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Blandly, my question is: "is the $j$-th coefficient" a continuous function on power series?

More specifically, here is the setting I am bumping into this question. I can prove the following formula, for a power series $f$ with coefficients $a_i$ and $a_0 \neq 0$: $$ \det ([x^j]f^i)_{1 \leqslant i,j \leqslant n} = a_1^{\frac{n(n+1)}{2}}, $$ where $[x^j]$ denotes the $j$-th coefficient. I would like to extend it to $a_0 = 0$ by a kind of continuity argument, and this amounts to know if $[x^j]$ is continuous "for this limit". I do not think $[x^j]$ is continuous for power series for all norms, since it is an obviously unbounded linear operator. But can this $a_0 \to 0$ limit still work in this case?

Wolker
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1 Answers1

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The value of $$ \det ([x^j]f^i)_{1 \le i,j \le n} $$ does not depend on $a_0$: If $g(x) = f(x) + C$ for some constant $C$ then each row of $$ \det ([x^j]g^i)_{1 \le i,j \le n} $$ is equal to the corresponding row of $ \det ([x^j]f^i)_{1 \le i,j \le n} $ plus multiples of the previous rows. By elementary rules for the determinant it follows that $$ \det ([x^j]f^i)_{1 \le i,j \le n} = \det ([x^j]g^i)_{1 \le i,j \le n} \, , $$ which means that you can assume $a_0 \ne 0$ for your calculations.

Alternatively, note that your determinant is equal to $$ g_n := W\left(\frac{f'}{1!}, \frac{(f^2)'}{2!}, \ldots, \frac{(f^n)'}{n!}\right) $$ evaluated at $x=0$. Here $W$ denotes the Wronskian determinant. We have $$ \begin{align} g_n &= W\left(f', \frac{f f'}{1!}, \ldots, \frac{f^{n-1} f'}{(n-1)!}\right) \\ &\overset{(*)}{=} (f')^n \cdot W\left(1, \frac{f }{1!}, \ldots, \frac{f^{n-1} }{(n-1)!}\right) \\ &= (f')^n \cdot W\left(\frac{f' }{1!}, \ldots, \frac{(f^{n-1})' }{(n-1)!}\right)\\ &= (f')^n \cdot g_{n-1} \, . \end{align} $$ At $(*)$ we used the “product rule” for Wronsktians, see for example Why does the Wronskian satisfy $W(yy_1,\ldots,yy_n)=y^n W(y_1,\ldots,y_n)$?.

Also $g_1 = f'$, so that $$ g_n = (f')^{n(n+1)/2} $$ by induction.

Yet another way is to use the “chain rule for Wronskians”: $$ \begin{align} g_n(x) &= W\left(1, \frac{f}{1!}, \frac{f^2}{2!}, \ldots, \frac{f^n}{n!}\right)(x) \\ &= W\left(1, \frac{x}{1!}, \frac{x^2}{2!}, \ldots, \frac{x^n}{n!}\right)(f(x)) \cdot (f'(x))^{n(n+1)/2} \\ &= (f'(x))^{n(n+1)/2} \end{align} $$ because $ W\left(1, \frac{x}{1!}, \frac{x^2}{2!}, \ldots, \frac{x^n}{n!}\right) = 1$.

Martin R
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