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I had a problem in my booked i tried to prove. Here is the problem

"Let $x_1,\ldots,x_n$ be different real numbers and $y_1,\ldots,y_n,s_1,\ldots,s_n$ some real numbers. Prove that there exists a polynomial $p(x)$ of a degree less than $2n$ such that $p(x_i)=y_i$ and $p'(x_i)=s_i$ for every $i=1,2,\ldots,n.$"

Here is my attempt:

Let $V= \{ p(x)\in \mathbb{R}[x] \mid \deg(p(x))<2n\}$ $\Rightarrow$ $\exists$ $h(x)\in V$ s.t $\deg(h(x))<n<2n$ from here we now can apply Lagrange Interpolation theorem. $\Rightarrow$ $h(x_i)=y_i$.

Let $Q= \{ g(x)\in \mathbb{R}[x] \mid g(x)=p'(x),\ p(x) \in V, \deg(p(x)) < n \}$

where $\dim(Q)=n-1$.

Now let $g(x)\in Q \Rightarrow \deg(g(x))<n-1$

Let $A: Q\rightarrow \mathbb{R}^n$

$g(x) \rightarrow (g(x_1),\ldots,g(x_n))$

Now assume $g(x)\in \ker(A) \Rightarrow g(x_i) = 0$, $n$ zeros $\Rightarrow \ g(x)=0 \ \Rightarrow \ker(A) = \{0\} \Rightarrow A$ injective $+$surjective $\Rightarrow g(x_i)=s_i$

But since $\deg(h(x))<n \Rightarrow h'(x) \in Q \Rightarrow g(x)=h'(x) \Rightarrow g(x_i)=h'(x_i)=s_i$

My Professor had i quick look and stated that this was a good attempt but it was not correct. He said that my proof does not with certainty show that this $h'(x_i)=s_i$ ill get in the end will fullfill $h(x_i)=y_i$

I didnt understand him, because in my opinion im sure that the derivitive of $h(x)$ which fullfill $h(x_i)=y_i$ lies in $Q$ and therefor i can let my $g(x)$ be equal to $h'(x)$

I would be grateful if someone can explain this?

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    $\begingroup$ I edited to include proper use of \dim, \deg, \ker,, \ldots, \mid, and some other things. At one point you manually added a small space before $\deg$. If you use \deg, then you automatically get proper spacing to the left and right in things like $a\deg b$ and to the left in things like $a\deg(b)$. The software is quite sophisticated on matters of when to add that space and when not to (and similarly with \ker, \dim, \cos, \log, \det, \max, \sup, etc.). You shouldn't keep alternating in and out of MathJax in things like $\deg(h(x))<n \Rightarrow h'(x)\in Q$ etc. ${}\qquad{}$ $\endgroup$ – Michael Hardy Oct 1 '15 at 19:27
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Define $T:\mathbb{R}[x]\rightarrow \mathbb{R}^{2n}$, $T(p(x))=(p(x_1),\ldots,p(x_n), p'(x_1),\ldots,p'(x_n))$.

Notice that if $p(x)\in \ker(T)$ then $x_1,\ldots,x_n$ are roots of $p(x)$ with multiplicity bigger or equal to $2$, since they are distinct the degree of $p(x)$ is bigger than 2n-1.

So if $P_{2n-1}$ is the subspace of polynomials with degree smaller or equal to 2n-1 then $T:P_{2n-1}\rightarrow \mathbb{R}^{2n}$ is injective and therefore surjective.

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  • $\begingroup$ You say that $p(x)$ has a degree bigger than $2n-1$. But my problem states that the degree of $p(x)$ is less then $2n$. So If I understand you correct. You conclude that if $p(x) \in \ker (A)$ then $\deg (p(x)) > 2n-1 \Rightarrow p(x) \not \in \ker (A)$ (Contradiction). Do I understand you correct? $\endgroup$ – Olba12 Oct 2 '15 at 6:34
  • $\begingroup$ @Olba12 If $p(x)\in \ker(T)$ then $p(x)$ does not belong to $P_{2n-1}$. Thus, if we restrict the domain of $T$ to $P_{2n-1}$, $T:P_{2n-1}\rightarrow \mathbb{R}^{2n}$, we obtain an injective map. Do you agree? Since $\dim(P_{2n-1})=2n$ then $T:P_{2n-1}\rightarrow \mathbb{R}^{2n}$ is surjective. Now, pick any vector $v=(y_1,\ldots, y_n,s_1,\ldots,s_n)\in\mathbb{R}^{2n}$. There is $q(x)\in P_{2n-1}$(so the degree of $q(x)$ is smaller than 2n) such that $T(q(x))=v$. This $q(x)$ is the required one. $\endgroup$ – Daniel Oct 2 '15 at 10:15
  • $\begingroup$ Yes I understand, thank you! $\endgroup$ – Olba12 Oct 4 '15 at 8:52
  • $\begingroup$ @Olba12 You are welcome $\endgroup$ – Daniel Oct 4 '15 at 13:43
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At first I didn't understand your proof at all, but here goes.

1) In the definition of $Q$, it is the set of all $G(x)$, but then you don't mention $G$ again, although I suppose we can assume it is a typo for $g(x)$. So essentially as far as I can tell, $Q = \{p'(x) : \deg p < n-1\}$. Since $V$ contains all polynomials of degree less than $2n$, if we say $\deg p < n-1$, then of course $p \in V$. But now notice that $p\mapsto p'$ has a one dimensional kernel, so that $Q$ has dimension $n-2$ not dimension $n-1$.

2) Then you define the map $A:Q \to \mathbb{R}^n$. You correctly show that the only polynomial in the kernel is 0, although you then say that $\ker A = \varnothing$, which is impossible, the kernel always contains the identity (0). Then you say that this implies $A$ is injective and surjective, but although $A$ is injective, it cannot be surjective, since $\dim Q=n-2< \dim\mathbb{R}^n=n$. Even if $\dim Q$ were $n-1$, this would still be impossible.

3) You then say that since $\deg h < n$, $h' \in Q$. Which makes me believe you meant $Q =\{ p' : \deg p < n\}$, which would be a vector space of dimension $n-1$, which does not invalidate the problem in (2). But you then claim that since $h'\in Q$, the $g(x)$ you found before must be $h'(x)$. This step makes no sense. Why on earth would these two different things be equal? Ah I see, your final question explains your thought process. The answer is that $g$ is a specific polynomial, not a variable. You have found a specific polynomial in $Q$ (or would have if that part had been correct) for which $g(x_i)=s_i$. Unfortunately since this is a specific polynomial, you cannot just let it equal something else.

Finally, if you would like a hint about how to approach the problem, think about Lagrange interpolation, and in particular think about the following polynomials: $$ E_i = \frac{\prod_{j=1}^n (x-x_j)^2}{x-x_i} = (x-x_1)^2(x-x_2)^2\cdots (x-x_{i-1})^2(x-x_i)(x-x_{i+1})^2\cdots (x-x_n)^2$$ What are their derivatives at the various $x_i$?

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  • $\begingroup$ How is $g(x)$ specific? Its an arbitrary polynomial which lies in $Q$? $\endgroup$ – Olba12 Oct 1 '15 at 21:27
  • $\begingroup$ @Olba12 It is not arbitrary, not all polynomials $q$ in $Q$ satisfy $q(x_i)=s_i$, it is only $g$ specifically that does that. $\endgroup$ – jgon Oct 1 '15 at 21:28
  • $\begingroup$ Ah, now I see where my thought process went wrong. Is it possible from where I am to prove that this $g$ must be $h$? Or is it a dead end? We were specificly told to take help from the proof of Lagrange Interpolation. Maybe one could instead consider the map $A: g(x) \rightarrow (g(x_1),\dots,g(x_n),g´(x_1),\dots,g´(x_n))$ $\endgroup$ – Olba12 Oct 1 '15 at 21:38
  • $\begingroup$ @Olba12 I'm pretty sure it's a dead end, since what you are trying to prove is that the Lagrange interpolation polynomial you get will have the correct derivatives, which is not likely to happen. $\endgroup$ – jgon Oct 1 '15 at 21:39

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