Suppose $f,g,h$ are functions from the set of positive real numbers into itself satisfying $$f(x)g(y)=h\left(\sqrt {x^2+y^2} \right)\ \ \forall \ x,y\in (0,\infty )$$ Show that the three functions $\frac {f(x)}{g(x)},\frac{g(x)}{h(x)},\frac{h(x)}{f(x)}$ are all constants. I only succeed in to show $f/g$ is constant function i,e, i got $f(x)=g(x)$. Anyone can help me to prove that $g/h$ is constant.
Asked
Active
Viewed 263 times
2
-
It is better if you show how you got $ f(x)=g(x)$. – Extremal Oct 11 '17 at 04:50
-
@Derivative divide $f(x)g(y)$ by $f(y)g(x)$ we will get $f(x)=g(y))$ for all x,y – Girish Kumar Chandora Oct 11 '17 at 04:53
-
That is not exactly correct. – Extremal Oct 11 '17 at 04:55
-
@Derivative can you explain it – Girish Kumar Chandora Oct 11 '17 at 04:56
-
See my answer below. – Extremal Oct 11 '17 at 04:58
-
Are you sure that here $x,y$ cannot be $0$? – Extremal Oct 11 '17 at 05:04
-
@Derivative yes the question is absloutely correct and this question was asked in M.sc entrance exam – Girish Kumar Chandora Oct 11 '17 at 05:06
-
Why not use that $f(x)g(0) = h(x)$ and $f(0)g(y) = h(y)$? – Matthew Cassell Oct 11 '17 at 06:19
-
Mattos: $x,y$ cannot be $0$ – Extremal Oct 11 '17 at 13:00
2 Answers
2
So, by dividing by $f(y)g(x)$, you get $f(x)/g(x)=f(y)/g(y)$ for all positive $x,y$. So then obviously $f/g$ is constant.
Ok, here is my try: $f(x)g(1)=h(\sqrt{x^2+1})$ and $f(x)g(\sqrt{2})=h(\sqrt{x^2+2})=f(\sqrt{x^2+1})g(1)$.
So,$f(x)g(\sqrt{2})=f(\sqrt{x^2+1})g(1)$. i.e. $f(x)=f(\sqrt{x^2+1})g(1)/g(\sqrt{2})$. Hence $f(\sqrt{x^2+1})/h(\sqrt{x^2+1})=g(\sqrt{2})/g(1)^2$. Now change variable $y=\sqrt{x^2+1}$. But we are done only for $y>1$.
Extremal
- 5,785
-
i also did the same thing. But can u prove h/g is constant – Girish Kumar Chandora Oct 11 '17 at 05:00
-
-
sorry it was $f=kg$ where is constant. The main aim is to prove h/g is constant – Girish Kumar Chandora Oct 11 '17 at 05:04
-
-
1you can change 1 and 2 to any positive number and then the equality hold for all y – vita nova Oct 11 '17 at 10:24
-
@novavita But not intependently, one must be the double of the other... – skyking Oct 12 '17 at 06:01
2
Assume that you have two positive numbers $u$ and $v$ and select $c$ smaller than both. Then you have that you can write $u=\sqrt{x^2+c^2}$ and $v=\sqrt{y^2+c^2}$ for some positive $x$ and $y$
$$f(x)g(c) = h\left(\sqrt{x^2+c^2}\right)$$ $$f(x)g(c\sqrt2) = h\left(\sqrt{x^2+2c^2}\right) = f\left(\sqrt{x^2+c^2}\right)g(c)$$
So
$${h(u)\over f(u)} = {h\left(\sqrt{x^2+c^2}\right)\over f\left(\sqrt{x^2+c^2}\right)} = {f(x)g(c)\over f(x)g(c\sqrt2)/g(c)} = g(c)^2/g(\sqrt c)$$
In similar way you get
$${h(v)\over f(v)} = {g(c)^2 \over \sqrt c} = {h(u)\over f(v)}$$
That is $h(v)/f(v) = h(u)/f(u)$ for any positive $u$ and $v$ which means that $h/f$ is constant.
skyking
- 16,654
-
0 is not positive real number and it us not included in our domain. If 0 is included then everyone can solve it – Girish Kumar Chandora Oct 11 '17 at 05:33
-