Decoding the function!

Algebra Level 3

{ f ( 1 + x ) = 1 + f ( x ) x 2 f ( 1 x ) = f ( x ) , x ≠ 0 \begin{cases} f(1+x) = 1 + f(x) \\ x^2f\left(\dfrac 1x\right) = f(x), \quad x \ne 0 \end{cases}

If f : R → R f: \mathbb {R \to R} is an odd function satisfying the system of equations above, find f ( 137 ) f(137) .


The answer is 137.

Mayank Jain by Mayank Jain , 22, India

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1 solution

Chew-Seong Cheong
Nov 10, 2016

Consider

f ( 1 + x ) = 1 + f ( x ) f ( 1 + ( 1 + x ) ) = 1 + f ( 1 + x ) f ( 2 + x ) = 2 + f ( x ) f ( 1 + ( 2 + x ) ) = 1 + f ( 2 + x ) f ( 3 + x ) = 3 + f ( x ) . . . = . . . ⟹ f ( n + x ) = n + f ( x ) Putting x = 1 f ( n + 1 ) = n + f ( 1 ) . . . ( 1 ) \begin{aligned} f(1+x) & = 1 + f(x) \\ f(1+(1+x)) & = 1 + f(1+x) \\ f(2+x) & = 2 + f(x) \\ f(1+(2+x)) & = 1 + f(2+x) \\ f(3+x) & = 3 + f(x) \\ ... \quad & = \quad ... \\ \implies f(n+x) & = n + f(x) & \small {\color{#3D99F6}\text{Putting }x=1} \\ f(n+1) & = n + f(1) & ...(1) \end{aligned}

Consider also

x = − 1 2 : f ( 1 − 1 2 ) = 1 + f ( − 1 2 ) Since f ( x ) is an odd function. f ( 1 2 ) = 1 − f ( 1 2 ) ⟹ f ( 1 2 ) = 1 2 x = 2 : 2 2 f ( 1 2 ) = f ( 2 ) ⟹ f ( 2 ) = 2 2 × 1 2 = 2 \begin{aligned} x = -\frac 12: \quad f \left(1-\frac 12\right) & = 1 + {\color{#3D99F6} f \left(-\frac 12\right)} & \small {\color{#3D99F6}\text{Since }f(x) \text{ is an odd function.}} \\ f \left(\frac 12\right) & = 1 {\color{#3D99F6} - f \left(\frac 12\right)} \\ \implies f \left(\frac 12\right) & = \frac 12 \\ x = 2: \quad 2^2 f \left(\frac 12\right) & = f(2) \\ \implies f(2) & = 2^2 \times \frac 12 = 2 \end{aligned}

( 1 ) : f ( n + 1 ) = n + f ( 1 ) Putting n = 1 f ( 2 ) = 1 + f ( 1 ) 2 = 1 + f ( 1 ) ⟹ f ( 1 ) = 1 ( 1 ) : ⟹ f ( n + 1 ) = n + 1 ⟹ f ( x ) = x ⟹ f ( 137 ) = 137 \begin{aligned} (1): \quad f(n+1) & = n + f(1) & \small {\color{#3D99F6}\text{Putting }n=1} \\ f(2) & = 1 + f(1) \\ 2 & = 1 + f(1) \\ \implies f(1) & = 1 \\ (1): \implies f(n+1) & = n + 1 \\ \implies f(x) & = x \\ \implies f(137) & = \boxed{137} \end{aligned}

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Your solution shows that f ( n ) = n f(n) = n for integers n n .

Are we able to conclude that f ( x ) = x f(x) = x for all real numbers x x ?

Calvin Lin Staff - 4 years, 7 months ago

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Tried but couldn't get it. Any reference?

Chew-Seong Cheong - 4 years, 7 months ago

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If you look at the maps x → 1 + x x \rightarrow 1 + x , x → 1 x x \rightarrow \frac{1}{x} and x → − x x \rightarrow - x , these help us define the types of values that we can get to by knowing the value of x x . From the value of 0, we can easily get from to all of the integers, and then from the integers to the reciprocals of the integers.

In fact, we can also get to all of the rational numbers, by simply using the continued fractions approach. E.g. if we want to get to 2 3 = 0 + 1 1 + 1 2 \frac{ 2}{3} = 0 + \frac{ 1 } { 1 + \frac{1}{2} } , we first obtain 1 \rightarrrow 2 → 1 2 → 1 + 1 2 → 1 1 + 1 [ 2 1 \rightarrrow 2 \rightarrow \frac{1}{2} \rightarrow 1 + \frac{1}{2} \rightarrow \frac{1}{1 + \frac{1}[2} .

Using this, we can show that for all rational numbers, we must have f ( p q ) = p q f( \frac{p}{q} ) = \frac{p}{q} .

However, we have less information about the irrational numbers. I believe that for with an algebraically independent irrational basis, we can define f ( α i ) = β i f(\alpha_i) = \beta_i , and then follow through the consequences.

Calvin Lin Staff - 4 years, 7 months ago

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