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I am wondering, is there some (general) theorem or lemma (that I am not aware of) which would state something about number of inflection points of a logarithm of a positive function if we know the number of inflection points of the function (or the other way)?

In other words, if we know that a positive function $f(x)$ has $n$ inflection points, can we say something about number of inflection points of $\log f(x)$? Also the other way. If we know that $\log f(x)$ has $m$ inflection points, can we say something about number of inflection points of $f(x)$? Thanks.

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  • $\begingroup$ Why would you think that log affects the number of inflection points of a function? $\endgroup$
    – Mohammad
    Jan 13, 2014 at 18:03
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    $\begingroup$ Well, for example even in simple case $f(x)=x^3+1$ there is one inflection point ($x=0$). But if we take the logarithm, i.e., $\log f(x)$ has two inflection points $x=0, x=\sqrt[3]{2}$. $\endgroup$
    – pisoir
    Jan 13, 2014 at 18:19
  • $\begingroup$ Of course one should restrict $f$ to be positive-valued in the first place, to make this a sensible question. $\endgroup$ Jan 13, 2014 at 18:23
  • $\begingroup$ Of course I meant positive function. Sorry for the error. I fixed the question. $\endgroup$
    – pisoir
    Jan 13, 2014 at 18:25

4 Answers 4

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In general, there is no relationship.

For example, take $f(x) = e^{x+(\sin x)/10}$. One can check that $f$ has no inflection points but $\log f$ has infinitely many inflection points.

On the other hand, let $h$ be a solution to the differential equation $h'(x) = (-1 + (\sin x)/2) h(x)^2$ (with say $h(0)=1$, on say $x\in[0,\infty)$). Let $g(x)$ be an antiderivative of $h(x)$, and let $f(x) = e^{g(x)}$. The inflection points of $f$ occur precisely where $g''(x)+g'(x)^2=0$ and thus where $h'(x) = -h(x)^2$, which are the infinitely many multiples of $\pi$. However, one can show that $h$ is positive and decreasing on $[0,\infty)$; therefore $g''(x)=h'(x)$ is never zero, hence $g=\log f$ has no inflection points.

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  • $\begingroup$ wow how did you come up with these examples? $\endgroup$
    – GPerez
    Jan 14, 2014 at 21:26
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Horizontal inflexions of the original curve are in one-to-one correspondence with horizontal inflexions of the log curve. Apart from that, there is no connexion.

Consider the regular curve parametrised by $\gamma(t) = (t,x(t))$. The curve has an inflexion if and only if the first two derivatives are linearly dependent, i.e. $\dot{\gamma}\parallel \ddot{\gamma}$. In otherwords: $\ddot{x}(t)=0$.

Next, consider the curve $\alpha(t)=(t,\ln x(t))$. We have $\dot{\alpha}(t) = (1,\dot{x}/x)$ and $$\ddot{\alpha}(t) = \left(0,\frac{x\ddot{x}-\dot{x}^2}{x^2}\right)$$.

Assuming that $x>0$ then $\dot{\alpha}\parallel \ddot{\alpha}$ if, and only if, $x\ddot{x}-\dot{x}^2=0$. If $\gamma(t_0)$ is an inflexion then we also need $\dot{x}(t_0)=0$ for $\alpha(t_0)$ to be an inflexion. If $\alpha(t_0)$ is an inflexion then we need $\dot{x}(t_0)=0$ for $\gamma(t_0)$ to be an inflexion.

Hence, for $x>0$, if $\dot{x}=0$ then $\gamma(t_0)$ is an inflexion if, and only if, $\alpha(t_0)$ is an inflexion.

The log curve $\alpha$ is free to have inflexions when the original $\gamma$ does not: we simply need $\dot{x}\neq 0$ at such points. Conversely, the original curve $\gamma$ is free to have inflexions when the log curve $\alpha$ does not: we just need $\dot{x} \neq 0$ at that point.

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  • $\begingroup$ This is a valid argument, but only for one very special case of inflection points. $\endgroup$ Jan 13, 2014 at 18:31
  • $\begingroup$ @GregMartin That's all the OP was interested in, the "...number of inflection points of a logarithm of a positive function." $\endgroup$ Jan 13, 2014 at 18:33
  • $\begingroup$ Well, no: this argument treats only points where the original function has derivative $0$. $\endgroup$ Jan 13, 2014 at 18:33
  • $\begingroup$ @GregMartin No it doesn't It give a condition for the original curve to have an inflexion: $\ddot{x}(t_0)=0$ and for the log curve to have an inflexion: $(x\ddot{x}-\dot{x}^2)(t_0)=0$. That only assumes the curve is a graph of a function. I then relate the inflexions on both curves. Feel free to add an answer if you can do any better. $\endgroup$ Jan 13, 2014 at 18:35
  • $\begingroup$ @Fly by Night I'm really not sure so correct me, but aren't the conditions you state necessary ones? The if and only if concers me. I thought that sufficiency was given by the first non-zero derivative being of odd order? $\endgroup$
    – GPerez
    Jan 13, 2014 at 18:51
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Write out the two successive derivatives (assuming $f\in\mathcal C^2$): $$\frac{\mathrm d^2}{\mathrm dx^2}(\log\circ f)=\frac{\mathrm d}{\mathrm dx}\frac{f'}{f}=\frac{f''f-(f')^2}{f^2}$$ We want this to be zero at some $x\in\mathbb R$: $$f''(x)f(x)-(f'(x))^2=0$$

So we can see that all the original (stationary) inflection points of $f$ restricted to $(0,\infty)$ are still inflection points, but there's no clear way of knowing whether there's more, without additional information about $f$. Notice that we may have lost any inflection points that were in the negative domain of $f$. We cannot conclude anything if $f'(x)\not = 0$.

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  • $\begingroup$ Actually, there's the caveat that there can be less if the values of $f$ are negative over a certain range, because the $\log$ can't be evaluated there. $\endgroup$ Jan 13, 2014 at 18:16
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    $\begingroup$ This is false, because there's no reason that $f'(x)$ should also equal $0$ at the inflection point. For example, take $f(x) = x^3+x+1$. Then $f(x)$ has an inflection point at $x=0$, while the inflection points of $\log f(x)$ are near $x=0.167056$ and $x=1.19858$. $\endgroup$ Jan 13, 2014 at 18:20
  • $\begingroup$ @Raskolnikov Well spotted! fixed it. $\endgroup$
    – GPerez
    Jan 13, 2014 at 18:24
  • $\begingroup$ @GregMartin, In the context I though we were talking about saddle type inflection points, but I've edited for full correction $\endgroup$
    – GPerez
    Jan 13, 2014 at 18:25
  • $\begingroup$ @GregMartin Now it should be ok $\endgroup$
    – GPerez
    Jan 13, 2014 at 18:32
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See here---- Actually the points of inflection of a function $y=f(x)$ means,where the function $y=f(x)$ has its 2nd order derivative equals to zero. or we can say mathematically where or for what values of $x$ the function has its 2nd order derivative = 0. OR $f '' ( x ) = 0 $.

the values of $x$ for which the equation $f'' (x) = 0$ is satisfied we can say those points in the domain of the function,it has the points of inflections or the $f '' (x)$ is $0$.

See where the function has its 2nd order derivative $f '' (x) = 0$ , at that values of $x$ the function $y = f(x)$ is changing its behavior from concave down to concave up or concave up to concave down.

Mathematically we can say if a function $y = f (x)$ has its $f '' (x)$ (2nd order derivative) less than $0$ or $f " (x) < 0$ or its $f '' (x)$ value is negative in a certain scale of $x$ then you can say that the function is just concave down or if we draw the curve of $y = f(x)$ in a co-ordinate system we will have a curve just having a concave down nature. And similarly if the function has its $f'' (x ) > 0$ in certain scale of $x$ then we say that the function has a concave up characteristics and if we draw the graph of $y= f(x)$ we will have just a concave up picture of the function $y = f(x)$.

please go to the link : coz I dont have the software to upload answer with pic. The link is the graph of a simple function to implement the concepts which I said earlier above. The link is a picture of a graph of $y = x^3$. You can go through the link or you can have a google search for it for better understanding along with my concepts....

graph

See here at the point $x=0$ the $f ''( x )$ of the function $f(x) = x^3$ that is $f'' (x) = 3(x^2)$ is equal to zero. See $f '' (x)$ for $x=0$ is $f'' (0) = 3(0^2) = 0$.

So you see there is the point of inflection at $x=0$ in the curve of the function $y = f(x)$. and see if $x>0$ the $f'' (x)= 3x^2$ is also greater than $0$. So it is concave up for $x>0$ and it is also true for $x<0$ and it is concave down when $x<0$.

But see here the log $f(x)$ will be: let us consider $g(x) = \log f(x) = \log {x^3} = 3\log x$.

Now if you have the second order derivative of $\log {x^3}$ or $3\log x$ , you will have $g'' (x) = \frac{- 3 }{ x^2}$.

see the function $g(x) = \log f(x)$ has no such points where the second order derivative is 0. So, There is no points of inflections for the function $g(x) = \log f(x)$.

I was trying to make your concept on points of inflections and see there is no such direct dependency between the number of inflection points for the function $f(x)$ and $\log f(x)$.

the function $\log f(x)$ will have the point of inflection at that values of $x$ for which the equation: $ \frac{( f(x) f " (x) - (f ' (x))^2 )}{ ((f(x))^2) } = 0$ has a solution for $x$. Or simply you can say if the equation :
$$ f(x) f ''(x) - (f ' (x) )^2 = 0$$ has any solution for $x$ or not.

If this equation has any solution then the function $g(x) = \log (f(x))$ will have the points of inflection. And you can also draw the curve of $y = f(x)$ or $g(x) = \log f(x)$ by applying the 2nd order derivative criteria.

Hope the answer will help you a bit.Best of luck.

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    $\begingroup$ Nice concepts, though I have to specify that a point of inflection is when $f''(x)$ changes sign, which implies that $f''(x)=0$, but $f''(x)=0$ alone isn't enough to say that there's a point of inflection! $\endgroup$
    – GPerez
    Jan 16, 2014 at 1:41
  • $\begingroup$ @arkadeep Thanks for explaining what an inflection point is;). Please, fix your errors in the explanation: $(x^3)^{''}\ne 3x^2$ but $(x^3)''=6x$. Then also your explanantion about the fact that $f''(x)>0$ for $x>0$ and $f''(x)<0$ for $x<0$ makes more sense. $\endgroup$
    – pisoir
    Jan 16, 2014 at 13:38

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