I was wondering whether there exists a function that escapes to infinity with a finite input. For a specific example, how about $f(0)=0$ and as $x$ tends to $10$, $f(x)$ tends to infinity. The use of this would be to produce unimaginably large numbers with imaginable inputs.

If there are many such functions, I would prefer a formula which is short and snappy. If need be, magical operators such as an infinite sum or product would suffice. I understand that the tangent function has many poles (is that what you call an un-definition?), but they are sort of expensive to compute. I am not adept in calculus, but if you must... and at the very least I would like the function to be computable.

I am just a bit curious, is all. Can anyone help me? Perhaps the answer is obvious.

  • $\begingroup$ What about things such as $\frac{1}{10-x}-\frac{1}{10}$ ? $\endgroup$ – Claude Leibovici Feb 1 '14 at 9:37
  • $\begingroup$ I thought a reciprocal would be in order. Nice and simple, an answer. So, I assume that replacing the constant $10$ with another number $n$ would generate a function that has a pole at $n$. So simple. I feel so dumb. $\endgroup$ – bimmo Feb 1 '14 at 9:42
  • $\begingroup$ In fact, you can select a function $f(x)$ which goes to infinity for $x=n$ and which has a finite value for $x=0$. Then use $g(x)=f(x)-f(0)$ $\endgroup$ – Claude Leibovici Feb 1 '14 at 9:51
  • $\begingroup$ consider $y=1/(n-x)-1/n$. I have realised that decreasing the numerator of the current reciprocals reduces the bend in the curve. (I have Graphmatica up). A numerator of 0 produces a flat line at y=0 and a hole in that line at y=n. The gradient at y=0 seems to dictate this bend. $\endgroup$ – bimmo Feb 1 '14 at 10:13
  • $\begingroup$ You can also use $y=(n-x)^{-a}-n^{-a}$ using $a>0$. If I amy ask, what do you plan to do ? If I can help, my pleasure. $\endgroup$ – Claude Leibovici Feb 1 '14 at 10:19

Look at $\tan(x)$. https://www.wolframalpha.com/input/?i=tan%28x%29. As $x \to \frac\pi2$ then $f(x) \to \infty$. Also true for any integer multiple of $\frac\pi2$.

In general, functions of the form $f(x) = \frac{g(x)}{h(x)}$ will tend to infinity as $h(x)$ tends to 0, which doesn't necessarily require $x$ to tend to $\infty$. Disclaimer: $g(x)$ and $h(x)$ have to be well defined on the same interval, continuous and $h$ doesn't divide $g$ for all $x$ for this to be the case.

Here's an example that fits your criteria $f(0) = 0 $ and $f(x) \to \infty$ as $x \to 10$: $f(x) = \tan(\frac{\pi x}{20})$ If you want to construct a function $g$ where $g(x) \to \infty$ as $x \to k$ for some non-zero real number $k$, you can simply use $g(x)=\frac1{x-k}$.

  • $\begingroup$ I have a graph up. It seem that using either @Claude's reciprocal function or this tangent function gives a non-zero gradient at $f(0)$. Interesting. This tangent function is 'smoother' or 'softer' than the reciprocals, presumably because of it's greater slope at $f(0)$. Is the non-zero slope necessary? Would pushing it to 0 produce a square 'kick' to infinity at f(k)? $\endgroup$ – bimmo Feb 1 '14 at 10:05
  • $\begingroup$ Graphmatica calls the 'kick' a hole. That makes sense. $\endgroup$ – bimmo Feb 1 '14 at 10:15

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.