Greater precision with regression?

I used my TI-83 to find the quadratic regression of two data columns. The accuracy wasn't close at all. So I tried cubic and then finally quartic regression. The accuracy still isn't close enough. Is there any way I can improve the accuracy? I can't add any more data, unfortunetly.

Edit 1:

To address Henry's comment, my plotted data looks similar to y=x^(1/3)

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It is difficult to comment without seeing the data. You could try plotting the points to see what the pattern looks like and whether that suggests anything. –  Henry May 20 '11 at 22:54
If it is in fact a cube root, that explains why polynomials don't fit well. The cube root gets too flat for a polynomial. As I suggested below, taking the logarithm of all the data will check this. y=x^(1/3) implies log(y)=(log(x))/3 and you should get a good linear fit. –  Ross Millikan May 20 '11 at 23:43

This may mean your data is poorly fit by a polynomial. Three ways it can fail is either to be exponential, to have poles, or be like a sine wave. Plotting the data with the fits you have overlayed may give you some ideas. If it is exponential, taking the logarithm of one will render it linear (or close to polynomial). If it has poles, using a rational function will help. Procedures are given in chapter 3.2 of Numerical Recipes. The obsolete versions are free. If it is a sine wave, an FFT can help-chapter 12 of NR.

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What do you mean 'take the logarithm of one'? Log the value of the equation? –  AedonEtLIRA May 20 '11 at 22:56
Take the logarithm of the data. If the data fits $y=exp(ax)$ your polynomial fitters will not do well, because polynomials can't increase that fast. But then $\log y=ax$ will be linear. –  Ross Millikan May 20 '11 at 23:01
I'm really sorry, I still don't follow. Say my function is y = -1.5E-8*x^4 + 1.8E-5*x^3 + .007*x^2 + 1.7*x + -15. Intead of raising each, I would lof each? –  AedonEtLIRA May 20 '11 at 23:19
I wasn't talking about the function, but the data itself. Instead of doing a regression of x and y, do one on log(x) and log(y). See if that helps. –  Ross Millikan May 20 '11 at 23:39
Ok thanks, that makes more sense. How do you predict something with that though? If I understood you right you take the log of both x and y, make a data chart and then do regression? –  AedonEtLIRA May 21 '11 at 19:51

If, as you say, it looks like $y=x^\frac{1}{3}$, then you might try a power regression, which will give you a formula of the form $y=ax^b$.

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