Mathematics Asked by Boby on January 3, 2021
Let me set up the question by stating something that is well-known. Suppose $f$ is an unknown function, but we are given $f’$ and the fact that
begin{align}
lim_{x to infty} f(x)=0.
end{align}
Then, by using fundamental theorem of calculus we have that
begin{align}
f(a)-f(x)= int_x^a f'(t) dt
end{align}
and by using that $f(a) to 0$ as $ato infty$ we arrive at
begin{align}
f(x)= int_x^infty -f'(t) dt.
end{align}
In other words, we can recover $f$ based on this information.
Here is the question: Suppose instead we know $frac{d}{dx} log f(x)=frac{f'(x)}{f(x)}$ and $lim_{x to infty} f(x)=0$. Can we still recover $f(x)$?
If we use, the previous approach we arrive at
begin{align}
frac{f(x)}{f(a)}= e^{-int_x^a frac{f'(t)}{f(t)} dt}.
end{align}
Clearly, we have a problem with this approach as we get a division by zero in the limit.
My guess is that in general, it is not possible to recover $f(x)$ based on this information. (See one of the answers) However, can we do this with some extra minimal assumptions on $f$?
One trivial assumption that I would like to avoid making, is that we know $f(a)$ at some point $a$ which would make the FTC approach work.
Given one of your functions $f$, we can always write it as $f(x) = e^{g(x)}$ where $f$ and $g$ determine each other uniquely. The condition $lim_{x to infty} f(x) = 0$ translates to $lim_{x to infty} g(x) = -infty$. Also, $frac{d}{dx} log(f(x)) = g'(x)$, so knowing the logarathmic derivative of $f$ equates to knowing the ordinary derivative of $g$.
Having made this translation, we see that the basic issue is that, given a function $g(x)$ with $lim_{x to infty} g(x) = -infty$, we cannot recover $g(x)$ from $g'(x)$. For example $g(x)=-x+c$ has $g'(x) = -1$ and $lim_{x to infty} g(x) = -infty$, regardless of the value of the constant $c$. Translating back to $f$, we can say that $f(x) = k e^{-x}$ has logarithmic derivative $-1$ and $lim_{x to infty} f(x) = 0$, regardless of the value of the constant $k$.
Correct answer by Mike F on January 3, 2021
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