Re: Comments needed for attempted proof of a simple integral
From: Lance Lamboy (lance.lamboy_at_lamboy.nospam.com)
Date: 06/16/04
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Date: Tue, 15 Jun 2004 22:18:02 -0400
On Tue, 15 Jun 2004 22:02:06 -0400, Chairman of the Ozzy Osbourne
Appreciation Society wrote:
> [adding sci.math for wider audience]
>
> No more comments?
>
> I'm self studying through Tom Apostol's Calculus and it's hard to know
> if I'm on the right track or if I've gone off track without feedback.
> Usenet is currently my only means of getting feedback.
>
> Apparently there are no answer books for this text. The standard "solve
> X" problems aren't so difficult because standard math education has been
> about solving problems of that nature and I can usually verify answers
> to those problems with software. It's proofs that are hardest type of
> exercise for me to verify, and there are a lot of those in this book.
> I've considered switching to Spivak just for access to answers to those
> kinds of problems although I feel that I've already made a large enough
> investment into Apostol's book to finish it out, even if that means
> sacrificing some degree of certainty about the subject matter.
>
> In exercise section 1.15 I'm asked to prove a few increasingly more
> difficult versions of the problem I posted here. Since I use a similar
> strategy to prove each of those that I used to prove the first, I
> figured that I should post that method and use the responses as a litmus
> test to see if it was general considered to be OK or generally
> considered to be NOT OK.
Looks OK to me.
<snip>
>>
>> My thinking is to translate the integral into a sum, and then use the
>> result of a previous excersize to show that the sum is equavalent to
>> n(n-1)/2, but I'm uncertain if I'm going about this the right way.
>>
>>
>> (1) INT[floor(t)] dt from 0 to n
>>
>> Let P = {0, 1, 2, ..., n} which satisfies the definitions above,
>> from the definition of integral we have:
>>
>> (2a) 0 (1 - 0) + 1 (2 - 1) + 2 (3 - 2) + ... + (n-1) (n - (n-1))
>>
>> noting that the length of each open subinterval is 1, gives the
>> following sum:
>>
>> (2) sum{k=0 to n-1} floor(k) * 1
>>
>> (3) As a consequence of the definition of floor, when x
>> is an integer, x = floor(x). Since each k in the sum up to n-1 is
>> an integer, we replace floor(k) with k. Also since 1 is a constant
>> multiplier, move it outside of the sum (and drop it), giving:
>>
>> (4) sum{k=0 to n-1} k
>>
>> (5*) since the 0th term is 0,
>>
>> sum{k=1 to n-1} k
>>
>> (6) From a previous result we know that
>>
>> sum{k=1 to n} k = n^2/2 + n/2
>>
>> (7) Subtracting the nth term we get a LHS that
>> matches the sum in step 5*.
>>
>> sum{k=1 to n-1} k = n^2/2 + n/2 - n
>>
>> sum{k=1 to n-1} k = n^2/2 + n/2 -2n/2
>>
>> sum{k=1 to n-1} k = n^2/2 - n/2
>>
>> sum{k=1 to n-1} k = n (n-1)/2
>>
>> Which completes the proof.
>>
>> Thanks!
>>
>>
>>
Lance Lamboy
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