computerages Posted May 25, 2008 Share Posted May 25, 2008 Hello! I was wondering if someone can give me a functions that increases gradually in the beginning, and then picks up an exponential rate after some time. It would be much similar to an exponential function such as x^2, but it would increase gradually for a certain time in the beginning. Thanks! Link to comment Share on other sites More sharing options...
Aeternus Posted May 26, 2008 Share Posted May 26, 2008 [math]x^2[/math] is not an exponential function or even a function exhibiting exponential growth. See http://en.wikipedia.org/wiki/Exponential_growth . I imagine the answer to your question is really going to depend on what exactly you want and what you mean by "in the beginning" and gradual growth. Can you go into a bit more detail about what it is you want to derive? Link to comment Share on other sites More sharing options...
Bignose Posted May 26, 2008 Share Posted May 26, 2008 I guess you don't want: [math]f(t) = \begin{cases} e t^2 \text{ if } t <= 1,\\ e^t \text{ if } t >1 \end{cases}[/math] Like Aeternus said, you probably need to provide more info to get better answers Link to comment Share on other sites More sharing options...
computerages Posted May 26, 2008 Author Share Posted May 26, 2008 please see the attachment to understand what i mean.. as u can see the graph suddently begins to exp. grow but in the begining just increases gradually (not exactly linearly though).... I hope that makes sense..... Link to comment Share on other sites More sharing options...
Cap'n Refsmmat Posted May 26, 2008 Share Posted May 26, 2008 So you mean something like [math]y = - \frac{1}{x-10}[/math] Link to comment Share on other sites More sharing options...
Aeternus Posted May 26, 2008 Share Posted May 26, 2008 If you don't mind it being symmetric about the y axis and only want strong growth not assymptotic like Capn's, then you might also consider something similar to what Bignose suggested - [math] y = e^{(\frac{x}{a})^b} - 1 [/math] Where you can change [math]a[/math] to change where it begins growing very quickly and change [math]b[/math] to change how quickly it grows at that point and how gradually it grows before that. To generalise what Capn said a bit, if you want assymptotic behaviour, similar to what he said you can do [math] y = \frac{1}{b \cdot x-b \cdot a} [/math] Where [math]a[/math] and [math]b[/math] have similar effects as above. Link to comment Share on other sites More sharing options...
YT2095 Posted May 26, 2008 Share Posted May 26, 2008 doesn`t n! do that also? Link to comment Share on other sites More sharing options...
ajb Posted May 29, 2008 Share Posted May 29, 2008 (edited) doesn`t n! do that also? For natural numbers yes. But looking at the plot given, he is looking for a function on the real numbers. But you could use the Gamma function on [math]\mathbb{R}^{+}[/math] Try, [math]y[x] = \Gamma[x+1][/math] Edited May 29, 2008 by ajb multiple post merged Link to comment Share on other sites More sharing options...
Riogho Posted May 29, 2008 Share Posted May 29, 2008 2^x Link to comment Share on other sites More sharing options...
ajb Posted May 30, 2008 Share Posted May 30, 2008 2^x Or indeed, [math] y[x] = a^{x} +b[/math] for [math] a > 1[/math] and [math]b \in \mathbb{R}[/math] The rate of growth is [math] y'[x] = a^{x}\log[a][/math]. Link to comment Share on other sites More sharing options...
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now