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Suppose the polynomial $f(x)$ is of degree $3$ and satisfies $f(3)=2$, $f(4)=4$, $f(5)=-3$, and $f(6)=8$. Determine the value of $f(0)$.

How would I solve this problem? It seems quite complicated...

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Let the polynomial be $f(x)=ax^3+bx^2+cx+d$. The numbers a,b,c,d are unknown.

Now, use f(3)=2, f(4)=4, f(5)=-3, and f(6)=8. You have 4 unknowns and 4 equations.

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I want to point out that, since $f(0) = d$, you only need to solve for one unknown. – Strants Jul 25 '14 at 22:03

If you just want to use subtraction, you could take first differences

               2   4  -3   8 
                 2  -7  11
                  -9  18

then expand this to the left

-328 -137 -36  2   4  -3   8 
   191  101  38  2  -7  11
     -90 -63 -36  -9  18
        27  27  27  27
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+1 Nice... this is probably the intended solution (it avoids actually calculating the interpolating polynomial, which is unnecessary). – NotNotLogical Jul 26 '14 at 1:19
Never seen this before.How does it work? – MonK Jul 26 '14 at 20:19
Yes an explanation of why this works would be a good thing to add @Henry – DavidButlerUofA Jul 26 '14 at 20:56
Taking first differences is like a discrete version of taking the derivative, so if the bottom row is constant (degree $0$) then the one above it is linear (degree $1$), the one above that is quadratic (degree $2$) and the top row is cubic (degree $3$). – Henry Jul 27 '14 at 9:36

Let $$L_3(x)=\frac{(x-4)(x-5)(x-6)}{(3-4)(3-5)(3-6)}$$ so notice that $L_3(3)=1$ and $L_3(k)=0$ for $k=4,5$ or $6$ and define by the same way $L_4, L_5$ and $L_6$ so we see that $$f(x)=f(3)L_3(x)+f(4)L_4(x)+f(5)L_5(x)+f(6)L_6(x)$$ and now the calculus of $f(0)$ isn't hard.

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Let $g(x)=f(x+3)$. Then the difference table for $g$ is given by

$2\;\;\; 4\;\; -3\;\; 8$

$\;\; 2 \;\; -7\;\; 11$

$\;\;\;-9\;\; 18$


so $f(0)=g(-3)=2+2\binom{-3}{1}-9\binom{-3}{2}+27\binom{-3}{3}=2+2(-3)-9(6)+27(-10)=-328.$

Alternatively, $f(x)=g(x-3)=2+2\binom{x-3}{1}-9\binom{x-3}{2}+27\binom{x-3}{3}=\frac{9}{2}x^3-\frac{117}{2}x^2+245x-328$ and $f(0)=-328$.

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