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

mathmari

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Apr 14, 2013
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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 $i$ for 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 be $T_k$, right? But how do we do that exactly?


:unsure:
 

mathmari

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Apr 14, 2013
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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:
 

Klaas van Aarsen

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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. 🤔
 

mathmari

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Apr 14, 2013
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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:
 

Klaas van Aarsen

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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. 🤔
 

mathmari

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Apr 14, 2013
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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 be $x_{1}\cdot \ldots x_{m}$ 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:
 

mathmari

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Or have I understood that wrongly how to use here the partial derivative? :unsure:
 

Klaas van Aarsen

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Should at the end of the second term be $x_{1}\cdot \ldots x_{m}$ 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 define $f_{i_1\ldots i_m}=\frac{\partial}{\partial x_{i_1}}\cdots\frac{\partial}{\partial x_{i_m}}f(x_0)$ 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? 🤔
 

mathmari

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Apr 14, 2013
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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:
 

Klaas van Aarsen

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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:
 

mathmari

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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:
 

Klaas van Aarsen

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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)
 

mathmari

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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:
 

Klaas van Aarsen

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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)
 

mathmari

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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:


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:
 

Klaas van Aarsen

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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)

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:

mathmari

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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:
 

Klaas van Aarsen

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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? :rolleyes:
 
Last edited:

mathmari

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Klaas van Aarsen

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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 become $0$.
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:
 
Last edited:

mathmari

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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 become $0$.
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:
 

Klaas van Aarsen

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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 $\frac{\partial}{\partial x_1}\frac{\partial}{\partial x_1}\frac{\partial}{\partial x_2} T_3(0,0)$? 🤔
 

mathmari

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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:
 

Klaas van Aarsen

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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. :rolleyes:

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? 🤔
 

mathmari

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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?



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: