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# Problem on Functional Equation | SMO, 2010 | Problem 31

Try this beautiful problem from Singapore Mathematics Olympiad, SMO, 2010 based on functional equation. You may use sequential hints.

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

But try the problem first…

Source

Challenges and Thrills – Pre College Mathematics

## Try with Hints

First hint

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 $$x \geq 2$$ .

Second Hint

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………..

Final Step

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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