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 \(p(8).\)

abcdefghijklmnopqrst Dec 9, 2021

#20**+8 **

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 \(p(8) = \frac{q(8) + 8}{8^2 - 1} = \boxed{\frac{3}{56}}.\)

abcdefghijklmnopqrst Dec 11, 2021

#1**+3 **

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**

--------------------------------------

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

Melody Dec 10, 2021

#3**+1 **

Geogebra is very useful for curve fitting, nice idea also with the matrix solver. Using wolframalpha isn't very intuitive (generic).

tinfoilhat
Dec 10, 2021

#2**+1 **

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...

tinfoilhat Dec 10, 2021

#4**+2 **

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.

geno3141 Dec 10, 2021

#5**+1 **

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?

Melody Dec 10, 2021

#6**0 **

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

--. .-

GingerAle
Dec 10, 2021

#7**+1 **

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.

Melody
Dec 10, 2021

#8**+1 **

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

--. .-

GingerAle
Dec 10, 2021

edited by
Guest
Dec 10, 2021

#11**+3 **

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

GA

--. .-

GingerAle Dec 11, 2021

#15

#19**+2 **

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.

Melody Dec 11, 2021

#20**+8 **

Best Answer

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 \(p(8) = \frac{q(8) + 8}{8^2 - 1} = \boxed{\frac{3}{56}}.\)

abcdefghijklmnopqrst Dec 11, 2021