sequence question

[Qwertyuiop[]]

a sequence is defined by [imath]u_o=1[/imath] and for all [imath]n \in N[/imath],[imath]u_{n+1}=\frac{1}{2}u_n+n-1[/imath].
1)Show that for n >= 3, u(n) is positive. And for n>=4, [imath]u_n\:\ge n-2[/imath]
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 [imath]u_{k}\ge 0[/imath] then we have to prove [imath]u_{k+1}\ge 0[/imath].

I made u(n) the subject in the formula and i know that [imath]u_{k}\ge 0[/imath] so setting that equation greater than or equal to 0 i get:
[math]u_k\ge 0 \\ 2\left(u_{k+1}-k+1\right)\ge 0\\ \left(u_{k+1}-k+1\right)\ge 0[/math]I have to get this expression in the form [imath]u_{k+1}\ge 0[/imath] 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 [imath]u_{k}\ge 0[/imath] then we have to prove [imath]u_{k+1}\ge 0[/imath].

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

[imath]u_0 = 1[/imath]

[imath]u_1 = \dfrac{1}{2} (1) + 0 - 1 = -\dfrac{1}{2}[/imath]

[imath]u_2 = \dfrac{1}{2} \left ( -\dfrac{1}{2} \right ) + 1 - 1 = -\dfrac{1}{4}[/imath]

[imath]u_3 = \dfrac{1}{2} \left ( -\dfrac{1}{4} \right ) + 2 - 1 = \dfrac{7}{8}[/imath]

The rest looks good. You left it with
[imath]u_{n+1} \geq n - 1[/imath]

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

-Dan

 

[Qwertyuiop[]]

[imath]u_0 = 1[/imath]

[imath]u_1 = \dfrac{1}{2} (1) + 0 - 1 = -\dfrac{1}{2}[/imath]

[imath]u_2 = \dfrac{1}{2} \left ( -\dfrac{1}{2} \right ) + 1 - 1 = -\dfrac{1}{4}[/imath]

[imath]u_3 = \dfrac{1}{2} \left ( -\dfrac{1}{4} \right ) + 2 - 1 = \dfrac{7}{8}[/imath]

The rest looks good. You left it with
[imath]u_{n+1} \geq n - 1[/imath]

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

-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 [imath]n\:\ge 4[/imath] , [imath]u_n\ge n-2[/imath].
Base: [imath]u_4=39/16[/imath] and [math]\frac{39}{16}\ge 2[/math] so true for n=4.
recurence hyp: if [imath]u_{k\:}\ge k-2[/imath] then show [imath]u_{k+1\:}\ge k+1-2\:=\:u_{k+1}=k-1[/imath]
[math]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[/math]Here it's not as obvious as it was for the first question,i don't get [imath]u_{k+1}\:\ge k-2[/imath]. What did I do wrong?

 

[topsquark]

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

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

not
[imath]u_{n+1} > (n + 1) - 2[/imath]

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

-Dan