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.

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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.

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

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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
yes, it is a double power
Does this make it clearer? \[3^{\left(-\left(\frac{x}{6}\right)^{2}\right)}\]
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.
So you want the decay to take longer time to asymptote 0 and also start the decay a little earlier?
@frx, exactly. That is a clearer explanation of what I'm trying to do.
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
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:\[2^{\left(-\left(\frac{x}{8}\right)^{2}\right)}\]
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Sorry, I'm equally unfamiliar with gnuplot and wolfram alpha, I guess I should just stick with one. :)
Hmm just an idea, what if you choose a couple of x and y values and construct the function backwards?
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
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.

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