MitPY 11: Asymptotes and Wolfram Alpha

January 31, 2021

6:23 pm

5 Comments on MitPY 11: Asymptotes and Wolfram Alpha

indices, Mathematical Methods, MitPY, roots, Specialist Mathematics, tests

This MitPY comes from frequent commenter, John Friend:

Dear colleagues,

I figured this was as good place as any to ask for help. I’m writing a small test on rational functions. One of my questions asks students to consider the function $\displaystyle f(x) = \frac{x^3 + x}{x^2 + ax - 2a}$ where $a \in R$ and to find the values of $a$ for which the function intersects its oblique asymptote.

The oblique asymptote is $y = x - a$ so they must first solve

$\displaystyle \frac{x^3 + x}{x^2 + ax - 2a} = x - a$ … (1)

for $x$ . The solution is $\displaystyle x = \frac{2a^2}{(a+1)^2}$ and there are no restrictions along the way to getting this solution that I can see. So obviously $a \neq -1$ .

It can also be seen that if $a = 0$ then equation (1) becomes $\displaystyle \frac{x^3 + x}{x^2} = x$ which has no solution. So obviously $a \neq 0$ .

When I solve equation (1) using Wolfram Alpha the result is also $\displaystyle x = \frac{2a^2}{(a+1)^2}$ . But here’s where I’m puzzled:

Wolfram Alpha gives the obvious restriction $a + 1 \neq 0$ but also the restriction $5a^3 + 4a^2 + a \neq 0$ .

$a \neq 0$ emerges naturally (and uniquely) from this second restriction and I really like that this happens as a natural part of the solution process. BUT ….

I cannot see where this second restriction comes from in the process of solving equation (1)! Can anyone see what I cannot?

Thanks.

5 Replies to “MitPY 11: Asymptotes and Wolfram Alpha”

Grant Cairns says:

January 31, 2021 at 7:23 pm

The solution $\displaystyle x = \frac{2a^2}{(a+1)^2}$ is fine as long as it doesn’t happen to equal one of the points where $f(x)$ isn’t defined; namely, $\frac{-a\pm \sqrt{a^2+8a}}2$ . A little exploration shows that this happens when $5a^3 + 4a^2 + a = 0$ .

Reply
Anonymous says:

January 31, 2021 at 7:25 pm

5a^3 + 4a^2 + a = 0, when solved over the reals, gives only a = 0.

If you type this into wolfram alpha: “(x^3 + x)/(x^2 + ax – 2a) = x – a, reals” , you won’t get the fancy second restriction.

As for why 5a^3 + 4a^2 + a =/= 0 over the complex numbers, when you sub x=2a^2 / (a+1)^2 into x^2 + ax – 2a, you get an expression that has the same complex zeros as 5a^3 + 4a^2 + a.

Wolfram alpha presumably converts a complex (pun intended) restriction into a polynomial where possible, so as not to have something like “(4 a^4)/(a + 1)^4 + (2 a^3)/(a + 1)^2 – 2 a =/= 0”.

Reply
1. John Friend says:
  
  January 31, 2021 at 7:43 pm
  
  Thanks very much GC and Anonymous. Much appreciated. I’d substituted x into all of f(x), not just the denominator and after simplifying a got something different. Of course, now that it’s been pointed out, it’s obvious to simply require the denominator of f(x) to not be zero.
  
  Thanks again.
  
  Reply
Steve R says:

February 8, 2021 at 10:30 pm

Hi,

GC and Anonymous have highlighted the shortcomings of Mathematica .

IMO it would not hurt to add a warning that 5a^2 + 4a + 1 = 0 has no real solutions as discriminant is negative

Steve R

Reply
1. John Friend says:
  
  February 9, 2021 at 6:59 am
  
  Thanks, Steve. I agree. I think Wolfram Alpha has assumed x and a are complex (this is the default assumption in Mathematica). Things look different when I specify reals:
  
  https://www.wolframalpha.com/input/?i=Solve+%28x%5E3+%2B+x%29%2F%28x%5E2+%2B+a*x+-+2a%29+%3D+x-a+where+x+is+a+real+number
  
  Reply

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5 Replies to “MitPY 11: Asymptotes and Wolfram Alpha”

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