Recurring Styles #5 - Really a unique Solution?
f n + 1 ( x ) = x 2 + 6 f n ( x ) , f 0 ( x ) = 8
Define a sequence of functions f 0 , f 1 , f 2 , … such that it satisfies the above conditions, for whole numbers n , and real x . For every positive integer n , solve the equation f n ( x ) = 2 x , for x .
The answer is 4.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
2 solutions
dude, you are so fetch, how are you so smart, please tell me the book you use for practicing and learning all such stuff.
Log in to reply
That's a weird question. Well, this is my profession, so to say. I've been taking interest in Mathematics since I was a baby, so!
We have to solve for n ≥ 1 , but it seems reasonable to assume that n = 0 gives the same solution.
Equating f 0 ( x ) = 8 = 2 x immediately gives x = 4 , and that solution works because 4 2 + 6 ⋅ 8 = 8 .
More rigorously, at the desired value for x all f n ( x ) should be equal, so that in particular f n + 1 ( x ) = f n ( x ) = 2 x . Then we have 2 x = x 2 + 6 ⋅ 2 x 4 x 2 = x 2 + 1 2 x 3 x 2 = 1 2 x x = 4
Since f n ( x ) is positive, f n ( x ) = 2 x has only positive solutions. We'll show that, for each n , f n ( x ) = 2 x has a solution x = 4 . Now, Since f 1 ( x ) = x 2 + 4 8 , x = 4 is a solution of f 2 ( x ) = 2 x . Now f_{n+1}(4) = \sqrt{4^2 + 6f_n(4)} = \sqrt{4^2 + 6 \cdot 8} = 8 = 2 \cdot 4 , which completes the inductive step. Next, induction on n gives us that for each n , x f n ( x ) decreases as x increases in ( 0 , ∞ ) . It follows that f n ( x ) = 2 x has the unique solution x = 4 .