# $f ' (x) = f(x - (x+1)^t + 1)$

Let $x > 0$ and $c$ a given real $> 0.$ Let $t$ be between $0$ and $1.$

How to find $f(x)$ or good asymptotics for $f(x)$ such that

$$f ' (x) = f(x - (x+1)^t + 1)$$

And $f(1) = 1 + c$. Also $f$ is a nonlinear function and twice differentiable for $x > 0$.

• Don't know what "good" means in this context, but in gerenal one would try numerical integration. – Rubi Shnol Apr 5 '15 at 13:38
• Im talking about closed forms as asymptotics. – mick Apr 5 '15 at 13:41
• Update : Some comments and a brute estimate by Tommy : math.eretrandre.org/tetrationforum/showthread.php?tid=981 – mick Apr 10 '15 at 15:29
• I don't know if this helps you, but I noticed that if we allow $0$ to be in the domain, the Taylor expansion about it yields $f(x)=\frac{1+c}{e}e^{x}$. – Archaick Apr 14 '15 at 23:35

I'm assuming you want asymptotics for large $x$. Changing variables $x \to (x+1)$ your equation can be written more simply as

$f'(x) =f( x- x^t +1)$

Let's write $f(x) = e^{g(x)}$.

Then:

$f'(x) = g'(x) e^{g(x)}$.

$f(x - x^t +1 ) = e^{ g(x)} e^{ g(x-x^t+1) - g(x)} \approx e^{g(x)} e^{ - (x^t-1) g'(x) }$

This approximation is accurate if $g''(t)$ is not very large.

So this gives us $g'(x) \approx e^{ - (x^t-1) g'(x) }$. I'm assuming that $x$ is large here, so we should be able to work with the simpler equation

$g'(x) \approx e^{ - x^t g'(x) }$

Now if we plug in $g'(x) = x^{-t} * ( t \log x - \log t - \log \log x)$, then we get

$$x^{-t} * ( t \log x - \log t - \log \log x) \approx e^{ -t \log x + \log t + \log \log x}= x^t (t \log x)$$

As long as $x$ is large, $t \log x$ is much larger than $\log t$ and $\log \log x$ and the approximation is pretty close.

So exponentiating an integral of this will give you an approximate asymptotic, although sadly I don't know how good it is. This integral is not even closed form, but the leading term is going to be $x^{1-t} \log x$ times a constant. So your asymptotic should be very, very roughly $x^{ x^{1-t} /(1-t) }$.

• The first line is confusing, shouldn't it be $f'(x-1) = f(x - x^t)$? are you defining a new $f$ by $f_{new}(x) = f_{old}(x-1)$ ??? – achille hui Apr 17 '15 at 22:37
• @achille hui exactly. I think that equation is slightly simpler. It doesn't really matter because I am using such a rough approximation. – Will Sawin Apr 17 '15 at 22:42
• Thanks for the effort, but to be honest it is a very weak estimate. Barely better than Tommy's i Linked in the comments. And actually i was aware of what you wrote, but i did not answer because the estimate is weak and - more importantly - im not even sure your steps are correct ?! There are Huge error terms not adequately handled imho. In it current state its An argument rather then An answer. – mick Apr 18 '15 at 14:28
• +1 i can give but i cannot accept in current form. Very sorry ! – mick Apr 18 '15 at 14:30