0
$\begingroup$

Suppose you're given function on a domain ([1,2] for instance) that exactly resembles $\frac{1}{x}$. Is this enough to know (assuming the function doesn't have a weird behavior, such as piecewise or jumps or anything) that the function is $\frac{1}{x}$?

My reasoning is that given two points, infinitesimally close, you can find it's derivative at the leftmost point. Given three of these, using polynomial approximations or some other method, you could find the second derivative, then the third derivative, and etc. The domain has to be finite, so that there are an "infinite" number of infinitesimally small points.

Then use a ramped up Euler's Method with infinitesimally small steps. Given a slope and point, using

$f(x+dx)=f(x)+dy=f(x)+f^\prime(x)dx$

you can find the next. With the second derivative

$f(x+2dx)=f(x+dx)+f^\prime(x+dx)dx=f(x)+f'(x)dx+(f'(x)+f''(x)dx)dx=f(x)+2f'(x)dx+f''(x)dx^2$

Given all the derivatives at a point and the point, shouldn't you be able to find the rest of the function using an infinite series?

Is this a method for rigorous induction?

$\endgroup$
  • $\begingroup$ I think all of this depends on if the function is smooth. In case, there is a point where the function is not smooth (but still continuous), your method may give a different function. $\endgroup$ – Aniruddha Deshmukh Jan 4 '18 at 17:16
  • 1
    $\begingroup$ The function $f(x)=e^{-1/x^2}$ for $x>0$, $f(x)=0$ for $x≤0$ coincides with $g(x)\equiv 0$ for $x≤0$ even though both are smooth. $\endgroup$ – lulu Jan 4 '18 at 17:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.