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One of my points is not a real number in my interval of convergence! This is how I calculated it:

$$\lim_{n\to\infty} \left|\frac{x^8-1}{3}\right|$$ Using root test

$\frac{x^8-1}{3} <1$

$x<4^{1/8}$

$\frac{x^8-1}{3} >-1$

$x>(-2)^{1/8}$

So you see this endpoint does not exist! what should I write for the interval?

Arpan
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Elsa
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1 Answers1

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EDIT: Sorry, the previous answer was not correct.

You have $$ \Bigg|\frac{x^8-1}{3}\Bigg| <1 \Leftrightarrow |{x^8}-1|<3 \Leftrightarrow |x|<4^{1/8} =\sqrt[4]{2}. $$ In these examples you typically want to work without getting rid of the absolute value, since the $x$ in your series may also be considered as complex number; and those have no relation of order, so you have to look at the modulus or absolute value in order to obtain the convergence radius, which will give you a disk in the complex case and an interval in the real case.

Alberto Debernardi
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  • What do the arrows mean in ur answer? why did u multiply by 3? so for the second one where i got a not real number, would it be (2)^1/8 since absolute value? – Elsa Apr 24 '15 at 07:45
  • You don't have to look at these extreme values without the absolute value, since the relation $x>(-2)^{1/8}$ does not exist (the complex numbers do not have a relation of order (see http://math.stackexchange.com/questions/312204/prove-that-field-of-complex-numbers-cannot-be-equipped-with-an-order-relation). – Alberto Debernardi Apr 24 '15 at 07:53
  • Did I do the math correct then? So I did |x^8-1|>|-3| and from then on it became 2^1/8 – Elsa Apr 24 '15 at 07:58
  • But my homework says 2^1/8 is incorrect endpoint. What did I do wrong? – Elsa Apr 24 '15 at 07:58
  • As I said, getting inequalities of complex numbers is bad. Also, it is not correct that if $-3<|x^8-1|$ then $|-3|<|x^8-1|$. You stop at the inequality $|x^8-1|<3$ and determine the radius at that point. If you take out the absolute value from that point you will probably get wrong things. – Alberto Debernardi Apr 24 '15 at 08:04
  • ok so i get |x^8-1|<3, but how do I move on from there? How do I get an actual point? – Elsa Apr 24 '15 at 08:06
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    If you don't see it directly from there then you can do some trick as considering the equation that takes the extreme value: $x^8-1=3$. Solve it to get $x=2^{1/4}$ and then prove that any $x$ with $|x|<2^{1/4}$ satisfies $|x^8-1|<3$ and that for $|x|=2^{1/4}$, the inequality is not satisfied. – Alberto Debernardi Apr 24 '15 at 08:10