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In solving a first order linear differential equation $(1-D)y=x^2$ where $D\equiv \frac{d}{dx}$ the way I learnt was that we proceed as $y=\frac{1}{1-D}x^2=(1-D)^{-1}x^2=(1+D+D^2+D^3+\cdots)x^2=x^2+2x+2+0+0+\cdots=x^2+2x+2.$

Now the question that comes to mind is that what justifies our saying that $$(1-D)^{-1}=1+D+D^2+D^3+\cdots$$

When I asked my teacher about this he said that it is not the case that the inverse operator of $1-D$ is $1+D+D^2+D^3+\cdots$ (whose meaning is unclear anyway). What is true instead is that $$(1-D)(1+D+D^2+\cdots+D^m)x^m=x^m$$ and that we are actually using this fact.

Although I understand this, but I am not entirely satisfied for two reasons. Firstly the resemblance with the expression for the geometric series must be there for a reason which I want to know. Secondly can an appropriate norm be given to an appropriate function space in which we can actually prove this geometric series to be true? (The answer to the second question also covers the first).

Thanks

  • To be careful, no differential operator with non-trivial differential part can be invertible. $L[y+c]=L[y]$ for any constant $c$. Therefore, $L$ is not injective hence is not invertible. Of course, if we narrow the scope to the appropriate space of equivalence classes of functions modulo homogeneous solutions then we may be able to find injectivity. This is hidden by the reliance on the "rest solution" which imposes a particular natural initial condition... but, really this is just a representative of the equivalence class. – James S. Cook Feb 21 '14 at 12:12

2 Answers2

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I think your prof is right, and this has nothing to do with series and/or norms. The space of polynomials of degree less than a fixed number is finite-dimensional and in such space the operator $D$ is nilpotent and satisfies your prof's formula.

As another way to see that your trick works only for polynomials, note that the solution to your equation is $y=x^2+2x+2+ce^x$, with the last term missed by the trick (which simply doesn't work in the homogenous case).

Martin Argerami
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If the functions are limited to polynomials, then if the degree of $f$ is $n$, $$ \mathrm{D}^{n+1}f=0 $$ Thus, $$ \left(\mathrm{I}+\mathrm{D}+\mathrm{D}^2+\dots\right)f $$ converges (it is actually a finite sum). In this way, $\mathrm{I}-\mathrm{D}$ is invertible on the set of polynomials.

Formally, this is one way to look at the Euler-Maclaurin Sum Formula. Notice that on polynomials, using Taylor's formula, $e^{\lambda\mathrm{D}}$ is the shift operator: $$ \begin{align} f(x+\lambda) &=f(x)+\lambda\mathrm{D}f(x)+\frac1{2!}\left(\lambda\mathrm{D}\right)^2f(x)+\frac1{3!}\left(\lambda\mathrm{D}\right)^3f(x)+\dots\\ &=e^{\lambda\mathrm{D}}f(x) \end{align} $$ Thus, $$ f(x)-f(x-1)=\left(1-e^{-\mathrm{D}}\right)f(x) $$ The idea of the Euler-Maclaurin Sum Formula is to invert $1-e^{-\mathrm{D}}$. Since the power series for $1-e^{-x}$ has no constant term, we need to use $\frac{x}{1-e^{-x}}\frac1x$. That is, formally, the Euler-Maclaurin Sum Formula is $$ \frac{\mathrm{D}}{1-e^{-\mathrm{D}}}\int f(x)\,\mathrm{d}x $$ and this is exact on polynomials (other functions have a remainder term that can be computed in a fashion similar to the remainder term for the Taylor series).

robjohn
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  • I agree that $I-D$ is invertible but is there any closed form expression for its inverse? –  Feb 22 '14 at 13:48
  • @Shahab: Since $(\mathrm{I}-\mathrm{D})e^x=0$, I wouldn't say that $\mathrm{I}-\mathrm{D}$ is invertible in general. Furthermore, $\dfrac1{1-x}=1+x+x^2+x^3+x^4+\dots$; however, I would say that $\dfrac1{1-x}$ is the closed form. Functionally, I would say that $\dfrac1{\mathrm{I}-\mathrm{D}}$ would be the closed form, but I don't think that is the answer you are looking for. – robjohn Feb 22 '14 at 15:08
  • I meant I-D is invertible on the set of polynomials. If you say $1/(1-D)$ I understand it as notation for $(1-D)^{-1}$ but I don't know how to give meaning to $\sum D^n$ without any notion of convergence of operators. –  Feb 23 '14 at 01:09
  • @Shahab: For a polynomial $p$, $\sum_{k=0}^\infty\mathrm{D}^kp$ is a finite sum. Convergence is trivial. Obviously, $\sum\limits_{n=0}^\infty\mathrm{D}^n$ cannot converge in any sense as an operator on $e^x$. To talk about the convergence of $\sum\limits_{n=0}^\infty\mathrm{D}^n$ as an operator, we need to say as an operator on what space. – robjohn Feb 23 '14 at 08:47
  • Sorry if I am misunderstanding, I know $\sum D^kp$ is a finite sum but by a closed form expression I am asking for an expression for the inverse of $(1-D)^{-1}$ and not of the "value" of this inverse at some polynomial. Also, whether there is an appropriate space equipped with a norm to make sense of convergence is part of the question. The set underlying the space is $Hom(\mathbb{R}^\infty,\mathbb{R}^\infty)$. –  Feb 23 '14 at 08:58
  • In general, there is no inverse for $\mathrm{I}-\mathrm{D}$, so I guess I don't understand what you are looking for. – robjohn Feb 23 '14 at 09:41
  • Let $\mathbb{R}^\infty$ be the space of all real polynomials, and $Hom(\mathbb{R}^\infty,\mathbb{R}^\infty)$ be the space of all linear operators on it. The operator $I-D\in Hom(\mathbb{R}^\infty,\mathbb{R}^\infty)$ is invertible. –  Feb 23 '14 at 11:48