6
$\begingroup$

Let $f\colon \mathbb{R}\to \mathbb{R}$ be a function with $k$ continuous derivatives. We want to find an expression for $$ S=f(1)+f(2)+f(3)+\ldots+f(n). $$ I'm currently reading Analysis by Its History by Hairer and Wanner. They first consider the shifted sum and arrive at the expression $$f(n)-f(0)=\sum_{i=1}^{n} f^{\prime}(i)-\frac{1}{2 !} \sum_{i=1}^{n} f^{\prime \prime}(i)+\frac{1}{3 !} \sum_{i=1}^{n} f^{\prime \prime \prime}(i)-\frac{1}{4 !} \sum_{i=1}^{n} f^{\prime \prime \prime \prime}(i)+\ldots$$

using Taylor series (provided that the Taylor series actually converges to $f$).

In order to turn this formula for $\sum f^{\prime}(i)$ into a formula for $\sum f(i)$, we replace $f$ by its primitive (again denoted by $f$ ): $$\sum_{i=1}^{n} f(i)=\int_{0}^{n} f(x) d x+\frac{1}{2 !} \sum_{i=1}^{n} f^{\prime}(i)-\frac{1}{3 !} \sum_{i=1}^{n} f^{\prime \prime}(i)+\frac{1}{4 !} \sum_{i=1}^{n} f^{\prime \prime \prime}(i)-\ldots$$ The second idea is to remove the sums $\sum f^{\prime}, \sum f^{\prime \prime}, \sum f^{\prime \prime \prime}$, on the right by using the same formula, with $f$ successively replaced by $f^{\prime}, f^{\prime \prime}, f^{\prime \prime \prime}$ etc.

I don't really understand the step which replaces $f$ by its primitive. Using $F$ for denoting the primitive of $f$ I obtain $$F(n)-F(0)=\sum_{i=1}^{n} F^{}(i)-\frac{1}{2 !} \sum_{i=1}^{n} F^{\prime}(i)+\frac{1}{3 !} \sum_{i=1}^{n} F^{\prime \prime}(i)-\frac{1}{4 !} \sum_{i=1}^{n} F^{\prime \prime \prime}(i)+\ldots$$ but I don't obtain any expression in terms of an integral. Clearly, $$ F(n) - F(0) = \int_0^n f(x) \textrm{d}x $$ but since the author mentions that he again denotes the primitive by $f$ this doesn't match up with the above formula. Can anyone explain me what my mistake is here?

$\endgroup$
4
  • $\begingroup$ Have a look at the Euler's summation formula. An example here. $\endgroup$
    – rtybase
    Aug 2 '21 at 9:39
  • $\begingroup$ @rtybase I know that it's the Euler-Mclaurin formula, but I want to understand its above derivation $\endgroup$
    – Sebastian
    Aug 2 '21 at 9:41
  • $\begingroup$ Then you need to add more context to the question, like specify the textbook and the formula they try to prove in the textbook. $\endgroup$
    – rtybase
    Aug 2 '21 at 9:45
  • $\begingroup$ @rtybase they try to give a short derivation of the euler mclaurin formula $\endgroup$
    – Sebastian
    Aug 2 '21 at 10:32
2
$\begingroup$

If you simply replace $f$ by $F$ with $F'=f$, you would get $$F(n)-F(0)=\sum_{i=1}^nF'(i)-\frac 1{2!}\sum_{i=1}^n F''(i)+\frac 1 {3!}\sum_{i=1}^n F'''(i)+\cdots,\qquad (1)$$ which implies $$\int_0^n f(x)~dx=\sum_{i=1}^n f(i)-\frac 1 {2!}\sum_{i=1}^n f'(i)+\frac 1{3!}\sum_{i=1}^n f''(i)+\cdots$$

$$\Leftrightarrow \sum_{i=1}^n f(i)=\int_0^n f(x)~dx+\frac 1 {2!}\sum_{i=1}^n f'(i)-\frac 1{3!}\sum_{i=1}^n f''(i)+\cdots.$$

Your mistake is on the right hand side of (1), where you didn't literally replace $f$ by $F$ as you did on the left hand side.

$\endgroup$
3
  • 1
    $\begingroup$ don't know what I did there honestly, thanks! $\endgroup$
    – Sebastian
    Aug 3 '21 at 10:48
  • $\begingroup$ why does the author then say that the primitive is denoted by $f$? Or did I just misunderstand this formulation? $\endgroup$
    – Sebastian
    Aug 3 '21 at 10:54
  • $\begingroup$ @Sebastian That might be the reason why the author got you confused. What he might mean is that denote its derivative again by $f$. Note that sometimes it is not good to read too much. When you read the line "in order to turn this formula for $\sum f'(i)$ into a formula for $\sum f(i)$", probably you can guess what the author is trying to do: shift the the order of differentiation. $\endgroup$
    – Pythagoras
    Aug 3 '21 at 11:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.