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Suppose that we have a function $f$ where $Q(\tau)$ is modified by multiplying $Q(\tau)$ by a real number. Let $f'$ be the modified function, and let $k \in \mathbb{R}$ or $k \in \mathbb{C}$. Take the ratio:

$\frac{f}{{f'}} = k$

$f = \exp \left( { - {{\int\limits_0^\tau {\frac{\omega }{{2Q(\tau ')}}\left( {\frac{\omega }{{{\omega _h}}}} \right)} }^{\frac{{ - 1}}{{\pi Q(\tau ')}}}}d\tau '} \right)\exp \left( {i\int\limits_0^\tau {\left( {{{\left( {\frac{\omega }{{{\omega _h}}}} \right)}^{\frac{1}{{\pi Q(\tau ')}}}} - 1} \right)\omega } d\tau '} \right)$

$f' = \exp \left( { - {{\int\limits_0^\tau {\frac{\omega }{{4Q(\tau ')}}\left( {\frac{\omega }{{{\omega _h}}}} \right)} }^{\frac{{ - 1}}{{2\pi Q(\tau ')}}}}d\tau '} \right)\exp \left( {i\int\limits_0^\tau {\left( {{{\left( {\frac{\omega }{{{\omega _h}}}} \right)}^{\frac{1}{{2\pi Q(\tau ')}}}} - 1} \right)\omega } d\tau '} \right)$

Note that the only thing changed between $f$ and $f'$ is that $Q(\tau)$ in $f$ has become $2Q(\tau)$ in $f'$.

Is it possible to find a constant value $k$ for this ratio? For example, $k = 1/2$. Why or why not?

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  • $\begingroup$ Not in general. Consider $f(x) = x^2 + x$. As for your given function, it is unclear what the variable is. $\endgroup$ – Karolis Juodelė Jul 24 '12 at 19:01
  • $\begingroup$ Thanks, Karolis. The variable is $Q(\tau)$, so that $Q(\tau)$ in $f$ has become $2Q(\tau)$ in $f'$. $\endgroup$ – Nicholas Kinar Jul 24 '12 at 19:17
  • $\begingroup$ $Q(\tau)$ cannot be a variable, as it does not have a constant meaning in the expression - it depends on $\tau'$ which itself is not a constant. $\endgroup$ – Karolis Juodelė Jul 24 '12 at 19:24
  • $\begingroup$ OK - then maybe the terminology is wrong. I've updated my question. $\endgroup$ – Nicholas Kinar Jul 24 '12 at 19:39
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Take for instance, for $f:\mathbb{R}\to\mathbb{R}$ and $f':\mathbb{R}\to\mathbb{R}$, let us also let our scale factor be $2$:

$$f(x)\equiv\int_{0}^{x}{\sin{x}\:dx}=\left[-\cos{x}\right]_{0}^{x}=-(\cos{x}-1)$$

And:

$$f'(x)\equiv\int_{0}^{2x}{\sin{x}\:dx}=\left[-\cos{x}\right]_{0}^{2x}=-(\cos{2x}-1)$$

So we have:

$$\frac{f(x)}{f'(x)}=\frac{\cos{x}-1}{\cos{2x}-1}=\frac{1}{2(1+\cos{x})}$$

Which is clearly dependent on $x$.

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  • $\begingroup$ Yes - I see that now, Shaktal. Thank you for pointing this out. $\endgroup$ – Nicholas Kinar Jul 24 '12 at 19:18

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