difference equations

[Knigh4321]

In short, can some one show me how this works:

the difference eqn is:

 

y(t+1)-2y(t)=1

therefore

 

y(t)=2^t-1

 

what i don't understand is the step to get the exponent (t) on the right side of the eqn....what happened to the y(t+1)?

any help would be nice. I'm going to go to my tech on Mon. to ask...but if I could get it before hand it'd be nice...

[JaKiri]

[oops misread the topic]

 

Looks like all you have to do is to put y(t+1) = 2y(t) + 1, and the final line follows directly.

[JaKiri]

To expand on that: the function is of the form y(t) = 2^t + c, because of the multiplication by 2 business, and putting that into the first given we get

 

2*2^t + c - 2 ( 2^t + c) = 1

2*2^t + c - 2*2^t - 2c = 1

- c = 1

c = -1

 

=> y(t) = 2^t - 1.

[Knigh4321]

that was an intermediate step...the exponent is killing me...I don't understand where it came from. Is there some properity that I'm forgetting about. I can't find it in any books. To be specific:

 

if

y(t+1)=y(t)+1+y(t)

then

y(t+1)=2y(t)+1

then

y(t+1)-2y(t)=1

(this is where things get screwy)

then

for y(1)=1 the eqn

y(t)=2^t-1....

 

there is a jump here that I am not understanding. Call it lack of knowlegde or whatever but I'm stumped.

[Knigh4321]

thank you sooo much

[JaKiri]

Remember that exponents are just shorthand for, say, 2*2*2*2*2*2*2*2*2*2 (2^10).

 

Lets ignore the = 1 from the original eqn for the moment (the conclusion is still valid, you just have to follow the procedure above to get any constant).

 

y(t+1) = 2y(t)

 

Let's say that y(0) = c.

 

y(1) = 2 * c

y(2) = 4c = 2^2 * c

y(3) = 8c = 2^3 * c

 

From this, it should be clear to see that y(t) = 2^t * c, which is where the exponent comes from; as you're multiplying by 2 every time, after t cycles, you must have multiplied by t 2s, which is equal to 2^t.

[JaKiri]

Originally posted by Knigh4321

thank you sooo much

 

'Beware the physics student awake at 6 in the morning with nothing to do.'