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If $f(x)=\frac{x}{\sqrt{1+x^2}}$ prove by mathematical induction that $f(f(...f(x)...))$ = $\frac{x}{\sqrt{1+nx^2}}$.

(Noting that there are n $f$'s in the LHS)

I have no idea where to start with this one. Which method is best suited for proving this?

Thanks

Jean Marie
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Brown
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    Hint: Let $g(x) = \frac{x}{\sqrt{1 + nx^2}}$. Then compute $g(f(x))$, i.e. $$\frac{\left[\frac{x}{\sqrt{1 + x^2}}\right]}{\sqrt{1 + n\left[\frac{x}{\sqrt{1 + x^2}}\right]^2}}.$$What do you notice? – Theo Bendit Jun 20 '19 at 02:11
  • Duplicate of this (https://math.stackexchange.com/q/581059) (nice answer by Lubin). – Jean Marie Apr 29 '20 at 09:39

2 Answers2

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Just do it by induction. You can set up a base case like showing that $f(f(x)) = \frac{x}{\sqrt{1+2x^2}}$ i.e. in the form above. Then assuming that $f^{n}(x)$ (composed n times) is of the form above, then apply $f$ again and show that we recover the function $\frac{x}{\sqrt{1+(n+1)x^2}}$ as required.

To start you off, $f(f(x)) = \frac{\frac{x}{\sqrt{1+x^2}}}{\sqrt{1+\frac{x^2}{1+x^2}}} = \frac{x}{\sqrt{1+x^2}\frac{\sqrt{1+2x^2}}{\sqrt{1+x^2}}}$

Theo Bendit
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fGDu94
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The case for $n=1$ comes from the definition $$ f(x) = \frac {x}{\sqrt {1+x^2}}$$

Noe if the statement if true for $n$, we need to show that it is true for $n+1$

That is if $$f^{n} (x) =\frac{x}{\sqrt{1+nx^2}}$$ then $$f^{n} (\frac {x}{\sqrt {1+x^2}}) = \frac {x}{\sqrt {1+(n+1)x^2}}$$

Note that $$f^{n}(\frac {x}{\sqrt {1+x^2}})= \frac{\frac{x}{\sqrt{1+x^2}}}{\sqrt{1+\frac{nx^2}{1+x^2}}} = \frac {x}{\sqrt {1+(n+1)x^2}}$$

Mohammad Riazi-Kermani
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