       Re: FindFit power law problem

• To: mathgroup at smc.vnet.net
• Subject: [mg117200] Re: FindFit power law problem
• From: Bill Rowe <readnews at sbcglobal.net>
• Date: Fri, 11 Mar 2011 04:34:01 -0500 (EST)

```On 3/10/11 at 6:06 AM, gwardall at gmail.com (Gary Wardall) wrote:

>I was wondering if you really meant to transform both the x and y
>data?

>Fit[Log[10, data], {1, x}, x]

>Does just that. Some software transforms just the y data when
>estimating the coefficients for the model y = a x^b.

>What happens is they transform just the y data and use the linear
>model

>yy=aa+bb*ln[x]

>Finding the coefficients aa and bb and state the results as

>y=Exp[aa]*x^bb

>Since:
>
>y=Exp[aa+bb*ln[x]]

>is equivalent to:

>y=Exp[aa]*Exp[bb*ln[x]]

>in turn is equivalent to:

>y=Exp[aa]*Exp[ln[x^bb]]

>which is also equivalent to:

>y=Exp[aa]*x^bb

>Note then that a=Exp[aa] and b=bb.

>I hope my Algebra is correct.

You start you post by asking about transforming both x and y.
You have a model where you transform y then seem to be
suggesting to do:

FindFit[xformedData, a Log[x] + b, {a,b}, x]

This is exactly the same as doing

FindFit[Log[data], a + b x, {a,b},x]

That is setting up the model as a linear function of Log[x] is
exactly the same as transforming both x and y then using a
linear model to fit the transformed data.

Note also, the difference in fit parameters comes about by
transforming y. For real data, errors are almost always
additive. Any non-linear transformation of measured y data with
additive error creates a fundamentally different model. The
point is f[y+error] does not map to f[y] + f[error] unless f is
a linear function. If the range for you data isn't too large,
doing a non-linear fit to the data will give approximately the
same result as doing a linear fit to appropriately transformed
data. A sufficiently small range in this context means the
deviation from true linearity is small. Or said differently, any
non-linear function restricted to a sufficiently small range
looks linear.

```

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