Invertible linear transformation on the space of polynomials

Question

Let $\mathbb{F}[x]$ be the polynomial ring over the field $\mathbb{F}$. Let us assume $\operatorname{char}\mathbb{F}$ is not $2$. Then the $\mathbb{F}$-linear transformation $$\mathscr{A}:\mathbb{F}[x]\rightarrow\mathbb{F}[x], f(x)\mapsto f(x)+f(x+1)$$ is invertible since its matrix with respect to the basis $1, x,\dotsc,x^n,\dotsc$ is upper-triangular with diagonal entries $2$. But is it possible to write down an explicit formula for $\mathscr{A}^{-1}(f(x))$?

It is not difficult to write down $\mathscr{A}^{-1}(f(x))$ if $f(x)$ is given. For example, if $f(x)=x^2$, then $\mathscr{A}^{-1}(x^2)=\frac{1}{2}x^2-\frac{1}{2}x$. But I can not find a formula working for all polynomials $f(x)$.

Answer 1

Let $D:f\mapsto f'$. Then $$A(f)_x=f(x)+f(x)+f'(x)+1/2f''(x)+\cdots=(I+e^D)(f)_x.$$ Then $A^{-1}=(I+e^D)^{-1}$. Suppose that $f$ is a polynomial of degree $20$. Then our preferred computer calculates the Taylor's development of $\dfrac{1}{e^x+1}$ until $O(x^{21})$.

and we are done.

For example, $f(x)=x^3$; $A^{-1}(f)=(1/2I-1/4D+1/48D^3)(f)$, that is, $1/2x^3-3/4x^2+1/8$.

EDIT. In fact, $\dfrac{1}{e^x+1}=1/2(1-\tanh(x/2))$. Then the coefficients of its Taylor' development are functions of Bernouilli's numbers cf.

https://en.wikipedia.org/wiki/Bernoulli_number

Then there exists a recurrence linking the successive coefficients.

Answer 2

Let $\Delta_x:\ f(x)\rightarrow f(x+1)-f(x)$ be the forward difference operator then $\mathscr{A}=2I+\Delta_x$ where $I$ is the identity operator. Using formal expansion of operators $$\mathscr{A}^{-1}=(2I+\Delta_x)^{-1}=\frac{1}{2}\sum_{n=0}^\infty(−\frac{1}{2})^n\Delta_x^n$$ where $\Delta_x^0=I$ and $\Delta_x^n=\Delta_x[\Delta_x^{n-1}]$, $n\in\mathbb N$. By expanding $\Delta_x^n[f(x)]$ as $$ \Delta_x^n[f(x)]=\sum_{k=0}^n\binom{n}{k}(-1)^{n-k}f(x+k)$$ we get by replacing $$\mathscr{A}^{-1}[f(x)]=\sum_{n=0}^\infty\frac{1}{2^{n+1}}\cdot\sum_{k=0}^n\binom{n}{k}(-1)^kf(x+k)$$ $$\mathscr{A}^{-1}[f(x)]=\sum_{k=0}^\infty(-1)^kf(x+k)\cdot\sum_{n=k}^\infty\binom{n}{k}\frac{1}{2^{n+1}}$$ Since $\Delta_x^n$ annihilates polynomials $p_\ell(x)$ of degree $\ell$ lower than $n$, i.e. $\Delta_x^n[p_\ell(x)]=0$ for $0\le\ell<n$, we get finite sums' upper limits by taking $f(x)=p_\ell(x)$. In fact $$\mathscr{A}^{-1}[p_\ell(x)]=\frac{1}{2}\sum_{n=0}^\ell(−\frac{1}{2})^n\Delta_x^n[p_\ell(x)]$$ $$\mathscr{A}^{-1}[p_\ell(x)]=\sum_{k=0}^\ell(-1)^k\,p_\ell(x+k)\cdot\sum_{n=k}^\ell\binom{n}{k}\frac{1}{2^{n+1}}$$ I think this is the formula you are looking for. Indeed, for single powers $p_\ell(x)=x^\ell$ we obtain using the following PARI/GP code

g(L,x='x)=for(l=0,L,print("p(",l,",x) = ",sum(k=0,l,(-1)^k*(x+k)^l*sum(n=k,l,binomial(n,k)/2^(n+1)))))

this output

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NOTE: If $D_x: f(x)\rightarrow f'(x)$ @loup_blanc answer can be set under the same context. In this case, by using $\tanh(z)$ Taylor's series the following formal odd order differential operator expansion is obtained $$\mathscr{A}^{-1}=(I+e^{D_x})^{-1}=\frac{1}{2}I-\sum_{n=1}^\infty\frac{(4^n-1)\,B_{2n}}{(2n)!}\cdot D_x^{2n-1}$$ where $B_{2n}$ are Bernoulli numbers. For polynomials $p_\ell(x)$ we have $$\mathscr{A}^{-1}[p_\ell(x)]=\frac{1}{2}p_\ell(x)-\sum_{n=1}^{\lfloor\frac{\ell+1}{2}\rfloor}\frac{(4^n-1)\,B_{2n}}{(2n)!}\cdot D_x^{2n-1}[p_\ell(x)]$$ For single powers $p_\ell(x)=x^\ell$ we obtain by using the following PARI/GP code

f(L,x='x)=for(l=0,L,print("p(",l,",x) = ",x^l/2-sum(n=1,(l+1)\2,(4^n-1)/(2*n)!*bernfrac(2*n)*derivn(x^l,2*n-1))))

the same output that has been got from difference operators.