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) $f(x-f(y))=f(f(y))+xf(y)+f(x)-1$

#Algebra #FunctionalEquations #RealNumbers

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Comments

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 - 6 years, 4 months ago

Thanks for the second! And I made a typo in the first one, I just edited it. Sorry

Siddharth G - 6 years, 4 months ago

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 - 6 years, 4 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$