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I am trying to solve the equation $f(x+t)=f(x)+f(t)+2\sqrt{f(x)}\sqrt{f(t)}$ - as in find a function that satisfies this equation. I notice that the RHS is $({\sqrt{f(x)}+\sqrt{f(t)}})^2$ but I am stuck after this.

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    If it is known that the answer is a polynomial, we can see that since f(0)=0, then $f(x)=ax^n$. Using the given expression we can prove that a=1.2012-01-01

3 Answers 3

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The presence of $\sqrt{f(x)}$ in your functional equation implies that the range of $f$ is nonnegative.

If you are looking for a continuous function, Simon's comment shows that $g$ must be linear. (With real functions, continuity and additivity imply full linearity. This is mentioned here although I'd prefer to show a link to an actual proof.)

Therefore $\sqrt{f(x)} = ax$

for some $a$.

  1. If $a=0$, then $f$ is the zero function.
  2. If $a>0$, then $f$ is only defined for $x\geq 0$, and $f(x)=a^2x^2$.
  3. If $a<0$, then $f$ is only defined for $x\leq 0$, and $f(x)=a^2x^2$.

That is, there are three families of solutions. They all have the same form $f(x)=a^2x^2$, but either you must restrict the domain to non-negatives, non-positives, or $a $ is necessarily $0$.

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Use the hint and solve in the usual step by step way for $g$ first. Assume wlog $g(1)= 1$.

  1. solve for $x, t$ natural numbers and $0$ (domain can't be negative) to see $g(n) = n$.

  2. solve for fractions with numerator being $1$ to see $n\cdot g(1/n) = 1$.

  3. now for rationals to see $q/p \cdot g(p/q) = 1$

  4. use limit argument for irrationals to get $g(r) = r$

  5. plug back into $f$ to conclude one solution is $f(n) = n^2$.

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    Reals, not integers, obviously.2011-12-31
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solution: $f(x) = a^2 x^2$ where a is any real number

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    Except that it won't work if $x$ and $t$ have opposite signs.2012-01-01