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"Loosely Approximating" a continuous curve

Posted 7 months ago
6 Replies
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I'm trying to save some CPU cycles in an application by finding a curve formula that's "very similar but simpler" than the ones I have. I'm wondering what the best way to do this in Wolfram Alpha is. I tried to split it up into a table and then run cubic fit on it but Alpha wasn't really having that.

Any suggestions?

Here are the two formulas I'm attempting to approximate:

plot 10^(((   (x*1.4+.3)^(.3)     )*128-128)/20) from x=0 to 1
plot 10^(((   (1-(x-.5))^.3     )*128-128)/20) from x=0 to 1

Here they are as tables:

Table[ 10^(((   (x*1.4+.3)^(.3)     )*128-128)/20) ,{x,0,1,.05}]
Table[ 10^(((   (1-(x-.5))^.3     )*128-128)/20)  ,{x,0,1,.05}]

And here's the syntax I tried.

quadratic fit[Table[ 10^(((   (1-(x-.5))^.3     )*128-128)/20)  ,{x,0,1,.05}]]

Here's what it looks like in Apple Grapher:


6 Replies

Please edit to add the actual Mathematica code (not a picture of code) for the function of interest.

Dan, Edited original post. Sorry for not being as rigorous as I could have been.

Using FindFormula function. See attached files.


Mariusz, I don't know how you generated that pdf or what a .nb file is but I now feel like I need to read a few books on how to use Wolfram Alpha. Either way, I see generally how you did this and I'm deeply appreciative!

There was a misunderstanding,I all done in Mathematica not wolframalfa. I don't know how to do it in portal of wolframalfa.

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