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Show that if x is small compared with unity, then $$f(x)=\frac{(1-x)^\frac{-2}{3}+(1-4x)^\frac{-1}{3}}{(1-3x)^\frac{-1}{3}+(1-4x)^\frac{-1}{4}}=1-\frac{7x^2}{36}$$

In my first attempt I expanded all four brackets up to second order of x, but this didn't lead me to something that could be expressed as the final result. In my second attempt I decided to find $f'(x)$ and $f''(x)$ and use these to find $f'(0)$ and $f''(0)$ to find the Maclaurin expansion of $f(x)$ but this was way too time consuming. Can someone lead me to right track and offer some assistance? Thank you

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  • $\begingroup$ Maybe instead of computing the derivatives of the LHS and RHS, you could multiply your equality through by the denominator and then compute taylor expansions. That might give you something! It's possible though that the exercice was --meant-- to be time consuming, in which case you need to compute the second order derivatives of your big fraction... $\endgroup$
    – Patrick Da Silva
    Nov 17, 2013 at 15:19
  • $\begingroup$ I'm certain that I'm not meant to be finding f''(x) or multiplying/dividing across the equality. There must be a simpler way of doing this... $\endgroup$
    – Vladimir Nabokov
    Nov 17, 2013 at 15:27
  • $\begingroup$ @StefanSmith I don't seem to be having that issue with my display... $\endgroup$
    – Vladimir Nabokov
    Nov 17, 2013 at 17:11
  • $\begingroup$ @VladimirNabokov : is the RHS of your equation supposed to be something like "$1 - 7x^2/36 + o(x^3)$"? $\endgroup$
    – Stefan Smith
    Nov 18, 2013 at 21:02

3 Answers 3

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Expand each piece in the numerator and denominator up to the second order in Taylor series (this is the minimum taking into account what you are supposed to find). Then develop the result again as a Taylor series ... and you will find it !

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  • $\begingroup$ I tired that before but ended up with a fraction of two quadratic equations, each with three terms! $\endgroup$
    – Vladimir Nabokov
    Nov 17, 2013 at 15:33
  • $\begingroup$ It is very simple. Just try using the small pieces. $\endgroup$
    – Claude Leibovici
    Nov 17, 2013 at 15:36
  • $\begingroup$ Will get back to you! Thanks $\endgroup$
    – Vladimir Nabokov
    Nov 17, 2013 at 15:36
  • $\begingroup$ Let me propose a simpler way : expand each of the four pieces, comine them to get the expansion of the numerator (call it A) and the expansion of the denominator (call it B). Now multiply the rhs by B and show that it is equal to A. This is similar to what Patrick Da Silva was proposing. $\endgroup$
    – Claude Leibovici
    Nov 17, 2013 at 16:35
  • $\begingroup$ I've ended up $f(x)=\frac{\frac{37x^2}{9}+2x+2}{\frac{9x^2}{2}+2x+2}$ How shall I proceed. Theres no way of simplifying this to $1-\frac{7x^2}{36}$ ? @ClaudeLeibovici, I don't think the method you suggested works.. $\endgroup$
    – Vladimir Nabokov
    Nov 17, 2013 at 19:12
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For two real functions $f,g$, we have $$ \left( \frac fg \right)' = \frac{f'g -fg'}{g^2}, \quad \left( \frac fg \right)'' = \frac{f''g -fg''}{g^2} - 2 \frac{f'g-fg'}{g^3}g'. $$ It follows that the Taylor expansion of your function to the second order depends only on the Taylor expansion of your numerator and denominator to the second order. Let $f(x)$ be your numerator and $g(x)$ your denominator. Compute $f(0), f'(0), f''(0), g(0), g'(0), g''(0)$ and plug them in the last equations ; they will give you your Taylor expansion for $f/g$ around $x=0$, thus giving you the polynomial on the right.

For instance, $$ f(x) = (1-x)^{-2/3} + (1-4x)^{-1/3} \\ f'(x) = \frac 23 (1-x)^{-5/3} + \frac 43 (1-4x)^{-4/3} \\ f''(x) = \frac{10}9 (1-x)^{-8/3} + \frac{64}{9}(1-4x)^{-7/3} $$ hence $f(0) = 2$, $f'(0) = 2$ and $f''(0) = 74/9$. Do the same for $g$, you'll get there!

Hope that helps,

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The development of the last fraction (ratio of the two quadratic polynomials) is
1 - 7 x^2 / 36 + 7 x^3 / 36 + 35 x^4 / 144.

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