Re: Radius of convergence

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Originally Posted by

**Educated** Attachment 27884
I can do part a) and b) but I'm not getting how to do part c)

So I'm guessing the part under the square root cannot be negative, and so $\displaystyle \dfrac{64b^2}{a^4}x^2 \le 1$

Your inequality is correct (though you may not necessarily have equality), but not the reason. The radius of convergence being one is the same as saying the expansion of $\displaystyle \sqrt{1+z}$ converges whenever $\displaystyle |z|<1$. In your case, $\displaystyle z=\frac{64b^2}{a^4}x^2$ which gives you $\displaystyle \frac{64b^2}{a^4}x^2<1$

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Solving that I would get:

$\displaystyle x^2 \le \dfrac{64b^2}{a^4}$

I suggest you take a look at this equation again. It should be $\displaystyle x^2<\frac{a^4}{64b^2}$. Then $\displaystyle |x|<\frac{a^2}{8b}$. When $\displaystyle x=\frac{a}{2}$, we have $\displaystyle \frac{a}{2}< \frac{a^2}{8b}$ (since a is positive, we drop the absolute value sign). This is true if and only if $\displaystyle b<\frac{a}{4}$ simply by manipulating the inequality.

Re: Radius of convergence

Quote:

Originally Posted by

**Gusbob** I suggest you take a look at this equation again. It should be $\displaystyle x^2<\frac{a^4}{64b^2}$. Then $\displaystyle |x|<\frac{a^2}{8b}$. When $\displaystyle x=\frac{a}{2}$, we have $\displaystyle \frac{a}{2}< \frac{a^2}{8b}$ (since a is positive, we drop the absolute value sign). This is true if and only if $\displaystyle b<\frac{a}{4}$ simply by manipulating the inequality.

Oh right, thank-you!

(I can't believe I made such a simple mistake by re-arranging incorrectly)