A very very hard college algebra problem

[nmego12345]

Homework Statement

Note: I'm saying it's very very hard because I still couldn't solve it and I've posted it in stackexchange and no answer till now.

I'm posting here the problem statement, all variables and known data in addition to my solving attempts. Because I'm posting an image of my question and that it would be hard to separate my solving attempts from the problem statement etc.., but they are obvious in the image. I'm posting an image because I'm not familiar to the formatting here yet, and I don't have time to format all this problem again ( took me 1.5 hours to format)

https://s3.amazonaws.com/diigo/thum...][/B][/B] [h2]The Attempt at a Solution[/h2]

 

[Stephen Tashi]

We need to state the problem correctly. Isn't the problem to find a polynomial $f_k (x)$ such that $f_k(x)- f_k(x-1) = x^k$ ?

 

[nmego12345]

I'm not sure If I understand the problem correctly, I found it on a textbook, I didn't invent it, but I think
that you're correct

 

[Stephen Tashi]

Let $p(x) = x^{k+1} + ...$ other terms. Suppose, for example, that $p(x) - p(x-1)$ has the term $Cx^3$.

By part 2) , if you form the new polynomial $q(x) = p(x) - C f_3(x)$ then

$q(x) - q(x-1) = p(x) - p(x-1) + C (f_3(x) - f_3(x-1)) = x^{k+1}+ ... Cx^3 + ... - C x^3$ so you can eliminate the $Cx^3$ term by subtracting a multiple of $f_3(x)$.

I think what the problem wants you to do is express $f_k(x)$ as $x^k$ minus multiples of $f_{k-1}, f_{k-2},...$.

After you do that, you may be able to find the numerical values of the coefficients, but perhaps the problem only wants you to write the "recursive" relation between $f_k$ and $f_{k-1}, f_{k-2} ...$.

 

[nmego12345]

First of all, what is Cx³?, is it combinations? but combinations are defined over 2 numbers not 1 number. Is it an arbitary constant?

 

[Stephen Tashi]

First of all, what is Cx³?, is it combinations? but combinations are defined over 2 numbers not 1 number. Is it an arbitary constant?

C is an arbitrary constant.

 

[nmego12345]

First of all, what is Cx³?, IIRC, aren't combinations have 2 numbers to begin with?

Oh Ok, I reread your answer, that makes sense.