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Find $f\circ f$ if $f(t)=\dfrac{t}{(1+t^2)^{1/2}},\ \ t\in \mathbb{R}$

$ a.)\ \dfrac{1}{(1+2t^2)^{1/2}} \\ \color{green}{ b.)\ \dfrac{t}{(1+2t^2)^{1/2}}}\\~\\ c.)\ (1+2t^2)\\~\\ d.)\ \text{none of these} \\$

I tried to put the value $t=1$ and after checking concluded that it's $d.)$ but the book is giving option $b.)$.

I look for a short and simple way.

I have studied maths up to 12th grade.

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  • $\begingroup$ $f(1) = \frac{1}{\sqrt 2}$ so $f \circ f (1) =f( \frac{1}{\sqrt 2})=\frac{1}{\sqrt 3}$ which matches both a) and b) $\endgroup$ – WW1 Jun 26 '15 at 23:15
  • $\begingroup$ f(0) = 0 eliminates a) and c). This doesn't prove that b) is correct, though. $\endgroup$ – Sjoerd Jun 26 '15 at 23:20
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Our formula for $f$ is $$f(\Box)=\frac{\Box}{\sqrt{1+\Box^2}}$$

With this in mind we have \begin{align*} f\bigl(f(t)\bigr) &=f\left(\fbox{$f(t)$}\right) \\ &=\frac{\fbox{$f(t)$}}{\sqrt{1+\fbox{$f(t)$}^2}} \\ &=\frac{\frac{t}{\sqrt{1+t^2}}}{\sqrt{1+\left(\frac{t}{\sqrt{1+t^2}}\right)^2}} \\ &=\frac{t}{\sqrt{1+\frac{t^2}{1+t^2}}\cdot\sqrt{1+t^2}} \\ &=\frac{t}{\sqrt{\left(1+t^2\right)+\frac{t^2}{1+t^2}\cdot\left(1+t^2\right)}} \\ &=\frac{t}{\sqrt{1+t^2+t^2}} \\ &= \frac{t}{\sqrt{1+2\,t^2}} \end{align*}

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  • 3
    $\begingroup$ +1 for the $\square$'s. I have always loved this notation to prove things like this. $\endgroup$ – Dmoreno Jun 26 '15 at 23:28
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$$ f(f(t))=\frac { \frac {t}{\sqrt {1+t^{ 2 }} } }{ \sqrt { 1+\frac { t^{ 2 } }{ 1+t^{ 2 } } } } =\frac { t }{ \sqrt { 1+2t^{ 2 } } }$$

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