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let $ f\in F[x] $ be a polynomial. and prove that the matrix $ f\left(J_{n}\left(\lambda\right)\right) $ satisfies

$ [f\left(J_{n}\left(\lambda\right)\right)]_{ij}=\begin{cases} \frac{1}{\left(j-i\right)!}f^{(j-i)}\left(\lambda\right) & 1\leq i\leq j\leq n\\ 0 & else \end{cases} $

when $ f^{(j-i)} $ is the (j-i) deriviative of $ f $.

Here's what i tried:

step 1: I proved that

$ [\left(J_{n}\left(0\right)\right)^{k}]=\begin{cases} 1 & j=i+k\\ 0 & else \end{cases} $

step 2: Using the binom formula, I proved that

$ \left(J_{n}\left(\lambda\right)\right)^{k}=\sum_{i=0}^{k}\binom{k}{i}\lambda^{k-i}\left(J_{n}\left(0\right)^{i}\right) $

Now assume $ f\left(x\right)=\sum_{j=0}^{k}a_{j}x^{j} $ then,

$ f\left(J_{n}\left(\lambda\right)\right)=\sum_{j=0}^{k}a_{j}\left(J_{n}\left(\lambda\right)^{j}\right)=\sum_{j=0}^{k}a_{j}\sum_{i=0}^{j}\binom{j}{i}\lambda^{j-i}\left(J_{n}\left(0\right)\right)^{i}=\sum_{j=0}^{k}\sum_{i=0}^{j}a_{j}\lambda^{j-i}\left(J_{n}\left(0\right)\right)^{i} $

Im not sure how to recognize the (j-i) deriviative out of the expression. And I'm not sure how to continue. Any ideas will help. Thanks in advance.

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  • $\begingroup$ Couldn't find the proof yet. A solution can be really helpful. $\endgroup$ – FreeZe May 31 at 14:14
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Hint

Your step 1 is OK.

Your step 2 also, but you're not using it completely. During the step 2, you're considering I imagine $f_k(x) = x^k$. Therefore

$$f_k^{(l)}(x)= \begin{cases} \frac{k!}{(k-l)!} x^{k-l} & \text{ for } 0 \le l \le k\\ 0 & \text{ else} \end{cases}$$

Use this and your step 1 to prove the expected formulae for the monomial $f_k(x) = x^k$.

Your last step could be a single sentence noticing that the requested formulae is linear in $f$ as the derivation is.

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  • $\begingroup$ Im sorry I dont understand how it is connected to the rows and columns of the matrix. the i,j term of the matrix should be equal to the (j-i) deriviative multiply by $ 1/{(j-i)!} $ and I cant see how to proceed with your hint. also I cant see why your hint is correct. I would say $ f_{k}^{(l)}\left(x\right)=\begin{cases} \frac{k!}{l!}x^{k-l} & 0\leq l\leq k\\ 0 & else \end{cases} $ rather than what you suggested $\endgroup$ – FreeZe May 30 at 15:15
  • $\begingroup$ I think that my formulae for the derivation of $f_k$ is correct. Take $l=0$ for example and then proceed by induction. Regarding the hint, use the formulae that you got at step 1 in your step 2 with the derivation formulae that I gave. $\endgroup$ – mathcounterexamples.net May 30 at 15:21
  • $\begingroup$ Do you think that for a general polynomial $ f_{k}\left(x\right)=\sum_{j=0}^{k}a_{j}x^{j} $ this formulae will work ? : $ f_{k}^{(l)}\left(x\right)=\begin{cases} \sum_{j=0}^{k}a_{j}\frac{j!}{(j-l)!}x^{k-l} & 0\leq j\leq l\leq k\\ 0 & else \end{cases}$ Also, the lower bound of the polynomial is 0 while the lower bound in the matrix would be 1 (the first row/column) how do you suggest fixing it? $\endgroup$ – FreeZe May 30 at 15:32
  • $\begingroup$ $ f\left(J_{n}\left(\lambda\right)\right)=\sum_{j=0}^{k}a_{j}\left(J_{n}\left(\lambda\right)\right)^{j}=\sum_{j=0}^{k}\sum_{i=0}^{j}\binom{j}{i}a_{j}\lambda^{j-i}J_{n}^{i}\left(0\right)=\sum_{i=0}^{j}\frac{1}{i!}f^{(i)}\left(\lambda\right)J_{n}^{i}\left(0\right) $ It does look closer to the form i want to get to. But still, I dont know how to proceed. can you elaborate ? $\endgroup$ – FreeZe May 30 at 15:57

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