Why is \( T_k \) the Unique Polynomial of Degree \( k \) with These Properties?

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Klaas van Aarsen said:
Anyway, consider $\frac{\partial}{\partial{x_{i_1}}}\ldots \frac{\partial}{\partial{x_{i_m}}}T_k(0)$.
So we differentiate $m$ times.
It means that all terms with an order lower than $m$ vanish due to differentiation.
And all terms with an order higher than $m$ that did not already vanish due to differentiation, will vanish when we substitute $0$.
That leaves the terms of order $m$, and only those that are a match with $x_{i_1}\cdot\ldots\cdot x_{i_m}$ will remain.
That is, all permutations of it.
And there are $m!$ permutations... 🤔
Ahh I see! And so the second property for $T_k$ follows!

For the uniqueness, we have to show that the coefficients of $T_k$ are uniquely defined, right? :unsure:
 
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mathmari said:
Ahh I see! And so the second property for $T_k$ follows!

For the uniqueness, we have to show that the coefficients of $T_k$ are uniquely defined, right?

Yep. (Nod)
 
Klaas van Aarsen said:
Yep. (Nod)

Great! Thank you! 🤩