For every positive integer n, consider all monic polynomials f(x) with integer coefficients, such that for some real number a
x(f(x+a)−f(x))=nf(x)
Find the largest possible number of such polynomials f(x) for a fixed n<1000.
11 years ago
11 years ago
"lin" also needs to learn how to spell "help" ... and that there is no class called "heeeeeeelp math" -- incredible inability to follow directions.
11 months ago
To find the largest possible number of polynomials f(x) for a fixed n<1000, we need to analyze the given equation:
x(f(x+a)−f(x))=nf(x)
First, let's rewrite the equation by expanding the brackets:
xf(x+a) - xf(x) = nf(x)
Simplifying further:
xf(x+a) - nf(x) = xf(x)
Rearranging the terms:
xf(x+a) - (n+1)f(x) = 0
Now, notice that this is a linear homogeneous differential equation of the form:
y' - (n+1)y = 0
To solve this differential equation, we can express it as a separable equation:
dy/y = (n+1)dx
Integrating both sides:
ln|y| = (n+1)x + C
Applying the exponential function to both sides:
|y| = e^((n+1)x + C)
Since we're looking for monic polynomials, we can discard the absolute value:
y = e^((n+1)x + C)
Now, let's consider the condition that the polynomial has integer coefficients. For this, the term inside the exponential function should be a real number. The expression, (n+1)x + C, should be equal to zero when x is an integer.
(n+1)x + C = 0
Solving for C:
C = - (n+1)x
Now, let's substitute this value of C back into the polynomial equation:
y = e^((n+1)x - (n+1)x)
Simplifying:
y = e^0
y = 1
Hence, the only polynomial that satisfies the given condition for any fixed n<1000 is:
f(x) = 1
Therefore, the largest possible number of such polynomials f(x) is 1.