-1

The equation of my set is: X(i) = 1.1 * X(i-1) + 10 where X(i) = 10

I want to know how to turn this into a equation into which I hope to insert a value of X(i-50) (for example) and find X(i) or graph on a graphic calculator.

rpgstar
  • 101
  • Perhaps you could give more details. An equation for a set is nonsense. Why do you expect a quadratic? – Michael Burr Mar 27 '18 at 03:25
  • @MichaelBurr each new number is the previous number times 1.1 + 10. sorry if my terminology is incorrect this is a new area for me. – rpgstar Mar 27 '18 at 03:30
  • @MichaelBurr Also the first number is always X(1) = 1 – rpgstar Mar 27 '18 at 03:34
  • I hate to be a stickler for terminology, but if you have the right terms you can make people understand you and search the internet. Sets do not have any order, so the elements do not have indices. Lines are straight, quadratics are not. You have a linear recurrence relation. It is easy to write a spreadsheet to generate the values you want. Presumably the initial condition is $X(1)=10$, not what you wrote in the first line. – Ross Millikan Mar 27 '18 at 03:47

2 Answers2

2

It looks like you are trying to solve a linear difference equation. These are of the form $$x_{n+1}=ax_n+b$$

There are two special cases:

  1. If $a=1$ then $x_n=x_0+nb$
  2. If $b=0$ then $x_n = a^nx_0$

To solve the general case we can transform our original equation into form $2$ by introducing a new set of variables $y_n$ and making a substitution.

For each $n$, let $y_n=x_n-c$ where $c$ is a constant we will set to suit our needs. Note that $c$ is the same for each $n$.

Then $x_{n+1}=ax_n+b$ becomes $y_{n+1}+c = a(y_n+c)+b = ay_n+ac+b$.

Here's the trick: Since we are free to set $c$ to be anything we want we will set it so that $c = ac+b\implies c=\frac{b}{1-a}$. Notice that we dealt with the special case $a=1$ above so we can assume $a\ne 1$ so we don't get a division by zero.

Now $y_{n+1}+c = ay_n+ac+b$ becomes $y_{n+1}=ay_n\implies y_n=a^ny_0$

Replacing $y_n$ with $x_n-c$ gives $$x_n-c = a^n(x_0-c)$$

Your numbers $a=1.1$, $b=10$ and $x_0 = 1$ give us that $c=\frac{10}{-.1}=-100$ and this gives us $$x_n=(1.1)^n(101) - 100$$

John Douma
  • 11,426
  • 2
  • 23
  • 24
1

You have an inhomogeneous linear recurrence relation. You can generate a table of values in a spreadsheet by making a column for $i$ and a column for $X(i)$, writing the equation into the $X(2)$ square and copying down. You get the below. The spreadsheet will graph it for you as well. John Douma has shown an analytic solution.
enter image description here

Ross Millikan
  • 374,822