Proof That $sum_{k=1}^{n-1}int_{k}^{k+1}left{xright}f´(x)dx=int_{1}^{n}left{xright}f´(x)dx$
I am Reading the following notes: Ramanujan summation of divergent series by B Candelpergher (https://hal.univ-cotedazur.fr/hal-01150208v2/document). There, the author derives the Euler-MacLaurin summation formula in the beginning of the first chapter.
In one of the steps (on page 18 of the file) he states that
$$sum_{k=1}^{n-1}int_{k}^{k+1}[x]f´(x)dx=int_{1}^{n}xf´(x)dx+int_{1}^{n}left{xright}f´(x)dx$$
$left{xright}$ is the fractional part of $x$.
The first integral on the right side I understand, but the second one involving the fractional part of $x$, I can´t really see how it´s true. In other words, he is saying that
$$sum_{k=1}^{n-1}int_{k}^{k+1}left{xright}f´(x)dx=int_{1}^{n}left{xright}f´(x)dx$$
Can someone kindly show the proof or the intuition behind it?
Thank you in advance.
integration summation euler-maclaurin ramanujan-summation
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I am Reading the following notes: Ramanujan summation of divergent series by B Candelpergher (https://hal.univ-cotedazur.fr/hal-01150208v2/document). There, the author derives the Euler-MacLaurin summation formula in the beginning of the first chapter.
In one of the steps (on page 18 of the file) he states that
$$sum_{k=1}^{n-1}int_{k}^{k+1}[x]f´(x)dx=int_{1}^{n}xf´(x)dx+int_{1}^{n}left{xright}f´(x)dx$$
$left{xright}$ is the fractional part of $x$.
The first integral on the right side I understand, but the second one involving the fractional part of $x$, I can´t really see how it´s true. In other words, he is saying that
$$sum_{k=1}^{n-1}int_{k}^{k+1}left{xright}f´(x)dx=int_{1}^{n}left{xright}f´(x)dx$$
Can someone kindly show the proof or the intuition behind it?
Thank you in advance.
integration summation euler-maclaurin ramanujan-summation
add a comment |
I am Reading the following notes: Ramanujan summation of divergent series by B Candelpergher (https://hal.univ-cotedazur.fr/hal-01150208v2/document). There, the author derives the Euler-MacLaurin summation formula in the beginning of the first chapter.
In one of the steps (on page 18 of the file) he states that
$$sum_{k=1}^{n-1}int_{k}^{k+1}[x]f´(x)dx=int_{1}^{n}xf´(x)dx+int_{1}^{n}left{xright}f´(x)dx$$
$left{xright}$ is the fractional part of $x$.
The first integral on the right side I understand, but the second one involving the fractional part of $x$, I can´t really see how it´s true. In other words, he is saying that
$$sum_{k=1}^{n-1}int_{k}^{k+1}left{xright}f´(x)dx=int_{1}^{n}left{xright}f´(x)dx$$
Can someone kindly show the proof or the intuition behind it?
Thank you in advance.
integration summation euler-maclaurin ramanujan-summation
I am Reading the following notes: Ramanujan summation of divergent series by B Candelpergher (https://hal.univ-cotedazur.fr/hal-01150208v2/document). There, the author derives the Euler-MacLaurin summation formula in the beginning of the first chapter.
In one of the steps (on page 18 of the file) he states that
$$sum_{k=1}^{n-1}int_{k}^{k+1}[x]f´(x)dx=int_{1}^{n}xf´(x)dx+int_{1}^{n}left{xright}f´(x)dx$$
$left{xright}$ is the fractional part of $x$.
The first integral on the right side I understand, but the second one involving the fractional part of $x$, I can´t really see how it´s true. In other words, he is saying that
$$sum_{k=1}^{n-1}int_{k}^{k+1}left{xright}f´(x)dx=int_{1}^{n}left{xright}f´(x)dx$$
Can someone kindly show the proof or the intuition behind it?
Thank you in advance.
integration summation euler-maclaurin ramanujan-summation
integration summation euler-maclaurin ramanujan-summation
edited Jan 4 at 14:31
Ricardo770
asked Jan 4 at 14:22
Ricardo770Ricardo770
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Just write out the sum: begin{align*}
sum_{k=1}^{n-1} int_k^{k+1} text{whatever}, dx &= int_1^2 text{whatever}, dx + int_2^3 text{whatever}, dx + cdots + int_{n-1}^n text{whatever}, dx \
&= int_1^n text{whatever}, dx.
end{align*}
1
...where $text{whatever}={x}f´(x)$ :)
– Wojowu
Jan 4 at 14:38
@Connor Harris I tried what you say, but I get the following: $int_{n-1}^{n} left{xright}f´(x)dx=int_{n-1}^{n}(x-left[xright])f´(x)dx$ integrating the first integral by parts I get $int_{n-1}^{n}(x-left[xright])f´(x)dx=f(n)-int_{n-1}^{n}f(x)dx$. summing up all the integrals I got $sum_{k=1}^{n-1}int_{k}^{k+1}left{xright}f´(x)dx= sum_{k=1}^{n-1}f(k)-int_{1}^{n} f(x)dx$
– Ricardo770
Jan 4 at 15:31
Why do you need to do any integrals?
– Connor Harris
Jan 4 at 16:22
add a comment |
Ok, I got it!
I took the reverse way to prove it.
$$int_{1}^{n}left{xright}f´(x)dx= int_{1}^{n}(x-left[xright])f´(x)dx$$
$$ =int_{1}^{n}x f´(x)dx -int_{1}^{n} left[xright]f´(x)dx$$
$$ =int_{1}^{n}x f´(x)dx -sum_{k=1}^{n-1}int_{k}^{k+1} left[xright]f´(x)dx$$
$$=sum_{k=1}^{n-1} int_{k}^{k+1}x f´(x)dx -sum_{k=1}^{n-1}int_{k}^{k+1} left[xright]f´(x)dx$$
$$=sum_{k=1}^{n-1} int_{k}^{k+1}(x - left[xright])f´(x)dx$$
$$=sum_{k=1}^{n-1} int_{k}^{k+1}left{xright}f´(x)dx$$
add a comment |
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2 Answers
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2 Answers
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Just write out the sum: begin{align*}
sum_{k=1}^{n-1} int_k^{k+1} text{whatever}, dx &= int_1^2 text{whatever}, dx + int_2^3 text{whatever}, dx + cdots + int_{n-1}^n text{whatever}, dx \
&= int_1^n text{whatever}, dx.
end{align*}
1
...where $text{whatever}={x}f´(x)$ :)
– Wojowu
Jan 4 at 14:38
@Connor Harris I tried what you say, but I get the following: $int_{n-1}^{n} left{xright}f´(x)dx=int_{n-1}^{n}(x-left[xright])f´(x)dx$ integrating the first integral by parts I get $int_{n-1}^{n}(x-left[xright])f´(x)dx=f(n)-int_{n-1}^{n}f(x)dx$. summing up all the integrals I got $sum_{k=1}^{n-1}int_{k}^{k+1}left{xright}f´(x)dx= sum_{k=1}^{n-1}f(k)-int_{1}^{n} f(x)dx$
– Ricardo770
Jan 4 at 15:31
Why do you need to do any integrals?
– Connor Harris
Jan 4 at 16:22
add a comment |
Just write out the sum: begin{align*}
sum_{k=1}^{n-1} int_k^{k+1} text{whatever}, dx &= int_1^2 text{whatever}, dx + int_2^3 text{whatever}, dx + cdots + int_{n-1}^n text{whatever}, dx \
&= int_1^n text{whatever}, dx.
end{align*}
1
...where $text{whatever}={x}f´(x)$ :)
– Wojowu
Jan 4 at 14:38
@Connor Harris I tried what you say, but I get the following: $int_{n-1}^{n} left{xright}f´(x)dx=int_{n-1}^{n}(x-left[xright])f´(x)dx$ integrating the first integral by parts I get $int_{n-1}^{n}(x-left[xright])f´(x)dx=f(n)-int_{n-1}^{n}f(x)dx$. summing up all the integrals I got $sum_{k=1}^{n-1}int_{k}^{k+1}left{xright}f´(x)dx= sum_{k=1}^{n-1}f(k)-int_{1}^{n} f(x)dx$
– Ricardo770
Jan 4 at 15:31
Why do you need to do any integrals?
– Connor Harris
Jan 4 at 16:22
add a comment |
Just write out the sum: begin{align*}
sum_{k=1}^{n-1} int_k^{k+1} text{whatever}, dx &= int_1^2 text{whatever}, dx + int_2^3 text{whatever}, dx + cdots + int_{n-1}^n text{whatever}, dx \
&= int_1^n text{whatever}, dx.
end{align*}
Just write out the sum: begin{align*}
sum_{k=1}^{n-1} int_k^{k+1} text{whatever}, dx &= int_1^2 text{whatever}, dx + int_2^3 text{whatever}, dx + cdots + int_{n-1}^n text{whatever}, dx \
&= int_1^n text{whatever}, dx.
end{align*}
answered Jan 4 at 14:34
Connor HarrisConnor Harris
4,350723
4,350723
1
...where $text{whatever}={x}f´(x)$ :)
– Wojowu
Jan 4 at 14:38
@Connor Harris I tried what you say, but I get the following: $int_{n-1}^{n} left{xright}f´(x)dx=int_{n-1}^{n}(x-left[xright])f´(x)dx$ integrating the first integral by parts I get $int_{n-1}^{n}(x-left[xright])f´(x)dx=f(n)-int_{n-1}^{n}f(x)dx$. summing up all the integrals I got $sum_{k=1}^{n-1}int_{k}^{k+1}left{xright}f´(x)dx= sum_{k=1}^{n-1}f(k)-int_{1}^{n} f(x)dx$
– Ricardo770
Jan 4 at 15:31
Why do you need to do any integrals?
– Connor Harris
Jan 4 at 16:22
add a comment |
1
...where $text{whatever}={x}f´(x)$ :)
– Wojowu
Jan 4 at 14:38
@Connor Harris I tried what you say, but I get the following: $int_{n-1}^{n} left{xright}f´(x)dx=int_{n-1}^{n}(x-left[xright])f´(x)dx$ integrating the first integral by parts I get $int_{n-1}^{n}(x-left[xright])f´(x)dx=f(n)-int_{n-1}^{n}f(x)dx$. summing up all the integrals I got $sum_{k=1}^{n-1}int_{k}^{k+1}left{xright}f´(x)dx= sum_{k=1}^{n-1}f(k)-int_{1}^{n} f(x)dx$
– Ricardo770
Jan 4 at 15:31
Why do you need to do any integrals?
– Connor Harris
Jan 4 at 16:22
1
1
...where $text{whatever}={x}f´(x)$ :)
– Wojowu
Jan 4 at 14:38
...where $text{whatever}={x}f´(x)$ :)
– Wojowu
Jan 4 at 14:38
@Connor Harris I tried what you say, but I get the following: $int_{n-1}^{n} left{xright}f´(x)dx=int_{n-1}^{n}(x-left[xright])f´(x)dx$ integrating the first integral by parts I get $int_{n-1}^{n}(x-left[xright])f´(x)dx=f(n)-int_{n-1}^{n}f(x)dx$. summing up all the integrals I got $sum_{k=1}^{n-1}int_{k}^{k+1}left{xright}f´(x)dx= sum_{k=1}^{n-1}f(k)-int_{1}^{n} f(x)dx$
– Ricardo770
Jan 4 at 15:31
@Connor Harris I tried what you say, but I get the following: $int_{n-1}^{n} left{xright}f´(x)dx=int_{n-1}^{n}(x-left[xright])f´(x)dx$ integrating the first integral by parts I get $int_{n-1}^{n}(x-left[xright])f´(x)dx=f(n)-int_{n-1}^{n}f(x)dx$. summing up all the integrals I got $sum_{k=1}^{n-1}int_{k}^{k+1}left{xright}f´(x)dx= sum_{k=1}^{n-1}f(k)-int_{1}^{n} f(x)dx$
– Ricardo770
Jan 4 at 15:31
Why do you need to do any integrals?
– Connor Harris
Jan 4 at 16:22
Why do you need to do any integrals?
– Connor Harris
Jan 4 at 16:22
add a comment |
Ok, I got it!
I took the reverse way to prove it.
$$int_{1}^{n}left{xright}f´(x)dx= int_{1}^{n}(x-left[xright])f´(x)dx$$
$$ =int_{1}^{n}x f´(x)dx -int_{1}^{n} left[xright]f´(x)dx$$
$$ =int_{1}^{n}x f´(x)dx -sum_{k=1}^{n-1}int_{k}^{k+1} left[xright]f´(x)dx$$
$$=sum_{k=1}^{n-1} int_{k}^{k+1}x f´(x)dx -sum_{k=1}^{n-1}int_{k}^{k+1} left[xright]f´(x)dx$$
$$=sum_{k=1}^{n-1} int_{k}^{k+1}(x - left[xright])f´(x)dx$$
$$=sum_{k=1}^{n-1} int_{k}^{k+1}left{xright}f´(x)dx$$
add a comment |
Ok, I got it!
I took the reverse way to prove it.
$$int_{1}^{n}left{xright}f´(x)dx= int_{1}^{n}(x-left[xright])f´(x)dx$$
$$ =int_{1}^{n}x f´(x)dx -int_{1}^{n} left[xright]f´(x)dx$$
$$ =int_{1}^{n}x f´(x)dx -sum_{k=1}^{n-1}int_{k}^{k+1} left[xright]f´(x)dx$$
$$=sum_{k=1}^{n-1} int_{k}^{k+1}x f´(x)dx -sum_{k=1}^{n-1}int_{k}^{k+1} left[xright]f´(x)dx$$
$$=sum_{k=1}^{n-1} int_{k}^{k+1}(x - left[xright])f´(x)dx$$
$$=sum_{k=1}^{n-1} int_{k}^{k+1}left{xright}f´(x)dx$$
add a comment |
Ok, I got it!
I took the reverse way to prove it.
$$int_{1}^{n}left{xright}f´(x)dx= int_{1}^{n}(x-left[xright])f´(x)dx$$
$$ =int_{1}^{n}x f´(x)dx -int_{1}^{n} left[xright]f´(x)dx$$
$$ =int_{1}^{n}x f´(x)dx -sum_{k=1}^{n-1}int_{k}^{k+1} left[xright]f´(x)dx$$
$$=sum_{k=1}^{n-1} int_{k}^{k+1}x f´(x)dx -sum_{k=1}^{n-1}int_{k}^{k+1} left[xright]f´(x)dx$$
$$=sum_{k=1}^{n-1} int_{k}^{k+1}(x - left[xright])f´(x)dx$$
$$=sum_{k=1}^{n-1} int_{k}^{k+1}left{xright}f´(x)dx$$
Ok, I got it!
I took the reverse way to prove it.
$$int_{1}^{n}left{xright}f´(x)dx= int_{1}^{n}(x-left[xright])f´(x)dx$$
$$ =int_{1}^{n}x f´(x)dx -int_{1}^{n} left[xright]f´(x)dx$$
$$ =int_{1}^{n}x f´(x)dx -sum_{k=1}^{n-1}int_{k}^{k+1} left[xright]f´(x)dx$$
$$=sum_{k=1}^{n-1} int_{k}^{k+1}x f´(x)dx -sum_{k=1}^{n-1}int_{k}^{k+1} left[xright]f´(x)dx$$
$$=sum_{k=1}^{n-1} int_{k}^{k+1}(x - left[xright])f´(x)dx$$
$$=sum_{k=1}^{n-1} int_{k}^{k+1}left{xright}f´(x)dx$$
answered Jan 4 at 16:33
Ricardo770Ricardo770
8017
8017
add a comment |
add a comment |
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