Knigh4321 Posted January 26, 2003 Share Posted January 26, 2003 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... Link to comment Share on other sites More sharing options...
JaKiri Posted January 26, 2003 Share Posted January 26, 2003 [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. Link to comment Share on other sites More sharing options...
JaKiri Posted January 26, 2003 Share Posted January 26, 2003 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. Link to comment Share on other sites More sharing options...
Knigh4321 Posted January 26, 2003 Author Share Posted January 26, 2003 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. Link to comment Share on other sites More sharing options...
Knigh4321 Posted January 26, 2003 Author Share Posted January 26, 2003 thank you sooo much Link to comment Share on other sites More sharing options...
JaKiri Posted January 26, 2003 Share Posted January 26, 2003 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. Link to comment Share on other sites More sharing options...
JaKiri Posted January 26, 2003 Share Posted January 26, 2003 Originally posted by Knigh4321 thank you sooo much 'Beware the physics student awake at 6 in the morning with nothing to do.' Link to comment Share on other sites More sharing options...
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