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

  • Thread starter Thread starter Albert1
  • Start date Start date
  • Tags Tags
    Degree Value
AI Thread Summary
The discussion revolves around finding the value of the polynomial function $f(102)$, where $f(x)$ is a degree 100 polynomial defined by $f(k) = \frac{1}{k}$ for integers $k$ from 1 to 101. Participants note that the initial solution provided is incorrect, indicating a misunderstanding or miscalculation. The problem emphasizes the need for careful evaluation of polynomial properties and constraints given the specific values at integer points. The hint reiterates the polynomial's degree and the conditions for $f(k)$. Ultimately, the correct approach to determine $f(102)$ remains unresolved in this thread.
Albert1
Messages
1,221
Reaction score
0
$f(x)$ 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:
Mathematics news on Phys.org
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
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.
Back
Top