For part b, we know the value of $x_0$ and $x_1$, how does that help with $x_j$? Or we don't need to worry about $x_j$ at all? And any thoughts on part c and part d?
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Part (b) just says to do it for $0\le j\le1$, so only $x_0$ and $x_1$ are involved. – Gerry Myerson Feb 20 '16 at 23:31
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Why do we need that? I still don't get it. What's gonna change if $j\ge 1$? – J.doe Feb 21 '16 at 00:09
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You're trying to find a polynomial, $H_p$, that, together with its first $p$ derivatives, agrees with the given function $f$ at $x_0$ and at $x_1$. So you want it to match $2p+2$ pieces of data, so it must have degree $2p+1$. If you increase $j$, you increase the amount of data to match, so you must increase the degree of $H_p$. E.g., if $j$ were 2, you'd need $H_p$ of degree $3p+2$. – Gerry Myerson Feb 21 '16 at 03:42
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Part (b). For $p=0$ we have Lagrange and for $p=1$ we have Hermite. How do we go about finding $Hp(x)$ for $p=2$? – J.doe Feb 21 '16 at 20:05
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So for part b, we are supposed to have several different formulas for the different value of p, correct? – J.doe Feb 21 '16 at 20:12
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Yes, there's one formula for each value of $p$. – Gerry Myerson Feb 21 '16 at 21:36