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

[Albert1]

$f(x)$ is a real polynominal function with degree 100,and $f(k)=\dfrac {1}{k} , \,\,(k=1,2,3,4,5,------,101),$
please find $f(102)=?$

 

[Albert1]

$f(x)$ 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:

Spoiler

set: $g(x)=xf(x)-1$

 

[kaliprasad]

$f(x)$ is a real polynominal function with degree 100,and $f(k)=\dfrac {1}{k} , \,\,(k=1,2,3,4,5,------,101),$
please find $f(102)=?$

Spoiler

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

Spoiler

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 $f(102) = \frac{1}{51}$

 

[Albert1]

Spoiler

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 $f(102) = 0$

Spoiler

please check again $g(102)=?$

 

[kaliprasad]

Spoiler

please check again $g(102)=?$

oops

Spoiler

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