Recurrence Relation: a(n) = 3a(n-1) + a(n-2)

[itaiezra8]

hi!
i have a question that i like to get answer.
it can be more than one solution in

Recurrence Relation that made the same answer?
for example : is the ​

a(n)=3a(n-1)+a(n-2) the same to a(n)=4a(n-1)-2a(n-2)-a(n-3) , a(0)=1 ,a(1)=4 ,a(2)=13
?
thanks...

 

[stapel]

i have a question that i like to get answer.
it can be more than one solution in

Recurrence Relation that made the same answer?
for example : is the ​

a(n)=3a(n-1)+a(n-2) the same to a(n)=4a(n-1)-2a(n-2)-a(n-3) , a(0)=1 ,a(1)=4 ,a(2)=13 ?

What is the full and exact text of the exercise, and its complete instructions? With respect to your notation, are the numbers in parentheses meant to into subscripts, so the above means the following?

. . . . .\(\displaystyle \mbox{Given }\, a_0\, =\, 1, a_1\, =\, 4,\,\mbox{ and }\, a_2\, =\, 13,\, \mbox{ is it true that the rela}\mbox{tion}\)

. . . . .\(\displaystyle a_n\, =\, 3a_{n-1}\, +\, a_{n-2}\, \mbox{ is equiv}\mbox{alent to }\, a_n\, =\, 4a_{n-1}\, -\, 2a_{n-2}\, -\, a_{n-3}\,?\)

When you reply, please include a clear listing of your thoughts and efforts so far. Thank you!

 

[itaiezra8]

What is the full and exact text of the exercise, and its complete instructions? With respect to your notation, are the numbers in parentheses meant to into subscripts, so the above means the following?

. . . . .\(\displaystyle \mbox{Given }\, a_0\, =\, 1, a_1\, =\, 4,\,\mbox{ and }\, a_2\, =\, 13,\, \mbox{ is it true that the rela}\mbox{tion}\)

. . . . .\(\displaystyle a_n\, =\, 3a_{n-1}\, +\, a_{n-2}\, \mbox{ is equiv}\mbox{alent to }\, a_n\, =\, 4a_{n-1}\, -\, 2a_{n-2}\, -\, a_{n-3}\,?\)

When you reply, please include a clear listing of your thoughts and efforts so far. Thank you!

the question, if it can be more than one solution for exercise?
in the exercise for given a(0)=1,a(1)=4,a(2)=13, the result is a(n)=3(n-1)+(n-2) but i claim that also a(n)=4a(n-1)-2a(n-2)-a(n-3) is the correct answer.
i used walfarm alpha but i am not sure.
how can i prove that?
thank you.

 

[ksdhart2]

Okay, now that's an entirely different story. To be able to say with any certainty whether a given problem has multiple solutions, we'd need to see the original problem you were given (as well as any and all work you did to arrive at your solutions). It's possible that both a(n)=3(n-1)+(n-2) and a(n)=4a(n-1)-2a(n-2)-a(n-3) are solutions, but it's also possible you made an error somewhere and only the former is a solution. We just can't say without seeing the problem. When you reply please include the original problem, quoted word-for-word if possible. Thank you.

 

[Ishuda]

hi!
i have a question that i like to get answer.
it can be more than one solution in

Recurrence Relation that made the same answer?
for example : is the ​

a(n)=3a(n-1)+a(n-2) the same to a(n)=4a(n-1)-2a(n-2)-a(n-3) , a(0)=1 ,a(1)=4 ,a(2)=13
?
thanks...

(1) a(n) = 3 a(n-1) + a(n-2)
(2) = 4 a(n-1) - [ a(n-1) ] + a(n-2)
(3) = 4 a(n-1) - [3 a(n-2) + a(n-3) ] + a(n-2)
(4) = 4 a(n-1) - 3 a(n-2) - a(n-3) + a(n-2)
(5) = 4 a(n-1) - 2 a(n-2) - a(n-3)

Of course the relationships are valid for a slightly different set of n values. For (1) and (2) the relationship is good for all n greater than or equal to 2. For (3), (4), and (5) the relationship is good for all n greater than or equal to 3.

 

[stapel]

the question, if it can be more than one solution for exercise?
in the exercise for given a(0)=1,a(1)=4,a(2)=13, the result is a(n)=3(n-1)+(n-2) but i claim that also a(n)=4a(n-1)-2a(n-2)-a(n-3) is the correct answer.

So (assuming that the parentheticals are meant to be subscripts) is the original exercise something along the lines of the following?

. . . . .\(\displaystyle \mbox{Given }\, a_0\, =\, 1,\, a_1\, =\, 4,\, \mbox{ and }\, a_2\, =\, 13,\, \mbox{ find at least}\)

. . . . .\(\displaystyle \mbox{one recursion formula that relates these first three terms.}\)

. . . . .\(\displaystyle \mbox{Must there be only one such relation? Prove or disprove.}\)

Thank you!