Do You See The Connection?
Let the sequence of real numbers x n be defined with
{ x 1 = k x n + 1 = 4 x n ( 1 − x n ) , n ≥ 1 .
Let there be N distinct values of k such that x 2 0 1 4 = 0 . Find the last three digits of N .
Note: For those interested, this question was partly inspired by the function f ( x ) = λ x ( 1 − x ) also know as the logistic function and is related to population growth.
The answer is 97.
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2 solutions
You made an off-by-one error. Since x 1 = k and you want x 2 0 1 4 = 0 , you are applying f 2013 times to k to get 0. Thus, the correct equation is f 2 0 1 3 ( k ) = 0 . This gives you an answer of 97.
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Thanks. I have udpated the answer to 97.
@Anqi Li Can you update the solution accordingly? Please also explain how to evaluate the final answer.
"Thus the roots of f^n(2014) = 0 precisely occurs when 2^n theta = k * pi". You don't want 2014 to be the argument of f. You should be substituting 2014 for n, not for sin^2(theta). Also, you're abusing notation to bring in a new meaning for k at the end. The k in k pi has no relationship to the k in k = sin^2(theta). Finally, you skip over the non-trivial task of finding the number of roots to 2^2014 theta = m*pi. The reader has no idea where 97 came from. What is the relationship between 97 and 2014?
S t a r t a t a 2 0 1 4 , have total number of solution is: 2 2 0 1 3 ) + 1 M O D ( 2 2 0 1 3 ) + 1 , 1 0 0 0 ) = 193]
x 2 0 1 4 = 4 x 2 0 1 3 ( 1 − x 2 0 1 3 ) = 0 ⇔ x 2 0 1 3 = 0 o r x 2 0 1 3 = 1 x 2 0 1 3 = 4 x 2 0 1 2 ( 1 − x 2 0 1 2 ) = 1 ⇔ 4 x 2 0 1 2 2 − 4 x 2 0 1 2 + 1 = 0 ( 1 ) Equation (1) has a root: x 2 0 1 2 = 2 1 Similarly: x 2 0 1 3 = 4 x 2 0 1 2 ( 1 − x 2 0 1 2 ) = 0 ⇔ x 2 0 1 2 = 0 o r x 2 0 1 2 = 1 x 2 0 1 2 = 4 x 2 0 1 1 ( 1 − x 2 0 1 1 ) = 2 1 ⇔ 4 x 2 0 1 1 2 − 4 x 2 0 1 1 + 2 1 = 0 ( 2 )
Equation (2) has two distinct roots: x 2 0 1 1 = 4 ± 2 + 2
Each value of x 2 0 1 1 will generate two distinct values of x 2 0 1 0 ... So x 2 0 1 3 = 1 will generate: 2 2 0 1 1 solutions. Similarly: x 2 0 1 2 = 1 will generate: 2 2 0 1 0 solutions. ... x 2 = 1 will generate: one solution x 2 = 0 will generate: two solutions Sum all possible solutions is as above. So result is: 1 9 3
can you please explain it a bit more
Um, I didn't understand why N is 2^2013 + 1, and not just 2^2013.
Actually, I had a IMHO quite elegant solution in mind.
The first thing is for notational convenience to define f ( x ) = 4 x ( 1 − x ) . So now we can rephrase the question to find distinct roots of f 2 0 1 4 ( k ) = 0
Now, as @AaaaaBbbbb noticed the image f ( x ) of x will be in the interval [ 0 , 1 ] if x ∈ ( 0 , 1 ) . To have any roots at all. x needs to be in [ 0 , 1 ] . In such a case, we can set k = sin 2 θ Some motivation so firstly, when we constrict x to intervals regarding 0 and 1 , we always opt for trigonometric substitutions , in this case we fumble around to find a nice thing to sub into f(x), the idea being that 1 − x resembles 1 − cos 2 θ which is easy to manipulate. Now notice that
f ( k ) = f ( sin 2 θ ) = 4 sin 2 θ ( 1 − sin 2 θ ) = ( 2 cos θ sin θ ) 2 = sin 2 ( 2 θ )
Iterating this,
f 2 ( k ) = f ( sin 2 ( 2 θ ) ) = sin 2 ( 4 θ )
Thus the roots of f n ( 2 0 1 4 ) = 0 precisely occurs when 2 n θ = k π . Just sub in n = 2 0 1 4 and you can get your answer.