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Square root of a fraction

asked 2021-12-09 10:32:05 +0200

keko gravatar image

updated 2021-12-09 10:34:32 +0200

I have the following variable and function:

f = function('f')(r)

with an equation given by:

eq = sqrt(-1/(f^2*sin(r)^2))-1/sqrt(-f^2*sin(r)^2)

which, in LaTeX expression, looks like:

\begin{equation} \sqrt{-\frac{1}{f\left(r\right)^{2} \sin\left(r\right)^{2}}} - \frac{1}{\sqrt{-f\left(r\right)^{2} \sin\left(r\right)^{2}}} \end{equation}

In principle, this should be equal to zero. However,


gives me False. I have tried with .simplify_full() and following the suggestion in here, I have used ._sympy_().simplify():

eq = eq._sympy_().simplify()

but I still get False. Am I missing something?

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answered 2021-12-09 12:51:49 +0200

cav_rt gravatar image

By doing

$\sqrt{-\frac{1}{f\left(r\right)^{2} \sin\left(r\right)^{2}}} - \frac{1}{\sqrt{-f\left(r\right)^{2} \sin\left(r\right)^{2}}} = \frac{i}{\sqrt{f\left(r\right)^{2} \sin\left(r\right)^{2}}}-\frac{1}{i\sqrt{f\left(r\right)^{2} \sin\left(r\right)^{2}}} = \frac{2i}{\sqrt{f\left(r\right)^{2} \sin\left(r\right)^{2}}}$

I wouldn't expect bool(eq==0) to return True.

You can deal with square roots using canonicalize_radical().


returns $\frac{2 i}{f\left(r\right) \sin\left(r\right)}$.

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answered 2021-12-15 17:08:05 +0200

keko gravatar image

I have realized that my problem here was of mathematical nature rather than computational. I was assuming that

$-\frac{a}{b} =\frac{a}{-b}$ $\implies$ $\sqrt{-\frac{a}{b}}=\sqrt{\frac{a}{-b}}$ $\implies$ $\frac{\sqrt{-a}}{\sqrt{b}}=\frac{\sqrt{a}}{\sqrt{-b}}$

However, if we consider only the positive branches of the square roots, the last equality does not hold.

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Asked: 2021-12-09 10:32:05 +0200

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Last updated: Dec 15 '21