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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.

MathMathCookie
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    Let $g(x) = \sqrt{f(x)}$ then you've noticed that $g(x+t) = g(x)+g(t)$. – Simon M Dec 31 '11 at 17:36
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    What domain and range? For example, is $f(x)$ defined and nonnegative for all real $x$, so that $\sqrt{f(x)}$ is real as well? Take this together with Simon's comment, and see if you can find the unique solution! – GEdgar Dec 31 '11 at 18:30
  • 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. – NoChance Jan 01 '12 at 20:41

3 Answers3

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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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gnometorule
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  • In the limit argument assume you want a continuous solution, as your task is to find "a" function, not the space of all functions. – gnometorule Dec 31 '11 at 18:57
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    $f(n) = n^2$ ?? $0 = 0^2 = f(0) = f((-1)+1) = f(1)+f(-1)+2\sqrt{f(1)}\sqrt{f(-1)} = 1^2+(-1)^2+2\sqrt{1^2}\sqrt{(-1)^2}= 1+1+2=4$, right? – GEdgar Dec 31 '11 at 21:12
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    Domain are the non-negative integers (see (1) in my answer). He needs one function, no general solution - domain isn't specified – gnometorule Dec 31 '11 at 21:16
  • Reals, not integers, obviously. – gnometorule Dec 31 '11 at 21:19
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solution: $f(x) = a^2 x^2$ where $a$ is any real number

Mohsen Shahriari
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wen
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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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