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Functional Equations(over reals)

A couple of doubts:
Q1)\(f(x+2y)=f(x)+2f(y)\) given that \(f(1)=2,f(0)=0\)
Q2)\(f(x-f(y))=f(f(y))+xf(y)+f(x)-1\)

Note by Siddharth G
1 year, 11 months ago

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Q1. Putting \( y' = 2y \), we get \( f(x+y') = f(x) + f(y') \), which is the Cauchy Functional Equation.

Q2 is IMO 1999/Q6 Siddhartha Srivastava · 1 year, 11 months ago

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@Siddhartha Srivastava Thanks for the second! And I made a typo in the first one, I just edited it. Sorry Siddharth G · 1 year, 11 months ago

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@Siddharth G It's almost the same anyways.

Putting \( x = 0 \), we get \( f(2y) = 2f(y) \).

Now we have \( f(x+y) = f(x+2\frac{y}{2}) = f(x) + 2f(\frac{y}{2}) = f(x) + f(y) \) , from the above statement. Cauchy's Functional Equation all over again. Siddhartha Srivastava · 1 year, 11 months ago

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