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I have a function $$f(x)=\frac{1}{x^2+1}$$ defined for $x\geq 0$. In one of the paper it is represented by its concave representation which is given below $$f'(x)\simeq -x^2+2\left(1-\frac{1}{(x_0^2+1)^2}\right)xx_0+\frac{1}{x_0^2+1}+\frac{2x_0^2}{(x_0^2+1)^2}-x_0^2$$ where the approximation is evaluated at $x=x_0$. I tried to get it through second order Taylor approximation but my expression does not match with it. Any help in this regard will be much appreciated.Thanks in advance.

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    $\begingroup$ You mention $a$ and $b$ are positive constants, but I don't see them anywhere in your question. $\endgroup$ – Toby Mak Mar 8 '18 at 8:57
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    $\begingroup$ @TobyMak thank you for your correction comment. I have removed them from my question. $\endgroup$ – Frank Moses Mar 8 '18 at 9:01
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    $\begingroup$ According to Wolfram Alpha, the Taylor series at $x=a$ is $\frac{1}{a^2+1} - \frac{2a(x-a)}{(a^2+1)^2}$. Subtracting this from the approximation gives $-(a-x)^2$, which approaches $0$ as $x$ approaches $a$. $\endgroup$ – Toby Mak Mar 8 '18 at 9:10
  • $\begingroup$ @TobyMak yes at $x=a$ they give same answer but at other points their answer are different. For example I tried to check for $x_0=2$ case. For this case the Taylor expansion provides a convex looking function while the above equation(in my post) provides a concave function over all values of $x$ $\endgroup$ – Frank Moses Mar 8 '18 at 9:16
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$$f(x)=\frac{1}{x^2+1}\quad\implies\quad f'(x)= -\frac{2x}{(x^2+1)^2}$$ The approximation evaluated at $x=x_0$ is proposed as : $$f'(x)\simeq -x^2+2\left(1-\frac{1}{(x_0^2+1)^2}\right)xx_0+\frac{1}{x_0^2+1}+\frac{2x_0^2}{(x_0^2+1)^2}-x_0^2$$ So, at $x=x_0$ the above formula gives : $$f'(x_0)\simeq -x_0^2+2\left(1-\frac{1}{(x_0^2+1)^2}\right)x_0x_0+\frac{1}{x_0^2+1}+\frac{2x_0^2}{(x_0^2+1)^2}-x_0^2$$ After simplification : $$f'(x_0)\simeq \frac{1}{x_0^2+1}$$

If the proposed approximation was true, we should have $\quad f'(x_0)=-\frac{2x_0}{(x_0^2+1)^2}$

So, the proposed approximation is obviously false.

It seems that this is an approximation for $f(x)=\frac{1}{x^2+1}$, not for $f'(x)$. Probably a typo.

NOTE :

Even supposing that the proposed formula is an approximation for $f(x)$, it doesn't correspond exactly with the first terms of the Taylor series.

The first terms of the Taylor series are : $$f(x)=\frac{1}{x^2+1}\simeq \frac{1}{x_0^2+1} -\frac{2x_0}{(x_0^2+1)^2}(x-x_0)-\frac{1-3x_0^2}{(x_0^2+1)^3}(x-x_0)^2+...$$ or, rearranged as : $$f(x)\simeq -\frac{1-3x_0^2}{(x_0^2+1)^3}x^2-\frac{8x_0^3}{(x_0^2+1)^3}x+\frac{1+3x_0^2+6x_0^4}{(x_0^2+1)^3}$$ This draw us to think that the textbook approximation doesn't comes from a Taylor series expansion, but from another method of fitting, for example least mean squares. Since the range of $x$ where such regression calculus was carried out isn't specified, it is impossible to reproduce it. Of course, the validity of the approximate formula is limited to a range of $x_0$ depending on the unknown range of the regression calculus. For example, if the approximation is bad in case of $x_0=2$, this means that $x_0=2$ is outside the range that was taken into account for the calculus of the formula.

All those approximate functions are valid only on a small range of $x$ around $x_0$ as one can see on the next figures :

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enter image description here

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  • $\begingroup$ you are right I just used $'$ to mean approximation I forgot that it is also used as to represent the first derivative $\endgroup$ – Frank Moses Mar 9 '18 at 8:17
  • $\begingroup$ @Frank Moses : See the comment in addition to my first answer. $\endgroup$ – JJacquelin Mar 9 '18 at 9:19

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