1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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

    User Avatar
    Staff Emeritus
    Science Advisor

    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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?