Show the following for $f(x)=\dfrac{1}{1+e^{-x}}$

Asked Jan 13 '21 at 13:40

Active Jan 15 '21 at 03:31

Viewed 200 times

$f(x)=\dfrac{1}{1+e^{-x}}$ is sigmoid function. Clearly $f(x)$ is bijective

Show that for an arbitrary interval $(a,b)$, by using the composition of $f$ with an appropriate function $g: (0,1) \to (a,b)$, show that $|\mathbb{R}|=|(a,b)|$ and use this to prove that $[a,b]\approx(a,b)$

This is an exercise in my book. But I couldnt solve it. Please help

Edited:

Considering $g: (0,1) \to (a,b), $ $g(x)=(b-a)x+a$ this function is bijective. So $gof$ is also bijective. So can we say that his shows $|\mathbb{R}|=|(a,b)|$

For the second part I think we need a function lets say $h$ such that $h:[a,b]→(a,b)$ is a bijection.

edited Jan 15 '21 at 03:31

Eric Wofsey

330,363

asked Jan 13 '21 at 13:40

Jalil Ahmad

Note that $f$ is a bijection between $\mathbb{R}$ and $(0,1)$. – Gary Jan 13 '21 at 14:32
@Gary can you please elaborate what this gives and how to proceed next ? – Jalil Ahmad Jan 13 '21 at 15:39
Take $g(x) = (b - a)x + a$. This is a bijection between $(0,1)$ and $(a,b)$. What can you say about $g(f(x))$? – Gary Jan 13 '21 at 16:20
@Gary so using $g(x)$ you proposed we have $g(f(x))=\dfrac{b-a}{1+e^{-x}}+a$ – Jalil Ahmad Jan 14 '21 at 20:46
2

@Gary as $f$ and $g$ are bijection so $gof$ will be bijection as well. so can we say that this shows $|\mathbb{R}| =|(a,b)|$ – Jalil Ahmad Jan 14 '21 at 23:28
1

@Gary for the second part I think we need a function lets say $h$ such that $h: [a,b] \to (a,b)$ is a bijection. – Jalil Ahmad Jan 14 '21 at 23:30
Your answer to the first part is correct. For the second part, see https://math.stackexchange.com/q/28570 – Gary Jan 15 '21 at 07:02
@Gary Thank you.. – Jalil Ahmad Jan 15 '21 at 07:06

Show the following for $f(x)=\dfrac{1}{1+e^{-x}}$

0 Answers0