#Functions
strange willowBOT
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vale gust
i think i got it
f(1) = 2^9 * 9! * f(m)
f(1) = 2 * f(3)
f(3)= 2 * 2 * f(5)
f(5)= 2 * 3 * f(7)
f(7)= 2 * 4 * f(9)
.
.
.
f(17) = 2 * 9 * f(19) --> m = 19
is it correct?
merry stratus
Hmm... Let's see.
f(x) = f(x + 2)(x + 1)
1/f(x + 2) = (x + 1)/f(x)
Let g(x) = 1/f(x).
g(x + 2) = (x + 1)g(x)
I noticed that it kinda looks like a recurrent relation for x!, but not quite. Let's tinker with it.
Recurrence relation for (mx + n)!: f(x + 1/m) = f(x)(mx + n + 1)
Our case of g(x) then seems close to m = 1/2, n = -1/2, so ((x - 1)/2)!. Let's try doing g(x) = h(x)((x - 1)/2)!.
h(x + 2)((x + 1)/2)! = h(x)((x - 1)/2)! (x + 1)
h(x + 2)((x + 1)/2) = h(x)(x + 1)
h(x + 2) = 2h(x)
Seems good! Solving this relation is easy by taking the logarithm, at which point it becomes a linear recurrence relation. I will omit that part as that isn't interesting. We get:
h(x) = 2^(x/2) (C1 + C2(-1)^x)
g(x) = 2^(x/2) ((x - 1)/2)! (C1 + C2(-1)^x)
f(x) = 2^(-x/2)/(((x - 1)/2)! (C1 + C2(-1)^x))
We also need f(1).
f(1) = 2^(-1/2)/(C1 - C2)
So:
f(1)/f(m) = 2^((m - 1)/2) ((m - 1)/2)! (C1 - C2)/(C1 + C2(-1)^m)
We are not given the initial conditions: that must mean that m is odd, as then the constants disappear:
f(1)/f(m) = 2^((m - 1)/2) ((m - 1)/2)!
And as this is equal to 2^9 9!, we must have (m - 1)/2 = 9, so m = 19.
merry stratus
Well, your approach is probably better for this situation 😅