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

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The discussion revolves around proving that the \( k \)-th Taylor polynomial \( T_k \) is the unique polynomial of degree \( k \) that satisfies specific conditions related to a continuously differentiable function \( f \). Participants explore how to compute partial derivatives of \( T_k \) and verify that they match those of \( f \) at a point \( x_0 \). The importance of proper notation and the application of the product rule in differentiation are emphasized to avoid confusion. The conversation also touches on the need for clarity in simplifications during derivative calculations. Ultimately, the group aims to establish the uniqueness of \( T_k \) through careful analysis of its properties.
  • #31
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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  • #32
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)
 
  • #33
Klaas van Aarsen said:
Yep. (Nod)

Great! Thank you! 🤩
 

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