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We know that $(a+b)^n=a^n+\binom{n}{1}a^{n-1}b+\cdots+b^n$ when $a,b$ commutes to each other.

Am I right?

My question is-

Can we expand $(\partial _x+f(x))^np(x)$ or $(\partial _x+f(x))^n$ in the above formula ?

i.e., can we have $(\partial _x+f(x))^n=\partial _x^n+\binom{n}{1} (\partial _x)^{n-1}f(x)+\cdots+(f(x))^n$ ?

Note that if the field is of characteristic $n$ (prime), then we can have equality.

So when $(\partial _x+f(x))^n$ can be expanded ?

Some hints is given here Formal series expansion of differential operator (d/dx + f(x))^n

MAS
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  • The way to see this is by induction. Check what happens when the operator is applied twice, thrice, and try to note the mechanism. – Allawonder Sep 17 '19 at 17:37
  • No, we cannot plug $a = \partial_x$ and $b = f\left(x\right)$ into the binomial formula, since $a$ and $b$ needs to commute ($ab=ba$) for the binomial formula to hold. – darij grinberg Sep 17 '19 at 17:39
  • @Allawonder, I think that way says $NO$. But is there any way to expand similarly? – MAS Sep 17 '19 at 17:40
  • @M.A.SARKAR If it's no, then that's that. What do you mean by expand similarly? – Allawonder Sep 17 '19 at 17:42
  • @darijgrinberg, So how to consider series like $\sum (\partial_x+f(x))^np(x)x^n$ and its convergence or divergence? – MAS Sep 17 '19 at 17:42
  • @darijgrinberg, I mean if we consider a field of characteristic $n$ (prime) then obviously the equality holds – MAS Sep 17 '19 at 17:43
  • @Allawonder,I mean if we consider a field of characteristic $n$ (prime) then obviously the equality holds – MAS Sep 17 '19 at 17:43
  • @M.A.SARKAR: I highly doubt that it holds for characteristic $n$. – darij grinberg Sep 17 '19 at 17:49

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