How can I deal with Functional Equations

When I came across this problem from PMO: f(a)+1/f(b) = f(1/a) + f(b) Where f is defined for all real numbers except zero. What are the possible values of f(1) - f(-1)?

Furthermore, how would I attack problems regarding functional equations, especially if the basic techniques may not work (eg. zeroing f(x))?

Note by John Ashley Capellan
7 years, 8 months ago

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Comments

Putting a=1,b=−1 a = 1, b = -1

f(1)+1f(−1)=f(1)+f(−1) f(1) + \frac{1}{f(-1)} = f(1) + f(-1)

1f(−1)=f(−1) \frac{1}{f(-1)} = f(-1)

f(−1)2=1 f(-1)^2 = 1

f(−1)=±1 f(-1) = \pm 1

Putting a=−1,b=1 a = -1, b = 1

f(−1)+1f(1)=f(−1)+f(1) f(-1) + \frac{1}{f(1)} = f(-1) + f(1)

1f(1)=f(1) \frac{1}{f(1)} = f(1)

f(1)2=1 f(1)^2 = 1

f(1)=±1 f(1) = \pm 1

Now, When f(1)=f(−1)=1, f(1) = f(-1) = 1,

f(1)−f(−1)=0 f(1) - f(-1) = 0

When f(1)=1,f(−1)=−1 f(1) = 1 , f(-1) = -1

f(1)−f(−1)=2 f(1) - f(-1) = 2

When f(1)=−1,f(−1)=1 f(1) = -1, f(-1) = 1

f(1)−f(−1)=−2 f(1) - f(-1) = -2

When f(1)=−1,f(−1)=−1 f(1) = -1 , f(-1) = -1

f(1)−f(−1)=0 f(1) - f(-1) = 0

Therefore possible values of f(1)−f(−1)=−2,0,2 f(1) - f(-1) = -2, 0 ,2

Siddhartha Srivastava - 7 years, 8 months ago

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Such elegant solution!

John Ashley Capellan - 7 years, 8 months ago
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