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Problem of the week #77 - September 16th, 2013

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Jameson

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Jan 26, 2012
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Given \(\displaystyle f(x) = \frac{x}{\sqrt{x^2 + 1}}\), evaluate \(\displaystyle \underbrace{f(f(f( \dots f}_{2013}(x) \dots ))).\)
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Jameson

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Jan 26, 2012
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Congratulations to the following members for their correct solutions:

1) MarkFL
2) johng
3) soroban
4) eddybob123
5) BAdhi
6) anemone

Solution (from eddybob123):
We observe the behavior of the first few compositions:
$$f(x)=\frac{x}{\sqrt{x^2+1}}$$
$$(f^2)(x)=\frac{f(x)}{\sqrt{(f(x))^2+1}}= \frac{\frac{x}{ \sqrt{x^2+1}}} {\sqrt{\frac{x^2}{x^2+1}+ \frac{x^2+1}{x^2+1}}}=\frac{x}{\sqrt{2x^2+1}}$$
$$(f^3)(x)=\frac{(f^2)(x)}{\sqrt{((f^2)(x))^2+1}}= \frac{\frac{x}{\sqrt{2x^2+1}}}{ \sqrt{\frac{x^2}{2x^2+1}+\frac{2x^2+1}{2x^2+1}}}= \frac{x}{\sqrt{3x^2+1}}$$
It appears that the following rule holds for all positive integers $n$:
$$(f^n)(x)=\frac{x}{\sqrt{nx^2+1}}$$
This is an inductive proof. We have already done the initial step. We need to show that if the above formula holds for an arbitrary positive integer $k$, then it holds for $k+1$:
$$(f^k)(x)=\frac{x}{\sqrt{kx^2+1}}$$
$$(f^{k+1})(x)=\frac{(f^k)(x)}{\sqrt{((f^k)(x))^2+1}}= \frac{\frac{x}{\sqrt{kx^2+1}}}{\sqrt{\frac{x^2}{kx^2+1}+\frac{kx^2+1}{kx^2+1}}}=\frac{x}{\sqrt{(k+1)x^2+1}}\,\,\,\,\boxed{}$$
Hence,
$$(f^{2013})(x)=\frac{x}{\sqrt{2013x^2+1}}$$
 
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