induction inequality question (1 Viewer)

[sasquatch]

Im not sure if i did this question correctly, could somebody check for me. Thanks.

If u₁ = 1 and u_n = root(2u_n-1) for n >= 2
a) Show that u_n < 2 for n >= 1.

Step 1. If n = 1,
u₁ = 1
***< 2
therefore true for n = 1.

Step 2. Assuming true for n = k,
u_k = root(2u_k-1) < 2

Step 3. Hence show true for n = k + 1,
root(2u_k) < 2, i.e. LHS - RHS < 0

LHS - RHS = root(2u_k) - 2
********< 2 - 2
********< 0

 

[airie]

Im not sure if i did this question correctly, could somebody check for me. Thanks.

If u₁ = 1 and u_n = root(2u_n-1) for n >= 2
a) Show that u_n < 2 for n >= 1.

Step 1. If n = 1,
u₁ = 1
***< 2
therefore true for n = 1.

Step 2. Assuming true for n = k,
u_k = root(2u_k-1) < 2

Step 3. Hence show true for n = k + 1,
root(2u_k) < 2, i.e. LHS - RHS < 0

LHS - RHS = root(2u_k) - 2
********< 2 - 2
********< 0

Where did you get root(2u_k) < 2 in your last step? You're supposed to show it, not sub it straight into the inequality

For the last step, I'd probably do:
From the inductive hypothesis, u_k = root(2u_k-1) < 2,
And u_k+1 = root(2u_k)
= root[2*root(u_k-1)]
< root[2*2] by inductive hypothesis,
= 2.

Therefore, the condition holds true for k+1 if it holds for k. And by induction...

EDIT: Btw, just wanted to mention that at the last step, u_k+1 < root[2*2] could be deduced from the inductive hypothesis as u₁ = 1 and u_n = root(2u_n-1) for n >= 2 ie. u_n > 0 for all positive integers n.

 

[sasquatch]

Where did you get root(2u_k) < 2 in your last step? You're supposed to show it, not sub it straight into the inequality

Erm now looking back.. i dunno what the hell i did.. my mistake..