Problem on Functional Equation | SMO, 2010 | Problem 31

Try this beautiful problem from Singapore Mathematics Olympiad, SMO, 2010 based on functional equation.

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 \(P_{2010} (-\frac {1}{2})\) .

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)

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