Suppose for all $x$, $$f’(x)/g’(x) = f(x)/g(x),$$ Then what can we say about the relationship between $f(x)$ and $g(x)$?
I think the only solution is $f(x) = cg(x)$ for some non-zero constant $c$. Is there any other possible relationship?
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3$\begingroup$ But $c$ could be $0$: there is no reason for $f$ not to be identically $0$ (as opposed to $g$, which can't be $0$ if we want the right side to be defined and can't be constant if we want the left side to be defined). $\endgroup$– Robert IsraelMay 3, 2019 at 3:13
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2 Answers
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We can obtain $f'/f=g'/g$ from the problem statement. Integrating both sides gives:
$$\ln(f(x))=\ln(g(x))+C_1,$$
where $C_1$ is our combined constant of integration. Exponentiating both sides gives us:
$$f(x)=cg(x),$$
where $c=e^{C_1}$. Your idea was correct.
EDIT: I saw JohnDoe's comment after posting my answer.
EDIT 2: Of course, this all only works when none of $g, g',$ or $f$ is zero.
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Then the numerator of quotient rule for derivative of $f/g$ vanishes and so $f/g$ constant. Need to do more work on it for proof...
answered May 3, 2019 at 2:39
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6$\begingroup$ Or rearrange to $f'/f=g'/g$, then integrate and exponentiate. Then $f=cg$ $\endgroup$– John DoeMay 3, 2019 at 2:40
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$\begingroup$ John-- I think that's better, avoids assumptions about derivatives... If you put as an answer I'd upvote it. $\endgroup$ May 3, 2019 at 2:46
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1$\begingroup$ Never mind, there is already an answer to that tune :) $\endgroup$– John DoeMay 3, 2019 at 2:48
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2$\begingroup$ @TheCount No, that's fine, keep the answer up - you also had the same idea and so posted it. I don't mind at all :) $\endgroup$– John DoeMay 3, 2019 at 2:53
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