Mixing
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Limits in the Linearity of Limit and Integral Operators
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I have the following function $$f(textbf{x})=frac{Q(textbf{x})}{P(textbf{x})}$$
where $P(textbf{x})$ is such that $P(textbf{x})=0iff textbf{x}=0$ and at the same time $Q(textbf{x})=0ifftextbf{x}=0$. The functions $P(textbf{x})$ and $Q(textbf{x})$ are such that: $lim limits _{textbf{x}rightarrow0}f(textbf{x})=0$ and $int limits _{Omega}f(textbf{x}) mathrm{d}textbf{x}=m$.
Suppose now that $Q(textbf{x})$ is a polynomial function and that I need to algebrically expand it. What one would get is: $$f(textbf{x})=frac{Q_1(textbf{x})+Q_2(textbf{x})+cdots+Q_n(textbf{x})}{P(textbf{x})}=sum limits _{i=1}^nfrac{Q_i(textbf{x})}{P(textbf{x})}$$
Unfortunately, the functions are such that $$lim limits _{textbf{x}rightarrow0} f(textbf{x})neq sum limits _{i=1}^nlim limits _{textbf{x}rightarrow0} frac{Q_i(textbf{x})}{P(textbf{x})}$$ and also $$int limits _{Omega}f(textbf{x}) mathrm{d}textbf{x}neq sum limits _{i=1}^nint limits _{Omega} frac{Q_i(textbf{x})}{P(textbf{x})} mathrm{d}textbf{x}$$
What I would like to know is, when one has to work with the "expanded" version of the function (meaning that one has to evaluate the integral or the limit forcibly in this form), how can he/she adapt the inequalities in order to be sure that he retrieves the results of the "compact" formulation?
I have read When can a sum and integral be interchanged?, but still find this situation obscure
A very simple example would be:
$$f(x)=frac{x-2}{x-2}$$ and $$g(x)=frac{x}{x-2}, h(x)=-frac{2}{x-2}$$ integrated over (for example) $I=[0,3]$
integration limits
asked Jan 18 at 12:08
RiccardoRiccardo
314
$endgroup$
add a comment |
$begingroup$
I have the following function $$f(textbf{x})=frac{Q(textbf{x})}{P(textbf{x})}$$
where $P(textbf{x})$ is such that $P(textbf{x})=0iff textbf{x}=0$ and at the same time $Q(textbf{x})=0ifftextbf{x}=0$. The functions $P(textbf{x})$ and $Q(textbf{x})$ are such that: $lim limits _{textbf{x}rightarrow0}f(textbf{x})=0$ and $int limits _{Omega}f(textbf{x}) mathrm{d}textbf{x}=m$.
Suppose now that $Q(textbf{x})$ is a polynomial function and that I need to algebrically expand it. What one would get is: $$f(textbf{x})=frac{Q_1(textbf{x})+Q_2(textbf{x})+cdots+Q_n(textbf{x})}{P(textbf{x})}=sum limits _{i=1}^nfrac{Q_i(textbf{x})}{P(textbf{x})}$$
Unfortunately, the functions are such that $$lim limits _{textbf{x}rightarrow0} f(textbf{x})neq sum limits _{i=1}^nlim limits _{textbf{x}rightarrow0} frac{Q_i(textbf{x})}{P(textbf{x})}$$ and also $$int limits _{Omega}f(textbf{x}) mathrm{d}textbf{x}neq sum limits _{i=1}^nint limits _{Omega} frac{Q_i(textbf{x})}{P(textbf{x})} mathrm{d}textbf{x}$$
What I would like to know is, when one has to work with the "expanded" version of the function (meaning that one has to evaluate the integral or the limit forcibly in this form), how can he/she adapt the inequalities in order to be sure that he retrieves the results of the "compact" formulation?
I have read When can a sum and integral be interchanged?, but still find this situation obscure
A very simple example would be:
$$f(x)=frac{x-2}{x-2}$$ and $$g(x)=frac{x}{x-2}, h(x)=-frac{2}{x-2}$$ integrated over (for example) $I=[0,3]$
integration limits
asked Jan 18 at 12:08
RiccardoRiccardo
314
$endgroup$
$begingroup$
I think it has to come down to the limits $Q_i(x)/P(x)$ for $xto 0$ not existing, otherwise those equalities would have to hold.
$endgroup$
– SvanN
Jan 18 at 12:21
$begingroup$
Yes I get some of the $Q_i(x)/P(x)$ tending to $infty $ for $xrightarrow 0$
$endgroup$
– Riccardo
Jan 18 at 12:26
add a comment |
$begingroup$
I have the following function $$f(textbf{x})=frac{Q(textbf{x})}{P(textbf{x})}$$
where $P(textbf{x})$ is such that $P(textbf{x})=0iff textbf{x}=0$ and at the same time $Q(textbf{x})=0ifftextbf{x}=0$. The functions $P(textbf{x})$ and $Q(textbf{x})$ are such that: $lim limits _{textbf{x}rightarrow0}f(textbf{x})=0$ and $int limits _{Omega}f(textbf{x}) mathrm{d}textbf{x}=m$.
Suppose now that $Q(textbf{x})$ is a polynomial function and that I need to algebrically expand it. What one would get is: $$f(textbf{x})=frac{Q_1(textbf{x})+Q_2(textbf{x})+cdots+Q_n(textbf{x})}{P(textbf{x})}=sum limits _{i=1}^nfrac{Q_i(textbf{x})}{P(textbf{x})}$$
Unfortunately, the functions are such that $$lim limits _{textbf{x}rightarrow0} f(textbf{x})neq sum limits _{i=1}^nlim limits _{textbf{x}rightarrow0} frac{Q_i(textbf{x})}{P(textbf{x})}$$ and also $$int limits _{Omega}f(textbf{x}) mathrm{d}textbf{x}neq sum limits _{i=1}^nint limits _{Omega} frac{Q_i(textbf{x})}{P(textbf{x})} mathrm{d}textbf{x}$$
What I would like to know is, when one has to work with the "expanded" version of the function (meaning that one has to evaluate the integral or the limit forcibly in this form), how can he/she adapt the inequalities in order to be sure that he retrieves the results of the "compact" formulation?
I have read When can a sum and integral be interchanged?, but still find this situation obscure
A very simple example would be:
$$f(x)=frac{x-2}{x-2}$$ and $$g(x)=frac{x}{x-2}, h(x)=-frac{2}{x-2}$$ integrated over (for example) $I=[0,3]$
integration limits
asked Jan 18 at 12:08
RiccardoRiccardo
314
$endgroup$
I have the following function $$f(textbf{x})=frac{Q(textbf{x})}{P(textbf{x})}$$
where $P(textbf{x})$ is such that $P(textbf{x})=0iff textbf{x}=0$ and at the same time $Q(textbf{x})=0ifftextbf{x}=0$. The functions $P(textbf{x})$ and $Q(textbf{x})$ are such that: $lim limits _{textbf{x}rightarrow0}f(textbf{x})=0$ and $int limits _{Omega}f(textbf{x}) mathrm{d}textbf{x}=m$.
Suppose now that $Q(textbf{x})$ is a polynomial function and that I need to algebrically expand it. What one would get is: $$f(textbf{x})=frac{Q_1(textbf{x})+Q_2(textbf{x})+cdots+Q_n(textbf{x})}{P(textbf{x})}=sum limits _{i=1}^nfrac{Q_i(textbf{x})}{P(textbf{x})}$$
Unfortunately, the functions are such that $$lim limits _{textbf{x}rightarrow0} f(textbf{x})neq sum limits _{i=1}^nlim limits _{textbf{x}rightarrow0} frac{Q_i(textbf{x})}{P(textbf{x})}$$ and also $$int limits _{Omega}f(textbf{x}) mathrm{d}textbf{x}neq sum limits _{i=1}^nint limits _{Omega} frac{Q_i(textbf{x})}{P(textbf{x})} mathrm{d}textbf{x}$$
What I would like to know is, when one has to work with the "expanded" version of the function (meaning that one has to evaluate the integral or the limit forcibly in this form), how can he/she adapt the inequalities in order to be sure that he retrieves the results of the "compact" formulation?
I have read When can a sum and integral be interchanged?, but still find this situation obscure
A very simple example would be:
$$f(x)=frac{x-2}{x-2}$$ and $$g(x)=frac{x}{x-2}, h(x)=-frac{2}{x-2}$$ integrated over (for example) $I=[0,3]$
integration limits
integration limits
asked Jan 18 at 12:08
RiccardoRiccardo
314
asked Jan 18 at 12:08
RiccardoRiccardo
314
edited Jan 18 at 12:23
Riccardo
asked Jan 18 at 12:08
RiccardoRiccardo
314
asked Jan 18 at 12:08
RiccardoRiccardo
314
asked Jan 18 at 12:08
RiccardoRiccardo
314
314
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I think it has to come down to the limits $Q_i(x)/P(x)$ for $xto 0$ not existing, otherwise those equalities would have to hold.
$endgroup$
– SvanN
Jan 18 at 12:21
$begingroup$
Yes I get some of the $Q_i(x)/P(x)$ tending to $infty $ for $xrightarrow 0$
$endgroup$
– Riccardo
Jan 18 at 12:26
add a comment |
$begingroup$
I think it has to come down to the limits $Q_i(x)/P(x)$ for $xto 0$ not existing, otherwise those equalities would have to hold.
$endgroup$
– SvanN
Jan 18 at 12:21
$begingroup$
Yes I get some of the $Q_i(x)/P(x)$ tending to $infty $ for $xrightarrow 0$
$endgroup$
– Riccardo
Jan 18 at 12:26
$begingroup$
I think it has to come down to the limits $Q_i(x)/P(x)$ for $xto 0$ not existing, otherwise those equalities would have to hold.
$endgroup$
– SvanN
Jan 18 at 12:21
$begingroup$
I think it has to come down to the limits $Q_i(x)/P(x)$ for $xto 0$ not existing, otherwise those equalities would have to hold.
$endgroup$
– SvanN
Jan 18 at 12:21
$begingroup$
Yes I get some of the $Q_i(x)/P(x)$ tending to $infty $ for $xrightarrow 0$
$endgroup$
– Riccardo
Jan 18 at 12:26
$begingroup$
Yes I get some of the $Q_i(x)/P(x)$ tending to $infty $ for $xrightarrow 0$
$endgroup$
– Riccardo
Jan 18 at 12:26
add a comment |
1 Answer
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I'm not sure why you'd want to split it up if you had it in such a nice form, as your example shows you'll run into problems. Let me address some of those problems though; hopefully they show why you wanna avoid breaking things up into 'bad-behaved' parts.
Whenever two functions are integrable on some interval, then we may reverse the order of the integral and the sum. But, the integrability condition is crucial here. Your example at the end is just such a case where this doesn't apply: $x/(x-2)$ has integral $-infty$ over $[0,3]$, while $-2/(x-2)$ has integral $+infty$ over $[0,3]$. (If you happen to know some measure theory, this is exactly the type of problems one wishes to avoid when defining the Lebesgue integral for a function.)
The other issue is similar: the limit $Q_i(x)/P(x)$ as $xto 0$ need not exist, so that we cannot even talk about the sum of these limits (let alone decide if it's equal to the limit of the sum).
answered Jan 18 at 16:13
SvanNSvanN
2,0661422
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
I'm not sure why you'd want to split it up if you had it in such a nice form, as your example shows you'll run into problems. Let me address some of those problems though; hopefully they show why you wanna avoid breaking things up into 'bad-behaved' parts.
Whenever two functions are integrable on some interval, then we may reverse the order of the integral and the sum. But, the integrability condition is crucial here. Your example at the end is just such a case where this doesn't apply: $x/(x-2)$ has integral $-infty$ over $[0,3]$, while $-2/(x-2)$ has integral $+infty$ over $[0,3]$. (If you happen to know some measure theory, this is exactly the type of problems one wishes to avoid when defining the Lebesgue integral for a function.)
The other issue is similar: the limit $Q_i(x)/P(x)$ as $xto 0$ need not exist, so that we cannot even talk about the sum of these limits (let alone decide if it's equal to the limit of the sum).
answered Jan 18 at 16:13
SvanNSvanN
2,0661422
$endgroup$
add a comment |
$begingroup$
I'm not sure why you'd want to split it up if you had it in such a nice form, as your example shows you'll run into problems. Let me address some of those problems though; hopefully they show why you wanna avoid breaking things up into 'bad-behaved' parts.
Whenever two functions are integrable on some interval, then we may reverse the order of the integral and the sum. But, the integrability condition is crucial here. Your example at the end is just such a case where this doesn't apply: $x/(x-2)$ has integral $-infty$ over $[0,3]$, while $-2/(x-2)$ has integral $+infty$ over $[0,3]$. (If you happen to know some measure theory, this is exactly the type of problems one wishes to avoid when defining the Lebesgue integral for a function.)
The other issue is similar: the limit $Q_i(x)/P(x)$ as $xto 0$ need not exist, so that we cannot even talk about the sum of these limits (let alone decide if it's equal to the limit of the sum).
answered Jan 18 at 16:13
SvanNSvanN
2,0661422
$endgroup$
add a comment |
$begingroup$
I'm not sure why you'd want to split it up if you had it in such a nice form, as your example shows you'll run into problems. Let me address some of those problems though; hopefully they show why you wanna avoid breaking things up into 'bad-behaved' parts.
Whenever two functions are integrable on some interval, then we may reverse the order of the integral and the sum. But, the integrability condition is crucial here. Your example at the end is just such a case where this doesn't apply: $x/(x-2)$ has integral $-infty$ over $[0,3]$, while $-2/(x-2)$ has integral $+infty$ over $[0,3]$. (If you happen to know some measure theory, this is exactly the type of problems one wishes to avoid when defining the Lebesgue integral for a function.)
The other issue is similar: the limit $Q_i(x)/P(x)$ as $xto 0$ need not exist, so that we cannot even talk about the sum of these limits (let alone decide if it's equal to the limit of the sum).
answered Jan 18 at 16:13
SvanNSvanN
2,0661422
$endgroup$
I'm not sure why you'd want to split it up if you had it in such a nice form, as your example shows you'll run into problems. Let me address some of those problems though; hopefully they show why you wanna avoid breaking things up into 'bad-behaved' parts.
Whenever two functions are integrable on some interval, then we may reverse the order of the integral and the sum. But, the integrability condition is crucial here. Your example at the end is just such a case where this doesn't apply: $x/(x-2)$ has integral $-infty$ over $[0,3]$, while $-2/(x-2)$ has integral $+infty$ over $[0,3]$. (If you happen to know some measure theory, this is exactly the type of problems one wishes to avoid when defining the Lebesgue integral for a function.)
The other issue is similar: the limit $Q_i(x)/P(x)$ as $xto 0$ need not exist, so that we cannot even talk about the sum of these limits (let alone decide if it's equal to the limit of the sum).
answered Jan 18 at 16:13
SvanNSvanN
2,0661422
answered Jan 18 at 16:13
SvanNSvanN
2,0661422
answered Jan 18 at 16:13
SvanNSvanN
2,0661422
answered Jan 18 at 16:13
SvanNSvanN
2,0661422
2,0661422
add a comment |
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$begingroup$
I think it has to come down to the limits $Q_i(x)/P(x)$ for $xto 0$ not existing, otherwise those equalities would have to hold.
$endgroup$
– SvanN
Jan 18 at 12:21
$begingroup$
Yes I get some of the $Q_i(x)/P(x)$ tending to $infty $ for $xrightarrow 0$
$endgroup$
– Riccardo
Jan 18 at 12:26