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For the moment, consider the corresponding problem involving integration. Let $s(x)$ be the explicit solution to the following integral.

$ \displaystyle s(x)=\int_a^x f(t) \, dt $

The function $s'(x)$ is equivalent to the derivative of the integral with respect to it's upper limit and may be expressed in integral form.

$ \displaystyle s'(x)=\partial _x\left(\int_a^x f(t) \, dt\right)=f(a)+\int_a^x f'(t) \, dt $

Now let $s(x)$ be the explicit solution to the following summation.

$ \displaystyle s(x)=\sum _{t=a}^x f(t) $

The function $s'(x)$ is equivalent to the derivative of the summation with respect to it's upper limit. What is the derivative of $s(x)$ expressed in summation form?

$ \displaystyle s'(x)=\partial _x\left(\sum _{t=a}^x f(t)\right)=\ ? $

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If $x$ is allowed to take real values, I'm expecting to see some floor functions. Else, your expression makes no sense. In the case you do introduce some floor function, you still need to see wether $s$ is differentiable or not. –  Pedro Tamaroff Nov 30 '12 at 19:01
    
Although the variable $x$ is restricted to be an integer within the summation, the derivative of $s(x)$ may still be considered without the floor function. It is assumed that $s(x)$ is differentiable. For example, if $s(x)=\frac{1}{2} x (x+1)=\sum _{t=1}^x t$, then $s'(x)=x+\frac{1}{2}=\frac{1}{2}+\sum _{t=1}^x \partial _t(t)$. –  cyclochaotic Dec 2 '12 at 3:02
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Still does not make sense to me. $\sum _{t=1}^x t$ is defined only for integer $x$. Your function on the LHS is just a real valued function that happens to coincide with the RHS for integer $x$. But that does not define $s(x)$, I could build other functions with the same property, eg: $x (x+1)/2 + sin(\pi x)$ –  leonbloy Dec 2 '12 at 3:53
    
@leonbloy, that is a fascinating difficulty you have raised. In addressing this point I think I have figured out a way to resolve the transition form x being a continuous real variable to a discrete integer variable. Define $s(x)$ by the continuous recurrence equation, $s'(x)-s'(x-1)=f'(x)$. I will post a second answer when I have completed the revision. –  cyclochaotic Dec 2 '12 at 22:07
    
In addressing the difficulty leonbloy raised I realized I was implicitly using an auxiliary function, which I define in my revised answer. Now, extraneous functions of the form, $g(x) \sin (\pi x)$ get filtered out. Also, in order to avoid any misinterpretation regarding the notation used for differentiation, I would like to clarify what the differential operator symbol $\partial _x$ means: $\partial _x\equiv D_x\equiv \frac{d}{dx}\equiv \frac{\partial }{\partial x}$. –  cyclochaotic Dec 6 '12 at 20:06

2 Answers 2

In order to express continuous summation, define an auxiliary function $s(x,f)$ equal to the $RHS$ of the Euler-Maclaurin formula with zero remainder, such that, the index variable $n$ is replaced with the real variable $x$. $$ \displaystyle s(x,f)\text{:=}\int_a^x f(t) \, dt+\frac{f(x)}{2}+\frac{f(a)}{2}+\sum _{k=1}^{\infty } \frac{B_{2 k} }{(2 k)!} \left(f^{(2 k-1)}(x)-f^{(2 k-1)}(a)\right) $$

Clearly, if $x$ is an integer $n$, then $s(n,f)$ is equal to the summation of $f(t)$.

$ \displaystyle s(n,f)=\sum _{t=a}^n f(t) $

We can now derive a differential equation for $s$. Differentiate $s$ with respect to $x$ and simplify.

$ \displaystyle \frac{\partial s(x,f)}{\partial x}=f(x)+\frac{f'(x)}{2}+\sum _{k=1}^{\infty } \frac{B_{2 k} f^{(2 k)}(x)}{(2 k)!} $

Substitute $f'(t)$ for $f(t)$.

$ \displaystyle s\left(x,f'\right)=f(x)-f(a)+\frac{f'(x)}{2}+\frac{f'(a)}{2}+\sum _{k=1}^{\infty } \frac{B_{2 k} }{(2 k)!} \left(f^{(2 k)}(x)-f^{(2 k)}(a)\right) $

Calculate the difference and simplify.

$ \displaystyle \frac{\partial s(x,f)}{\partial x}-s\left(x,f'\right)=\sum _{k=0}^{\infty } \frac{B_k}{k!} f^{(k)}(a) $

Observe that the term $\frac{f^{(k)}(a)}{k!}$ is the coefficient for the Taylor expansion of $f(t)$, hence.

$ \displaystyle f(t)=\sum _{k=0}^{\infty } \frac{f^{(k)}(a)}{k!} (t-a)^k=\sum _{k=0}^{\infty } c(k) (t-a)^k $

$ \displaystyle \frac{\partial s(x,f)}{\partial x}-s\left(x,f'\right)=\sum _{k=0}^{\infty } c(k) B_k $

Consider what happens if we restrict $x$ to be an integer $n$.

$ \displaystyle \left.\frac{\partial s(x,f)}{\partial x}\right|_{x=n}=\sum _{k=0}^{\infty } c(k) B_k+\sum _{t=a}^n f'(t) $

Now, let's get creative and allow the following symbolic equivalence.

$ \displaystyle \left.\frac{\partial s(x,f(t))}{\partial x}\right|_{x=n}\equiv \partial _n\sum _{t=a}^n f(t) $

The derivative of a summation with respect to it's upper limit may then be expressed as,

$$ \displaystyle \partial _n\sum _{t=a}^n f(t)=\sum _{k=0}^{\infty } c(k) B_k+\sum _{t=a}^n f'(t) $$

Example

As an example consider the following summation identity with integer $m\geq 1$.

$ \displaystyle \sum _{t=0}^n t^m=\frac{B_{m+1}(n+1)-B_{m+1}}{m+1} $

Define the functions $f$ and $s$.

$ \displaystyle f(t)=t^m $

$ \displaystyle s(x,f)=\frac{B_{m+1}(x+1)-B_{m+1}}{m+1} $

Verify the differential equation.

$ \displaystyle \partial _n\sum _{t=a}^n f(t)=\sum _{k=0}^{\infty } c(k) B_k+\sum _{t=a}^n f'(t) $

$ \displaystyle \left.\partial _x\left(\frac{B_{m+1}(x+1)-B_{m+1}}{m+1}\right)\right|_{x=n}=B_m+\sum _{t=0}^n m t^{m-1} $

$ \displaystyle B_m(n+1)=B_m+m\left(\frac{(B_m(n+1)-B_m)}{m}\right) $

$ \displaystyle B_m(n+1)=B_m(n+1) $

Application

As an application consider the following summation identity that uses the incomplete gamma function, $\Gamma (a,z)$.

$ \displaystyle \sum _{t=0}^n \frac{1}{\Gamma (t+1)}=\frac{e \Gamma (n+1,1)}{n!} $

Define the functions $f$ and $s$.

$ \displaystyle f(t)=\frac{1}{\Gamma (t+1)} $

$ \displaystyle s(x,f)=\frac{e \Gamma (x+1,1)}{\Gamma (x+1)} $

Let the constant $C$ represent the Bernoulli sum.

$ \displaystyle C=\sum _{k=0}^{\infty } c(k) B_k $

Because $C$ is a constant, we may choose any valid $n$ to solve for $C$, choose $n=0$.

$ \displaystyle C=-e \text{Ei}(-1) $

In conclusion, assemble the differential equation and simplify. The variable $t$ is replaced with the standard index variable $k$. The result is a summation identity that makes use of these additional functions: (1) $\, _2\tilde{F}_2$, (2) $\text{Ei}$, and (3) $\psi ^{(0)}$.

$$ \displaystyle \frac{-1}{e}\sum _{k=0}^n \frac{1}{k!}\psi ^{(0)}(k+1)=\text{Ei}(-1)+\psi ^{(0)}(n+1) \left(1-\frac{1}{n!}\Gamma (n+1,1)\right)\\ +n! \, _2\tilde{F}_2(n+1,n+1;n+2,n+2;-1) $$

Peace

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Consider the following integral equation with constant $C$.

$ \displaystyle \partial _x\left(\int_a^x f(t) \, dt\right)-\int_a^x f'(t) \, dt=C $

Is there a corresponding summation equation?

$ \displaystyle \partial _x\left(\sum _{t=a}^x f(t)\right)-\sum _{t=a}^x f'(t)=C $

The proof of this result is based on the Euler-Maclaurin formula assuming zero remainder.

$ \displaystyle \sum _{t=a}^x f(t)=\int_a^x f(t) \, dt+\frac{f(x)}{2}+\frac{f(a)}{2}+\sum _{k=1}^{\infty } \frac{B_{2 k} }{(2 k)!} \left(f^{(2 k-1)}(x)-f^{(2 k-1)}(a)\right) $

Differentiate with respect to x.

$ \displaystyle \partial _x\left(\sum _{t=a}^x f(t)\right)=f(x)+\frac{f'(x)}{2}+\sum _{k=1}^{\infty } \frac{B_{2 k}}{(2 k)!} f^{(2 k)}(x) $

Apply the formula to $f'(t)$.

$ \displaystyle \sum _{t=a}^x f'(t)=f(x)-f(a)+\frac{f'(x)}{2}+\frac{f'(a)}{2}+\sum _{k=1}^{\infty } \frac{B_{2 k}}{(2 k)!} \left(f^{(2 k)}(x)-f^{(2 k)}(a)\right) $

Calculate the difference and simplify.

$ \displaystyle \partial _x\left(\sum _{t=a}^x f(t)\right)-\sum _{t=a}^x f'(t)=\sum _{k=0}^{\infty } \frac{B_k}{k!} f^{(k)}(a)\text{ }\ \square $

Observe that the term $\frac{f^{(k)}(a)}{k!}$ is the coefficient for the Taylor expansion of $f(t)$, hence.

$ \displaystyle f(t)=\sum _{k=0}^{\infty } \frac{f^{(k)}(a)}{k!} (t-a)^k=\sum _{k=0}^{\infty } c(k) (t-a)^k $

The derivative of a summation with respect to it's upper limit is therefore, $$ \displaystyle \partial _x\left(\sum _{t=a}^x f(t)\right)=\sum _{k=0}^{\infty } c(k)B_k+\sum _{t=a}^x f'(t) $$

Example

As an example let $f(t)=e^{r t}$ and $a=0$. The summation of $e^{r t}$ is just the geometric series in $e^{r}$, hence.

$ \displaystyle \sum _{t=0}^x f(t)=\sum _{t=0}^x e^{r t}=\sum _{t=0}^x \left(e^r\right)^t=\frac{\left(e^r\right)^{x+1}-1}{e^r-1}=\frac{e^{r (x+1)}-1}{e^r-1} $

$ \displaystyle \partial _x\left(\sum _{t=0}^x f(t)\right)=\partial _x\left(\sum _{t=0}^x e^{r t}\right)=\partial _x\left(\frac{e^{r (x+1)}-1}{e^r-1}\right)=\frac{r e^{r (x+1)}}{e^r-1} $

$ \displaystyle \sum _{t=0}^x f'(t)=\sum _{t=0}^x r e^{r t}=\frac{r\left(e^{r (x+1)}-1\right)}{e^r-1} $

The Bernoulli sum in terms of the coefficient function $c(k)$.

$ \displaystyle e^{r t}=\sum _{k=0}^{\infty } \frac{r^k}{k!} t^k $

$ \displaystyle c(k)=\frac{r^k}{k!} $

$ \displaystyle \sum _{k=0}^{\infty } c(k) B_k=\sum _{k=0}^{\infty } \frac{r^k}{k!} B_k $

Notice that this is the generating form for the Bernoulli numbers.

$ \displaystyle \frac{r}{e^r-1}=\sum _{k=0}^{\infty } \frac{B_k}{k!} r^k $

Now assemble the equation and simplify.

$ \displaystyle \partial _x\left(\sum _{t=a}^x f(t)\right)=\sum _{k=0}^{\infty } c(k)B_k+\sum _{t=a}^x f'(t) $

$ \displaystyle \frac{r e^{r (x+1)}}{e^r-1}=\frac{r}{e^r-1}+\frac{r\left(e^{r (x+1)}-1\right)}{e^r-1} $

$ \displaystyle r=r $

Peace

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I cannot make sense of the question or the answer. 1) Peter's comment 2) your second formula in the question only applies if $f$ is function of $x$ and $t$, and the derivative is respect with $x$ 3) (to jump to the end) you dont' prove anything by asumming an equality, doing valid operations on it, and reaching an equality, the way is the reverse –  leonbloy Dec 1 '12 at 21:33
    
I recently addressed Peter's comment above. The partial derivative may be applied to any suitable expression of one or more variables, for example: $\partial _x\left(x^2\right)=2 x$ and $\partial _t\left(x^2\right)=0$. The Euler-Maclaurin formula is an identity not an assumption. However, assuming that the remainder term is zero simply restricts f(t) to a proper class of functions, refer to the link. –  cyclochaotic Dec 2 '12 at 3:38
    
Sorry, neither your response to Peter, nor this makes sense to me. If $f$ depends on $x$ and $t$, then write that explicitly, and write explicitly the derivatives, don't write $f'(t)$ (is that derivative with respect to... what?) –  leonbloy Dec 2 '12 at 3:57
    
For this discussion $f(t)$ depends only on $t$ and is independent of $x$. Also to clarify, $f'(t)$ is equivalent to $\frac{df}{dt}$. –  cyclochaotic Dec 2 '12 at 22:08
    
"f(t) depends only on t and is independent of x." Then your second formula in the question is just wrong. –  leonbloy Dec 2 '12 at 22:23

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