Problem :

If $f(x)$ is a polynomial of degree $n$ and if $f(k) = \frac{k}{k+1}$ where $k =0,1,2,\ldots,n$, find $f(x)$.

Can we go like this : Let the polynomial be $a(x-\alpha_1)(x-\alpha_2)(x-\alpha_3)......(x-\alpha_n)$ where $\alpha_1, \alpha_2....\alpha_n$ are the roots of the $f(x)$ and $a$ is the leading coefficient.

Please guide how to proceed in such question.

  • $\begingroup$ It might be easier not to factor the polynomial, and just solve the problem for n=0, n=1, n=2, and n=3 and see if you spot some sort of theme/pattern. $\endgroup$ – Mark S. May 12 '13 at 16:42

Hint: Consider the polynomial $g(x) = (x+1) f(x) - x$. What is the degree of $g(x)$? What are the zeros of $g(x)$? What is the value of $g(-1)$?

More details added.

The degree of $g(x)$ is equal to the degree of $f(x)$ plus 1 (since we multiplied by $x+1$), which is equal to $n+1$.

The zeros of $g(x)$ are (at least) $0, 1, 2, \ldots n$, so by the Remainder-Factor Theorem,

$$g(x) = A(x) \prod_{i=0}^n (x-i)$$

Since $g$ has degree $n+1$, it follows that $A(x)$ is a constant polynomial, which we will denote by $A$.

The value of $g(-1)$ is $g(-1) = (-1 + 1) f(-1) - (-1) = 1 $. Hence

$$1 = g(-1) = A \prod_{i=0}^n (-1-i)$$

This allows us to determine that $A = \frac{1}{(n+1)!}$.

Hint: $\frac{ g(x) + x) } { x+1}$ is indeed a polynomial, and is our $f(x)$.

  • $\begingroup$ Hi,Could u brief me about this concept. how you reach at : g(x) =(x+1)f(x) -x, I will be greatful to you.. thanks.. again.. $\endgroup$ – Sachin May 14 '13 at 2:32
  • $\begingroup$ @Sultan It based off looking at $f(k) = \frac{k}{k+1} $. We want to write it as $f(x) = \frac{x}{x+1} $, however that is not a polynomial. The 'approximation' that we can get, is that $(x+1) f(x) = x $, and that is what we consider. $\endgroup$ – Calvin Lin May 14 '13 at 12:54
  • $\begingroup$ Hi, Thanks for your clarification what I got is as follows : (x+1)f(x) -x=0 . Let g(x) be the polynomial where g(x) = (x+1)f(x) -x therefore g(-1) = 1 since when g(x) is divided by f(x) it is giving remainder x $\Rightarrow $ g(x) is second degree polynomial. therefore let it be $ax^2+bx+c $ and putting x =-1 we get $a -b+c = 1$...Please guide now from where to get value of a,b,c thanks for all your hints given.... $\endgroup$ – Sachin May 16 '13 at 7:46
  • $\begingroup$ when g(x) is divided by f(x) it is giving remainder x Sorry this line is : when g(x) is divided by f(x)(x+1) it is giving remainder x $\endgroup$ – Sachin May 16 '13 at 8:15
  • $\begingroup$ @sultan I added more details. I don't understand why you're looking at quadratics. Please work through the initial hints, and complete the rest of the question. $\endgroup$ – Calvin Lin May 16 '13 at 16:01

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