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Simple Convergence

  1. Jan 29, 2007 #1
    Just out of curiosity, does
    [tex]\sum\limits_{n = 1}^\infty {\sin \left( {x^n } \right)} [/tex]
    converge [itex]\forall x \in \left( -1 , 1 \right)[/itex] ?
     
    Last edited: Jan 29, 2007
  2. jcsd
  3. Jan 29, 2007 #2

    quasar987

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    I would be tempted to say yes, because of the inequality |sin(x)|<|x|, which implies |sin(x^n)|<|x^n|, which implies

    [tex]\sum^m|\sin(x^n)|\leq \sum^m|x^n|[/tex]

    But for -1<x<1, the sum on the right is geometric, which converges. So the sine series converges absolutely.
     
  4. Jan 30, 2007 #3
    can you please tell me how i can prove that |sin(x)| < |x|?
     
  5. Jan 30, 2007 #4

    HallsofIvy

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    Strictly speaking, it's not true- sin(x)= x when x= 0!

    At x=0 sin(0)= 0. Let f(x)= x- sin(x). Then f'(x)= 1- cos(x) which is always greater than or equal to 0. That is, x- sin(x) is an increasing function. For any x> 0 x> sin(x). For x< 0 use the fact that both x and sin(x) are odd functions: if x is negative, then -x is positive and so -x> sin(-x)= -sin(x). But x is negative and (for [itex]0> x> -\pi[/itex]) so is sin(x). |sin(x)|= -x> |sin(x)|= -sin(x). For |x|> 1, it is obvious that |x|> |sin(x)|.
     
  6. Feb 2, 2007 #5
    I see, thank you :smile:

    A similar question (but like the previous, related to a larger problem):
    [tex]\forall n \in \mathbb{N},{\text{ does }}\exists x > 1:\frac{1}
    {n}\sum\limits_{k = 1}^n {\sin \left( {x^k } \right)} \leqslant \sin 1{\text{ ?}}[/tex]
     
  7. Feb 6, 2007 #6
    In particular, I'm looking at the function
    [tex]f_n \left( x \right) = \frac{1}{n}\sum\limits_{k = 1}^n {\sin \left( {x^k } \right)} [/tex]

    where
    [tex]\forall n \geqslant 1,\; - 1 \leqslant f_n \left( x \right) \leqslant 1[/tex]
    and
    [tex]\forall x \in \left[ 0,1 \right], \; \frac{{df_n }}{{dx}} > 0 [/tex]

    From inspection at n=2,3,..., up to n=25 (my graphing program's constraints!),
    the first critical point (i.e., dfn(x)/dx=0) that fn(x) approaches (for positive x) is always a maximum at some x>1

    Furthermore, if we let xn be the x-value of this maximum for fn(x)
    (i.e., the smallest x>0 such that dfn(x)/dx=0), then we notice
    [tex]x_2 > \cdots > x_n > 1[/tex]
    --------------------------------------------------------------------------------
    *But, what is
    [tex] \lim \limits_{n \to \infty } x_n \; ? [/tex]

    Clearly, [itex] \lim \limits_{n \to \infty } x_n \ne 1 [/itex], since
    [tex]f'_n \left( 1 \right) = \left( {\cos 1 + 2\cos 1 + 3\cos 1 + \cdots } \right)/n > 0[/tex]

    Also, since
    [tex]\frac{{df_n }}{{dx}} = \frac{1}{n}\sum\limits_{k = 1}^n {kx^{k - 1} \cos \left( {x^k } \right)} [/tex]

    my question is simply
    [tex]{\text{What is the smallest }}x > 0\;{\text{such that }}\frac{d}
    {{dx}}\sum\limits_{k = 1}^{\infty} {\sin \left( {x^k } \right)} = 0\;? [/tex]
     
    Last edited: Feb 6, 2007
  8. Feb 10, 2007 #7
    No replies? :redface:
    In particular, before asking
    [tex]\text{What is } \lim \limits_{n \to \infty } x_n ?[/tex]

    someone might ask
    [tex]{\text{How do we know if }}\forall n \in \mathbb{N},\;\exists x_n > 1: f_n {\kern 1pt} ' \left( {x_n } \right) = 0\;?[/tex]

    which means I must
    [tex]{\text{Prove/disprove that }}\forall n \in \mathbb{N},\;\exists x_n > 1:\frac{d}{{dx}}\sum\limits_{k = 1}^n {\sin \left( {x_n^k } \right)} = 0[/tex]

    or, equivalently (due to Mean & Intermediate Value Theorems),
    [tex]{\text{Prove/disprove that }}\forall n \in \mathbb{N},\;\exists x > 1: \frac{1}{n} \sum\limits_{k = 1}^n {\sin \left( {x^k } \right)} \leqslant \sin 1[/tex]

    (I tried induction, but showing n[itex]\to[/itex]n+1 wasn't quite as easy as I hoped...)

    *So, does anyone have any ideas how I may prove (or disprove :rolleyes:) that
    [tex]\forall n \in \mathbb{N},\;\exists x > 1: \frac{1}{n} \sum\limits_{k = 1}^n {\sin \left( x^k \right)} \leqslant \sin 1[/tex]
    ?
     
  9. Feb 20, 2007 #8
    Same as before,
    [tex]\begin{gathered}
    f_n \left( x \right) = \frac{1}
    {n}\sum\limits_{k = 1}^n {\sin \left( {x^k } \right)} \Rightarrow f'_n \left( x \right) = \frac{1}
    {n}\sum\limits_{k = 1}^n {kx^{k - 1} \cos \left( {x^k } \right)} \hfill \\
    x_n = \min \left\{ {x > 0:f'_n \left( x \right) = 0} \right\} = \min \left\{ {x > 0:\frac{d}
    {{dx}}\sum\limits_{k = 1}^n {\sin \left( {x^k } \right)} = 0} \right\} \hfill \\ \end{gathered} [/tex]

    From inspection, it appears that
    [tex]\frac{d}{{dx}}\sum\limits_{k = 1}^{n + 1} {\sin \left( {x_n^k } \right)} < 0[/tex]

    though...
    [tex]\begin{gathered}
    \cos x_n + 2x_n \cos x_n^2 + 3x_n^2 \cos x_n^3 + \cdots + nx_n^{n - 1} \cos x_n^n = 0 \Rightarrow \hfill \\
    \cos x_n + 2x_n \cos x_n^2 + 3x_n^2 \cos x_n^3 + \cdots + \left( {n + 1} \right)x_n^n \cos x_n^{n + 1} < 0 \hfill \\
    \end{gathered} [/tex]

    ~or equivalently,
    [tex]\begin{gathered}
    \cos x_n + 2x_n \cos x_n^2 + 3x_n^2 \cos x_n^3 + \cdots + nx_n^{n - 1} \cos x_n^n = 0 \Rightarrow \hfill \\
    \cos x_n + 2x_n \cos x_n^2 + 3x_n^2 \cos x_n^3 + \cdots + nx_n^{n - 1} \cos x_n^n + \left( {n + 1} \right)x_n^n \cos x_n^{n + 1} < 0 \hfill \\ \end{gathered} [/tex]

    implies (as π/2=x1>x2>...)
    [tex]\left( {n + 1} \right)x_n^n \cos x_n^{n + 1} < 0 \Rightarrow \cos x_n^{n + 1} < 0 \Rightarrow \sqrt[{n + 1}]{{\pi /2}} < x_n < \sqrt[{n + 1}]{{3\pi /2}}[/tex]

    which, by the Squeeze Theorem, implies
    [tex]\lim \limits_{n \to \infty } \sqrt[{n + 1}]{{\pi /2}} < \lim \limits_{n \to \infty } x_n < \lim \limits_{n \to \infty } \sqrt[{n + 1}]{{3\pi /2}} \Rightarrow \mathop {\lim }\limits_{n \to \infty } x_n = 1[/tex]

    *Which is strange, considering that
    [tex]\begin{gathered}
    \forall n \in \mathbb{N},\;f'_n \left( x \right) > 0\;{\text{for }}0 \leqslant x \leqslant 1,\;{\text{and}} \hfill \\
    \mathop {\lim }\limits_{n \to \infty } f'_n \left( 1 \right) = \frac{{\cos 1 + 2\cos 1 + 3\cos 1 + \cdots + n\cos 1}}
    {n} = \mathop {\lim }\limits_{n \to \infty } \frac{{n + 1}}
    {2}\cos 1,\;{\text{which diverges}} \hfill \\
    \end{gathered} [/tex]
     
    Last edited: Feb 20, 2007
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