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How can I convert this polynomial into a linear equation and then convert the answer back into the original form? $$ Y=-2.2083 X^{5} + 39.875 X^{4} - 270.71 X^{3} + 846.12 X^{2} - 1184.1 X + 626 $$ Can I use $\log$'s to convert it into linear form ?. If so, how ?.

Mike

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  • $\begingroup$ What makes you think that might work? And what does it have to do with statistics? $\endgroup$ – Henry Aug 11 '14 at 21:53
  • $\begingroup$ I thought this might work because of this pdf: "Non-Linear Regression by Samuel L. Baker", "Polynomial Forecasting by Moore Tech LLc", etc. It is a polynomial (a form of regression) and statisticians do regression. $\endgroup$ – Michael Aug 11 '14 at 22:39
  • $\begingroup$ @Michael, is there a particular reason that this has to be a fifth degree polynomial? From what I understand, you have Excel fit this line. Are the results for the 5th degree any better than those for the 4th degree? What range of X values are you working in? $\endgroup$ – BeaumontTaz Aug 12 '14 at 1:36
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First, rewrite your polynomial as $Y=\displaystyle\prod_{k=1}^5\Big(X-x_k\Big)$, where $x_{_{1-5}}$ are its five roots. Then take the logarithm of both sides, and use the fact that the logarithm of a product is the sum of logarithms.

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  • $\begingroup$ How could this be implemented into excel? $\endgroup$ – Michael Aug 11 '14 at 22:51
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    $\begingroup$ @Michael: I'm sorry, I don't use Excel. $\endgroup$ – Lucian Aug 11 '14 at 22:55
  • $\begingroup$ @Michael, for your particular case, you only have one real root and 4 imaginary roots: $x=6.07324,1.36806\pm0.45817i,4.62375\pm1.02227i$. This doesn't make a nice linear function: $\ln(y)=\ln(x-6.07324)+\ln(x-1.36806+0.45817i)+\ln(x-1.36806-0.45817i)+\ln(x-4.62375-1.02227i)+\ln(x-4.62375+1.02227i)$ $\endgroup$ – BeaumontTaz Aug 12 '14 at 1:52
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Not the way you are hoping. If you take the log of a sum, you don't get the sum of the logs, so taking the log of both sides won't help. If you had only one term on the right, it could be useful: $$y=2.2803 x^5\\\log y = \log 2.2803 + 5 \log x\\$$ but with all those right side terms you are out of luck .

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  • $\begingroup$ I was hoping I could use logs like in this article:"Non-Linear Regression by Samuel L. Baker". This is a polynomial I got from excel and I want to forecast one period ahead but there is no way that I know of yet to extrapolate using a polynomial. I thought if I could convert the polynomial into a linear equation I could get a decent estimation. $\endgroup$ – Michael Aug 11 '14 at 22:31
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    $\begingroup$ You can just plug in larger $X$ values to make a forecast. Unfortunately, using polynomials (or any functional form that you don't have a theoretical basis for) usually fails. There are delicate cancellations to make the function fit the data. When $X$ gets outside the range of the data points, $Y$ tends to blow up or down unrealistically. $\endgroup$ – Ross Millikan Aug 11 '14 at 22:37
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    $\begingroup$ Note that he is using logs to convert a functional form like $y=a \exp(bx)$ to a straight line, not to linearize a fit he already has. $\endgroup$ – Ross Millikan Aug 11 '14 at 22:40
  • $\begingroup$ You are correct. When I used an x outside the range y went unrealistic. I also thought of trying this: mhhe.com/math/calc/smithminton2e/cd/folder_structure/text/…; to convert the polynomial shape into a straight line. Don't know if it would work. $\endgroup$ – Michael Aug 11 '14 at 22:54
  • $\begingroup$ If you want a straight line, you can just fit one to the data. For extrapolation, that is often better than a higher degree polynomial. Lowering the degree smooths the data and reduces the impact of noise on your extrapolation. Excel will do it for you. $\endgroup$ – Ross Millikan Aug 12 '14 at 0:16
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Oddly enough, I have to do something like this from time to time (I've done it for 4th and 6th order polynomials, never 5, though). I solve it numerically for my x- and y-value range of interest:

  • Create an (x,y) array of solutions to your polynomial equation (in my case I can use steps of 0.5).
  • Plot using Excel
  • Fit a log function to your data using the the Trendline overlay graphing feature
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See https://onlinelibrary.wiley.com/doi/pdf/10.1002/we.1676, Appendix A, Equation A.6.

Cheers, Fraser

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  • $\begingroup$ Welcome to MSE. This should be a comment, not an answer. $\endgroup$ – José Carlos Santos May 31 '18 at 13:12

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