I am reading a textbook, and it talks about doing synthetic division in order to rewrite a function into the quotient

$$R(x)=\frac{p(x)}{q(x)}= f(x) + \frac{r(x)}{q(x)}$$ Since $\frac{r(x)}{q(x)}$ approaches 0 for large x's, R(x) basically approaches f(x)

If f(x) = C, then you have a HA @ y=C

If f(x) = mx+b, then you have a HA @ y=mx+b

What if f(x) is a quadratic or a cubic, when the degrees of p(x) and q(x) are more than 1 apart. Does this yield a parabolic asymptote, or is that not recognized?

  • $\begingroup$ 'Parabolic asymptote' is actually mentioned on Wikipedia's page on asymptotes: en.wikipedia.org/wiki/Asymptote_Curvilinear_asymptotes. $\endgroup$ – Semiclassical Nov 21 '14 at 14:27
  • $\begingroup$ You may also be interested to read about en.wikipedia.org/wiki/Big_O_notation which essentially is used to describe the asymptotic behavior of functions in general (up to multiplication by some constant). (also @Semiclassical, your link is broken. There should be a hashtag after Asymptote instead of an underscore) $\endgroup$ – JMoravitz Nov 21 '14 at 15:08
  • $\begingroup$ Corrected link: Curvilinear asymptotes $\endgroup$ – Semiclassical Nov 21 '14 at 15:28
  • $\begingroup$ Thanks! Feel free to post a formal answer for approval. $\endgroup$ – JackOfAll Nov 21 '14 at 23:33
  • $\begingroup$ Here is an answer. If this is not proper I will change it. $\endgroup$ – Arbuja Oct 19 '15 at 20:46

Like they said this has to do with curvilinear asymptotes. For most curvilinear asymptotes, they can be in the form or a parabola, cubic or any function continuous at infinity.

As the above a simple functions. Laurent series at infinity helps a taylor series that doesn't converge to infinity converge.

Lets say you take the function such as $$\left({{x}^{\left(3/4\right)}+{x}^{(2/5)}+3}\right)^{(2/3)}$$ $$\left({\left(x^{1/20}\right)}^{15}+{\left(x^{1/20}\right)}^{8}+3\right)$$ Substitute $a={x}^{(1/10)}$ $${\left({\left(a\right)}^{15}+{\left(a\right)}^{8}+3\right)}^{2/3}$$ $$\left(\frac{{1}+{\left(\frac{1}{\left(a\right)}\right)}^{7}+\frac{3}{{a}^{15}}}{\frac{1}{a^{15}}}\right)^{2/3}$$ Replace with a with $1/a$ $$\left(\frac{{1}+{\left({a}\right)}^{7}+{3}{{a}^{15}}}{{a^{15}}}\right)^{2/3}$$ $$\left(\frac{{1}+{\left({a}\right)}^{7}+{3}{{a}^{15}}}{{a^{15}}}\right)^{2/3}$$ $${\left(\frac{1}{{a}^{15}}\right)}^{2/3}\left({{1}+{\left({a}\right)}^{7}+{3}{{a}^{15}}}\right)^{2/3}$$ Take the taylor series at $x=0$ for the second bracket. $${\left(\frac{1}{{a}^{10}}\right)}\left({1+\frac{2{a}^{7}}{3}-\frac{{a}^{14}}{9}+\frac{2{a}^{15}}{3}}...\right)$$ Now to substitue $a$ for $\frac{1}{20x}$ $${\left({x}^{1/2}\right)}\left({1+\frac{2}{{3x}^{7/20}}-\frac{1}{9{x}^{14/20}}+\frac{2}{{3}{x}^{15/20}}}...\right)$$ $$\lim_{x\to\infty}\left({{x}^{1/2}+\frac{2{x^{3/20}}}{{3}}-\frac{1}{9{x}^{4/20}}+\frac{2}{{3}{x}^{5/20}}}...\right)\approx{{x}^{1/2}+\frac{2{x^{3/20}}}{{3}}}$$

Use this for $\frac{x^3+5}{x+5}$ and you will approximately get $x^2-5x+25$.

Now lets say we have a rational function inside but something very complicated like .... Taking, for example, $${\left(\frac{x^3+3x^2+5}{x+3}\right)}^{1/3}\left({\frac{x^4+2x^3+5}{x+5}}\right)^{2/5}$$ using a shortcut technique I found you could divide both the polynomials in the bracket.

If you take the brackets and divide inside you get $(x^2)^{1/3}(x^3-3x^2+15x-75)^{2/5}$.

Now you can take this as an asymptote instead taking another laurent series since that would be painful. However all terms on the donimantor must be filled.

Now taking if you take the highest exponent number inside the brackets of ${\left(x^2\right)}^{1/3}$ multiply by x and so the same thing with $\left({x^3-3x^2+15x-75}\right)^{2/5}$ except you multiply with y. Then you replace x and y with the exponents that are lesser than the lowest high exponent number.

$$2(1/3)+3(2/5)<3$$ $$(2/3)+(6/5)<3$$ $$10/15+18/15<3$$ $$33/15<3$$ Thus $(x^2)^{1/3}(x^3-3x^2+15x-75)^{2/5}$ can be an asymptote. However, these asymptotes can only be used at a specific range. And they're are several rules behind this when using sine functions and adding up polynomials....which I'm still not done with plus the proofs for certain rules are too long.


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