Euler-Maclaurin Formula Definition Confusion
$begingroup$
I am confused about these $2$ definitions of the Euler-Maclaurin formula.
I read the following here:
The full Euler-Maclaurin formula with no remainder term (for infinitely differentiable $f$ ) is given in Concrete Mathematics [2, p. 471]:
$sum_{j=0}^{m-1}{f(j)}=int_{0}^{m}{f(x)dx}+sum_{k=1}^{infty}{frac{B_k}{k!}left(f^{(k-1)}(m)-f^{(k-1)}(0)right)}$.
But the wikipedia article found here says this:
Explicitly, for $p$ a positive integer and a function $f(x)$ that is $p$ times continuously differentiable in the interval $[m,n]$, we have
$S-I=sum_{k=1}^{p}{frac{B_k}{k!}left(f^{(k-1)}(n)-f^{(k-1)}(m)right)}+R_p$
My question is, if a function is infinitely differentiable over a given interval, how can the wikipedia definition hold since infinity is not an integer?
I could not find a reference that gives the definition the way Wikipedia does. Can anyone explain the discrepancy between the two definitions? What am I misunderstanding?
real-analysis calculus bernoulli-numbers euler-maclaurin
asked Jan 23 at 16:11
GnumbertesterGnumbertester
668113
$endgroup$
add a comment |
$begingroup$
I am confused about these $2$ definitions of the Euler-Maclaurin formula.
I read the following here:
The full Euler-Maclaurin formula with no remainder term (for infinitely differentiable $f$ ) is given in Concrete Mathematics [2, p. 471]:
$sum_{j=0}^{m-1}{f(j)}=int_{0}^{m}{f(x)dx}+sum_{k=1}^{infty}{frac{B_k}{k!}left(f^{(k-1)}(m)-f^{(k-1)}(0)right)}$.
But the wikipedia article found here says this:
Explicitly, for $p$ a positive integer and a function $f(x)$ that is $p$ times continuously differentiable in the interval $[m,n]$, we have
$S-I=sum_{k=1}^{p}{frac{B_k}{k!}left(f^{(k-1)}(n)-f^{(k-1)}(m)right)}+R_p$
My question is, if a function is infinitely differentiable over a given interval, how can the wikipedia definition hold since infinity is not an integer?
I could not find a reference that gives the definition the way Wikipedia does. Can anyone explain the discrepancy between the two definitions? What am I misunderstanding?
real-analysis calculus bernoulli-numbers euler-maclaurin
asked Jan 23 at 16:11
GnumbertesterGnumbertester
668113
$endgroup$
1
$begingroup$
If a function is infinitely differentiable, then it is $p$ times differentiable for every positive integer $p$. Also, if you look at the paper more closely, you'll see that it's the formula with remainder that is used to prove the theorems about sums of integer powers.
$endgroup$
– saulspatz
Jan 23 at 16:23
$begingroup$
@saulspatz, if a function is infinitely differentiable though, what value would I plug in for the upper bound, $p$, since the wikipedia formula gives $p$ explicitly as an integer? Also, is a function $f(x)=x^3$ is $4$ times differentiable or infinitely times differentiable in the case of this formula?
$endgroup$
– Gnumbertester
Jan 23 at 16:38
1
$begingroup$
If it's infinitely differentiable you can use any $p$ you like. Just as in Taylor's formula, you get different estimates using different numbers of terms. The difference is, however, that the Euler-MacLaurin formula is not necessarily convergent, and then one tries to pick the value of $p$ that will give the smallest (or at least an acceptably small) error. $f(x)=x^3$ is infinitely differentiable, and a fortiori it's also $4$ times differentiable.
$endgroup$
– saulspatz
Jan 23 at 16:45
$begingroup$
@saulspatz, Thank you, that clears it up for me, I did not realize that $p$ is arbitrary depending on how much error is acceptable.
$endgroup$
– Gnumbertester
Jan 23 at 17:14
add a comment |
$begingroup$
I am confused about these $2$ definitions of the Euler-Maclaurin formula.
I read the following here:
The full Euler-Maclaurin formula with no remainder term (for infinitely differentiable $f$ ) is given in Concrete Mathematics [2, p. 471]:
$sum_{j=0}^{m-1}{f(j)}=int_{0}^{m}{f(x)dx}+sum_{k=1}^{infty}{frac{B_k}{k!}left(f^{(k-1)}(m)-f^{(k-1)}(0)right)}$.
But the wikipedia article found here says this:
Explicitly, for $p$ a positive integer and a function $f(x)$ that is $p$ times continuously differentiable in the interval $[m,n]$, we have
$S-I=sum_{k=1}^{p}{frac{B_k}{k!}left(f^{(k-1)}(n)-f^{(k-1)}(m)right)}+R_p$
My question is, if a function is infinitely differentiable over a given interval, how can the wikipedia definition hold since infinity is not an integer?
I could not find a reference that gives the definition the way Wikipedia does. Can anyone explain the discrepancy between the two definitions? What am I misunderstanding?
real-analysis calculus bernoulli-numbers euler-maclaurin
asked Jan 23 at 16:11
GnumbertesterGnumbertester
668113
$endgroup$
I am confused about these $2$ definitions of the Euler-Maclaurin formula.
I read the following here:
The full Euler-Maclaurin formula with no remainder term (for infinitely differentiable $f$ ) is given in Concrete Mathematics [2, p. 471]:
$sum_{j=0}^{m-1}{f(j)}=int_{0}^{m}{f(x)dx}+sum_{k=1}^{infty}{frac{B_k}{k!}left(f^{(k-1)}(m)-f^{(k-1)}(0)right)}$.
But the wikipedia article found here says this:
Explicitly, for $p$ a positive integer and a function $f(x)$ that is $p$ times continuously differentiable in the interval $[m,n]$, we have
$S-I=sum_{k=1}^{p}{frac{B_k}{k!}left(f^{(k-1)}(n)-f^{(k-1)}(m)right)}+R_p$
My question is, if a function is infinitely differentiable over a given interval, how can the wikipedia definition hold since infinity is not an integer?
I could not find a reference that gives the definition the way Wikipedia does. Can anyone explain the discrepancy between the two definitions? What am I misunderstanding?
real-analysis calculus bernoulli-numbers euler-maclaurin
real-analysis calculus bernoulli-numbers euler-maclaurin
asked Jan 23 at 16:11
GnumbertesterGnumbertester
668113
asked Jan 23 at 16:11
GnumbertesterGnumbertester
668113
edited Jan 23 at 16:16
Gnumbertester
asked Jan 23 at 16:11
GnumbertesterGnumbertester
668113
asked Jan 23 at 16:11
GnumbertesterGnumbertester
668113
asked Jan 23 at 16:11
GnumbertesterGnumbertester
668113
668113
1
$begingroup$
If a function is infinitely differentiable, then it is $p$ times differentiable for every positive integer $p$. Also, if you look at the paper more closely, you'll see that it's the formula with remainder that is used to prove the theorems about sums of integer powers.
$endgroup$
– saulspatz
Jan 23 at 16:23
$begingroup$
@saulspatz, if a function is infinitely differentiable though, what value would I plug in for the upper bound, $p$, since the wikipedia formula gives $p$ explicitly as an integer? Also, is a function $f(x)=x^3$ is $4$ times differentiable or infinitely times differentiable in the case of this formula?
$endgroup$
– Gnumbertester
Jan 23 at 16:38
1
$begingroup$
If it's infinitely differentiable you can use any $p$ you like. Just as in Taylor's formula, you get different estimates using different numbers of terms. The difference is, however, that the Euler-MacLaurin formula is not necessarily convergent, and then one tries to pick the value of $p$ that will give the smallest (or at least an acceptably small) error. $f(x)=x^3$ is infinitely differentiable, and a fortiori it's also $4$ times differentiable.
$endgroup$
– saulspatz
Jan 23 at 16:45
$begingroup$
@saulspatz, Thank you, that clears it up for me, I did not realize that $p$ is arbitrary depending on how much error is acceptable.
$endgroup$
– Gnumbertester
Jan 23 at 17:14
add a comment |
1
$begingroup$
If a function is infinitely differentiable, then it is $p$ times differentiable for every positive integer $p$. Also, if you look at the paper more closely, you'll see that it's the formula with remainder that is used to prove the theorems about sums of integer powers.
$endgroup$
– saulspatz
Jan 23 at 16:23
$begingroup$
@saulspatz, if a function is infinitely differentiable though, what value would I plug in for the upper bound, $p$, since the wikipedia formula gives $p$ explicitly as an integer? Also, is a function $f(x)=x^3$ is $4$ times differentiable or infinitely times differentiable in the case of this formula?
$endgroup$
– Gnumbertester
Jan 23 at 16:38
1
$begingroup$
If it's infinitely differentiable you can use any $p$ you like. Just as in Taylor's formula, you get different estimates using different numbers of terms. The difference is, however, that the Euler-MacLaurin formula is not necessarily convergent, and then one tries to pick the value of $p$ that will give the smallest (or at least an acceptably small) error. $f(x)=x^3$ is infinitely differentiable, and a fortiori it's also $4$ times differentiable.
$endgroup$
– saulspatz
Jan 23 at 16:45
$begingroup$
@saulspatz, Thank you, that clears it up for me, I did not realize that $p$ is arbitrary depending on how much error is acceptable.
$endgroup$
– Gnumbertester
Jan 23 at 17:14
1
1
$begingroup$
If a function is infinitely differentiable, then it is $p$ times differentiable for every positive integer $p$. Also, if you look at the paper more closely, you'll see that it's the formula with remainder that is used to prove the theorems about sums of integer powers.
$endgroup$
– saulspatz
Jan 23 at 16:23
$begingroup$
If a function is infinitely differentiable, then it is $p$ times differentiable for every positive integer $p$. Also, if you look at the paper more closely, you'll see that it's the formula with remainder that is used to prove the theorems about sums of integer powers.
$endgroup$
– saulspatz
Jan 23 at 16:23
$begingroup$
@saulspatz, if a function is infinitely differentiable though, what value would I plug in for the upper bound, $p$, since the wikipedia formula gives $p$ explicitly as an integer? Also, is a function $f(x)=x^3$ is $4$ times differentiable or infinitely times differentiable in the case of this formula?
$endgroup$
– Gnumbertester
Jan 23 at 16:38
$begingroup$
@saulspatz, if a function is infinitely differentiable though, what value would I plug in for the upper bound, $p$, since the wikipedia formula gives $p$ explicitly as an integer? Also, is a function $f(x)=x^3$ is $4$ times differentiable or infinitely times differentiable in the case of this formula?
$endgroup$
– Gnumbertester
Jan 23 at 16:38
1
1
$begingroup$
If it's infinitely differentiable you can use any $p$ you like. Just as in Taylor's formula, you get different estimates using different numbers of terms. The difference is, however, that the Euler-MacLaurin formula is not necessarily convergent, and then one tries to pick the value of $p$ that will give the smallest (or at least an acceptably small) error. $f(x)=x^3$ is infinitely differentiable, and a fortiori it's also $4$ times differentiable.
$endgroup$
– saulspatz
Jan 23 at 16:45
$begingroup$
If it's infinitely differentiable you can use any $p$ you like. Just as in Taylor's formula, you get different estimates using different numbers of terms. The difference is, however, that the Euler-MacLaurin formula is not necessarily convergent, and then one tries to pick the value of $p$ that will give the smallest (or at least an acceptably small) error. $f(x)=x^3$ is infinitely differentiable, and a fortiori it's also $4$ times differentiable.
$endgroup$
– saulspatz
Jan 23 at 16:45
$begingroup$
@saulspatz, Thank you, that clears it up for me, I did not realize that $p$ is arbitrary depending on how much error is acceptable.
$endgroup$
– Gnumbertester
Jan 23 at 17:14
$begingroup$
@saulspatz, Thank you, that clears it up for me, I did not realize that $p$ is arbitrary depending on how much error is acceptable.
$endgroup$
– Gnumbertester
Jan 23 at 17:14
add a comment |
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1
$begingroup$
If a function is infinitely differentiable, then it is $p$ times differentiable for every positive integer $p$. Also, if you look at the paper more closely, you'll see that it's the formula with remainder that is used to prove the theorems about sums of integer powers.
$endgroup$
– saulspatz
Jan 23 at 16:23
$begingroup$
@saulspatz, if a function is infinitely differentiable though, what value would I plug in for the upper bound, $p$, since the wikipedia formula gives $p$ explicitly as an integer? Also, is a function $f(x)=x^3$ is $4$ times differentiable or infinitely times differentiable in the case of this formula?
$endgroup$
– Gnumbertester
Jan 23 at 16:38
1
$begingroup$
If it's infinitely differentiable you can use any $p$ you like. Just as in Taylor's formula, you get different estimates using different numbers of terms. The difference is, however, that the Euler-MacLaurin formula is not necessarily convergent, and then one tries to pick the value of $p$ that will give the smallest (or at least an acceptably small) error. $f(x)=x^3$ is infinitely differentiable, and a fortiori it's also $4$ times differentiable.
$endgroup$
– saulspatz
Jan 23 at 16:45
$begingroup$
@saulspatz, Thank you, that clears it up for me, I did not realize that $p$ is arbitrary depending on how much error is acceptable.
$endgroup$
– Gnumbertester
Jan 23 at 17:14