Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
Not in general. Consider $f(x) = x^2 + x$. As for your given function, it is unclear what the variable is. – Karolis Juodelė Jul 24 '12 at 19:01
Thanks, Karolis. The variable is $Q(\tau)$, so that $Q(\tau)$ in $f$ has become $2Q(\tau)$ in $f'$. – Nicholas Kinar Jul 24 '12 at 19:17
$Q(\tau)$ 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. – Karolis Juodelė Jul 24 '12 at 19:24
OK - then maybe the terminology is wrong. I've updated my question. – Nicholas Kinar Jul 24 '12 at 19:39
up vote 1 down vote accepted

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 $2$:




So we have:


Which is clearly dependent on $x$.

share|cite|improve this answer
Yes - I see that now, Shaktal. Thank you for pointing this out. – Nicholas Kinar Jul 24 '12 at 19:18

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.