- #1
Battlemage!
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Homework Statement
The problem asks to use a substitution y(x) to turn a series dependent on a real number x into a power series and then find the interval of convergence.
[tex]
\sum_{n=0}^\infty (
\sqrt{x^2+1})^n
\frac{2^n
}{3^n + n^3}[/tex]
Homework Equations
After making a substitution, the book asks you to find pn, then take the limit as n approaches infinity. Basically just the ratio test.
The Attempt at a Solution
I got the right numerical answer for the interval of convergence, but I encountered a pretty bad logical inconsistency. I am pretty sure I've made an algebra mistake, but I can't find anything online that has roots, rational inequalities and absolute values all in the same problem, so I have turned to the lovely people at physicsforums.com.
Just to be thorough I'll include every step of the entire problem in case (a) the mistake is in the series manipulation or (b) some other error was made that I missed. I suspect, however, that I have a fundamental algebra misunderstanding.The most obvious substitution here for me was
[tex]y=\sqrt{x^2+1}[/tex]
so putting that in I have:
[tex]
\sum_{n=0}^\infty
y^n
\frac{2^n
}{3^n + n^3}[/tex]
so then pn
[tex]p_n = \left |
\frac {y^{n+1} \text{ }
2^{n+1}
}{3^{n+1} + (n+1)^3} \frac {3^n + n^3}
{y^n \text{ }
2^n} \right |[/tex]
[tex]p_n = \left |
\frac {y^n \text{ }y \text{ }
2^n \text{ }2
}{3^n \text{ }3 + n^3 + 3n^2 + 3n + 1} \frac {3^n + n^3}
{y^n \text{ }
2^n} \right |[/tex]
I always do that ^ so it's easier to see what cancels.
[tex]p_n = \left |
\frac {2y(3^n + n^3)
}{3^n \text{ }3 + n^3 + 3n^2 + 3n + 1} \right |[/tex]
Next I divided everything by 3n.
[tex]p_n = \left |
\frac {2y(1 + \frac{n^3}{3^n})
}{3 + \frac{n^3}{3^n} + \frac{3n^2}{3^n} + \frac{3n}{3^n} + \frac{1}{3^n}} \right |[/tex]So then I have that p = the limit of the above as n goes to infinity, which should be:
[tex]p = \left |
\frac {2}{3}y \right |[/tex]
since an exponential increases way faster than a polynomial.
So, that means the interval of convergence is going to be, if I understand it correctly, [tex] \left |
\frac {2}{3}y \right |<1[/tex]
(not including the endpoints, which must be tested separately.
This is where I think I'm the most lost. As I understand it correctly, normally this would mean:
[tex]
-1< \frac {2}{3}y <1[/tex]
Putting x back in gives
[tex]
-1< \frac {2}{3} \sqrt{x^2+1} <1[/tex]Moving the 2/3 over:
[tex]
-\frac {3}{2} < \sqrt{x^2+1} < \frac {3}{2} [/tex]Squaring everything:
[tex]
\frac {9}{4} < x^2+1 < \frac {9}{4} [/tex]Then getting the 1 on the other side:
[tex]
\frac {5}{4} < x^2 < \frac {5}{4} [/tex]and lastly taking the root to solve for x:
[tex]
\frac {\sqrt{5}}{2} < x < \frac {\sqrt{5}}{2} [/tex]And what is this madness? How can x be both less than and greater than the same number?Incidentally, according to the textbook answers, the interval of convergence is
[tex] |x|<\frac {\sqrt{5}}{2}[/tex]
so I have the right number. But somewhere along the line I lost logical consistency.Hopefully someone can see where I messed up and enlighten me about the mistake I'm making. Thanks for reading and responding!