what y(2) ,y(3), y(4) , y(5) ,by using matlab ?

y(0)=1 y(2)=5

y(n+2)=3y(n+1)-5y(n)

what y(2) ,y(3), y(4) , y(5) ,by using matlab ?

[Walter Roberson]

This is straight forward looping too.

Whenever you see equations like this, if there is a finite endpoint and it is less than a a billion or so, it is usually easier just to loop.

With very large numbers or with symbolic endpoints then it might be worth going through the math to calculate the formula for the terms.

[Walter Roberson]

Are you sure you are given y0 and y2 but not y1? If so then you can indeed calculate y1 but it will be 10/3 not an integer.

[Stephen23]

Original question by majed alharbi retrieved from Google Cache:

"what y(2) ,y(3), y(4) , y(5) ,by using matlab ?"

y(0)=1 y(2)=5

y(n+2)=3y(n+1)-5y(n)

what y(2) ,y(3), y(4) , y(5) ,by using matlab ?

[Rena Berman]

(Answers Dev) Restored edit

Answer 1

[Alan Stevens]

You need to start with y(1) = 1, not y(0) = 1. Then with y(2) = 5, you can use a for loop:

for n=1:3, y(n+2) = 3*y(n+1)-5*y(n); end

[majed alharbi]

still

[majed alharbi]

[Walter Roberson]

you did not loop like I suggested. You tried to vectorize.

The function can only be vectorized after a bunch of mathematical calculations to figure out the general form. I recommend looping, it is much easier.