Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Could someone point me to any function ${ f(x) }$ which is continuous at some interval ${ x \in [x_0; x_1] }$ and can be represented by formula, so that it rises until some point and then quickly "falls" like on image below?

enter image description here

What may cause such behavior?

share|improve this question
1  
Do you need it to be differentiable at that "point"? –  Ian Mateus Jan 21 '13 at 17:57
    
@IanMateus That is not necessary. –  Edward Ruchevits Jan 21 '13 at 21:46
add comment

3 Answers

up vote 3 down vote accepted

How about the function $f(x)=e^{-|x|}$, which has the following graph:

enter image description here

If you don't want it to be symmetric, you could use $f(x)=e^{-|1-e^x|}$, which has the following graph:

enter image description here

If you need the base line to be the same (the left asymptote of the above graph is $\frac{1}{e}$), you could use $$f(x)=e^{-|e^{-ax}-e^x|}$$ where $0<a<1$ is a constant you can vary. For example, with $a=1/10$, we get

enter image description here

share|improve this answer
    
Thank you! The second modification is what I wanted. –  Edward Ruchevits Jan 21 '13 at 21:54
add comment

Try something like a log-normal, $$f(x) = \frac{1}{\sqrt{2\pi}\sigma x}\exp\bigg({-\frac{(\log(x) - \mu)^2}{2\sigma^2}\bigg)}$$ For example,

enter image description here

share|improve this answer
1  
Thank you, that's interesting! –  Edward Ruchevits Jan 21 '13 at 21:53
add comment

How quickly? Piecewise linear functions satisfy this. In general, for real valued functions you can essentially get as nice bumps as necessary by considering $e^{-x^n}$, where the speed of decay can be acquired by increasing $n$. You can also construct bumps which are supported on compact sets and satisfy the above by playing with functions of the form $e^{-\frac{1}{x^2}}$.

share|improve this answer
1  
This was my first thought too, but the OP asks for the function to be "represented by formula". Of course, this is a vague notion, but I don't think piecewise-defined functions are what the OP is after. –  Zev Chonoles Jan 21 '13 at 17:59
    
Thank you for the answer! I meant single function, not piecewise. –  Edward Ruchevits Jan 21 '13 at 21:51
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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