+483 Let \(p(x)\) be a polynomial of degree 5 such that \(p(n) = \frac{n}{n^2 - 1} \) for \(n = 2, 3, 4, \dots, 7\). Find \(p(8).\)
+483 Thank you all so much for the helpful answers!!!!
Here is the solution given:
Let \(q(x) = (x^2 - 1) p(x) - x.\) Then \(q(n)\) has degree 7, and \(q(n) = 0\) for n=2, 3, 4, ... 7, so \(q(x) = (ax + b)(x - 2)(x - 3) \dotsm (x - 7)\)
for some constants \(a\) and \(b.\)
We know that \(q(1) = (1^2 - 1)p(1) - 1 = -1.\) Setting \(x=1 \) in the equation above, we get \(q(1) = 720(a + b),\)
so \(a+b=-\frac{1}{720}.\)
We also know that \(q(-1) = ((-1)^2 - 1)p(-1) + 1 = 1.\) Setting \(x=-1\) in the equation above, we get \(q(-1) = 20160(-a + b),\)
so \(-a + b = \frac{1}{20160}.\) Solving for \(a\) and \(b,\) we find \(a = -\frac{29}{40320}\) and \(b = -\frac{3}{4480}.\) Hence, \(\begin{align*} q(x) &= \left( -\frac{29}{40320} x - \frac{3}{4480} \right) (x - 2)(x - 3) \dotsm (x - 7) \\ &= -\frac{(29x + 27)(x - 2)(x - 3) \dotsm (x - 7)}{40320}. \end{align*}\)
In particular, \(q(8) = -\frac{(29 \cdot 8 + 27)(6)(5) \dotsm (1)}{40320} = -\frac{37}{8}, \)
so \(p(8) = \frac{q(8) + 8}{8^2 - 1} = \boxed{\frac{3}{56}}.\)
+118687 At first glance many would say that p(8)=8/63 which is approx 0.127 but that is not going to be correct.
I solved it graphically, though the answer may not be exact.
Here is the link, It says the polynomial is of degree 4 not 5. (Even though I specifically asked for degree 5)
I can't check it easily because the coefficients are rounded to 2 decimal places. 
https://www.geogebra.org/classic/fesnhtkm
And here is the graph.
So this is the approx answer (or exactly) f(8)=1/20 = 0.05
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I also tried to plug the values into a matrix solver calculator but it said matrix A was singular. To be honest I do't quite know what that means but it didn't give me an answer. Maybe it did not understand my input. idk.
I keep being given a polynomial of degree 4 not 5 maybe this is the problem.
https://www.mathstools.com/section/main/system_equations_solver#.YbKPR71BxnI
+133 Geogebra is very useful for curve fitting, nice idea also with the matrix solver. Using wolframalpha isn't very intuitive (generic).
+133 I wonder if curve fitting may be useful. I'm using a brute force method, so if there is a better solution please answer!
Let the polynomial be p(x) = ax^5 + bx^4 + cx^3 + dx^2 + ex + f.
We will set up a system of equations now.
2/(2^2-1) = 2/3 = 32a + 16b + 8c + 4d + 2e + f
3/8 = 243a + 81b + 27c + 9d + 3e + f
4/15 = 4^5(a) + 4^4(b) + etc...
You probably see the idea by now. I'm certain this is the tedious method since one wouldn't want to calculate 7^5 by hand, let alone solve a system of equations with monstrous coefficients like these.
Even wolframalpha doesn't accept the equations...
+23252 I used the first 6 values (n = 2, ... n = 7) to find the values for a, ... f in the equation:
y = ax5 + bx4 + cx3 + dx2 + ex + f
to get these values:
a = -7.192460317 x 10-4
b = 0.01875
c = -0.1948164682
d = 1.01875
e = -2.750892857
f = 3.375
If these are correct, they do show that 8/63 is not correct for n = 8.
+118687 If I had to, I would do it by matrix elimination.
Can anyone point out to me why the matrix calculator would not give me an answer?
What is the significance of a matrix being singular?
+2511 Hi Melody,
A square matrix is singular if its determinant is zero (0). (There is no inverse for this matrix.)
Note: This definition applies only if the matrix is square.
GA
--. .-
+118687 Thanks Ginger,
I read that but why does it stop it from being used for a simultaneous solution?
I thought it might be because the coeff of x^5 was zero.
So I tried repeating it by not using one of the equations.
This way the degree wold be 4, not 5
But exactly the same thing happened.
+2511 The determinate of a square matrix [A] is equal to zero if and only if the determinate of its reduced row echelon form [B] is equal to zero. It’s important to note that matrix [A] and reduced-form [B] do not have the same determinates, but they are zero at the same time. Because they are zero at the same time its undefined.
A graph of the \(y =\frac{n}{n^2-1}\) equation shows two asymptotes and a root of zero. The inverse of this equation \(y= \frac{n^2-1}{n}\) will give two roots (1) and (-1) (the singularities of its inversion) and an asymptote at zero. This suggests the fifth-degree polynomial equation that represents a finite portion of that equation has one real root and four complex roots. Creating this polynomial equation by using (7) points of \(y =\frac{n}{n^2-1}\), and then extending (interpolating) it linearly yields a point that is notably different from the base equation. That may be (one of) the point of the exercise –an example of catastrophe ...
My attempt at using Gauss-Jordan elimination was a catastrophe...
I’m interested in the official AoPS solution. I’m sure it will be over my head, but it will give me something to look up to.... 
GA
--. .-
+2511 Note that Geno’s data set plots y ≈ 0.0535715 for P(8). This value is very close to 0.05 for P(8) on Melody’s graph, so it’s probably correct. So, the answer to your question is, The polynomial at P(8) has y ≈ 0.0535715
This math problem demonstrates how extrapolating (infer by extending known information) a point from a polynomial equation created from a limited and narrow data set will have an error –sometimes a significant error –for the location of a nearby data point. The reason is the created polynomial does not include data points for the change in the curve.
Using Melody’s Geobraga graph, overlay g(n)=((n)/(n^(2)-1)). This will show the points on the original plot compared to the polynomial based on the sample points. Note that the graphs align perfectly until after the seventh point, where they begin to diverge. By the eighth point, the polynomial graph shows a sudden divergence down toward the x-axis, instead of the asymptotic glide toward the x-axis. So, for the polynomial, at P(8) y ≈ 0.0535715 compared to y =8/63 (≈ 0.12698) for the original equation.
GA
--. .-
+118687 Here is Alan's answer graphed, from Desmos.
I wouldn't guess that this was a degree 5 polynomial. (without doing lots of work)
Although it would have to be an odd degree. That makes 5 reasonable I guess.
+483 Thank you all so much for the helpful answers!!!!
Here is the solution given:
Let \(q(x) = (x^2 - 1) p(x) - x.\) Then \(q(n)\) has degree 7, and \(q(n) = 0\) for n=2, 3, 4, ... 7, so \(q(x) = (ax + b)(x - 2)(x - 3) \dotsm (x - 7)\)
for some constants \(a\) and \(b.\)
We know that \(q(1) = (1^2 - 1)p(1) - 1 = -1.\) Setting \(x=1 \) in the equation above, we get \(q(1) = 720(a + b),\)
so \(a+b=-\frac{1}{720}.\)
We also know that \(q(-1) = ((-1)^2 - 1)p(-1) + 1 = 1.\) Setting \(x=-1\) in the equation above, we get \(q(-1) = 20160(-a + b),\)
so \(-a + b = \frac{1}{20160}.\) Solving for \(a\) and \(b,\) we find \(a = -\frac{29}{40320}\) and \(b = -\frac{3}{4480}.\) Hence, \(\begin{align*} q(x) &= \left( -\frac{29}{40320} x - \frac{3}{4480} \right) (x - 2)(x - 3) \dotsm (x - 7) \\ &= -\frac{(29x + 27)(x - 2)(x - 3) \dotsm (x - 7)}{40320}. \end{align*}\)
In particular, \(q(8) = -\frac{(29 \cdot 8 + 27)(6)(5) \dotsm (1)}{40320} = -\frac{37}{8}, \)
so \(p(8) = \frac{q(8) + 8}{8^2 - 1} = \boxed{\frac{3}{56}}.\)
