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## Directrix 4 years ago There are an infinite number of polynomials P for which P(x+5) - P(x) = 2 for all x. What is the least possible value of P(4) - P(2)?

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1. perl

P(x+5) - P(x) -2 = 0

2. anonymous

..

3. anonymous

Let $P(x) = ax^{n}, .... P(x+5) = a(x+5)^{n}$ Then $a(x+5)^{n} -ax^{n} = 2$ $\rightarrow a = \frac{2}{(x+5)^{n} -x^{n}}$ since it applies for all x, let x=0 $a = \frac{2}{5^{n}}$ Evaluate P(4)-P(2) $\rightarrow \frac{2}{5^{n}}4^{n} - \frac{2}{5^{n}}2^{n} = \frac{2(4^{n}-2^{n})}{5^{n}}$ Take limit as n-> infinity $\lim_{n \rightarrow \infty}\frac{2(4^{n}-2^{n})}{5^{n}} = 0$ smallest possible value goes to 0

4. anonymous

0

5. anonymous

@dumbcow: why the initial assumption of $$P(x) = ax^{n}$$

6. anonymous

$P(x+5)-P(x)=2$$P(x+5)-P(x)-2=0$ if this is the function, given two value of x, 4 and 2, to be subtracted. . $[P(4+5)-P(4)-2] -[P(2+5)-P(2)-2]=0$$[P(9)-P(4)-2]-[P(7)-P(2)-2]=0$$P(5)-2 -P(5)-2=0$$p(0)-4=0$$P(0)=4$ kindly check my solution. i got 0 as the least value of x, and 4 is the result. .

7. anonymous

@FoolFormath because it was the simplest polynomial to work with and since it doesn't matter what polynomial it is

8. anonymous

@dumbcow, how about my solution? is it acceptable. .

9. anonymous

in one of your steps(3rd line) what happened to P(9) and P(7)

10. anonymous

oh i got it but how can you assume P(9) -P(4) = P(5) ?

11. anonymous

they're functions P

12. Directrix

No correct answer yet. Hint: Consider the value(s) of n a polynomial P of degree n would have in order for P(x+5) - P(x) to be constant.

13. anonymous

It seems that I have understood the solution (with some help) the polynomial should be for the form $$P(x)=\frac{2}{5}x+c$$ so $$P(4)-P(2)=\frac{4}{5}$$

14. Directrix

4/5 is correct

15. anonymous

i see what i did wrong now. isn't the question a little misleading though, there really is only 1 possible value for P(4)-P(2)

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