# How to show that T_k is the only polynomial of degree k with the specific properties?

• MHB
• mathmari
In summary, the conversation discusses the properties of the $k$-th Taylor polynomial of a $k$-times continuously differentiable function $f$ in an open set $U \subset \mathbb{R}^n$, with a fixed point $x_0 \in U$. The polynomial is defined as $T_k(x_1, \ldots, x_n) = \sum_{m=0}^k \frac{1}{m!} \sum_{i_1=1}^n \ldots \sum_{i_m=1}^n \frac{\partial}{\partial{x_{i_1}}} \ldots \frac{\partial}{\partial{x_{i_m}}} mathmari Gold Member MHB Hey! :giggle: Let$U\subset \mathbb{R}^n$be an open set and$f:U\rightarrow \mathbb{R}$is a$k$-times continusouly differentiable function. Let$x_0\in U$be fixed. The$k$-th Taylor polynomial of$f$in$x_0$is $$T_k(x_1,\ldots ,x_n)=\sum_{m=0}^k\frac{1}{m!}\sum_{i_1=1}^n \ldots \sum_{i_m=1}^n \frac{\partial}{\partial{x_{i_1}}}\ldots \frac{\partial}{\partial{x_{i_m}}}f(x_0)\cdot x_{i_1}\cdot \ldots x_{i_m}$$ Show that$T_k(x_1,\ldots ,x_n)$is the only polynomial of degree$k$such that $$T_k(0)=f(x_0) \\ \frac{\partial}{\partial{x_{i_1}}}\ldots \frac{\partial}{\partial{x_{i_m}}}T_k(0)=\frac{\partial}{\partial{x_{i_1}}}\ldots \frac{\partial}{\partial{x_{i_m}}}f(x_0)$$ for all$m\in\{1,\ldots , k\}$,$i_\ell\in \{1, \ldots ,n\}$and$\ell \in \{1, \ldots , m\}$. To show that$T_k$satisfies these properties we just have to replace$x_i=0$for all$ifor the first property and for the second one we have to calculate these partial derivatives, right? I am trying to show that $$\frac{\partial}{\partial{x_{i_1}}}\ldots \frac{\partial}{\partial{x_{i_m}}}T_k(0)=\frac{\partial}{\partial{x_{i_1}}}\ldots \frac{\partial}{\partial{x_{i_m}}}f(x_0)$$ Is it maybe as follows? \begin{align*}\frac{\partial}{\partial{x_{i_1}}}T_k(x_1,\ldots ,x_n)&=\frac{\partial}{\partial{x_{i_1}}}\left [\sum_{m=0}^k\frac{1}{m!}\sum_{i_1=1}^n \ldots \sum_{i_m=1}^n \frac{\partial}{\partial{x_{i_1}}}\ldots \frac{\partial}{\partial{x_{i_m}}}f(x_0)\cdot x_{i_1}\cdot \ldots x_{i_m}\right ]\\ & =\sum_{m=0}^k\frac{1}{m!}\sum_{i_1=1}^n \ldots \sum_{i_m=1}^n \frac{\partial}{\partial{x_{i_1}}}\ldots \frac{\partial}{\partial{x_{i_m}}}f(x_0)\cdot \frac{\partial}{\partial{x_{i_1}}}\left (x_{i_1}\cdot \ldots x_{i_m}\right )\\ & =\sum_{m=0}^k\frac{1}{m!}\sum_{i_1=1}^n \ldots \sum_{i_m=1}^n \frac{\partial}{\partial{x_{i_1}}}\ldots \frac{\partial}{\partial{x_{i_m}}}f(x_0)\cdot x_{i_2}\cdot \ldots x_{i_m}\end{align*} Or how do we calculate the partial derivatives? To show the uniqueness, we have toconsider that there is also an other polynomial with the specific properties and then we have toshow that this polynomial must beT_k$, right? But how do we do that exactly? :unsure: Or do we use the product rule at the calculation of$\frac{\partial}{\partial{x_{i_1}}}T_k(x_1,\ldots ,x_n)$? :unsure: Hey mathmari! First off, we should not use the same symbol twice if they are not the same. The$i_1$in the condition is different from the one that is an index in the summation. Perhaps we should start with$\frac\partial{\partial x_1}$. Then we have to indeed apply the product rule. Klaas van Aarsen said: First off, we should not use the same symbol twice if they are not the same. The$i_1$in the condition is different from the one that is an index in the summation. Perhaps we should start with$\frac\partial{\partial x_1}. Then we have to indeed apply the product rule. Do you mean it as follows? \begin{align*}&\frac{\partial}{\partial{x_1}}T_k(x_1,\ldots ,x_n)=\frac{\partial}{\partial{x_1}}\left [\sum_{m=0}^k\frac{1}{m!}\sum_{i_1=1}^n \ldots \sum_{i_m=1}^n \frac{\partial}{\partial{x_{i_1}}}\ldots \frac{\partial}{\partial{x_{i_m}}}f(x_0)\cdot x_{i_1}\cdot \ldots x_{i_m}\right ]\\ & =\frac{\partial}{\partial{x_1}}\left [\sum_{m=0}^k\frac{1}{m!}\sum_{i_1=1}^n \ldots \sum_{i_m=1}^n \frac{\partial}{\partial{x_{i_1}}}\ldots \frac{\partial}{\partial{x_{i_m}}}f(x_0)\right ]\cdot x_{i_1}\cdot \ldots x_{i_m}+\sum_{m=0}^k\frac{1}{m!}\sum_{i_1=1}^n \ldots \sum_{i_m=1}^n \frac{\partial}{\partial{x_{i_1}}}\ldots \frac{\partial}{\partial{x_{i_m}}}f(x_0)\cdot \frac{\partial}{\partial{x_1}}\left [x_{i_1}\cdot \ldots x_{i_m}\right ]\end{align*} :unsure: The summation in the first term is a constant, so its derivative is 0. We can still apply the product rule in the second term. Klaas van Aarsen said: The summation in the first term is a constant, so its derivative is 0. We can still apply the product rule in the second term. Should at the end of the second term bex_{1}\cdot \ldots x_{m}$instead of$x_{i_1}\cdot \ldots x_{i_m}? Do you mean to be as follows? \begin{align*}\frac{\partial}{\partial{x_1}}T_k(x_1,\ldots ,x_n)&=\frac{\partial}{\partial{x_1}}\left [\sum_{m=0}^k\frac{1}{m!}\sum_{i_1=1}^n \ldots \sum_{i_m=1}^n \frac{\partial}{\partial{x_{i_1}}}\ldots \frac{\partial}{\partial{x_{i_m}}}f(x_0)\cdot x_{1}\cdot x_2\cdot \ldots x_{m}\right ]\\ & =\sum_{m=0}^k\frac{1}{m!}\sum_{i_1=1}^n \ldots \sum_{i_m=1}^n \frac{\partial}{\partial{x_{i_1}}}\ldots \frac{\partial}{\partial{x_{i_m}}}f(x_0)\cdot \frac{\partial}{\partial{x_1}}\left [x_{1}\cdot x_2\cdot \ldots x_{m}\right ] \\ & =\sum_{m=0}^k\frac{1}{m!}\sum_{i_1=1}^n \ldots \sum_{i_m=1}^n \frac{\partial}{\partial{x_{i_1}}}\ldots \frac{\partial}{\partial{x_{i_m}}}f(x_0)\cdot \left [x_{2}\cdot \ldots x_{m}\right ]\end{align*} :unsure: Or have I understood that wrongly how to use here the partial derivative? :unsure: mathmari said: Should at the end of the second term bex_{1}\cdot \ldots x_{m}$instead of$x_{i_1}\cdot \ldots x_{i_m}? No. Otherwise the terms would consist of only a small subset of the possible factorizations. (Shake) Instead it should be like \begin{align*}\frac{\partial}{\partial{x_1}}T_k(x_1,\ldots ,x_n) & =\sum_{m=0}^k\frac{1}{m!}\sum_{i_1=1}^n \ldots \sum_{i_m=1}^n \frac{\partial}{\partial{x_{i_1}}}\ldots \frac{\partial}{\partial{x_{i_m}}}f(x_0)\cdot \frac{\partial}{\partial{x_1}}\left [x_{i_1}\cdot x_{i_2}\cdot \ldots \cdot x_{i_m}\right ] \\ & =\sum_{m=0}^k\frac{1}{m!}\sum_{i_1=1}^n \ldots \sum_{i_m=1}^n \frac{\partial}{\partial{x_{i_1}}}\ldots \frac{\partial}{\partial{x_{i_m}}}f(x_0)\cdot \left [ \delta_{1i_1}\cdot x_{i_2}\cdot \ldots \cdot x_{i_m} + x_{i_1}\cdot \delta_{1i_2}\cdot \ldots\cdot x_{i_m} + \ldots + x_{i_1}\cdot x_{i_2}\cdot \ldots \cdot \delta_{1i_m} \right ] \\ \end{align*} Anyway, let's take a look at an example. First, let's definef_{i_1\ldots i_m}=\frac{\partial}{\partial x_{i_1}}\cdots\frac{\partial}{\partial x_{i_m}}f(x_0)$to make the expressions a bit shorter. Then: $$T_3(x_1,x_2)=\frac 1{0!}f(x_0) + \frac 1{1!}(f_1 x_1 + f_2 x_2) + \frac 1{2!}(f_{11}x_1x_1+f_{12}x_1x_2 + f_{21}x_2x_1+f_{22}x_2x_2) +\frac 1{3!}(f_{111}x_1x_1x_1 + f_{112}x_1x_1x_2+\ldots + f_{222}x_2x_2x_2)$$ yes? Now what will$\frac{\partial}{\partial x_1}\frac{\partial}{\partial x_1}\frac{\partial}{\partial x_2} T_3(0)$be? Klaas van Aarsen said: Then: $$T_3(x_1,x_2)=\frac 1{0!}f(x_0) + \frac 1{1!}(f_1 x_1 + f_2 x_2) + \frac 1{2!}(f_{11}x_1x_1+f_{12}x_1x_2 + f_{21}x_2x_1+f_{22}x_2x_2) +\frac 1{3!}(f_{111}x_1x_1x_1 + f_{112}x_1x_1x_2+\ldots + f_{222}x_2x_2x_2)$$ yes? Now what will$\frac{\partial}{\partial x_1}\frac{\partial}{\partial x_1}\frac{\partial}{\partial x_2} T_3(0)$be? It will be equal to$0$, or not? :unsure: mathmari said: It will be equal to$0$, or not? Nope. (Shake) According to the problem statement, we're supposed to prove that it is equal to$f_{112}$. :sneaky: Klaas van Aarsen said: Nope. (Shake) According to the problem statement, we're supposed to prove that it is equal to$f_{112}$. :sneaky: Is the first partial derivative as for$x_2$the following? $$\frac{\partial}{\partial x_2}T_3= (f_{12} x_1 + f_{22} x_2+f_2) + \frac 1{2}(f_{112}x_1x_1+f_{122}x_1x_2+f_{12}x_1 + f_{212}x_2x_1+f_{21}x_1+f_{222}x_2x_2+2f_{22}x_2) +\frac 1{6}(f_{1112}x_1x_1x_1 + f_{1122}x_1x_1x_2+f_{112}x_1x_1+\ldots + f_{2222}x_2x_2x_2+3f_{222}x_2x_2)$$ :unsure: mathmari said: Is the first partial derivative as for$x_2$the following? $$\frac{\partial}{\partial x_2}T_3= (f_{12} x_1 + f_{22} x_2+f_2) + \frac 1{2}(f_{112}x_1x_1+f_{122}x_1x_2+f_{12}x_1 + f_{212}x_2x_1+f_{21}x_1+f_{222}x_2x_2+2f_{22}x_2) +\frac 1{6}(f_{1112}x_1x_1x_1 + f_{1122}x_1x_1x_2+f_{112}x_1x_1+\ldots + f_{2222}x_2x_2x_2+3f_{222}x_2x_2)$$ No. (Shake) Suppose we have the function$f$given by$f(x_1,x_2) = x_1x_2$and we have some point$(a_1,a_2)$. Let$g$be the function given by$g(x_1,x_2)=f(a_1,a_2)$. Then what is$\frac{\partial}{\partial x_2}f(a_1, a_2)$? (Wondering) And what is$\frac{\partial}{\partial x_2}g(a_1, a_2)$? (Wondering) Klaas van Aarsen said: No. (Shake) Suppose we have the function$f$given by$f(x_1,x_2) = x_1x_2$and we have some point$(a_1,a_2)$. Let$g$be the function given by$g(x_1,x_2)=f(a_1,a_2)$. Then what is$\frac{\partial}{\partial x_2}f(a_1, a_2)$? (Wondering) And what is$\frac{\partial}{\partial x_2}g(a_1, a_2)$? (Wondering) The second one is the derivative of a constant function, so$0$. The first one first we calculate the derivative then we substitute the point. :unsure: mathmari said: The second one is the derivative of a constant function, so$0$. The first one first we calculate the derivative then we substitute the point. Indeed. (Nod) So what do we get when we apply that to$\frac{\partial}{\partial x_2}T_3(x_1,x_2)$? (Wondering) And what do we get for$\frac{\partial}{\partial x_2}T_3(0,0)$? (Wondering) Klaas van Aarsen said: So what do we get when we apply that to$\frac{\partial}{\partial x_2}T_3(x_1,x_2)$? (Wondering) Isn't this what I have done in #11 ? :unsure: Klaas van Aarsen said: And what do we get for$\frac{\partial}{\partial x_2}T_3(0,0)$? (Wondering) If$\frac{\partial}{\partial x_2}T_3(x_1,x_2)$is correct in #11, then $$\frac{\partial}{\partial x_2}T_3(0,0)=f_2$$ , right? :unsure: mathmari said: Isn't this what I have done in #11 ? In #11 you've been taking the partial derivative of$f_{i_1\ldots i_m}$with respect to$x_2$. But we defined$f_{i_1\ldots i_m}=\frac{\partial}{\partial x_{i_1}}\cdots\frac{\partial}{\partial x_{i_m}}f(x_0)$, which is a constant and independent of$x_2$. (Worried) mathmari said: If$\frac{\partial}{\partial x_2}T_3(x_1,x_2)$is correct in #11, then $$\frac{\partial}{\partial x_2}T_3(0,0)=f_2$$ , right? It is at least the correct result. (Wink) Last edited: Klaas van Aarsen said: In #11 you've been taking the partial derivative of$f_{i_1\ldots i_m}$with respect to$x_2$. But we defined$f_{i_1\ldots i_m}=\frac{\partial}{\partial x_{i_1}}\cdots\frac{\partial}{\partial x_{i_m}}f(x_0)$, which is a constant and independent of$x_2$. (Worried)It is at least the correct result. (Nod) I got stuck right now. If$f_{i_1\ldots i_m}=\frac{\partial}{\partial x_{i_1}}\cdots\frac{\partial}{\partial x_{i_m}}f(x_0)$is a constant how can the result$f_2$be correct? :unsure: mathmari said: I got stuck right now. If$f_{i_1\ldots i_m}=\frac{\partial}{\partial x_{i_1}}\cdots\frac{\partial}{\partial x_{i_m}}f(x_0)$is a constant how can the result$f_2$be correct? Can we evaluate$\frac{\partial}{\partial x_2}T_3(x_1,x_2)$again while treating$f_{i_1\ldots i_m}as a constant? Oh, and can we do simplifications in a different step than the one where we find the partial derivative? Last edited: Klaas van Aarsen said: Oh, and can we do simplifications in a different step than the one where we find the partial derivative? What do you mean? :unsure: mathmari said: What do you mean? I meant like: \begin{align*}\frac{\partial}{\partial x_2} T_3(x_1,x_2)&=\frac 1{1!}(f_2) + \frac 1{2!}(f_{12}x_1 + f_{21}x_1+2f_{22}x_2) +\frac 1{3!}(f_{112}x_1x_1+ f_{121}x_1x_1+ f_{211}x_1x_1+ 2f_{122}x_1x_2+ 2f_{212}x_1x_2+2f_{221}x_1x_2+ 3f_{222}x_2^2) \\ &=f_2 + f_{12}x_1 + f_{22}x_2 +\frac 1{2}f_{112}x_1^2+ f_{122}x_1x_2+ \frac 12 f_{222}x_2^2 \end{align*} The first step takes the partial derivative and only that. I've only left out the terms that become0$. That way it's easier to spot any patterns that apply to the more general formula. Not to mention that it makes it easier to check/correct where mistakes are if there are any. :geek: It's also easier for a reader to follow what is happening. The second step takes care of the simplifications. In this case that includes that e.g.$f_{12}=f_{21}$, which follows from the fact that$f$is$k=3times continuously differentiable. See how we get: $$\frac{\partial}{\partial x_2} T_3(0,0)=f_2$$ now? :unsure: Last edited: Klaas van Aarsen said: I meant like: \begin{align*}\frac{\partial}{\partial x_2} T_3(x_1,x_2)&=\frac 1{1!}(f_2) + \frac 1{2!}(f_{12}x_1 + f_{21}x_1+2f_{22}x_2) +\frac 1{3!}(f_{112}x_1x_1+ f_{121}x_1x_1+ f_{211}x_1x_1+ 2f_{122}x_1x_2+ 2f_{212}x_1x_2+2f_{221}x_1x_2+ 3f_{222}x_2^2) \\ &=f_2 + f_{12}x_1 + f_{22}x_2 +\frac 1{2}f_{112}x_1^2+ f_{122}x_1x_2+ \frac 12 f_{222}x_2^2 \end{align*} The first step takes the partial derivative and only that. I've only left out the terms that become0$. That way it's easier to spot any patterns that apply to the more general formula. Not to mention that it makes it easier to check/correct where mistakes are if there are any. :geek: It's also easier for a reader to follow what is happening. The second step takes care of the simplifications. In this case that includes that e.g.$f_{12}=f_{21}$, which follows from the fact that$f$is$k=3$times continuously differentiable. See how we get: $$\frac{\partial}{\partial x_2} T_3(0,0)=f_2$$ now? :unsure: I understand now this specific example! But I haven't really understood how to get a general formula for the initial exercise statement. :unsure: mathmari said: I understand now this specific example! But I haven't really understood how to get a general formula for the initial exercise statement. How is it that we got only$f_2$from all those terms of$\frac{\partial}{\partial x_2} T_3(x_1,x_2)$? What is special about the term that we got$f_2$from? What are the reasons that all the other terms vanished? And getting back to my original question about the example I gave, what is$\frac{\partial}{\partial x_1}\frac{\partial}{\partial x_1}\frac{\partial}{\partial x_2} T_3(0,0)$? Klaas van Aarsen said: How is it that we got only$f_2$from all those terms of$\frac{\partial}{\partial x_2} T_3(x_1,x_2)$? What is special about the term that we got$f_2$from? What are the reasons that all the other terms vanished? All other terms except$f_2$are multipliedby some$x_i-term and if we calculate that in the point $(0,0)$ these ones vanish, right? :unsure:

mathmari said:
All other terms except $f_2$ are multipliedby some $x_i$-term and if we calculate that in the point $(0,0)$ these ones vanish, right?
That is part of it.

Let's go back to the expression:
$$T_3(x_1,x_2)=\frac 1{0!}f(x_0) + \frac 1{1!}(f_1 x_1 + f_2 x_2) + \frac 1{2!}(f_{11}x_1x_1+f_{12}x_1x_2 + f_{21}x_2x_1+f_{22}x_2x_2) +\frac 1{3!}(f_{111}x_1x_1x_1 + f_{112}x_1x_1x_2+\ldots + f_{222}x_2x_2x_2)$$
It has a term $\frac 1{0!}f(x_0)$. What happened to it? Why did it vanish?
It also had 2nd and higher order terms in it. Those vanished because after differentiating once, they still had at least some $x_i$ in them.
What happened to the 1st order terms? They also had an $x_i$ in them, didn't they?

Klaas van Aarsen said:
That one had a term $\frac 1{0!}f(x_0)$. What happened to it? Why did it vanish?

This one vanished because this is a constant, isn't it?
Klaas van Aarsen said:
It also had 2nd and higher order terms in it. Those vanished because after differentiating once, they still had at least some $x_i$ in them.
What happened to the 1st order terms? They also had an $x_i$ in them, didn't they?

These vanished after the calculation at the zero-vector because these are still multiplied by $x_i$ terms, right?
:unsure:

mathmari said:
This one vanished because this is a constant, isn't it?

In this case, yes.
We still need a more specific pattern if we evaluate for instance $\frac{\partial}{\partial x_1}\frac{\partial}{\partial x_1}\frac{\partial}{\partial x_2} T_3(0,0)$.

mathmari said:
These vanished after the calculation at the zero-vector because these are still multiplied by $x_i$ terms, right?
This applies to the 2nd and higher order terms that were still there after differentiation.
But what happened to the 1st order terms? Those did not have an $x_i$ in them after differentiation, did they?
Some of the other 2nd and higher order terms vanished for a different reason as well, didn't they?

Klaas van Aarsen said:
In this case, yes.
We still need a more specific pattern if we evaluate for instance $\frac{\partial}{\partial x_1}\frac{\partial}{\partial x_1}\frac{\partial}{\partial x_2} T_3(0,0)$. This applies to the 2nd and higher order terms that were still there after differentiation.
But what happened to the 1st order terms? Those did not have an $x_i$ in them after differentiation, did they?
Some of the other 2nd and higher order terms vanished for a different reason as well, didn't they?

\begin{align*}T_3(x_1,x_2)=\frac 1{0!}f(x_0) + \frac 1{1!}(f_1 x_1 + f_2 x_2) + \frac 1{2!}(f_{11}x_1x_1+f_{12}x_1x_2 + f_{21}x_2x_1+f_{22}x_2x_2)
+\frac 1{3!}(f_{111}x_1x_1x_1 + f_{112}x_1x_1x_2+\ldots + f_{222}x_2x_2x_2)
\end{align*}

After the differentiation :
- The term $f(x_0)$ vanishes because it is a constant.
- The term $f_1$ vanishes because it is multiplied by a constant (since it is not the differentiation variable). The same holds for $f_{11}$ and $f_{111}$. :unsure:

mathmari said:
After the differentiation :
- The term $f(x_0)$ vanishes because it is a constant.
- The term $f_1$ vanishes because it is multiplied by a constant (since it is not the differentiation variable). The same holds for $f_{11}$ and $f_{111}$.
That works for $\frac{\partial}{\partial x_2} T_3(0,0)$. (Nod)
So some terms vanish due to the differentiation, and some terms vanish because we substitute $(0,0)$.

Now let's take a look at, say, $\frac{\partial}{\partial x_1}\frac{\partial}{\partial x_2} T_3(0,0)$.
Terms with $x_1$ or $x_2$ in them will still vanish even though they are not constants - at least not for both partial differentiations.
Can we categorize them? Find more general patterns?

Note that we are working towards the patterns with respect to $\frac{\partial}{\partial{x_{i_1}}}\ldots \frac{\partial}{\partial{x_{i_m}}}T_k(0)$.

Klaas van Aarsen said:
Now let's take a look at, say, $\frac{\partial}{\partial x_1}\frac{\partial}{\partial x_2} T_3(0,0)$.
Terms with $x_1$ or $x_2$ in them will still vanish even though they are not constants - at least not for both partial differentiations.
Can we categorize them? Find more general patterns?

Ah do we write for this reason the expression with $\delta_{ij}$ as you wrote in #8 ? I mean if the differentiation variable is multiplied with the f-term or not?

:unsure:

mathmari said:
Ah do we write for this reason the expression with $\delta_{ij}$ as you wrote in #8 ? I mean if the differentiation variable is multiplied with the f-term or not?

Not really.
I was merely trying to demonstrate what happens when we take a partial derivative of $T_k(x_1,\ldots,x_n)$ and more specifically of $x_{i_1}\cdot\ldots\cdot x_{i_m}$. (Tauri)

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...

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:

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!

## 1. How do you prove that Tk is the only polynomial of degree k with the specific properties?

To prove that Tk is the only polynomial of degree k with the specific properties, you need to show that it satisfies all the necessary conditions and that no other polynomial of degree k can satisfy those conditions. This can be done through mathematical induction or by using the properties of Tk to derive a contradiction.

## 2. What are the specific properties of Tk that make it unique?

The specific properties of Tk include being a polynomial of degree k, having k distinct roots, and having a leading coefficient of 1. These properties set Tk apart from other polynomials of degree k.

## 3. Can you use a different method to show that Tk is the only polynomial of degree k with the specific properties?

Yes, there are other methods that can be used to show that Tk is the only polynomial of degree k with the specific properties. Some possible methods include using the fundamental theorem of algebra, using the properties of roots and coefficients of polynomials, or using the properties of Tk to prove uniqueness.

## 4. Why is it important to show that Tk is the only polynomial of degree k with the specific properties?

Showing that Tk is the only polynomial of degree k with the specific properties is important because it helps to establish the uniqueness of Tk and its properties. This can be useful in various mathematical proofs and applications where Tk is used.

## 5. Are there any real-life applications for proving that Tk is the only polynomial of degree k with the specific properties?

Yes, there are real-life applications for proving that Tk is the only polynomial of degree k with the specific properties. For example, in engineering and physics, Tk can be used to model various phenomena, and proving its uniqueness can help ensure the accuracy and reliability of these models.

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