How to solve Stieltjes integral $int_0^n f(x) d lfloor x rfloor$?
$begingroup$
I'm trying to solve the following
$$
int_0^n f(x) d lfloor x rfloor
$$
where $lfloor x rfloor$ is the floor function, $f : [0,n] rightarrow mathbb{R}$ is continuous, and $n in mathbb{N}$.
I know $frac{d}{dx} lfloor x rfloor$ itself cannot be secured (i.e. not differentiable). I cannot proceed further.
Can anyone give some hints?
calculus integration stieltjes-integral
asked Aug 31 '18 at 8:01
MoreblueMoreblue
8931217
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add a comment |
$begingroup$
I'm trying to solve the following
$$
int_0^n f(x) d lfloor x rfloor
$$
where $lfloor x rfloor$ is the floor function, $f : [0,n] rightarrow mathbb{R}$ is continuous, and $n in mathbb{N}$.
I know $frac{d}{dx} lfloor x rfloor$ itself cannot be secured (i.e. not differentiable). I cannot proceed further.
Can anyone give some hints?
calculus integration stieltjes-integral
asked Aug 31 '18 at 8:01
MoreblueMoreblue
8931217
$endgroup$
1
$begingroup$
The answer is $f(1)+...+f(n-1)$. Just write down Riemann - Steiltjes sums and take the limit.
$endgroup$
– Kavi Rama Murthy
Aug 31 '18 at 8:06
3
$begingroup$
@KaviRamaMurthy Shouldn't it be $$int_0^n,f(x),text{d}lfloor xrfloor = f(1)+f(2)+ldots+f(n),?$$ I think we have $$int_0^n,f(x),text{d}lceil xrceil=f(0)+f(1)+ldots+f(n-1),.$$
$endgroup$
– Batominovski
Aug 31 '18 at 8:08
add a comment |
$begingroup$
I'm trying to solve the following
$$
int_0^n f(x) d lfloor x rfloor
$$
where $lfloor x rfloor$ is the floor function, $f : [0,n] rightarrow mathbb{R}$ is continuous, and $n in mathbb{N}$.
I know $frac{d}{dx} lfloor x rfloor$ itself cannot be secured (i.e. not differentiable). I cannot proceed further.
Can anyone give some hints?
calculus integration stieltjes-integral
asked Aug 31 '18 at 8:01
MoreblueMoreblue
8931217
$endgroup$
I'm trying to solve the following
$$
int_0^n f(x) d lfloor x rfloor
$$
where $lfloor x rfloor$ is the floor function, $f : [0,n] rightarrow mathbb{R}$ is continuous, and $n in mathbb{N}$.
I know $frac{d}{dx} lfloor x rfloor$ itself cannot be secured (i.e. not differentiable). I cannot proceed further.
Can anyone give some hints?
calculus integration stieltjes-integral
calculus integration stieltjes-integral
asked Aug 31 '18 at 8:01
MoreblueMoreblue
8931217
asked Aug 31 '18 at 8:01
MoreblueMoreblue
8931217
asked Aug 31 '18 at 8:01
MoreblueMoreblue
8931217
asked Aug 31 '18 at 8:01
MoreblueMoreblue
8931217
asked Aug 31 '18 at 8:01
MoreblueMoreblue
8931217
8931217
1
$begingroup$
The answer is $f(1)+...+f(n-1)$. Just write down Riemann - Steiltjes sums and take the limit.
$endgroup$
– Kavi Rama Murthy
Aug 31 '18 at 8:06
3
$begingroup$
@KaviRamaMurthy Shouldn't it be $$int_0^n,f(x),text{d}lfloor xrfloor = f(1)+f(2)+ldots+f(n),?$$ I think we have $$int_0^n,f(x),text{d}lceil xrceil=f(0)+f(1)+ldots+f(n-1),.$$
$endgroup$
– Batominovski
Aug 31 '18 at 8:08
add a comment |
1
$begingroup$
The answer is $f(1)+...+f(n-1)$. Just write down Riemann - Steiltjes sums and take the limit.
$endgroup$
– Kavi Rama Murthy
Aug 31 '18 at 8:06
3
$begingroup$
@KaviRamaMurthy Shouldn't it be $$int_0^n,f(x),text{d}lfloor xrfloor = f(1)+f(2)+ldots+f(n),?$$ I think we have $$int_0^n,f(x),text{d}lceil xrceil=f(0)+f(1)+ldots+f(n-1),.$$
$endgroup$
– Batominovski
Aug 31 '18 at 8:08
1
1
$begingroup$
The answer is $f(1)+...+f(n-1)$. Just write down Riemann - Steiltjes sums and take the limit.
$endgroup$
– Kavi Rama Murthy
Aug 31 '18 at 8:06
$begingroup$
The answer is $f(1)+...+f(n-1)$. Just write down Riemann - Steiltjes sums and take the limit.
$endgroup$
– Kavi Rama Murthy
Aug 31 '18 at 8:06
3
3
$begingroup$
@KaviRamaMurthy Shouldn't it be $$int_0^n,f(x),text{d}lfloor xrfloor = f(1)+f(2)+ldots+f(n),?$$ I think we have $$int_0^n,f(x),text{d}lceil xrceil=f(0)+f(1)+ldots+f(n-1),.$$
$endgroup$
– Batominovski
Aug 31 '18 at 8:08
$begingroup$
@KaviRamaMurthy Shouldn't it be $$int_0^n,f(x),text{d}lfloor xrfloor = f(1)+f(2)+ldots+f(n),?$$ I think we have $$int_0^n,f(x),text{d}lceil xrceil=f(0)+f(1)+ldots+f(n-1),.$$
$endgroup$
– Batominovski
Aug 31 '18 at 8:08
add a comment |
1 Answer
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$begingroup$
Note that there is a difference between integrating $[0,n]$, $[0,n)$, $(0,n]$ and $(0,n)$:
$$int_{[0,n]}f(x)dlfloor xrfloor =sum_{i=0}^nf(i)$$
$$int_{[0,n)}f(x)dlfloor xrfloor =sum_{i=0}^{n-1}f(i)$$
$$int_{(0,n]}f(x)dlfloor xrfloor =sum_{i=1}^nf(i)$$
$$int_{(0,n)}f(x)dlfloor xrfloor =sum_{i=1}^{n-1}f(i)$$
since the associated measure for the floor-function is
$$mu_{lfloorcdotrfloor}=sum_{k=-infty}^infty delta_k$$
answered Jan 12 at 23:20
user408858user408858
482213
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add a comment |
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1 Answer
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$begingroup$
Note that there is a difference between integrating $[0,n]$, $[0,n)$, $(0,n]$ and $(0,n)$:
$$int_{[0,n]}f(x)dlfloor xrfloor =sum_{i=0}^nf(i)$$
$$int_{[0,n)}f(x)dlfloor xrfloor =sum_{i=0}^{n-1}f(i)$$
$$int_{(0,n]}f(x)dlfloor xrfloor =sum_{i=1}^nf(i)$$
$$int_{(0,n)}f(x)dlfloor xrfloor =sum_{i=1}^{n-1}f(i)$$
since the associated measure for the floor-function is
$$mu_{lfloorcdotrfloor}=sum_{k=-infty}^infty delta_k$$
answered Jan 12 at 23:20
user408858user408858
482213
$endgroup$
add a comment |
$begingroup$
Note that there is a difference between integrating $[0,n]$, $[0,n)$, $(0,n]$ and $(0,n)$:
$$int_{[0,n]}f(x)dlfloor xrfloor =sum_{i=0}^nf(i)$$
$$int_{[0,n)}f(x)dlfloor xrfloor =sum_{i=0}^{n-1}f(i)$$
$$int_{(0,n]}f(x)dlfloor xrfloor =sum_{i=1}^nf(i)$$
$$int_{(0,n)}f(x)dlfloor xrfloor =sum_{i=1}^{n-1}f(i)$$
since the associated measure for the floor-function is
$$mu_{lfloorcdotrfloor}=sum_{k=-infty}^infty delta_k$$
answered Jan 12 at 23:20
user408858user408858
482213
$endgroup$
add a comment |
$begingroup$
Note that there is a difference between integrating $[0,n]$, $[0,n)$, $(0,n]$ and $(0,n)$:
$$int_{[0,n]}f(x)dlfloor xrfloor =sum_{i=0}^nf(i)$$
$$int_{[0,n)}f(x)dlfloor xrfloor =sum_{i=0}^{n-1}f(i)$$
$$int_{(0,n]}f(x)dlfloor xrfloor =sum_{i=1}^nf(i)$$
$$int_{(0,n)}f(x)dlfloor xrfloor =sum_{i=1}^{n-1}f(i)$$
since the associated measure for the floor-function is
$$mu_{lfloorcdotrfloor}=sum_{k=-infty}^infty delta_k$$
answered Jan 12 at 23:20
user408858user408858
482213
$endgroup$
Note that there is a difference between integrating $[0,n]$, $[0,n)$, $(0,n]$ and $(0,n)$:
$$int_{[0,n]}f(x)dlfloor xrfloor =sum_{i=0}^nf(i)$$
$$int_{[0,n)}f(x)dlfloor xrfloor =sum_{i=0}^{n-1}f(i)$$
$$int_{(0,n]}f(x)dlfloor xrfloor =sum_{i=1}^nf(i)$$
$$int_{(0,n)}f(x)dlfloor xrfloor =sum_{i=1}^{n-1}f(i)$$
since the associated measure for the floor-function is
$$mu_{lfloorcdotrfloor}=sum_{k=-infty}^infty delta_k$$
answered Jan 12 at 23:20
user408858user408858
482213
edited Jan 16 at 22:01
answered Jan 12 at 23:20
user408858user408858
482213
answered Jan 12 at 23:20
user408858user408858
482213
answered Jan 12 at 23:20
user408858user408858
482213
482213
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$begingroup$
The answer is $f(1)+...+f(n-1)$. Just write down Riemann - Steiltjes sums and take the limit.
$endgroup$
– Kavi Rama Murthy
Aug 31 '18 at 8:06
3
$begingroup$
@KaviRamaMurthy Shouldn't it be $$int_0^n,f(x),text{d}lfloor xrfloor = f(1)+f(2)+ldots+f(n),?$$ I think we have $$int_0^n,f(x),text{d}lceil xrceil=f(0)+f(1)+ldots+f(n-1),.$$
$endgroup$
– Batominovski
Aug 31 '18 at 8:08