My Math Forum Help understanding/putting together this type of problem

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 March 6th, 2014, 12:23 AM #11 Senior Member   Joined: Jan 2014 Posts: 196 Thanks: 3 Re: Help understanding/putting together this type of problem I see this just plugging in values..... I see the middle does seem to be the lowest value. Just not understanding why, or the general approach. Do I set this = 0 an solve for c? Also something was mentioned about finding the min between an interval. how would I find the function for this?
March 6th, 2014, 02:31 AM   #12
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Re: Help understanding/putting together this type of problem

Quote:
 Originally Posted by WWRtelescoping I see this just plugging in values..... I see the middle does seem to be the lowest value. Just not understanding why, or the general approach.
Why not try it out with sets of numbers. For example:

2, 6, 21

1, 5, 10, 12, 13

0, 8, 9, 20, 25, 30, 33

Try the middle and see what happens when you move either side of it.

 March 6th, 2014, 06:19 AM #13 Senior Member   Joined: Jan 2014 Posts: 196 Thanks: 3 Re: Help understanding/putting together this type of problem I did something like this. e.g. $|2-2| + |6-2| + |21-2|= 23$ $|2-6| + |6-6| + |21-6|= 19$ $|2-21| + |6-21| + |21-21|=34$ And with other random numbers, have seen the same thing, that when $c$ is the middle term, w/ an odd number of terms, it minimizes more than the others. Though I have seen that this works w/ some examples, is there a way to show this works for all cases? How do I know there isn't another $c$; $c\neq \{2,6,21} \}$ that minimizes this further? Thanks for all the help with this!
March 6th, 2014, 06:30 AM   #14
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Re: Help understanding/putting together this type of problem

Quote:
 Originally Posted by WWRtelescoping Though I have seen that this works w/ some examples, is there a way to show this works for all cases? How do I know there isn't another $c$; $c\neq \{2,6,21} \}$ that minimizes this further?
It's called logic!

Sometimes you prove things with equations and formulas; sometimes, you can prove them with a logical argument, as I did a few posts above. In a nutshell:

As you move away from the middle term (in either direction) you are moving away from more terms than you are moving towards. You should try to see this with the examples you've done.

E.g. if you move from 6 to 5.9, you move 0.1 away from 6 and 21 and 0.1 towards 2. So, the sum of differences increases by 0.1. And, it must increase by any movement away from the middle term.

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1,2,3,5,15,25,50

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