4

Simple Question: If two functions $f(x)$ and $g(x)$ are provided, Wolfram Alpha suggests that $\sqrt{f(x)\cdot g(x)}$ is only equal to $\sqrt{f(x)}\cdot\sqrt{g(x)}$ when $x > 0$.

I searched "Is (f(x)^(1/2*g(x)^(1/2)) = ((f(x)*g(x))^(1/2)) ?"

May you please explain this condition?

Thank you.

Michael Hardy
  • 1
  • Wolfram Alpha suggests that ... Please quote what you literally typed into WA. – dxiv Aug 26 '17 at 03:59
  • 1
    Sure thing: I searched "Is (f(x)^(1/2)g(x)^(1/2)) = ((f(x)g(x))^(1/2)) ?" –  Aug 26 '17 at 03:59
  • Assuming you meant Is (f(x)^(1/2)*g(x)^(1/2)) = ((f(x)*g(x))^(1/2)) ? the WA answer I see is "$\sqrt{f(x)} \sqrt{g(x)}$ is not always equal to $\sqrt{f(x) g(x)}$" which is not the same as what you quoted. – dxiv Aug 26 '17 at 04:03
  • 2
    @dxiv if you wait for it to load, it says it is true when $x>0$ ... probably a bug in WA – Zubin Mukerjee Aug 26 '17 at 04:03
  • @ZubinMukerjee All I get after the wait is "Standard computation time exceeded...". If it does indeed say it's true for $x \gt 0$ then it must be a WA bug, indeed. – dxiv Aug 26 '17 at 04:06
  • @OswaldChisala For real functions what is true is that $\sqrt{f(x) \cdot g(x)} = \sqrt{,|f(x)|,} \cdot \sqrt{,|g(x)|,}$. – dxiv Aug 26 '17 at 04:10
  • 3
    Rather than speaking of whether $x>0$ or $x\ge0,$ one should speak of whether $f(x)\ge0$ and $g(x)\ge0. \qquad$ – Michael Hardy Aug 26 '17 at 04:11
  • 1
    @dxiv See the screenshot. I think if you refresh a few times you might get the same output ... @ MichaelHardy Yes, that is probably what they meant to write. – Zubin Mukerjee Aug 26 '17 at 04:13
  • 2
    @ZubinMukerjee +1 for that. As a side thought, I would add a note to the complex case making it plenty clear that $,\sqrt{\cdot},$ denotes the principal value of the complex square root in that case. – dxiv Aug 26 '17 at 04:16

1 Answers1

4

The condition is not correct: for example, let

$$f(x) = -2$$

$$g(x) = -8$$

Then, when $x=1$, which satisfies $x>0$, we get

$$\sqrt{f(x)g(x) }= \sqrt{-2 \cdot -8} = \sqrt{16} = 4$$

but

$$\sqrt{f(x)}\sqrt{g(x)} = \sqrt{-2}\sqrt{-8} = \left(i\sqrt{2} \right)\left(i\sqrt{8}\right)= -4$$

assuming principal square roots, as opposed to $-i\sqrt{2}$ and $-i\sqrt{8}$

For what it's worth, I put in the same query and got the same erroneous result:

Wolfram|Alpha output

Alternate form assuming $x>0$: $\sqrt{f(x)}\sqrt{g(x)}$ is always equal to $\sqrt{f(x)g(x)}$

It's probably a bug in Wolfram|Alpha.

As Michael Hardy has pointed out, if $f(x) \geq 0$ and $g(x) \geq 0$, then it is always true that $$\sqrt{f(x)}\sqrt{g(x)} = \sqrt{f(x) g(x)}$$

I think it's likely that this positive-radicand case is what Wolfram|Alpha is intended to output for the "alternate form" section.

Zubin Mukerjee
  • 17,832
  • 1
    Thanks Zubin, there's a good chance you're right about it being a bug. Kudos. –  Aug 26 '17 at 04:16