This is from an A-Level Maths paper.

Show that $\frac{x}{(1-x)^3} = x + 3x^2 + 6x^3 + O(x^4)$ (Only first three terms of infinite series expansion are asked for) Use the result to find an approximate value of $\frac{100}{729}$.

I can do the first part without a problem. On the second part, I know substituting $x = 0.1$ is what I need to do. But how do I go about figuring out that $x = 0.1$ is indeed what I need.

Setting $\frac{x}{(1-x)^3}=\frac{100}{729}$ and solving the resulting cubic does give $x=0.1$, but I feel there must be something simpler that I'm missing.



closed as off-topic by Jack D'Aurizio, user91500, Claude Leibovici, R_D, user223391 Jun 24 '16 at 22:26

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    $\begingroup$ Your first equality cannot hold: the RHS is a polynomial, the LHS is a meromorphic function with a triple pole at $x=1$. If you are talking about Taylor expansions, please state it clearly, and have a look at meta.math.stackexchange.com/questions/5020/… $\endgroup$ – Jack D'Aurizio Jun 22 '16 at 12:01
  • $\begingroup$ @NumberCruncher You have probably never heard of meromorphic functions, but just multiply across: the LHS is just $x$ the RHS is a polynomial of degree 6. You have miscopied the question. $\endgroup$ – almagest Jun 22 '16 at 12:04
  • $\begingroup$ Thanks for your replies. Yes, it is an infinite series, it's meant to be LHS is approximately equal to RHS, or LHS = RHS + O(x^4) $\endgroup$ – NumberCruncher Jun 22 '16 at 13:31
  • $\begingroup$ "how do I go about figuring out that $x = 0.1$ is indeed what I need." (No idea how the other comments address this, which seems to be your main, and probably only, question.) The idea is to hope that $x/(1-x)^3=100/729$ has a rational solution $x=a/b$ with $1\leqslant a<b$ relatively prime integers, or equivalently, that $3^6ab^2=10^2(b-a)^3$. Since $ab^2$ and $b-a$ are relatively prime, we know that $ab\mid 10$, and suddenly we are left with very few cases (recall that $1\leqslant a\leqslant b-1$), namely, $(a,b)=(1,2)$ or $(1,5)$ or $(1,10)$ or $(2,5)$. ... $\endgroup$ – Did Jun 23 '16 at 7:31
  • $\begingroup$ ... Likewise, $3^2\mid b-a$ hence $(a,b)=(1,2)$, $(1,5)$ and $(2,5)$ are impossible. Finally, if $(a,b)=(1,10)$ works, we are done, otherwise there is no rational solution. And $(a,b)=(1,10)$ works... :-) $\endgroup$ – Did Jun 23 '16 at 7:31