let p(x) be a polynomial function of (n−1)^(th) degree and p(k)=k for k=1,2,3,...,n find p(0) and p(n+1). example: n=10


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Question Number 98268 by mr W last updated on 16/Jun/20

$l e t p (x) b e a p o l y n o m i a l f u n c t i o n o f$ ${(n - 1)}^{t h} d e g r e e a n d$ $p (k) = k f o r k = 1, 2, 3, . . ., n$ $f i n d p (0) a n d p (n + 1) .$ $e x a m p l e : n = 10$
Commented by MJS last updated on 13/Jun/20

$I think this is impossible for degree n - 1 .$ $for n^{th} degree we have$ $p (x) = x + \underset{i = 1}{\prod^{n}} (x - i)$ $p (0) = {(- 1)}^{n} n!$ $p (n + 1) = n! + n + 1$
Commented by mr W last updated on 13/Jun/20

$t h a n k s s i r!$ $f o r {(n - 1)}^{t h} d e g r e e w e h a v e n e q n .$ $f o r n c o e f f i c i e n t s, s o t h e f u n c t i o n$ $s h o u l d b e u n i q u e . c a n y o u e x p l a i n$ $w h y t h i s i s n o t p o s s i b l e ?$ $f o r n^{t h} d e g r e e, t h e f u n c t i o n$ $p (x) = x + \underset{i = 1}{\prod^{n}} (x - i)$ $f u l f i l l s a l l c o n d i t i o n s, b u t i t i s n o t$ $u n i q u e, i t h i n k .$
Commented by MJS last updated on 13/Jun/20

$n = 3$ ${a x}^{2} + b x + c = y$ $x = y = 1 : a + b + c = 1$ $x = y = 2 : 4 a + 2 b + c = 2$ $x = y = 3 : 9 a + 3 b + c = 3$ $solving leads to a = c = 0 \land b = 1$ $p (x) = x$ $for n = 4$ ${a x}^{3} + {b x}^{2} + c x + d = y$ $. . .$ $a = b = d = 0 \land c = 1 \Rightarrow p (x) = x$ $. . .$
Commented by mr W last updated on 13/Jun/20

$n o w c l e a r s i r, t h a n k s!$ $a l l p o i n t s (k, k) w i t h k = 1 . . n l i e o n$ $t h e l i n e y = x, s o p (x) = x i s i n d e e d$ $a l w a y s a s u i t a b l e p o l y n o m i a l f o r$ $t h e s o l u t i o n .$ $w h a t d o y o u t h i n k ? i f p (k) = {(- 1)}^{k + 1} k,$ $d o e s t h e {(n - 1)}^{t h} d e g r e e p o l y n o m i a l$ $f u n c t i o n e x i s t ?$
Commented by MJS last updated on 14/Jun/20

$the function does exist .$ $p_{1} (x) = 1$ $p_{2} (x) = - 3 x + 4$ $p_{3} (x) = 4 x^{2} - 15 x + 12$ $p_{4} (x) = - \frac{10}{3} x^{3} + 24 x^{2} - \frac{155}{3} x + 32$ $p_{5} (x) = 2 x^{4} - \frac{70}{3} x + 94 x^{2} - \frac{455}{3} x + 80$ $p_{6} (x) = - \frac{14}{15} x^{5} + 16 x^{4} - \frac{308}{3} x^{3} + 304 x^{2} - \frac{2037}{5} x + 192$ $p_{7} (x) = \frac{16}{45} x^{6} - \frac{42}{5} x^{5} + \frac{704}{9} x^{4} - 364 x^{3} + \frac{39664}{45} x^{2} - \frac{5173}{5} x + 448$ $the sequence of p_{n} (0) is$ $a_{n} = ⟨ 1, 4, 12, 32, 80, 192, 448, . . . ⟩$ $a_{1} = 1$ $a_{2} = a_{1} \times 4$ $a_{3} = a_{2} \times 4 \times \frac{2^{2} - 1}{2^{2}}$ $a_{4} = a_{3} \times 4 \times \frac{2^{2} - 1}{2^{2}} \times \frac{3^{2} - 1}{3^{2}}$ $a_{5} = a_{4} \times 4 \times \frac{2^{2} - 1}{2^{2}} \times \frac{3^{2} - 1}{3^{2}} \times \frac{4^{2} - 1}{4^{2}}$ $. . .$ $I seem to be unable to find a_{n} in terms of n$ $the sequence of p_{n} (n + 1) is$ $b_{n} = ⟨ 1, - 5, 16, - 43, 106, - 249, 568, . . . ⟩$ $no idea to this . . .$
Commented by mr W last updated on 14/Jun/20

$t h a n k s v e r y m u c h s i r!$ $i s f o l l o w i n g t h e r i g h t p a t h t o s o l v e ?$ $s a y p_{n} (x) = \underset{k = 0}{\sum^{n - 1}} c_{n, k} x^{k}$ $p_{n} (r) = \underset{k = 0}{\sum^{n - 1}} c_{n, k} r^{k} = {(- 1)}^{r + 1} r$ $[\begin{matrix} 1 & 1 & 1 & . . . & 1 \\ 1 & 2 & 2^{2} & . . . & 2^{n - 1} \\ 1 & 3 & 3^{2} & . . . & 3^{n - 1} \\ . . . & . . . & . . . & . . . & . . . \\ 1 & n & n^{2} & . . . & n^{n - 1} \end{matrix}] (\begin{matrix} c_{n, 0} \\ c_{n, 1} \\ c_{n, 2} \\ . . . \\ c_{n, n - 1} \end{matrix}) = (\begin{matrix} 1 \\ - 2 \\ 3 \\ . . . \\ {(- 1)}^{n + 1} n \end{matrix})$ $\Rightarrow (\begin{matrix} c_{n, 0} \\ c_{n, 1} \\ c_{n, 2} \\ . . . \\ c_{n, n - 1} \end{matrix}) = {[\begin{matrix} 1 & 1 & 1 & . . . & 1 \\ 1 & 2 & 2^{2} & . . . & 2^{n - 1} \\ 1 & 3 & 3^{2} & . . . & 3^{n - 1} \\ . . . & . . . & . . . & . . . & . . . \\ 1 & n & n^{2} & . . . & n^{n - 1} \end{matrix}]}^{- 1} (\begin{matrix} 1 \\ - 2 \\ 3 \\ . . . \\ {(- 1)}^{n + 1} n \end{matrix})$
Commented by MJS last updated on 15/Jun/20

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