Algebraic Illusion
Let f be an injective function such that f ( n ) = 2 f ( m ) for every pair of real numbers ( m , n ) satisfying the equation m 2 + n 2 m − 1 = 0 Also, f ( 0 . 0 6 2 5 8 ) = 1 6 1 What is the value of ⌊ 1 0 0 f − 1 ( 1 ) ⌋ ?
The answer is 155.
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1 solution
First I have tried tan(1) in degrees and then in radian.
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I thought beforehand about that. People will more likely fall in this trap because most calculators have degrees as the default input.
Very Sneaky! f(1) = pi/4 should've been a dead give away. By the way, I had some strange things happen when I tried some iterative techniques in Matlab. For instance, repeatedly substituting 2m/(1-m^2) into itself jumps around back and forth (starting the iteration at pi/4), but the function value should be repeatedly doubling. I'm not sure why that seems to be happening.
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This problem is not well-written. We're actually finding p such that f ( p ) = 1 .
We can go two ways. (1) f ( n ) = 2 f ( m ) (2) f ( m ) = 2 1 f ( n )
However, the values that this function would produce from f ( 1 ) = 4 π can be written as 2 k π . If we want to equate it to 1 , we will get k ≈ − 1 . 6 5 . If k doesn't turn out to be an integer, then we cannot get there. I'll probably change the condition f ( 1 ) = 4 π to accommodate for a more useful condition.
By the way, I am interested in knowing your iterative method in detail.
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Here is some Matlab code I just wrote.for i = 1:12 x = pi/4; for n = 1:10000 i x = 2 x/(1-x^2); end x end As you can see, x jumps around, but f(x (n+1)) = 2*f(x n) with x_0 = pi/4. So the function values are doubling each time but x goes back and forth.
How about starting the iteration with f ( 0 . 0 6 2 5 8 ) = 1 6 1 and f ( 1 − m 2 2 m ) = 2 f ( m ) ?
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That seems to jump around as well, indicating that the function is extremely wavy with tall waves. I'm confused as to why that happens.
How can we ensure that no other function satisfies this functional equation?
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Actually, I couldn't ensure that. So, I changed the condition f ( 1 ) = 4 π to f ( 0 . 0 6 2 5 8 ) = 1 6 1 .
The before-edit approach
From the equation, n 2 m = 1 − m 2 or, n = 1 − m 2 2 m So, f ( 1 − m 2 2 m ) = 2 f ( m ) Hence, f is the arctangent function [ 2 t a n − 1 x = t a n − 1 1 − x 2 2 x ].
Therefore, f − 1 ( 1 ) = t a n ( 1 ) which implies the the answer is ⌊ 1 0 0 t a n ( 1 ) ⌋ = 1 5 5 .
The after-edit approach
We are trying to find such p which satisfies f ( p ) = 1 . Let, x 0 = 0 . 0 6 2 5 8 and, x i + 1 = 1 − x i 2 2 x i
Hence, f ( x i + 1 ) = 2 f ( x i ) because f ( 1 − x i 2 2 x i ) = 2 f ( x i )
Using this recursive equation, we get f ( x 4 ) = 2 f ( x 3 ) = 4 f ( x 2 ) = 8 f ( x 1 ) = 1 6 f ( x 0 ) = 1
So, p = x 4 = f − 1 ( 1 ) ≈ 1 . 5 5 7
P . S .
The condition f ( 1 ) = 4 π was later changed into f ( 0 . 0 6 2 5 8 ) = 1 6 1 because the functional equation doesn't necessarily lead to inverse tangent function.