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- Jan 26, 2012

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- Thread starter Jameson
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- Thread starter
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- #1

- Jan 26, 2012

- 4,055

--------------------

Remember to read the POTW submission guidelines to find out how to submit your answers!

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- #2

- Jan 26, 2012

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1) MarkFL

2) johng

3) soroban

4) eddybob123

5) BAdhi

6) anemone

Solution (from eddybob123):

$$f(x)=\frac{x}{\sqrt{x^2+1}}$$

$$(f^2)(x)=\frac{f(x)}{\sqrt{(f(x))^2+1}}= \frac{\frac{x}{ \sqrt{x^2+1}}} {\sqrt{\frac{x^2}{x^2+1}+ \frac{x^2+1}{x^2+1}}}=\frac{x}{\sqrt{2x^2+1}}$$

$$(f^3)(x)=\frac{(f^2)(x)}{\sqrt{((f^2)(x))^2+1}}= \frac{\frac{x}{\sqrt{2x^2+1}}}{ \sqrt{\frac{x^2}{2x^2+1}+\frac{2x^2+1}{2x^2+1}}}= \frac{x}{\sqrt{3x^2+1}}$$

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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