Let $P_k(x)$ denote the space of polynomials of at most degree $k$. Let $D$ denote differentiation with respect to $x$. Regard the differential operator $L: P_k\rightarrow P_k$ such that $L=\frac{1}{n!}D^n+\frac{1}{(n-1)!}D^{n-1}+...+D+I$ . If $k\leq n$, find the dimension of the kernel of $L-T$ where $T:P_k\rightarrow P_k$ is given by $T(p(x))=p(x+1)$.

To minimize the amount of calculation, I start with finding the matrix representation of $D$ w.r.t $\{1,x,x^2,...,\}$ basis, which is a matrix with $1,2,3,..,n$ on the super diagonal and 0 everywhere else. Then should I find $D^k$ for each $k$? The computation seems to be insane. Are there any easier way? Any shortcuts?

  • $\begingroup$ well, these are polynomials and a bunch of differentiation. Basically, use calculus and attack directly if I had to guess. $\endgroup$ Jul 24, 2015 at 3:36
  • $\begingroup$ It might be better to use $I$ in place of $1$ (which might be confused with the constant function $1$) in the definition of $L$. $\endgroup$ Jul 24, 2015 at 3:39
  • $\begingroup$ The set of polynomials of degree $k$ is not a vector space. Did you mean polynomials of degree at most $k$? $\endgroup$ Jul 24, 2015 at 4:53

2 Answers 2


Hint: Taylor's formula tells us that when $p(x)$ is a polynomial: $$p(x+1)=p(x)+p'(x)+\frac{1}{2}p''(x)+\cdots+\frac{1}{n!}p^{(n)}(x)+\cdots$$ since the remainder (Lagrange, etc.) will eventually become zero. Write this formula in the form of operators on $p$.


Hint Since $D: P_k \to P_k$ is nilpotent of order $k + 1$, that is, $D^l = 0$ for $l \geq k + 1$, the series defining its exponential truncates at that order. For $n \geq k + 1$ we have: $$ L = I + D + \frac{1}{2} D + \cdots + \frac{1}{n!} D^n = \sum_{i = 0}^{\infty} \frac{1}{i!} D^i =: \exp D. $$ For the matrix representation $[D]$ of $D$ w.r.t. the given basis $\{1, x, \ldots, x^k\}$, it's tractable to work out manually a change-of-basis matrix that puts $D$ in Jordan Normal Form, after which it is easy to exponentiate.

Additional hint The matrix representation for $L$ with respect to the given basis has $(i, j)$ entry $j \choose i$ for $j \geq i$ and $0$ for $j < i$, that is, it is upper triangular with nonzero entries given by (the transpose of) the first $k + 1$ rows of Pascal's Triangle.


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