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Linear approximation around a point through Taylor series requires the first order derivative to be non-zero unless you want to get only the value at that point. However this is only true when you are extremely close to the referred point. Is there a better way to linearly approximate such a function.

For example, using Taylor Series to evaluate linear approximation of the function, $f(x) = (1+x^2)^{1/2}$, yields $1$ which is really true if you are extremely close to $x=0$. Also it is not a first order approximation since its constant.

Is there another way to approximate, lets say, if I know the lower and upper bounds on $x$?

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    $\begingroup$ If you use 0 for the 1st order term you have a bound on the error (Lagrange's remainder). If you use something else for the 1st order term you don't know, in general. BTW, a constant is a particular case of a linear approximation $\endgroup$ – Luis Mendo Mar 29 '16 at 15:00
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    $\begingroup$ There's nothing wrong with the first-order approximation being constant. It just means the function is rather flat near that point. After all, how else would you approximate a function that actually is constant? Or what about the function $x\mapsto x^{100}+ 7$ near $x=0$? $\endgroup$ – MPW Mar 29 '16 at 15:01
  • $\begingroup$ @LuisMendo sorry I didn't get the first part...the only better way I can think of is to use the middle point of my bounds and approximate there...As for the constant being a particular case of linear approximation I am not sure since using this argument a lower order polynomial is always a particular case of a higher order polynomial $\endgroup$ – Zero Mar 29 '16 at 15:03
  • $\begingroup$ @MPW I can understand what you mean but I was wondering if there is another way to approximate if your margin of error is relatively large $\endgroup$ – Zero Mar 29 '16 at 15:05
  • $\begingroup$ If you require linear approximation, then move the center to another point which is closer to the point at which you need to evaluate the approximation. Approximations are implicitly local approximations, which means they are only valid for small neighborhoods, not globally. $\endgroup$ – MPW Mar 29 '16 at 15:16
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For simplicity I will use your function as an example. The pointwise error of linear approximation can be defined as $$dE=\big(\sqrt{1+x^2}-ax-b \big)^2$$ If you integrate the error over your domain (let's say from $-x_1$ to $x_1$) you find the total error over the domain $$E=\int_{-x_1}^{x_1}dE=\int_{-x_1}^{x_1}\big(\sqrt{1+x^2}-ax-b \big)^2dx$$ Completing the integration you get the closed form for total error $$E= \frac23(1+a^2)x_1^3+2x_1\bigg(1+b^2-b\sqrt{1+x_1^2}\bigg)-2b\text{ArcSinh(}x_1\text{)}$$ Now you can minimize the error by taking the partial derivatives wrt a and b; equating the equations to zero.

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  • $\begingroup$ appreciate the comment but it is a least square fit... and since the function as well as the bounds are symmetric, it would give you a constant...I was wondering more in the sense of different kinds of approximation algorithms like the Taylor series approximation... :) $\endgroup$ – Zero Mar 30 '16 at 11:20
  • $\begingroup$ @Zero Taylor series approximation is a local one. It only approximates the function close to your evaluation point irrespective of the bounds. You can vary the algorithm by increasing the norm and/or adding constraints for slope. $\endgroup$ – AnilB Mar 30 '16 at 16:15

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