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Homework Help: Using substitution to turn a series into a power series.

  1. Nov 1, 2016 #1
    1. The problem statement, all variables and given/known data
    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.

    \sum_{n=0}^\infty (
    }{3^n + n^3}[/tex]

    2. Relevant 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.

    3. 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
    so putting that in I have:
    }{3^n + n^3}[/tex]

    so then pn

    [tex]p_n = \left |
    \frac {y^{n+1} \text{ }
    }{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:

    -1< \frac {2}{3}y <1[/tex]

    Putting x back in gives

    -1< \frac {2}{3} \sqrt{x^2+1} <1[/tex]

    Moving the 2/3 over:

    -\frac {3}{2} < \sqrt{x^2+1} < \frac {3}{2} [/tex]

    Squaring everything:

    \frac {9}{4} < x^2+1 < \frac {9}{4} [/tex]

    Then getting the 1 on the other side:
    \frac {5}{4} < x^2 < \frac {5}{4} [/tex]

    and lastly taking the root to solve for x:
    \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 text book 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!
  2. jcsd
  3. Nov 1, 2016 #2

    Stephen Tashi

    User Avatar
    Science Advisor

    You don't preserve the solutions to an inequality when you square both sides.

    (-2) < (-1) doesn't imply (-2)(-2) < (-1)(-1)

    could be written as [itex]0 < \frac {2}{3} \sqrt{x^2+1} <1[/itex] or simply as [itex] \frac {2}{3} \sqrt{x^2+1} <1 [/itex] since the square root must be non-negative.
  4. Nov 1, 2016 #3
    Thanks (Weird that I couldn't find anything on this online).

    But that still leaves one small conundrum. What is the interval of convergence?
    [tex]0 \text{ to } \frac {\sqrt{5}}{2}?[/tex] But as I said the solution for this problem in Boas 3rd ed. according to a solutions book says that the interval of convergence is still an absolute value, namely
    [tex]|x| <\frac {\sqrt{5}}{2}[/tex]

    EDIT- A thought just occurred to me. When solving for x here, I eventually end up with

    [tex]x^2 <\frac {5}{4}[/tex]

    In middle school we learned that the solution for x2 = 4 was x = 2 or x = -2. Is it something that simple? Would taking the square root give me
    [tex]x <\frac {\sqrt{5}}{2}[/tex]
    [tex]x>-\frac {\sqrt{5}}{2}[/tex]

    and could that save me?
  5. Nov 1, 2016 #4

    Stephen Tashi

    User Avatar
    Science Advisor

    You have the right idea, but aren't expressing it precisely. The square root of a quantity (if that square root exists) is a non-negative number. It isn't "plus or minus" some number.

    The solutions to an equation or an inequality are often found by "doing the same thing to both sides of it", but this is not a completely reliable process. You have to think before you act and you can't always find the solutions by "doing the same thing to both sides".

    A correct way to solve the equation ##x^2 = 25 ## is think about the possibilities and realize that this equation is has the same solutions as the equation ##|x|^2 = 25##. You can take the square root of both sides of that equation without losing track of the solution x = -5.

    If you try to solve ##x^2 = 25## by taking the square root of both sides, you can still save the situation if you realize that ##\sqrt{x^2} = |x| ##.

    The misconception that a square root can be "plus or minus" is just a garbled way of describing the above process.
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