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I constructed with a degree 6 polynomial but apparently even a degree 5 would suffice.

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$$\array{f(x)&=&ax^6&+&bx^5&+&cx^4&+&dx^3+ex^2+fx+g \\ f'(x)&=& &&6ax^5&+&&&\cdots \\ f''(x)&=&&&&&30ax^4&+&\cdots}$$

So it is now just a matter of substitution in the values from the table and solving a set of linear equations?

By making the degree higher than it needs to be, am I going to just make the algebra unnecessarily longer or will it lead to a system of equations with no unique solutions?

stackdsewew
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    Substituting the values and solving the set of linear equations is right, but you don't need the second derivative anywhere; the resulting interpolation polynomial will have function values and first derivatives equal to $f$ on all three points $0,\frac12,1$. – Mees de Vries May 03 '16 at 10:15
  • Is that true generally or just in this case? – stackdsewew May 03 '16 at 10:16
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    The general case is something like: given $n$ points and $k$ values for each point, you can make a polynomial $f$ of degree $n*k$, such that for each $0 \leq i < k$ the $i$th derivative of $f$ is at each point is equal to the $i$th value for that point. See also e.g. the Wikipedia page on Hermite interpolation. – Mees de Vries May 03 '16 at 10:19
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    Perhaps you only want a degree $5$ polynomial. As written, you have $7$ variables $a,\cdots,g$. – Michael Burr May 03 '16 at 10:21
  • How would I know what the degree the polynomial should be before I begin the problem? Would having a higher degree polynomial just make the algebra unnecessarily longer or would it give me a system of equations that do not have a unique solution? – stackdsewew May 03 '16 at 10:23
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    Good question! My latest question http://math.stackexchange.com/questions/1770738/is-the-ideal-of-a-variety-the-annihilator-of-a-subspace-of-the-symmetric-algebra was inspired after reading yours. – user062295 May 04 '16 at 02:25
  • Your analysis is correct. In fact, the ambiguity in the solutions that you will get from assuming your solution is a polynomial in any number of variables will be the ideal of polynomials of the form $f(x)(x(x-1)(x-0.5))^2$ – user062295 May 04 '16 at 02:33
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    You asked another question about the smallest degree you can assume your polynomial to be when doing your calculations. Suppose you are interpolating through $n$ different values in $x$. My analysis below shows that the minimum degree polynomial in which there are polynomials which have the same values for $f$ and $f'$, is $2n$: namely in this degree there is an explicit example. It is always possible to assume your polynomial will be of degree $2n-1$(because there exists a degree where there is only one polynomial that interpolates the values for $f$ and $f'$ at the given points.) – user062295 May 04 '16 at 03:47
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    That is why you know in this example that there will exist an interpolating polynomial of degree $2*3-1=5$. – user062295 May 04 '16 at 03:48

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$\newcommand{\R}{\mathbb{R}}$.

Let $W=Sym({\R^*})$ be the algebra of polynomials in one variable with real coefficients.

Let $e_1$ be a basis for $\R$ and $x$ be the function $\R \to \R$ given by $x(e_1)=1$.

Thus if you didn't like the description above, $W$ is just the vector space generated by the monoid generated by $x$.

In any event if you add any element in the ideal in $W$ of polynomials that vanish, and whose first derivative vanishes on the set $\{0,0.5,1\}$, you will get another polynomial that interpolates the values in the table. These are in fact all of the polynomials that interpolate the values in the table.

The surprising part is that this set is an ideal in $W$. Here is why

The set of such polynomials will be a subspace of polynomials of the form $g(x)(x)(x-0.5)(x-1)$ by remainder theorem. To find the subspace we use the following table

$\begin{array} x & 0 & 0.5 & 1\\ g'(x)& g(x)(x-1)(x-0.5)|_{0} & g(x)x(x-0.5)|_1 & g(x)x(x-1)|_0.5 \\ g'(x) & g(0)0.5 & g(1)0.5 & -g(0.5)0.25 \end{array} $ so the condition that a polynomial must have its derivatives 0 at the given x values is that $g(0)=0, g(1)=0, g(2)=0$.

Thus the set of all polynomials that interpolates that values will need to have factors of $x(x-0.5)(x-1)$ in $g(x)$. Thus we know we got all of the polynomials that interpolate the values given above and they are of the form $f(x)(x(x-0.5)(x-1))^2$ so in particular it is an ideal.

This is a very interesting question because it gives one a reason to consider ideals that are not radical like $\langle (x(x-0.5)(x-1))^2 \rangle$ in $W$.

Keep up the good work!

(If you want a more direct answer to the question, then in the special case where you assume your polynomial is of degree 6, the kernel of the LHS of the linear system that you create will be the coefficients of $(x(x-0.5)(x-1))^2$, $d/dx((x(x-0.5)(x-1))^2)$ when these are expanded out.)

user062295
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