This is true, but I’m looking for an explicit, non-recursive formula that needs to handle the general case of the kth anti-derivative (instead of just the first).
The solution involves doing something funny with formal power series, like in this post.
Promote f from a function to a linear operator on the space of functions, F. The action of this operator is just “multiply by f”. We’ll similarly define F∼,F∼2 meaning to multiply by the first, second integral of f, etc.
Observe:
IF=F∼−IF∼D
IF=F∼−F∼2D+F∼3D2−⋯
Now we can calculate what we get when applying k times. The calculation simplifies when we note that all terms are of the form F∼a(−D)(a−k). Result:
IkF=∞∑j=k(j−1k−1)F∼j(−D)j−k
Now we apply the above operator to p:
IkFp=∞∑j=k(j−1k−1)F∼j(−D)j−kp
Ik(fp)=∞∑j=k(j−1k−1)(Ijf)(−D)j−kp
The sum terminates because a polynomial can only have finitely many derivatives.
Very nice! Notice that if you write r=j−k,I as D−1, and play around with binomial coefficients a bit, we can rewrite this as:
D−k(fp)=∑∞r=0(−kr)(D−k−rf)(Drp)
which holds for k<0 as well, in which case it becomes the derivative product rule. This also matches the formal power series expansion of (x+y)−k, which one can motivate directly
Use integration by parts:
I(pf)=pIf−I((Dp)(If))
Then Dp is another polynomial (of smaller degree), and If is another “nice” function, so we recurse.
This is true, but I’m looking for an explicit, non-recursive formula that needs to handle the general case of the kth anti-derivative (instead of just the first).
The solution involves doing something funny with formal power series, like in this post.
Heh, sure.
Promote f from a function to a linear operator on the space of functions, F. The action of this operator is just “multiply by f”. We’ll similarly define F∼,F∼2 meaning to multiply by the first, second integral of f, etc.
Observe:
IF=F∼−IF∼D
IF=F∼−F∼2D+F∼3D2−⋯
Now we can calculate what we get when applying k times. The calculation simplifies when we note that all terms are of the form F∼a(−D)(a−k). Result:
IkF=∞∑j=k(j−1k−1)F∼j(−D)j−k
Now we apply the above operator to p:
IkFp=∞∑j=k(j−1k−1)F∼j(−D)j−kp
Ik(fp)=∞∑j=k(j−1k−1)(Ijf)(−D)j−kp
The sum terminates because a polynomial can only have finitely many derivatives.
Very nice! Notice that if you write r=j−k, I as D−1, and play around with binomial coefficients a bit, we can rewrite this as:
D−k(fp)=∑∞r=0(−kr)(D−k−rf)(Drp)
which holds for k<0 as well, in which case it becomes the derivative product rule. This also matches the formal power series expansion of (x+y)−k, which one can motivate directly
(By the way, how do you spoiler tag?)
Oh, very cool, thanks! Spoiler tag in markdown is: