Does $sin ^n x$ converge uniformly on $[0,fracpi2)$?

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Does $f_n(x)=sin ^n x$ converge uniformly on $[0,fracpi2)$ ?




I know $f_n(x) rightarrow 0$ pointwise, since $vert sin ^n x vert< 1$. How about the uniform convergence?



Any hint?




real-analysis uniform-convergence




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    up vote
    1
    down vote

    favorite













    Does $f_n(x)=sin ^n x$ converge uniformly on $[0,fracpi2)$ ?




    I know $f_n(x) rightarrow 0$ pointwise, since $vert sin ^n x vert< 1$. How about the uniform convergence?



    Any hint?




    real-analysis uniform-convergence




    share|cite|improve this question























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite












      Does $f_n(x)=sin ^n x$ converge uniformly on $[0,fracpi2)$ ?




      I know $f_n(x) rightarrow 0$ pointwise, since $vert sin ^n x vert< 1$. How about the uniform convergence?



      Any hint?




      real-analysis uniform-convergence




      share|cite|improve this question














      Does $f_n(x)=sin ^n x$ converge uniformly on $[0,fracpi2)$ ?




      I know $f_n(x) rightarrow 0$ pointwise, since $vert sin ^n x vert< 1$. How about the uniform convergence?



      Any hint?





      real-analysis uniform-convergence





      share|cite|improve this question












      share|cite|improve this question




      share|cite|improve this question








      edited Jul 25 at 8:11









      Bernard

      110k635103




      110k635103









      asked Jul 25 at 7:41









      Learning Mathematics

      479213




      479213




















          2 Answers
          2






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          up vote
          5
          down vote



          accepted










          Let $x_n=sin ^-1 (1-frac 1 n)$. Then $sin^n (x_n))=(1-frac 1 n )^n to 1/e$. So the convergence is not uniform.






          share|cite|improve this answer





















          • Oh..! $x_n=sin ^-1 Bigl(1-frac1n Bigr) $ is the sequence in $[0,fracpi2)$ with $f_n(x_n)-f(x_n)$ does not tend to zero, so the convergence is not uniform. Thanks!
            – Learning Mathematics
            Jul 25 at 7:48


















          up vote
          1
          down vote













          Uniform convergence of $(f_n)$ to $f(x) = 0$ would require that
          $$
          M_n = sup
          $$
          converges to zero. However, for each $n$
          $$
          M_n ge lim_x to pi /2 f_n(x) = 1 , .
          $$






          share|cite|improve this answer





















          • Martin.You were faster+1).Shall delete mine.
            – Peter Szilas
            Jul 25 at 10:31










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          2 Answers
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          active

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          2 Answers
          2






          active

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          active

          oldest

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          up vote
          5
          down vote



          accepted










          Let $x_n=sin ^-1 (1-frac 1 n)$. Then $sin^n (x_n))=(1-frac 1 n )^n to 1/e$. So the convergence is not uniform.






          share|cite|improve this answer





















          • Oh..! $x_n=sin ^-1 Bigl(1-frac1n Bigr) $ is the sequence in $[0,fracpi2)$ with $f_n(x_n)-f(x_n)$ does not tend to zero, so the convergence is not uniform. Thanks!
            – Learning Mathematics
            Jul 25 at 7:48















          up vote
          5
          down vote



          accepted










          Let $x_n=sin ^-1 (1-frac 1 n)$. Then $sin^n (x_n))=(1-frac 1 n )^n to 1/e$. So the convergence is not uniform.






          share|cite|improve this answer





















          • Oh..! $x_n=sin ^-1 Bigl(1-frac1n Bigr) $ is the sequence in $[0,fracpi2)$ with $f_n(x_n)-f(x_n)$ does not tend to zero, so the convergence is not uniform. Thanks!
            – Learning Mathematics
            Jul 25 at 7:48













          up vote
          5
          down vote



          accepted







          up vote
          5
          down vote



          accepted






          Let $x_n=sin ^-1 (1-frac 1 n)$. Then $sin^n (x_n))=(1-frac 1 n )^n to 1/e$. So the convergence is not uniform.






          share|cite|improve this answer













          Let $x_n=sin ^-1 (1-frac 1 n)$. Then $sin^n (x_n))=(1-frac 1 n )^n to 1/e$. So the convergence is not uniform.







          share|cite|improve this answer













          share|cite|improve this answer



          share|cite|improve this answer











          answered Jul 25 at 7:44









          Kavi Rama Murthy

          20.1k2829




          20.1k2829











          • Oh..! $x_n=sin ^-1 Bigl(1-frac1n Bigr) $ is the sequence in $[0,fracpi2)$ with $f_n(x_n)-f(x_n)$ does not tend to zero, so the convergence is not uniform. Thanks!
            – Learning Mathematics
            Jul 25 at 7:48

















          • Oh..! $x_n=sin ^-1 Bigl(1-frac1n Bigr) $ is the sequence in $[0,fracpi2)$ with $f_n(x_n)-f(x_n)$ does not tend to zero, so the convergence is not uniform. Thanks!
            – Learning Mathematics
            Jul 25 at 7:48
















          Oh..! $x_n=sin ^-1 Bigl(1-frac1n Bigr) $ is the sequence in $[0,fracpi2)$ with $f_n(x_n)-f(x_n)$ does not tend to zero, so the convergence is not uniform. Thanks!
          – Learning Mathematics
          Jul 25 at 7:48





          Oh..! $x_n=sin ^-1 Bigl(1-frac1n Bigr) $ is the sequence in $[0,fracpi2)$ with $f_n(x_n)-f(x_n)$ does not tend to zero, so the convergence is not uniform. Thanks!
          – Learning Mathematics
          Jul 25 at 7:48











          up vote
          1
          down vote













          Uniform convergence of $(f_n)$ to $f(x) = 0$ would require that
          $$
          M_n = sup
          $$
          converges to zero. However, for each $n$
          $$
          M_n ge lim_x to pi /2 f_n(x) = 1 , .
          $$






          share|cite|improve this answer





















          • Martin.You were faster+1).Shall delete mine.
            – Peter Szilas
            Jul 25 at 10:31














          up vote
          1
          down vote













          Uniform convergence of $(f_n)$ to $f(x) = 0$ would require that
          $$
          M_n = sup
          $$
          converges to zero. However, for each $n$
          $$
          M_n ge lim_x to pi /2 f_n(x) = 1 , .
          $$






          share|cite|improve this answer





















          • Martin.You were faster+1).Shall delete mine.
            – Peter Szilas
            Jul 25 at 10:31












          up vote
          1
          down vote










          up vote
          1
          down vote









          Uniform convergence of $(f_n)$ to $f(x) = 0$ would require that
          $$
          M_n = sup
          $$
          converges to zero. However, for each $n$
          $$
          M_n ge lim_x to pi /2 f_n(x) = 1 , .
          $$






          share|cite|improve this answer













          Uniform convergence of $(f_n)$ to $f(x) = 0$ would require that
          $$
          M_n = sup
          $$
          converges to zero. However, for each $n$
          $$
          M_n ge lim_x to pi /2 f_n(x) = 1 , .
          $$







          share|cite|improve this answer













          share|cite|improve this answer



          share|cite|improve this answer











          answered Jul 25 at 9:36









          Martin R

          23.8k32743




          23.8k32743











          • Martin.You were faster+1).Shall delete mine.
            – Peter Szilas
            Jul 25 at 10:31
















          • Martin.You were faster+1).Shall delete mine.
            – Peter Szilas
            Jul 25 at 10:31















          Martin.You were faster+1).Shall delete mine.
          – Peter Szilas
          Jul 25 at 10:31




          Martin.You were faster+1).Shall delete mine.
          – Peter Szilas
          Jul 25 at 10:31












           

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