Question

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)?

Answer 1

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

Answer 2

..

Answer 3

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

Answer 4

0

Answer 5

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

Answer 6

\[P(x+5)-P(x)=2\]\[P(x+5)-P(x)-2=0\]
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. .

Answer 7

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

Answer 8

@dumbcow,

how about my solution? is it acceptable. .

Answer 9

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

Answer 10

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

Answer 11

they're functions P

Answer 12

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.

Answer 13

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 \( P(4)-P(2)=\frac{4}{5} \)

Answer 14

4/5 is correct

Answer 15

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)