# Uncertain summation of polynomials

I've been trying out the summation of polynomials. My line of strike is to deal with the subject the means I would certainly for calculus, yet not making use of restrictions.

Using a really straightforward instance, intend I desire to add the all numbers in between $10$ and also $20$ comprehensive, and also locate a polynomial which I can connect the numbers right into to get my solution. I believe its some kind of polynomial with level $2$. So I do a integer 'distinction': $$ \mathrm{diff}\left(x^{2}\right)=x^{2}-\left(x-1\right)^{2}=2x-1 $$

I can see from this that I virtually have my solution, so thinking an inverted 'assimilation' procedure and also re-arranging: $$ \frac{1}{2}\mathrm{diff}\left(x^{2}+\mathrm{int}\left(1\right)\right)=x $$

Currently, I recognize that the 'uncertain indispensable' of 1 is simply x, from 'setting apart' $x-(x-1) = 1$. So inevitably: $$ \frac{1}{2}\left(x^{2}+x\right)=\mathrm{int}\left(x\right) $$

So to get my solution I take the 'precise' indispensable: $$ \mathrm{int}\left(x\right):10,20=\frac{1}{2}\left(20^{2}+20\right)-\frac{1}{2}\left(9^{2}+9\right)=165 $$ (the lower bound demands lowering by one)

My inquiry is, exists a basic means I can 'incorporate' any kind of polynomial, this way?

Please excuse my absence of rigour and also the weird symbols.

Your "diff" is in fact called (backward) finite difference.

\begin{align} \nabla_1 [ P ](x) &= P(x) - P(x-1) \\ &= P(x-1+1) - P(x-1) \\ &= \Delta_1[ P ](x-1) \end{align}

The inverted the forward limited distinction is called indefinite sum. Removed from Wikipedia, the valuable solutions for polynomials is :

\begin{align} \Delta^{-1}_1 x^n &= \frac{B_{n+1}(x)}{n+1} + C \\ \Delta^{-1}_1 af(x) &= a \Delta^{-1}_1 f(x) \end{align}

(B _{n +1 } (x) is the Bernoulli polynomial.)

The Δ ^{ - 1 } can be transformed back to your "int" by replacing $x \mapsto x + 1$.

As various other solutions have actually kept in mind, you will uncover the calculus of finite differences.

For sensible estimations, below a most valuable reality : the regulation

$\frac{d}{dx} x^n = n x^{n-1}$

represents

$\Delta_1 x(x-1)(x-2)\cdots(x-(n-1)) = n x(x-1)(x-2)\cdots(x-(n-2))$

You appear to be grabbing the calculus of finite differences, as soon as a well - recognized subject yet instead antiquated nowadays. The response to your inquiry is of course : offered a polynomial $f(x)$ there is a polynomial $g(x)$ (of level one more than $f$) such that $$f(x)=g(x)-g(x-1).$$ This polynomial $g$ (like the indispensable of $f$) is one-of-a-kind save for its constant term. As soon as one has $g$ after that certainly $$f(a)+f(a+1)+\cdots+f(b)=g(b)-g(a-1).$$

When $f(x)=x^n$ is a monomial, the coefficients of $g$ entail the constantly remarkable Bernoulli numbers.

For any kind of certain polynomial there is a less complicated means to do uncertain summation than making use of the Bernoulli numbers, going off of Greg Graviton's solution. Below we'll make use of the forward distinction $\Delta f(x) = f(x+1) - f(x)$. After that

$\displaystyle \Delta {x \choose n} = {x \choose n-1}.$

This indicates that we can execute a "Taylor development" on any kind of polynomial to write it in the kind $f(x) = \sum a_n {x \choose n}$ by reviewing the limited distinctions $\Delta^n f(0)$ at absolutely no. For any kind of certain polynomial $f$ it is very easy to write these limited distinctions down by creating a table. As a whole, the formula is

$\displaystyle a_n = \Delta^n f(0) = \sum_{k=0}^{n} (-1)^{n-k} {n \choose k} f(k)$

as one can conveniently confirm by creating $\Delta = S - I$ where $S$ is the change driver $S f(x) = f(x+1)$ and also $I$ is the identification driver $I f(x) = f(x)$. After that the uncertain amount of $f$ is simply $\sum a_n {x \choose n+1}$. This is the most convenient means I recognize just how to do such calculations by hand, and also it additionally brings about a rather very easy method for polynomial interpolation offered the values of a polynomial at successive integers.