Find the Value of $f(102)$ for $f(x)$ of Degree 100

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Discussion Overview

The discussion revolves around finding the value of the polynomial function $f(x)$ of degree 100 at the point $f(102)$, given that $f(k) = \frac{1}{k}$ for integer values of $k$ from 1 to 101. The scope includes mathematical reasoning and problem-solving related to polynomial functions.

Discussion Character

  • Mathematical reasoning

Main Points Raised

  • Participants present the problem of determining $f(102)$ based on the given conditions of the polynomial function.
  • Some participants reiterate the problem statement without providing additional insights or solutions.
  • One participant suggests that an earlier proposed solution is incorrect, indicating a need for further clarification or correction.

Areas of Agreement / Disagreement

There is no consensus on the solution to the problem, as participants have not reached an agreement on the correctness of the proposed solutions or methods.

Contextual Notes

The discussion lacks detailed exploration of the methods to derive $f(102)$, and there are unresolved aspects regarding the polynomial's behavior outside the specified range of $k$ values.

Albert1
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$f(x)$ is a real polynominal function with degree 100,and $f(k)=\dfrac {1}{k} , \,\,(k=1,2,3,4,5,------,101),$
please find $f(102)=?$
 
Last edited:
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Albert said:
$f(x)$ is a real polynominal function with degree 100,and $f(k)=\dfrac {1}{k} \,\,(k=1,2,3,4,5,------,101),$
please find $f(102)=?$
hint:
set: $g(x)=xf(x)-1$
 
Albert said:
$f(x)$ is a real polynominal function with degree 100,and $f(k)=\dfrac {1}{k} , \,\,(k=1,2,3,4,5,------,101),$
please find $f(102)=?$

take $g(x) = x f(x) - 1$
it is polynomial of degree 101 and zero for x = 1 through 101
so $g(x) = A (x-1)(x-2)\cdots(x-101)$
the constant term = $- A * 101! = -1$ so $A = \frac{1}{101!}$

so $g(x) = \frac{1}{101!} (x-1)(x-2)(x-3)\cdots(x-101)$
or $g(102) = 102 f(102) - 1 = -1 $ or $f(102) = 0$

above solution is incorrect the solution is

take $g(x) = x f(x) - 1$
it is polynomial of degree 101 and zero for x = 1 through 101
so $g(x) = A (x-1)(x-2)\cdots(x-101)$
the constant term = $- A * 101! = -1$ so $A = \frac{1}{101!}$

so $g(x) = \frac{1}{101!} (x-1)(x-2)(x-3)\cdots(x-101)$
or $g(102) = 102 f(102) - 1 = 1 $ or $f(102) = \frac{1}{51}$
 
Last edited:
kaliprasad said:
take $g(x) = x f(x) - 1$
it is polynomial of degree 101 and zero for x = 1 through 101
so $g(x) = A (x-1)(x-2)\cdots(x-101)$
the constant term = $- A * 101! = -1$ so $A = \frac{1}{101!}$

so $g(x) = \frac{1}{101!} (x-1)(x-2)(x-3)\cdots(x-101)$
or $g(102) = 102 f(102) - 1 = -1 $ or $f(102) = 0$
please check again $g(102)=?$
 
Albert said:
please check again $g(102)=?$

oops
g(102) = 1 and I shall update the solution above
 

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