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

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SUMMARY

The polynomial function \( f(x) \) is of degree 100, defined such that \( f(k) = \frac{1}{k} \) for \( k = 1, 2, \ldots, 101 \). To find \( f(102) \), one must utilize polynomial interpolation techniques, specifically Lagrange interpolation or Newton's divided differences. The correct value of \( f(102) \) can be determined by evaluating the polynomial at this point, ensuring that the conditions for the given values are satisfied.

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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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