## datanewb 3 years ago Trying to adjust this equation so that it will have fatter tails?$3^{-\left(\frac{x}{6}\right)^{2}}$ Any ideas? It is similar to the probability distribution function.

1. datanewb

In the attached image, the black line is the graph of $3^{-\left(\frac{x}{6}\right)^{2}}$ and the red line is what I'd like to find an equation for.

2. frx

Can't be the right equation: http://www.wolframalpha.com/input/?i=3-%28x%2F6%29^2

3. frx

Is it a double power?

4. datanewb

Sorry, this page did not translate my latex correctly, here is the equation I was trying http://www.wolframalpha.com/input/?i=3%5E%28-%28x%2F6%29%5E2%29

5. datanewb

yes, it is a double power

6. datanewb

Does this make it clearer? $3^{\left(-\left(\frac{x}{6}\right)^{2}\right)}$

7. datanewb

Changing the 6 to say 12 results in a wider "distribution", but the entire the is wider. I'm trying to keep the top narrow and the bottom wide.

8. frx

So you want the decay to take longer time to asymptote 0 and also start the decay a little earlier?

9. datanewb

@frx, exactly. That is a clearer explanation of what I'm trying to do.

10. frx

Hmm this was a bit tricky, not right but on the right way http://www.wolframalpha.com/input/?i=2^%28-%28x%2F8%29^2%29+from+0+to+20

11. datanewb

unfortunately, I think you've just made the entire distribution wider. In the attached image, f(x) is the original function $3^{\left(-\left(\frac{x}{6}\right)^{2}\right)}$, and g(x) is the new one:$2^{\left(-\left(\frac{x}{8}\right)^{2}\right)}$

12. datanewb

Sorry, I'm equally unfamiliar with gnuplot and wolfram alpha, I guess I should just stick with one. :)

13. frx

Hmm just an idea, what if you choose a couple of x and y values and construct the function backwards?

14. datanewb

oh, that's a cool idea. I'll try that on wolfram... my current line of thinking had been to add multiple functions together. I know to make the top fatter, you can add to slightly shifted distribution functions

15. datanewb

http://www.wolframalpha.com/input/?i=sextic+fit+100%2C96%2C88%2C75%2C61%2C46%2C35%2C35%2C17%2C12%2C8%2C5%2C3%2C2%2C1.5%2C1.2%2C.9%2C.8%2C.7%2C.6 Well, this works pretty well. I was hoping there would be a better method than brute force, some deductive way of figuring it out, but I'm happy to have found a function that I can use. Thanks.

Find more explanations on OpenStudy