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Suppose I have an infinite series, $$0= a_0 x^p + a_1 x^{p+1} + a_2 x^{p+2} + \mathcal{O}(x^{p+3})$$ where $x<1$ and $\mathbb{N}\ni p\geq 1$.

Under what conditions does it hold that $a_0=0$? In a proof, it was stated that $a_0=0$ because terms of order $\mathcal{O}(x^{p+1})$ cannot correct for this, otherwise i.e. $a_0 \neq 0$ would imply the series does not converge to $0$.

I don't immediately see why. Maybe it is not true in general, but under certain conditions?!

DoubleKeys
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  • Does the identity mean that the "series" is $0$ for that specific value of $x$, or does it mean that the sum is the zero function? –  May 19 '20 at 10:28
  • @Gae.S. In the application, $x$ varies, but the sum always converges to zero. – DoubleKeys May 19 '20 at 10:31
  • $0$ is an analytic function, and analytic functions have a unique power series expansion. Thus all coefficients $a_i=0$. – Calvin Khor May 19 '20 at 11:39
  • Uniqueness here https://math.stackexchange.com/questions/3433492/is-there-a-proof-that-all-analytic-functions-only-have-one-unique-taylor-series – Calvin Khor May 19 '20 at 11:45
  • I see $a_0, a_1,a_2, x$ and $p$ but I do not see any infinite series. – DanielWainfleet May 19 '20 at 12:07

1 Answers1

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The problem is oddly stated. Where is the infinite series? I see four terms added together. Nevertheless, we can show $a_0=0.$ On the interval $(-1,1)$ we are given

$$a_0x^{p}+a_1x^{p+1}+a_2x^{p+2} + O(x^{p+3})=0$$

(where you should say $O(x^{p+3})$ as $x\to 0$). Rewrite the above as

$$x^p(a_0+ a_1x+a_2x^{2} + O(x^{3}))=0.$$

For $x\ne0,$ we can divide by $x^p$ to get

$$a_0+ a_1x+a_2x^{2} + O(x^{3})=0.$$

Take the limit of this as $x\to 0.$ On the left we get $a_0.$ On the right we get $0.$ Therefore $a_0=0.$

zhw.
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  • The infinite series is implicit in the term $\mathcal{O}(x^{p+3})$ which is supposed to mean that there is an infinite number of terms of order $p+3$ or higher following the same pattern. You solution works in either case, I think. – DoubleKeys May 21 '20 at 15:27