Proving if a linear part of a function exists using algebra.(proof of the second derivative being zero at some point/points)

Question

I know that the following question can be resolved using derivatives but nonetheless, I would like to hear a more fluid and lucid approach. Any hand waving explanation and approximations are welcome.

The question is:-

Prove that there exists a linear part of the function $f($x$)$ $=$ $\frac{1}{\sqrt{(x^2+1)^3}}$ at x= $-1/2$ and x=$1/2$

Now initially I thought, we could use the binomial expression to expand the quantity, then I would find the quantity around the said $x$ values and pray to almighty that it would resemble a linear function. But could get nowhere. Thanks in Advance!!!

Do you just mean that the second derivative is $0$? If so, I suggest editing your post to state that. — lulu, Feb 13 '24 at 12:12
Looking quickly at functions of the form $F_n(x)=\frac 1{(1+x^2)^n}$ we see that they have inflection points at $\pm \frac 1{\sqrt{2n+1}}$. That gives you want you want. Of course, you still need to argue that the form is correct but maybe there's an easy way to see that. The functional equation $F\left(\frac 1x\right)=x^{2n}F_n(x)$ might help. — lulu, Feb 13 '24 at 12:31

score 1 · Accepted Answer · answered Feb 13 '24 at 13:33

1

Let $\varepsilon=\pm1$.

$$\begin{align}f\left(\frac\varepsilon2+h\right)&=\left(\left(\frac\varepsilon2+h\right)^2+1\right)^{-3/2}\\ &=\left(\frac54+\varepsilon h+h^2\right)^{-3/2}\\ &=\left(\frac54\right)^{-3/2}\left(1+\frac{4\varepsilon}5h+\frac45h^2\right)^{-3/2}\\ &=\left(\frac54\right)^{-3/2}\left(1-\frac32\left(\frac{4\varepsilon}5h+\frac45h^2\right)+\frac{\left(-\frac32\right)\left(-\frac52\right)}2\left(\frac{4\varepsilon}5h\right)^2+o(h^2)\right)\\ &=\left(\frac54\right)^{-3/2}\left(1-\frac{6\varepsilon}5h+o(h^2)\right). \end{align} $$

answered Feb 13 '24 at 13:33

Anne Bauval

34,650

what is ‘O’………… – TheCuriousOne Feb 13 '24 at 14:33
I wrote $o$ (little o, defined here), not $O$ (big O). $o(h^2)$ means: $h^2e(h)$, where $e$ is a function such that $\lim_{h\to0}e(h)=0$. – Anne Bauval Feb 13 '24 at 14:36

Proving if a linear part of a function exists using algebra.(proof of the second derivative being zero at some point/points)

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