac{1}{3} {\left(- 2\right)}^{n} + \frac{5}{3}$

Then:

${x}_{n + 1} = f \left({x}_{n}\right)$

$\textcolor{w h i t e}{{x}_{n + 1}} = - 2 {x}_{n} + 5$

$\textcolor{w h i t e}{{x}_{n + 1}} = - 2 \left(\frac{1}{3} {\left(- 2\right)}^{n} + \frac{5}{3}\right) + 5$

$\textcolor{w h i t e}{{x}_{n + 1}} = \frac{1}{3} {\left(- 2\right)}^{n + 1} - \frac{10}{3} + 5$

$\textcolor{w h i t e}{{x}_{n + 1}} = \frac{1}{3} {\left(- 2\right)}^{n + 1} + \frac{5}{3}$

So our formula is correct.