sequence question

[Qwertyuiop[]]

a sequence is defined by $u_o=1$ and for all $n \in N$,$u_{n+1}=\frac{1}{2}u_n+n-1$.
1)Show that for n >= 3, u(n) is positive. And for n>=4, $u_n\:\ge n-2$
I'm stuck at the first part of the question(show that for n>=3, u(n) is positive), I am not sure how to proceed.

EDIT: I wanted to use induction to prove but the sequence is defined by a recursive formula and we have u(0)=1 but we have to show for n>=3.

 

[Harry_the_cat]

Use the recursive formula to find $\displaystyle u_1, u_2$ and $\displaystyle u_3$ and then start your first induction proof from there.

 

[Qwertyuiop[]]

Use the recursive formula to find $\displaystyle u_1, u_2$ and $\displaystyle u_3$ and then start your first induction proof from there.

Here's what i get after using induction:
Base: u(3) = 39/16>=0 so true.
Recurrence : if $u_{k}\ge 0$ then we have to prove $u_{k+1}\ge 0$.

I made u(n) the subject in the formula and i know that $u_{k}\ge 0$ so setting that equation greater than or equal to 0 i get:
$$u_k\ge 0 \\ 2\left(u_{k+1}-k+1\right)\ge 0\\ \left(u_{k+1}-k+1\right)\ge 0$$I have to get this expression in the form $u_{k+1}\ge 0$ but not sure how. Is my working correct and is there another way to do it? ?

 

[topsquark]

Here's what i get after using induction:
Base: u(3) = 39/16>=0 so true.
Recurrence : if $u_{k}\ge 0$ then we have to prove $u_{k+1}\ge 0$.

I made u(n) the subject in the formula and i know that $u_{k}\ge 0$ so setting that equation greater than or equal to 0 i get:
$$u_k\ge 0 \\ 2\left(u_{k+1}-k+1\right)\ge 0\\ \left(u_{k+1}-k+1\right)\ge 0$$I have to get this expression in the form $u_{k+1}\ge 0$ but not sure how. Is my working correct and is there another way to do it? ?

$u_0 = 1$

$u_1 = \dfrac{1}{2} (1) + 0 - 1 = -\dfrac{1}{2}$

$u_2 = \dfrac{1}{2} \left ( -\dfrac{1}{2} \right ) + 1 - 1 = -\dfrac{1}{4}$

$u_3 = \dfrac{1}{2} \left ( -\dfrac{1}{4} \right ) + 2 - 1 = \dfrac{7}{8}$

The rest looks good. You left it with
$u_{n+1} \geq n - 1$

So is $u_{n+1}$ positive? (Hint: What do you know about n - 1 for $n \geq 3$?)

-Dan

 

[Qwertyuiop[]]

$u_0 = 1$

$u_1 = \dfrac{1}{2} (1) + 0 - 1 = -\dfrac{1}{2}$

$u_2 = \dfrac{1}{2} \left ( -\dfrac{1}{2} \right ) + 1 - 1 = -\dfrac{1}{4}$

$u_3 = \dfrac{1}{2} \left ( -\dfrac{1}{4} \right ) + 2 - 1 = \dfrac{7}{8}$

The rest looks good. You left it with
$u_{n+1} \geq n - 1$

So is $u_{n+1}$ positive? (Hint: What do you know about n - 1 for $n \geq 3$?)

-Dan

oh, n-1 is positive for n>=3 because 3-1=2>0. And it is positive for all values of k>=3. correct?

 

[khansaheb]

oh, n-1 is positive for n>=3 because 3-1=2>0. And it is positive for all values of k>=3. correct?

 

[Qwertyuiop[]]

I used the same method for the second part that says for $n\:\ge 4$ , $u_n\ge n-2$.
Base: $u_4=39/16$ and $$\frac{39}{16}\ge 2$$ so true for n=4.
recurence hyp: if $u_{k\:}\ge k-2$ then show $u_{k+1\:}\ge k+1-2\:=\:u_{k+1}=k-1$
$$u_k\ge k-2\\ 2\left(u_{k+1}-k+1\right)\ge k-2\\ 2\cdot u_{k+1}\:\ge 3k-4\\ u_{k+1}\:\ge \frac{3}{2}k-2$$Here it's not as obvious as it was for the first question,i don't get $u_{k+1}\:\ge k-2$. What did I do wrong?

 

[topsquark]

I used the same method for the second part that says for $n\:\ge 4$ , $u_n\ge n-2$.
Base: $u_4=39/16$ and $$\frac{39}{16}\ge 2$$ so true for n=4.
recurence hyp: if $u_{k\:}\ge k-2$ then show $u_{k+1\:}\ge k+1-2\:=\:u_{k+1}=k-1$
$$u_k\ge k-2\\ 2\left(u_{k+1}-k+1\right)\ge k-2\\ 2\cdot u_{k+1}\:\ge 3k-4\\ u_{k+1}\:\ge \frac{3}{2}k-2$$Here it's not as obvious as it was for the first question,i don't get $u_{k+1}\:\ge k-2$. What did I do wrong?

You are trying to show that $u_{n+1}$ is positive. So you have
$u_{n+1} > n - 2$

not
$u_{n+1} > (n + 1) - 2$

You are trying to advance n by 1, but you still need to use n in your formula. If you want to talk about $u_{n+2}$, then you would have
$u_{n+2} > (n + 1) - 2$

-Dan