Something about Fourier expansion












1














Suppose f is continuity on $[-pi,+pi],period ~T=2pi$
$$f(x)sim frac{a_0}{2}+sum_{n=1}^{+infty}a_n cos nx +b_nsin nx$$
$$F(x)=int_0^x[f(t)-frac{a_0}{2}]dt$$



1)Calculate the F(x) Fourier expansion



2)proof
$$int_a^bf(t)dt=int_a^b frac{a_0}{2}dt +sum_{n=1}^{infty}int_a^b a_n cos nx +b_nsin nx dt$$



My attempt



1)$a_n=int_{-pi}^{pi}f(t)cos ntdt$



$b_n=int_{-pi}^{pi}f(t)sin nt dt$
Then I don’t know what to do next
2)
I think it’s a little like uniformly convergence’character
But I don’t have idea to proof this






real-analysis calculus integration functional-analysis analysis







share|cite|improve this question



























    1














    Suppose f is continuity on $[-pi,+pi],period ~T=2pi$
    $$f(x)sim frac{a_0}{2}+sum_{n=1}^{+infty}a_n cos nx +b_nsin nx$$
    $$F(x)=int_0^x[f(t)-frac{a_0}{2}]dt$$



    1)Calculate the F(x) Fourier expansion



    2)proof
    $$int_a^bf(t)dt=int_a^b frac{a_0}{2}dt +sum_{n=1}^{infty}int_a^b a_n cos nx +b_nsin nx dt$$



    My attempt



    1)$a_n=int_{-pi}^{pi}f(t)cos ntdt$



    $b_n=int_{-pi}^{pi}f(t)sin nt dt$
    Then I don’t know what to do next
    2)
    I think it’s a little like uniformly convergence’character
    But I don’t have idea to proof this






    real-analysis calculus integration functional-analysis analysis







    share|cite|improve this question

























      1












      1








      1


      1





      Suppose f is continuity on $[-pi,+pi],period ~T=2pi$
      $$f(x)sim frac{a_0}{2}+sum_{n=1}^{+infty}a_n cos nx +b_nsin nx$$
      $$F(x)=int_0^x[f(t)-frac{a_0}{2}]dt$$



      1)Calculate the F(x) Fourier expansion



      2)proof
      $$int_a^bf(t)dt=int_a^b frac{a_0}{2}dt +sum_{n=1}^{infty}int_a^b a_n cos nx +b_nsin nx dt$$



      My attempt



      1)$a_n=int_{-pi}^{pi}f(t)cos ntdt$



      $b_n=int_{-pi}^{pi}f(t)sin nt dt$
      Then I don’t know what to do next
      2)
      I think it’s a little like uniformly convergence’character
      But I don’t have idea to proof this






      real-analysis calculus integration functional-analysis analysis







      share|cite|improve this question













      Suppose f is continuity on $[-pi,+pi],period ~T=2pi$
      $$f(x)sim frac{a_0}{2}+sum_{n=1}^{+infty}a_n cos nx +b_nsin nx$$
      $$F(x)=int_0^x[f(t)-frac{a_0}{2}]dt$$



      1)Calculate the F(x) Fourier expansion



      2)proof
      $$int_a^bf(t)dt=int_a^b frac{a_0}{2}dt +sum_{n=1}^{infty}int_a^b a_n cos nx +b_nsin nx dt$$



      My attempt



      1)$a_n=int_{-pi}^{pi}f(t)cos ntdt$



      $b_n=int_{-pi}^{pi}f(t)sin nt dt$
      Then I don’t know what to do next
      2)
      I think it’s a little like uniformly convergence’character
      But I don’t have idea to proof this






      real-analysis calculus integration functional-analysis analysis




      real-analysis calculus integration functional-analysis analysis






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 2 days ago









      jacksonjackson

      758




      758






















          1 Answer
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          3














          Let $$F(x) sim frac{c_0}{2}+sum_{n=1}^{infty}c_n cos nx +d_nsin nx.$$ Since $F$ is $2pi$-periodic $C^1$, note that the Fourier series of $F$ converges uniformly to $F$ and therefore equality holds ($alpha$-Holder continuity is one of the sufficient conditions for uniform convergence, see e.g. this earlier post). Now, we can find $c_n$ and $d_n$ as follows.



          $$
          pi c_n = int_{-pi}^pi F(t)cos nt dt = frac{sin nt}{n}F(t)|^{pi}_{-pi}-int_{-pi}^pi (f(t)-frac{a_0}{2})frac{sin nt}{n}dt = -pi frac{b_n}{n}, quad nneq 0,
          $$
          $$
          pi d_n = int_{-pi}^pi F(t)sin nt dt = frac{-cos nt}{n}F(t)|^{pi}_{-pi}+int_{-pi}^pi (f(t)-frac{a_0}{2})frac{cos nt}{n}dt = pi frac{a_n}{n}.
          $$           This gives
          $$
          F(x) = frac{c_0}{2}+sum_{n=1}^{infty}left(-frac{b_n}{n}right)cos nx +frac{a_n}{n}sin nxtag{*}.
          $$           where $frac{c_0}{2} = sum_{n=1}^infty frac{b_n}{n}$ from $F(0) = 0$. It can be also calculated explicitly:
          $$begin{eqnarray}
          pi c_0 = int_{0}^{2pi} F(t) dt& =& (t-pi)F(t)|^{2pi}_{0}-int_{0}^{2pi}(t-pi) (f(t)-frac{a_0}{2})dt \&=&int_{0}^{2pi}left(sum_{n=1}^infty frac{2}{n}sin ntright) (f(t)-frac{a_0}{2})dt \
          &=&2pisum_{n=1}^infty frac{b_n}{n}.
          end{eqnarray}$$



          For (b), note that the stament is equivalent to
          $$
          F(b) - F(a) = sum_{n=1}^infty int_a^b left(a_n cos nx +b_nsin nxright)dx.
          $$           But this is obvious from $(*)$:
          $$begin{eqnarray}
          sum_{n=1}^infty int_a^b left(a_n cos nx +b_nsin nxright)dx &=&sum_{n=1}^{infty}frac{a_n}{n}(sin nb-sin na)+left(-frac{b_n}{n}right)(cos nb-cos na)\& =& F(b) - F(a).
          end{eqnarray}$$






          share|cite|improve this answer























          • i don’t understand this why F is $in C^1$then F is uniformly convergence
            – jackson
            2 days ago












          • And the five line integration by part is wrong
            – jackson
            2 days ago












          • It should be$pi c_0=tF(t)|_{-pi}^{pi}-int_{-pi}^{+pi}(f(t)-frac{a_0}{2}) dt$ and I don’t know how to proof it equal to 0
            – jackson
            2 days ago












          • The five line the left of $c_n$ should be $pi$?because $c_n=frac{1}{pi}int_0^{2pi}F(t)cos ntdt$ so do the next line ,am I right?
            – jackson
            yesterday












          • You are absolutely right ... Sorry for the confusion.
            – Song
            yesterday











          Your Answer





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          1 Answer
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          1 Answer
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          active

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          3














          Let $$F(x) sim frac{c_0}{2}+sum_{n=1}^{infty}c_n cos nx +d_nsin nx.$$ Since $F$ is $2pi$-periodic $C^1$, note that the Fourier series of $F$ converges uniformly to $F$ and therefore equality holds ($alpha$-Holder continuity is one of the sufficient conditions for uniform convergence, see e.g. this earlier post). Now, we can find $c_n$ and $d_n$ as follows.



          $$
          pi c_n = int_{-pi}^pi F(t)cos nt dt = frac{sin nt}{n}F(t)|^{pi}_{-pi}-int_{-pi}^pi (f(t)-frac{a_0}{2})frac{sin nt}{n}dt = -pi frac{b_n}{n}, quad nneq 0,
          $$
          $$
          pi d_n = int_{-pi}^pi F(t)sin nt dt = frac{-cos nt}{n}F(t)|^{pi}_{-pi}+int_{-pi}^pi (f(t)-frac{a_0}{2})frac{cos nt}{n}dt = pi frac{a_n}{n}.
          $$           This gives
          $$
          F(x) = frac{c_0}{2}+sum_{n=1}^{infty}left(-frac{b_n}{n}right)cos nx +frac{a_n}{n}sin nxtag{*}.
          $$           where $frac{c_0}{2} = sum_{n=1}^infty frac{b_n}{n}$ from $F(0) = 0$. It can be also calculated explicitly:
          $$begin{eqnarray}
          pi c_0 = int_{0}^{2pi} F(t) dt& =& (t-pi)F(t)|^{2pi}_{0}-int_{0}^{2pi}(t-pi) (f(t)-frac{a_0}{2})dt \&=&int_{0}^{2pi}left(sum_{n=1}^infty frac{2}{n}sin ntright) (f(t)-frac{a_0}{2})dt \
          &=&2pisum_{n=1}^infty frac{b_n}{n}.
          end{eqnarray}$$



          For (b), note that the stament is equivalent to
          $$
          F(b) - F(a) = sum_{n=1}^infty int_a^b left(a_n cos nx +b_nsin nxright)dx.
          $$           But this is obvious from $(*)$:
          $$begin{eqnarray}
          sum_{n=1}^infty int_a^b left(a_n cos nx +b_nsin nxright)dx &=&sum_{n=1}^{infty}frac{a_n}{n}(sin nb-sin na)+left(-frac{b_n}{n}right)(cos nb-cos na)\& =& F(b) - F(a).
          end{eqnarray}$$






          share|cite|improve this answer























          • i don’t understand this why F is $in C^1$then F is uniformly convergence
            – jackson
            2 days ago












          • And the five line integration by part is wrong
            – jackson
            2 days ago












          • It should be$pi c_0=tF(t)|_{-pi}^{pi}-int_{-pi}^{+pi}(f(t)-frac{a_0}{2}) dt$ and I don’t know how to proof it equal to 0
            – jackson
            2 days ago












          • The five line the left of $c_n$ should be $pi$?because $c_n=frac{1}{pi}int_0^{2pi}F(t)cos ntdt$ so do the next line ,am I right?
            – jackson
            yesterday












          • You are absolutely right ... Sorry for the confusion.
            – Song
            yesterday
















          3














          Let $$F(x) sim frac{c_0}{2}+sum_{n=1}^{infty}c_n cos nx +d_nsin nx.$$ Since $F$ is $2pi$-periodic $C^1$, note that the Fourier series of $F$ converges uniformly to $F$ and therefore equality holds ($alpha$-Holder continuity is one of the sufficient conditions for uniform convergence, see e.g. this earlier post). Now, we can find $c_n$ and $d_n$ as follows.



          $$
          pi c_n = int_{-pi}^pi F(t)cos nt dt = frac{sin nt}{n}F(t)|^{pi}_{-pi}-int_{-pi}^pi (f(t)-frac{a_0}{2})frac{sin nt}{n}dt = -pi frac{b_n}{n}, quad nneq 0,
          $$
          $$
          pi d_n = int_{-pi}^pi F(t)sin nt dt = frac{-cos nt}{n}F(t)|^{pi}_{-pi}+int_{-pi}^pi (f(t)-frac{a_0}{2})frac{cos nt}{n}dt = pi frac{a_n}{n}.
          $$           This gives
          $$
          F(x) = frac{c_0}{2}+sum_{n=1}^{infty}left(-frac{b_n}{n}right)cos nx +frac{a_n}{n}sin nxtag{*}.
          $$           where $frac{c_0}{2} = sum_{n=1}^infty frac{b_n}{n}$ from $F(0) = 0$. It can be also calculated explicitly:
          $$begin{eqnarray}
          pi c_0 = int_{0}^{2pi} F(t) dt& =& (t-pi)F(t)|^{2pi}_{0}-int_{0}^{2pi}(t-pi) (f(t)-frac{a_0}{2})dt \&=&int_{0}^{2pi}left(sum_{n=1}^infty frac{2}{n}sin ntright) (f(t)-frac{a_0}{2})dt \
          &=&2pisum_{n=1}^infty frac{b_n}{n}.
          end{eqnarray}$$



          For (b), note that the stament is equivalent to
          $$
          F(b) - F(a) = sum_{n=1}^infty int_a^b left(a_n cos nx +b_nsin nxright)dx.
          $$           But this is obvious from $(*)$:
          $$begin{eqnarray}
          sum_{n=1}^infty int_a^b left(a_n cos nx +b_nsin nxright)dx &=&sum_{n=1}^{infty}frac{a_n}{n}(sin nb-sin na)+left(-frac{b_n}{n}right)(cos nb-cos na)\& =& F(b) - F(a).
          end{eqnarray}$$






          share|cite|improve this answer























          • i don’t understand this why F is $in C^1$then F is uniformly convergence
            – jackson
            2 days ago












          • And the five line integration by part is wrong
            – jackson
            2 days ago












          • It should be$pi c_0=tF(t)|_{-pi}^{pi}-int_{-pi}^{+pi}(f(t)-frac{a_0}{2}) dt$ and I don’t know how to proof it equal to 0
            – jackson
            2 days ago












          • The five line the left of $c_n$ should be $pi$?because $c_n=frac{1}{pi}int_0^{2pi}F(t)cos ntdt$ so do the next line ,am I right?
            – jackson
            yesterday












          • You are absolutely right ... Sorry for the confusion.
            – Song
            yesterday














          3












          3








          3






          Let $$F(x) sim frac{c_0}{2}+sum_{n=1}^{infty}c_n cos nx +d_nsin nx.$$ Since $F$ is $2pi$-periodic $C^1$, note that the Fourier series of $F$ converges uniformly to $F$ and therefore equality holds ($alpha$-Holder continuity is one of the sufficient conditions for uniform convergence, see e.g. this earlier post). Now, we can find $c_n$ and $d_n$ as follows.



          $$
          pi c_n = int_{-pi}^pi F(t)cos nt dt = frac{sin nt}{n}F(t)|^{pi}_{-pi}-int_{-pi}^pi (f(t)-frac{a_0}{2})frac{sin nt}{n}dt = -pi frac{b_n}{n}, quad nneq 0,
          $$
          $$
          pi d_n = int_{-pi}^pi F(t)sin nt dt = frac{-cos nt}{n}F(t)|^{pi}_{-pi}+int_{-pi}^pi (f(t)-frac{a_0}{2})frac{cos nt}{n}dt = pi frac{a_n}{n}.
          $$           This gives
          $$
          F(x) = frac{c_0}{2}+sum_{n=1}^{infty}left(-frac{b_n}{n}right)cos nx +frac{a_n}{n}sin nxtag{*}.
          $$           where $frac{c_0}{2} = sum_{n=1}^infty frac{b_n}{n}$ from $F(0) = 0$. It can be also calculated explicitly:
          $$begin{eqnarray}
          pi c_0 = int_{0}^{2pi} F(t) dt& =& (t-pi)F(t)|^{2pi}_{0}-int_{0}^{2pi}(t-pi) (f(t)-frac{a_0}{2})dt \&=&int_{0}^{2pi}left(sum_{n=1}^infty frac{2}{n}sin ntright) (f(t)-frac{a_0}{2})dt \
          &=&2pisum_{n=1}^infty frac{b_n}{n}.
          end{eqnarray}$$



          For (b), note that the stament is equivalent to
          $$
          F(b) - F(a) = sum_{n=1}^infty int_a^b left(a_n cos nx +b_nsin nxright)dx.
          $$           But this is obvious from $(*)$:
          $$begin{eqnarray}
          sum_{n=1}^infty int_a^b left(a_n cos nx +b_nsin nxright)dx &=&sum_{n=1}^{infty}frac{a_n}{n}(sin nb-sin na)+left(-frac{b_n}{n}right)(cos nb-cos na)\& =& F(b) - F(a).
          end{eqnarray}$$






          share|cite|improve this answer














          Let $$F(x) sim frac{c_0}{2}+sum_{n=1}^{infty}c_n cos nx +d_nsin nx.$$ Since $F$ is $2pi$-periodic $C^1$, note that the Fourier series of $F$ converges uniformly to $F$ and therefore equality holds ($alpha$-Holder continuity is one of the sufficient conditions for uniform convergence, see e.g. this earlier post). Now, we can find $c_n$ and $d_n$ as follows.



          $$
          pi c_n = int_{-pi}^pi F(t)cos nt dt = frac{sin nt}{n}F(t)|^{pi}_{-pi}-int_{-pi}^pi (f(t)-frac{a_0}{2})frac{sin nt}{n}dt = -pi frac{b_n}{n}, quad nneq 0,
          $$
          $$
          pi d_n = int_{-pi}^pi F(t)sin nt dt = frac{-cos nt}{n}F(t)|^{pi}_{-pi}+int_{-pi}^pi (f(t)-frac{a_0}{2})frac{cos nt}{n}dt = pi frac{a_n}{n}.
          $$           This gives
          $$
          F(x) = frac{c_0}{2}+sum_{n=1}^{infty}left(-frac{b_n}{n}right)cos nx +frac{a_n}{n}sin nxtag{*}.
          $$           where $frac{c_0}{2} = sum_{n=1}^infty frac{b_n}{n}$ from $F(0) = 0$. It can be also calculated explicitly:
          $$begin{eqnarray}
          pi c_0 = int_{0}^{2pi} F(t) dt& =& (t-pi)F(t)|^{2pi}_{0}-int_{0}^{2pi}(t-pi) (f(t)-frac{a_0}{2})dt \&=&int_{0}^{2pi}left(sum_{n=1}^infty frac{2}{n}sin ntright) (f(t)-frac{a_0}{2})dt \
          &=&2pisum_{n=1}^infty frac{b_n}{n}.
          end{eqnarray}$$



          For (b), note that the stament is equivalent to
          $$
          F(b) - F(a) = sum_{n=1}^infty int_a^b left(a_n cos nx +b_nsin nxright)dx.
          $$           But this is obvious from $(*)$:
          $$begin{eqnarray}
          sum_{n=1}^infty int_a^b left(a_n cos nx +b_nsin nxright)dx &=&sum_{n=1}^{infty}frac{a_n}{n}(sin nb-sin na)+left(-frac{b_n}{n}right)(cos nb-cos na)\& =& F(b) - F(a).
          end{eqnarray}$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited yesterday

























          answered 2 days ago









          SongSong

          6,050318




          6,050318












          • i don’t understand this why F is $in C^1$then F is uniformly convergence
            – jackson
            2 days ago












          • And the five line integration by part is wrong
            – jackson
            2 days ago












          • It should be$pi c_0=tF(t)|_{-pi}^{pi}-int_{-pi}^{+pi}(f(t)-frac{a_0}{2}) dt$ and I don’t know how to proof it equal to 0
            – jackson
            2 days ago












          • The five line the left of $c_n$ should be $pi$?because $c_n=frac{1}{pi}int_0^{2pi}F(t)cos ntdt$ so do the next line ,am I right?
            – jackson
            yesterday












          • You are absolutely right ... Sorry for the confusion.
            – Song
            yesterday


















          • i don’t understand this why F is $in C^1$then F is uniformly convergence
            – jackson
            2 days ago












          • And the five line integration by part is wrong
            – jackson
            2 days ago












          • It should be$pi c_0=tF(t)|_{-pi}^{pi}-int_{-pi}^{+pi}(f(t)-frac{a_0}{2}) dt$ and I don’t know how to proof it equal to 0
            – jackson
            2 days ago












          • The five line the left of $c_n$ should be $pi$?because $c_n=frac{1}{pi}int_0^{2pi}F(t)cos ntdt$ so do the next line ,am I right?
            – jackson
            yesterday












          • You are absolutely right ... Sorry for the confusion.
            – Song
            yesterday
















          i don’t understand this why F is $in C^1$then F is uniformly convergence
          – jackson
          2 days ago






          i don’t understand this why F is $in C^1$then F is uniformly convergence
          – jackson
          2 days ago














          And the five line integration by part is wrong
          – jackson
          2 days ago






          And the five line integration by part is wrong
          – jackson
          2 days ago














          It should be$pi c_0=tF(t)|_{-pi}^{pi}-int_{-pi}^{+pi}(f(t)-frac{a_0}{2}) dt$ and I don’t know how to proof it equal to 0
          – jackson
          2 days ago






          It should be$pi c_0=tF(t)|_{-pi}^{pi}-int_{-pi}^{+pi}(f(t)-frac{a_0}{2}) dt$ and I don’t know how to proof it equal to 0
          – jackson
          2 days ago














          The five line the left of $c_n$ should be $pi$?because $c_n=frac{1}{pi}int_0^{2pi}F(t)cos ntdt$ so do the next line ,am I right?
          – jackson
          yesterday






          The five line the left of $c_n$ should be $pi$?because $c_n=frac{1}{pi}int_0^{2pi}F(t)cos ntdt$ so do the next line ,am I right?
          – jackson
          yesterday














          You are absolutely right ... Sorry for the confusion.
          – Song
          yesterday




          You are absolutely right ... Sorry for the confusion.
          – Song
          yesterday


















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          The Binding of Isaac: Rebirth/Afterbirth

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          Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! The Binding of Isaac: Rebirth/Afterbirth From The Cutting Room Floor < The Binding of Isaac: Rebirth Jump to: navigation, search This is a sub-page of The Binding of Isaac: Rebirth . Afterbirth is the first expansion pack for Rebirth and features a variety of improvements. Contents 1 Unused Graphics 1.1 Old Man Keeper 1.2 Skinless Hush 1.3 Miscellaneous Unused Hush Sprites 1.4 Generosity achievement 1.5 Afterbirth+ Room Editor 1.6 Bearded Ultra Greed 1.7 Early Ultra Greed 1.8 Unused Brownie Expression 1.9 Frail's Spinning Head 1.10 Mom's Dead Hand Alt 2 Unused Audio 2.1 Unused Dead Womb Layer 2.2 Unused Hush Battle Outro 2.3 The Stain Grunt Unused Graphics Old Man Keeper Image file: gfxuimain me...
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          What does “Dominus providebit” mean?

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          5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1 ...
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