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Suppose
y = x^3 + 3x^2 for x < 0
and
y = 4x^4 + 4x for x >= 0
~~~~
This can be written as a single equation
y = (x^2 + 3x)(|x|-x)/2 + (4x^3 + 4)(|x|+x)/2
~~~~~~
the piece times (|x|-x)/2x is 1 when x<0 and 0 when x>0
the piece times (|x|+x)/2x is 1 when x>0 and 0 when x<0
The only problem is when x = 0, because both pieces in the sample equation all of at least one x in there term, we just factored this out to avoid the issue. Let's say we don't have this option, can we still avoid the problem?
try
y=5x for x < 0
y = x^2 + 3x for x >= 0
~~~~~~~~
Is it then possible to avoid every possible piecewise defined function by turning it into a single equation. (Granted |x| is piecewise defined, but we ignore that as the only acceptable one)
A good number of fields of mathematics accept 1/0 = ∞. And it's actually meaningful for solutions.
It's called the Riemann sphere "R union {∞}" and has a few number of other names too.
Whereby -∞ and ∞ are the same number as ∞.
It makes sense if I have one apple and make groups of 0 apples, how many groups can I make. Well here's your group of zero apples and here's your group of 0... so ∞ many people can get a group of 0 apples!
Granted you have to be careful when using this idea of infinity, but so-what, you have to be careful when using 0 as well. If you accept the notion of a historical infinity (one where you keep track of where it's derived from) math works quite well with it.
For example, Normally
2 * ∞ = ∞, but instead keep the history of the infinity such that 2 * ∞ = 2∞
1/0 = ∞
2/0 = 2 ∞
By keeping the history of it, we can see later on what is possible to cancel out of it.
This also helps a lot because multiplication is finally a closed group like addition.
The concept of an identity and an inverse exist for all elements.
Also if you graph something like y = 1/x as x -> 0, y -> ∞. We might as well say when x = 0, y = ∞ where it's not a positive or negative, it's both. they are the same. It just makes sense.
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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