So I am stuck in a rut. I have a function, and I am taking the Taylor Series.

$$f(x+\ell)=f(x)+\ell f'(x)+\frac{\ell^2}{2!}f''(x)+\dots$$

Unfortunately, I know in advance that the second derivative is undesirable but the first derivative is good. Is there an approximation preferably involving the first derivative for the second derivative? I'm trying to avoid central difference, and other similar approximation methods.

Ideally I'd like something that looks kinda like (Note I'm making this expression up)

$$f''(x)\approx f(x)+\frac{f'(x)}{f(x)}$$

This way I can avoid the horrid second derivative and also avoid "h" terms that come up from other approximations(e.g. Central Difference). Until I can check it, I am going to assume that a second derivative approximation i.t.o. the function and it's derivative is better than the second derivative itself.

Note: I have also looked into using other approximations such as Chebyshev and Padé.

If someone with more experience can recommend an alternative to the Taylor series (That I haven't seen) or a second derivative approximation that would be extremely helpful.

Many Thanks


EDIT: @Dylan requested that my question have better context so I will provide some.

Problem: The analytic solution exhibits a discontinuity in $\frac{\partial^2 \bar{f}}{\partial x^2}$ instead of predicting an expected asymptotic approach.

Context: A fluid lump comes from above at $x+\ell$ and has some velocity $f(x+\ell)$ which is displaced over a transverse distance. The fluctuation is given by $\Delta f = \bar{f}(x+\ell)-\bar{f}(x)$ .$\bar{f}(x+\ell)$ is expanded in a Taylor series up to the linear term, resulting in $\Delta f = \ell \frac{\partial \bar{f}}{\partial x}$ Assuming that the transverse fluctuations of $g$ are of the same order of magnitude as $f$ we have $g=cf$ , $c$ some constant. The product of these fluctuations is therefore $\Delta g\Delta f = c (\Delta f)^2 $

$$= c \ell^2 \left(\frac{\partial \bar{f}}{\partial x}\right)\left|\frac{\partial \bar{f}}{\partial x}\right|$$

In order to include $\frac{\partial^2 \bar{f}}{\partial y^2}$, we let fluid lumps from the side of the greater velocity entertain higher values of velocity f, those from the side of smaller velocities, smaller values , with the result that more momentum is transported in one direction than in the other. Approximated $$f=\bar{f} + \ell \frac{\partial \bar{f}}{\partial x}$$ and $$g =\bar{f} - \ell \frac{\partial \bar{f}}{\partial x}$$

Taking a statistically average value of $$\left|\frac{\partial f}{\partial x}\right|$$ that is proportional to $$\sqrt{\left(\frac{\partial f}{\partial x}\right)^2}$$

Again assuming that $f$ and $g$ are of the same order of magnitude we have that $$\sqrt{\left(\frac{\partial f}{\partial x}\right)^2}=\sqrt{\left(\frac{\partial f}{\partial x}\right)\left(\frac{\partial g}{\partial x}\right)}=\sqrt{\left(\frac{\partial \bar{f}}{\partial x} + \ell \frac{\partial^2 \bar{f}}{\partial x^2}\right)\left(\frac{\partial \bar{f}}{\partial x} - \ell \frac{\partial^2 \bar{f}}{\partial x^2}\right)}$$ $$\approx\sqrt{\left(\frac{\partial \bar{f}}{\partial x}\right)^2 + \ell^2 \left(\frac{\partial^2 \bar{f}}{\partial x^2}\right)^2}$$

As before $\Delta g\Delta f = c (\Delta f)^2 $

$$= c \ell^2 \left(\frac{\partial \bar{f}}{\partial x}\right)\sqrt{\left(\frac{\partial \bar{f}}{\partial x}\right)^2 + \ell^2 \left(\frac{\partial^2 \bar{f}}{\partial x^2}\right)^2}$$

Thus when $\frac{\partial^2 \bar{f}}{\partial x^2}=0$, we have our previous result.

I wish to approximate $\frac{\partial^2 \bar{f}}{\partial x^2}$ in terms of $f(x)$or $f'(x)$ using another method i.e. $f''(x)$ in some other way, which was previously asked briefly before the edit.

  • 1
    $\begingroup$ What is the context of your problem? Are there any hypotheses on $f$? Why do you consider $f''(x)$ as "undesirable"? $\endgroup$ – TZakrevskiy May 24 '18 at 14:01
  • 1
    $\begingroup$ aren't Taylor series $f(x+\ell)=f(\ell)+(x+\ell)f'(\ell)+\frac{(x+\ell)^2}{2!}f''(\ell)+\dots$? $\endgroup$ – Jepsilon May 24 '18 at 14:04
  • $\begingroup$ Why do you not want to use the second derivative? $\endgroup$ – Allawonder May 24 '18 at 14:10
  • 1
    $\begingroup$ This is too vague. More context is needed. What is the actual function? What is the full problem? What kind of approximation do you want? Why is only the first derivative suitable? $\endgroup$ – Dylan May 24 '18 at 14:10
  • $\begingroup$ @Dylan The Taylor series is expanded up to the linear term only, for a particular problem I'm working on. They defined $$\mathit{\Delta} f = \bar{f}(x+\ell)-\bar{f}(x)$$ so that after the Taylor series expansion wo obtain $$\mathit{\Delta} f = \ell \frac{\partial \bar{f}}{\partial x}$$. I would like to work through the problem but including the nonlinear term from the second derivative but only an approximate of this second derivative. $\endgroup$ – KennyB May 24 '18 at 14:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.