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# Discrete Cauchy-Schwarz Problem watch

1. I'm wondering if someone could help me with part (c), I'm not sure how to start it at all. Also I would be thankful if someone could also quickly verify my solutions to (a) and (b), especially (a) as I just assumed the ds changes to dx at the end but I don't know if this is correct..?
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2. Anyone?
3. (Original post by BahamutXV)
Anyone?
The given result doesn't seem to be true.

Take N=1, v0 = 1, v1=0. Then the 2nd sum in the rhs product is zero, the LHS isn't.

This isn't just nitpicking; I think it's clear the difficulty in proving this is what happens at 0 and N in the RHS sums.
4. (Original post by DFranklin)
The given result doesn't seem to be true.

Take N=1, v0 = 1, v1=0. Then the 2nd sum in the rhs product is zero, the LHS isn't.

This isn't just nitpicking; I think it's clear the difficulty in proving this is what happens at 0 and N in the RHS sums.
You're meant to assume v_0 = 0.

Anyway, bound the |v_k + v_{k-1}| term by |v_k| + |v_{k-1}|. This is where the two will come from.
5. (Original post by IrrationalNumber)
You're meant to assume v_0 = 0.

Anyway, bound the |v_k + v_{k-1}| term by |v_k| + |v_{k-1}|. This is where the two will come from.
This is what I ended up getting but I'm still not sure where the 2 comes from? I assumed the 2 would just be in the continuous case as in part (a), not the discrete.

6. (Original post by IrrationalNumber)
You're meant to assume v_0 = 0.

Anyway, bound the |v_k + v_{k-1}| term by |v_k| + |v_{k-1}|. This is where the two will come from.
Oops, thanks!

(Original post by BahamutXV)
This is what I ended up getting but I'm still not sure where the 2 comes from? I assumed the 2 would just be in the continuous case as in part (a), not the discrete.
If you write out the sum for small N, you'll see you end up with a lot of repeated terms. This is where the 2 is going to come from. There's still a little fiddling you'll need to do to deal with the end conditions.
7. (Original post by DFranklin)
Oops, thanks!

If you write out the sum for small N, you'll see you end up with a lot of repeated terms. This is where the 2 is going to come from. There's still a little fiddling you'll need to do to deal with the end conditions.

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