Problem on Functional Equation | SMO, 2010 | Problem 31
Try this beautiful problem from Singapore Mathematics Olympiad, SMO, 2010 based on functional equation.
Problem - Functional Equation (SMO Entrance)
Consider the identity \(1+2+......+n = \frac {1}{2}n(n+1)\). If we set \(P_{1}(x) = \frac{1}{2}x(x+1)\) , then it is the unique polynomials such that for all positive integer n,\(p_{1}(n) = 1+2+..............+n\) . In general, for each positive integer k, there is a unique polynomial \(P_{k} (x) \) such that :
\(P_{k} (n) = 1^k + 2^ k+3^k +..................+n^k\) for each n =1,2,3...............
Find the value of \(P_{2010} (-\frac {1}{2})\) .
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Key Concepts
Polynomials
Functional Equation
Check the Answer
Answer : 0
Singapore Mathematics Olympiad
Challenges and Thrills - Pre College Mathematics
Try with Hints
If you got stuck in this question we definitely can start from here:
In the question given above say k is the positive even number :
so let \(f(x) = P_{k} - P_(x-1)\)
Then \(f(n) = n^k \) for all integer \(n\geq 2\) (when f is polynomials)
Like this then \(f(x) = x^k\)(again for all \( x \geq 2\) .
If you got stuck after first hint try this one
\(P_{k} (-n + 1) - P_{k}(-n) = f(-n +1) = (n-1)^k\)..................................(1)
Again, \(P_{k} (-n + 2) - P_{k}(-n+1) = f(-n +2) = (n-2)^k\).......................................(2)
Now taking n = 1;The \(eq^n\)(1) becomes, \(P_{k}(0) - P_{k}(-1) = f(0) = 0^{k}\),
And for \(eq^(n)\) (2) ; \(P_{k}(1) - P_{k}(0) = f(1) = 1^{k}\).
Now sum these equation and try to solve the rest...........
Summing this two equation we get , \(P_{k}(1)-P_{k}(-n) = 1^{k} + 0^{k}+1^{k}+.......+(n-1)^k\).
so , \(P_{k}(-n)+P_{k}(n-1)=0\)
Again if \(g(x) = (P_{k}(-x)+P_{k}(x-1)\)
Then g(n) is equal to 0 for all integer \(n\geq2\)
As g is polynomial, g(x) =0;
So , \(P_{k}(-\frac {1}{2}) + P_{k}(-\frac {1}{2}) = 0\)
so \(p_{k}(-\frac {1}{2}) = 0\) ............(Answer)
Other useful links
- https://cheenta.com/power-mean-inequality-for-math-olympiad/
- https://www.youtube.com/watch?v=PfRqs9W8nPQ&t=2s
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