# how to repeated evalution of the recurrence relation

• May 2nd 2011, 12:07 PM
hazeleyes
how to repeated evalution of the recurrence relation
$u_{r}=4u_{r}+3\\u_{1}=6$
Test the answer by finding the value of $u_{4}$ in 2 different ways:
by substitution in your solution and by repeated evaluation of the equation.

by substitution I have found u2=27,u3=111 and u4=447
I'm wondering how to find these values by repeated evaluation of the equation?
Cheers

Moderator edit: Please use tex tags NOT math tags. See Questions etc. subforum.
• May 2nd 2011, 12:18 PM
hazeleyes
Find the solution to the equation Ur+1= 4Ur + 3?
U1 = 6

Test your answer by finding the value of u4 in 2 different ways: by substitution in your solution and by repeated evaluation of the equation.

• May 2nd 2011, 12:37 PM
TheEmptySet
Quote:

Originally Posted by hazeleyes
$u_{r}=4u_{r}+3\\u_{1}=6$
Test the answer by finding the value of $u_{4}$ in 2 different ways:
by substitution in your solution and by repeated evaluation of the equation.

by substitution I have found u2=27,u3=111 and u4=447
I'm wondering how to find these values by repeated evaluation of the equation?
Cheers

I think you are trying to solve the difference equation

$u_{r+1}-4u_{r}=3$ I think there is a typo and you are missing the +1

So to solve we do it in two parts:

First we solve the homogenous equation

$u_{r+1}-4u_{r}=0$

we use the "guess"

$u_{r}=n^{r} \implies u_{r+1}=n^{r+1}$

Plugging this in gives

$n^{r+1}-4n^{r}=0 \iff n^{r}(n-4)=0$

this gives $n=4$

so the solution has the form

$u_{r}=C4^{r}$

Since the right hand side is a constant we guess that the particular solution is also a constant

$u_r=C4^r+A$

$C4^{r+1}+A-4(C4^r+A)=3 \iff A-4A=3 \iff A=-1$

So the solution is

$u_{r}=C4^r-1$

Now using the initial condition gives

$u_{1}=4C-1=6 \iff C=\frac{7}{4}$

So the solution is

$u_{r}=7(4)^{r-1}-1$
Now if we plug in

$r=4$

we get

$u_{4}=7\cdot 4^3-1=447$
• May 2nd 2011, 12:42 PM
Plato
Quote:

Originally Posted by hazeleyes
Find the solution to the equation Ur+1= 4Ur + 3?
U1 = 6

Look at this.
• May 2nd 2011, 01:51 PM
hazeleyes
thanks so much for your super clear explanation!!!