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#1 2008-10-29 17:41:20

neonash7777
Member
Registered: 2008-10-29
Posts: 3

Riemann Sum Calc Law??

I'll provide proof below but then a possible problem later
~~~~~~~~~
I propose the following rule:
Given:
∑ f(x)/g(x)
where m≤f(x)≤n & g(x) > 0

then there exist some k such that
m≤k≤n

k ∑ 1/g(x) = ∑ f(x)/g(x)
~~~~~~~~~~

Suppose for the Riemann sum has 3 elements for starters, we'll see that no matter how many elements it has my proof will hold true.

a/x + b/y + c/z = some value

Now we know f(x) spawns values a,b,c which are between m and n
and that x,y, and z are all positive from our given g(x) > 0

the total is (ayz + bxz + cxy)/xyz
now I'm saying that there exists some k within m to n such that:
(ayz + bxz + cxy)/xyz = k * (yz + xz + xy)/xyz

k is like a common factor to a, b, and c that we pulled out. Only it's more so an average common factor so-to-speak.

Now the question is how can we prove k exists between m and n?
Let's get rid of the denominator since it's the same on both sides. So we have:
(ayz + bxz + cxy) = k * (yz + xz + xy)

Now x,y,z are positive constants formed by g(x)
and a,b,c have a min value of m and max value of n. So the smallest possible value would be when a,b,c all equal m and max when all equal n.

Min
(myz + mxz + mxy) = k * (yz + xz + xy)
Obviously k = m then

Max
(nyz + nxz + nxy) = k * (yz + xz + xy)
Obviously k = n then

So k will never need to exceed n nor be less than m.
~~~~~~~~~~
My proof seems to justify what I've stated, the only problem is when g(x) can be < 0 too.
Let's say that x = 2, y = -2, z = 2 could come from: [g(x) = 2*(-1)^x]

a(-4) + b(4) + c(-4) = k (-4 + 4 + -4)

a(-4) + b(4) + c(-4) = k (-4)

now if a,c = m and b = n
-8m + 4n = k(-4)
for convenience we'll say n=2, m = 10
-80 + 8 = k(-4)
-72 = k(-4)
k = 18
but k is not between 2 and 10!

Now this doesn't currently disprove my proof above because I give the stipulation that g(x) > 0.
However the problem comes from the fact that you're allowed to move a negative sign in and out of a Reimann sum
-∑f(x) = ∑ -f(x)
So I would have to accept the fact that g(x) should be able to be < 0 but I clearly can't...

???
Help anyone, is there an error in my proof?

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