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Find a polynomial $q(a)$ of degree less than equal to $2$ that saitsifies the condition $q(a_0)=b_0, q'(a_0)=b'_0, \ \text{and} \ q'(a_1)=b'_1,$ where $a_0,a_1,b_0,b'_0,b'_1\in \mathbb{R}$, where $a_0\ne a_1$. And give a formula of the form $q(a)=b_0k_0(a)+b'_0k_1(a)+b'_1k_2(a).$

How can I do this question? I am self teaching numerical analysis and this question is in the book An Introduction to Numerical Analysis, by Atkinson but I don't know how to do it.

Tom
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A polynomial $q(a)$ of degree less than $2$ is linear, so you can write $q(a)=ca+d$. You will fail in this unless $b_0'=b_1'$. Probably you are asked for a polynomial of degree less than or equal to $2$ and should write $q(a)=ca^2+da+e$ Now plug in the restrictions you have on $q$ and solve for $c,d,e$

Ross Millikan
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  • You are right, it did say less than equal to 2. – Tom Apr 30 '13 at 00:50
  • @Tom: the statement that you will fail with a linear polynomial shows well what is happening here. You don't have enough parameters to solve the equations. A linear equation has only two parameters, but you have three restrictions. – Ross Millikan Apr 30 '13 at 05:34
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Let $q(x) = k_0 + k_1 x + k_2 x^2 + \cdots + k_n x^n$.

Since the polynomial is of degree less than or equal to 2, $n \le 2$, as $n$ represents the degree of the polynomial.

Thus, $q(x) = k_0 + k_1x + k_2x^2$.

If $q(a_0) = b_0$, then $k_0+k_1 a_0 +l_2a_0^2 = b_0$. Proceed from here.

Emily
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  • I made an error, I forgot to put less than or equal to 2. – Tom Apr 30 '13 at 00:50
  • @Tom Edited to reflect the change. – Emily Apr 30 '13 at 00:55
  • Thank you for editing, but I am having trouble continuing. How can I plug in the conditions I have and solve for $$q(x) = k_0 + k_1x + k_2x^2?$$ – Tom Apr 30 '13 at 04:19
  • Take $x = a_0$. Then $q(a_0) = k_0 + k_1 a_0 + k_2a_0^2$. Since $q(a_0) = b_0$, then you have a linear equation for $k_0, k_1, k_2$. Use the other conditions to get two more equations. – Emily Apr 30 '13 at 15:30