1. ## Plotting Difference Equation

Hey guys, I am currently trying to plot the graph of the equation
N(t+1) = Lambda * N(t)

Where Lambda is a constant.
It is the basic unlimitted growth equation, and ideally i would like to plot it for different values of N(0) and different Lambda

Only problem is, I have no idea how I plot this, do I need to solve it? if so how do i solve such an equation, i seem to remember somewhere that the solution would be N(t) = N(0) exp lambda*t

is that correct?

I am using the computer program Autograph, which lets me enter any equation (apart from ones like this :P) Hence I guess i need to solve it and plot the solution? Im kinda confusing myself.

Many thanks for any help!

Tom

2. Originally Posted by MWG_Thomas
Hey guys, I am currently trying to plot the graph of the equation
N(t+1) = Lambda * N(t)

Where Lambda is a constant.
It is the basic unlimitted growth equation, and ideally i would like to plot it for different values of N(0) and different Lambda

Only problem is, I have no idea how I plot this, do I need to solve it? if so how do i solve such an equation, i seem to remember somewhere that the solution would be N(t) = N(0) exp lambda*t

is that correct?

I am using the computer program Autograph, which lets me enter any equation (apart from ones like this :P) Hence I guess i need to solve it and plot the solution? Im kinda confusing myself.

Many thanks for any help!

Tom
The solution to the functional equation $\displaystyle N(t+1) = \lambda N(t)$ is $\displaystyle N(t) = a \lambda^{t+1}$.

Edit: Should be $\displaystyle N(t) = a \lambda^{t}$ - thanks for the catch, topsquark.

3. Thanks, how do i know what a is? and then once i have this a can i just plot it as you would any y=mx+c equation?

Thanks again,

Tom

4. Originally Posted by MWG_Thomas
Thanks, how do i know what a is? and then once i have this a can i just plot it as you would any y=mx+c equation?

Thanks again,

Tom
a = N(0).
Yes.

5. $\displaystyle N(t+1) = \lambda N(t)$ is $\displaystyle N(t) = N(0) \lambda^{t+1}$.

So if that is my solution, why is it t+1 and not simply t?

As say my N(0) is 1000 and lambda is 5 this equation would give when t=0
is $\displaystyle N(t) = 1000* 5^{1}$. which = 5000, so then N(0) would be 5000? but i know that N(0) is 1000, so that can't be right?

Isnt the solution
$\displaystyle N(t) = N(0) \lambda^{t}$. This would then give me the correct answer for N(0)

Thanx!!

Tom

6. Originally Posted by MWG_Thomas
N(t+1) = Lambda * N(t)
$\displaystyle N(t + 1) - \lambda N(t) = 0$

The characteristic equation for this is
$\displaystyle m - \lambda = 0$

$\displaystyle m = \lambda$

Thus the solutions are of the form
$\displaystyle N(t) = a \lambda ^t$

where a = N(0).

-Dan