A function whose reciprocal is an inverse

Hi! Can anyone think of any continuous function that satisfies the following? ‘-1’ denotes the ‘inverse’ of the function.

f1(x)=1f(x) f^{-1}(x)= \frac{1}{f(x)}

Note by Inquisitor Math
7 months, 2 weeks ago

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A close solution is f(x) = (1+x)/(1-x) which results in f(f(x)) = -1/x. Looking at the function by substituting x = tan θ, we can easily see that f simply adds π/4 to the argument θ, which when done twice in a row gives -cot θ or -1/x.

Trying to find a function in the reals which satisfies f(f(x)) = 1/x may ,thus , be impossible as simply adding some φ to the argument θ can never result in cot θ , although I don't have a proof for this.

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If you try to find functions of the form of a Möbius transformation, we get a^2 + b^2 =0, a = d, b = c. So, we get two solutions

        f(x) = ( ix + 1)/(x + i )        and       f(x) = (-ix + 1)/(x - i )

where i is the square root of -1.

A Former Brilliant Member - 4 months, 1 week ago

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Bingo! f(x)=xif(x)=x^{i }

Jeff Giff - 7 months, 1 week ago

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Wow, I was thinking it would be hard to get a real function but you have taken yours to complex

And btw replace 1\sqrt{-1} with iota because 1\sqrt{-1} is equal to iota and -iota

And one more you have to restrict the domain of the function so that the inverse exists, one restriction could be x(0,e2π]\forall x \in (0,e^{2π}]

Jason Gomez - 7 months, 1 week ago

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I think I heard of one... note that 1f(f(x))=x.\dfrac{1}{f(f(x))}=x.

Jeff Giff - 7 months, 1 week ago

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Just think about this, xf(f(x))=1, if we put x=0, we have 0=1, which is absurd, thus f(f(x)) is discontinuous at x=0, but then that means f(x) must also be discontinuous at some point. Thus we cannot find a continuous function, whose reciprocal = inverse.

Kushal Dey - 4 months, 1 week ago

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You can restrict the domain

Jason Gomez - 4 months, 1 week ago

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f(x)=x1f(x) = x^{-1}

f1(x)=x1f^{-1}(x) = x^{-1}

Yajat Shamji - 7 months, 1 week ago

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Here f(x)=f1(x)=1f(x)f(x)=f^{-1}(x)\cancel{=}\frac{1}{f(x)}

Wasi Husain - 7 months, 1 week ago

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There may be no function such that:

f1(x)=1f(x)f^{-1}(x) = \frac{1}{f(x)}

Yajat Shamji - 7 months, 1 week ago

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@Yajat Shamji f:{1}{1},f(x)=xf:\{1\} ➝ \{1\} ,f(x)=x satisfies

Jason Gomez - 7 months, 1 week ago

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@Jason Gomez Continuity cannot be defined for such a function.

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