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Ask for constructing function from coordinate(x,y) points

bundit

New member
May 30, 2013
2
Hello,

I'm new in this forum. I would like to know how to calculate to obtain the function expression of some given coordinate points(x,y). For example, if I have 6 coordinate points which are

(x,y)
(0.23, 4.5)
(0.25, 6.5)
(0.35, 8.8)
(0.43, 15)
(0.45, 17)
(0.5, 25)

The (x,y) relation are apparently not linearized. It seems to by y=a.exp(b.x). How to calculate to derive the equation of y in term of y=f(x)?

Thank you.
Bundit
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,191
My recommendation: Excel. Plot the points up in Excel, and try fitting trendlines of various types to your data. Warning: six points isn't going to give you that much information. Do you need to interpolate or extrapolate?

The best fit I found (without going to very high-order polynomials) was
$$y=303.31x^{2}-151.14x+24.221,$$
with an $R^{2}$-value of $0.9895$. You can do a bit better with a cubic:
$$y=1357.5x^{3}-1205.6x^{2}+387.32x-36.859,$$
with an $R^{2}$ value of $0.9968$.
 

bundit

New member
May 30, 2013
2
My recommendation: Excel. Plot the points up in Excel, and try fitting trendlines of various types to your data. Warning: six points isn't going to give you that much information. Do you need to interpolate or extrapolate?

The best fit I found (without going to very high-order polynomials) was
$$y=303.31x^{2}-151.14x+24.221,$$
with an $R^{2}$-value of $0.9895$. You can do a bit better with a cubic:
$$y=1357.5x^{3}-1205.6x^{2}+387.32x-36.859,$$
with an $R^{2}$ value of $0.9968$.
Thank you,Ackbach.
By the way, can you brief me the definition of R^2 in your quote.?

Thanks.
Bundit