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Discrete Cauchy-Schwarz Problem

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..?
cauchy3.png257.1KB
cauchy3a.jpg75.5KB
Reply 1
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BahamutXV
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Anyone?
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.
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.
Reply 4
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BahamutXV
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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.

cauchy3c1.jpg
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 ∑k=1N∣vk∣+∣vk−1∣\sum_{k=1}^N |v_k| + |v_{k-1}| 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.
Reply 6
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BahamutXV
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Original post by DFranklin
Oops, thanks!

If you write out the sum ∑k=1N∣vk∣+∣vk−1∣\sum_{k=1}^N |v_k| + |v_{k-1}| 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.


Thank you for your help! :smile:

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