What are the types of rational algebraic fractions?

floymdiT
2021-11-07
Answered

What are the types of rational algebraic fractions?

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Brighton

Answered 2021-11-08
Author has **103** answers

Step 1

There are two types of rational algebraic fractions namely proper algebraic fractions and improper algebraic fractions.

A rational algebraic fraction$\frac{f\left(x\right)}{g\left(x\right)}$ consisting of the numerator with lower degree than the denominator is defined as proper algebraic fractions.

Example:$\frac{{x}^{2}}{{x}^{3}+1}$

Step 2

A rational algebraic fraction$\frac{f\left(x\right)}{g\left(x\right)}$ consisting of the numerator with higher degree than the denominator is defined as improper algebraic fractions.

Example:$\frac{{x}^{2}+1}{x}$

There are two types of rational algebraic fractions namely proper algebraic fractions and improper algebraic fractions.

A rational algebraic fraction

Example:

Step 2

A rational algebraic fraction

Example:

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I have a measure $\mu $ that is supported on $[-3,3]\times \mathbb{R}$. What we are given is that, if we fix the first component $i$, then $\mu (i,\cdot )$ is a probability measure. Formally (maybe it should integrate to $\mathrm{d}i$ or similar, I cannot define this very well):

${\int}_{x\in \mathbb{R}}\mathrm{d}\mu (i,x)=1\phantom{\rule{1em}{0ex}}\mathrm{\forall}i\in [-3,3].$

I find it very hard to understand this tuple-indexing notation. My question is the following: What kind of assumptions do we need to have a result similar to:

${\int}_{(i,x)\in [-3,3]\times \mathbb{R}}\mathrm{d}\mu (i,x)={\int}_{i\in [-3,3]}\mathrm{d}i$

I just think that since for any fixed $i$ the measure $\mu $ integrates to 1 (on the second dimension -- apologies for my poor terminology), then integrating over all $(i,x)\in [-3,3]\times \mathbb{R}$ should also give simply an 'iterated integral' where we first integrate wrt $x\in \mathbb{R}$ for a fixed $i$, which will integrate to 1 , and then integrate over $i\in [-3,3]$. But of course, we cannot define

${\int}_{(i,x)\in [-3,3]\times \mathbb{R}}\mathrm{d}\mu (i,x)={{\int}_{i\in [-3,3]}{\int}_{x\in \mathbb{R}}\mathrm{d}\mu (i,x)={\int}_{i\in [-3,3]}\mathrm{d}y}$

where in the red parts I make an abuse of integartion rules.

${\int}_{x\in \mathbb{R}}\mathrm{d}\mu (i,x)=1\phantom{\rule{1em}{0ex}}\mathrm{\forall}i\in [-3,3].$

I find it very hard to understand this tuple-indexing notation. My question is the following: What kind of assumptions do we need to have a result similar to:

${\int}_{(i,x)\in [-3,3]\times \mathbb{R}}\mathrm{d}\mu (i,x)={\int}_{i\in [-3,3]}\mathrm{d}i$

I just think that since for any fixed $i$ the measure $\mu $ integrates to 1 (on the second dimension -- apologies for my poor terminology), then integrating over all $(i,x)\in [-3,3]\times \mathbb{R}$ should also give simply an 'iterated integral' where we first integrate wrt $x\in \mathbb{R}$ for a fixed $i$, which will integrate to 1 , and then integrate over $i\in [-3,3]$. But of course, we cannot define

${\int}_{(i,x)\in [-3,3]\times \mathbb{R}}\mathrm{d}\mu (i,x)={{\int}_{i\in [-3,3]}{\int}_{x\in \mathbb{R}}\mathrm{d}\mu (i,x)={\int}_{i\in [-3,3]}\mathrm{d}y}$

where in the red parts I make an abuse of integartion rules.