• Apr 10th 2013, 03:21 AM
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 $\dfrac{64b^2}{a^4}x^2 \le 1$

Solving that I would get:

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

$|x| \le \dfrac{8b}{a^2}$

And so the radius of convergence is: $\dfrac{8b}{a^2}$ ?

And how do I show that for the series to converge, $b < \dfrac{a}{4}$ ?
• Apr 10th 2013, 04:21 AM
Gusbob
Quote:

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 $\dfrac{64b^2}{a^4}x^2 \le 1$

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 $\sqrt{1+z}$ converges whenever $|z|<1$. In your case, $z=\frac{64b^2}{a^4}x^2$ which gives you $\frac{64b^2}{a^4}x^2<1$
Quote:

Solving that I would get:

$x^2 \le \dfrac{64b^2}{a^4}$
I suggest you take a look at this equation again. It should be $x^2<\frac{a^4}{64b^2}$. Then $|x|<\frac{a^2}{8b}$. When $x=\frac{a}{2}$, we have $\frac{a}{2}< \frac{a^2}{8b}$ (since a is positive, we drop the absolute value sign). This is true if and only if $b<\frac{a}{4}$ simply by manipulating the inequality.
• Apr 10th 2013, 04:54 AM
Educated
I suggest you take a look at this equation again. It should be $x^2<\frac{a^4}{64b^2}$. Then $|x|<\frac{a^2}{8b}$. When $x=\frac{a}{2}$, we have $\frac{a}{2}< \frac{a^2}{8b}$ (since a is positive, we drop the absolute value sign). This is true if and only if $b<\frac{a}{4}$ simply by manipulating the inequality.