Series expansion upto linear order












1












$begingroup$


I want to do a series expansion of the function given below around $a=0$ and keep the terms only upto $O(a)$. The function is



$$
f(a)=frac{a pi^2sin^2theta}{a^2-pi^2cos^2theta}
$$
Using $f(0)+f'(0)a$, I would get the expansion as $f(a)=-tan^2theta ~a$. I suspect whether this expansion would be correct near $theta$ equal to $pi/2$. Any help, how should I proceed in that case. Should I take $lim~thetarightarrow pi/2$ first and then do the expansion or should I do something else?






calculus mathematical-physics







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    I want to do a series expansion of the function given below around $a=0$ and keep the terms only upto $O(a)$. The function is



    $$
    f(a)=frac{a pi^2sin^2theta}{a^2-pi^2cos^2theta}
    $$
    Using $f(0)+f'(0)a$, I would get the expansion as $f(a)=-tan^2theta ~a$. I suspect whether this expansion would be correct near $theta$ equal to $pi/2$. Any help, how should I proceed in that case. Should I take $lim~thetarightarrow pi/2$ first and then do the expansion or should I do something else?






    calculus mathematical-physics







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      I want to do a series expansion of the function given below around $a=0$ and keep the terms only upto $O(a)$. The function is



      $$
      f(a)=frac{a pi^2sin^2theta}{a^2-pi^2cos^2theta}
      $$
      Using $f(0)+f'(0)a$, I would get the expansion as $f(a)=-tan^2theta ~a$. I suspect whether this expansion would be correct near $theta$ equal to $pi/2$. Any help, how should I proceed in that case. Should I take $lim~thetarightarrow pi/2$ first and then do the expansion or should I do something else?






      calculus mathematical-physics







      share|cite|improve this question











      $endgroup$




      I want to do a series expansion of the function given below around $a=0$ and keep the terms only upto $O(a)$. The function is



      $$
      f(a)=frac{a pi^2sin^2theta}{a^2-pi^2cos^2theta}
      $$
      Using $f(0)+f'(0)a$, I would get the expansion as $f(a)=-tan^2theta ~a$. I suspect whether this expansion would be correct near $theta$ equal to $pi/2$. Any help, how should I proceed in that case. Should I take $lim~thetarightarrow pi/2$ first and then do the expansion or should I do something else?






      calculus mathematical-physics




      calculus mathematical-physics






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 17 '18 at 20:15







      Parveen

















      asked Dec 17 '18 at 20:01









      ParveenParveen

      275




      275






















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$

          There being two variables $a$ and $theta$, you have to be careful with what $O(a)$ means, and you should probably write $f(a,theta)$ rather than just $f(a)$. It is true that $f(a,theta) sim -frac{a}{pi} tan^2(theta)$ if $theta$ is fixed and $cos(theta) ne 0$. But if $cos(theta) = 0$, $f(a,theta) = pi/a$. Various
          other possibilities exist if $cos(theta) to 0$ as $a to 0$. For example, if $theta = pi/2 + c a$ for constant $c ne 1/pi$,
          $$ f(a,pi/2 + c a) sim frac{pi}{1 - pi^2 c^2} a^{-1}$$
          while if $theta = pi/2 + a/pi$
          $$ f(a, pi/2 + a/pi) sim frac{3 pi^3}{a^3} $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            If I say that $theta=frac{pi}{2}pm epsilon$ where $epsilon$ is a very small number so $costheta$ is a small number but not exactly zero, then would $f(a)=-a~tan^2theta$ be true or not?
            $endgroup$
            – Parveen
            Dec 17 '18 at 20:35










          • $begingroup$
            If $epsilon$ is a nonzero constant, then indeed $f(a) = -frac{a}{pi} tan^2(theta) + O(a^3)$.
            $endgroup$
            – Robert Israel
            Dec 17 '18 at 20:51



















          0












          $begingroup$

          We know that near $X=0$,



          $$frac{1}{1-X}=1+X+X^2+...$$
          $$=1+O(X)$$



          thus



          $$f(a)=-frac{api^2sin^2(theta)}{pi^2cos^2(theta)}frac{1}{1-frac{a^2}{pi^2cos^2(theta)}}$$
          $$=-atan^2(theta)Bigl(1+O(a^2)Bigr)$$
          $$=-atan^2(theta)+O(a)$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            How are treating the denominator in $1-X$ form in this case?
            $endgroup$
            – Parveen
            Dec 17 '18 at 20:12










          • $begingroup$
            @Parveen Please look again.
            $endgroup$
            – hamam_Abdallah
            Dec 17 '18 at 20:16










          • $begingroup$
            This doesnt address the case $cos(theta) to 0$.
            $endgroup$
            – Robert Israel
            Dec 17 '18 at 20:17










          • $begingroup$
            You got the same answer as I mentioned in my post. As $tan^2theta$ diverges as $theta$ goes to $pi/2$, I doubt whether this is correct thing to do!
            $endgroup$
            – Parveen
            Dec 17 '18 at 20:20










          • $begingroup$
            If $thetato pi/2$ then $f(a)=frac{pi^2}{a}$ goes to infty when $ato 0$.
            $endgroup$
            – hamam_Abdallah
            Dec 17 '18 at 20:23











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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

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          active

          oldest

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          active

          oldest

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          1












          $begingroup$

          There being two variables $a$ and $theta$, you have to be careful with what $O(a)$ means, and you should probably write $f(a,theta)$ rather than just $f(a)$. It is true that $f(a,theta) sim -frac{a}{pi} tan^2(theta)$ if $theta$ is fixed and $cos(theta) ne 0$. But if $cos(theta) = 0$, $f(a,theta) = pi/a$. Various
          other possibilities exist if $cos(theta) to 0$ as $a to 0$. For example, if $theta = pi/2 + c a$ for constant $c ne 1/pi$,
          $$ f(a,pi/2 + c a) sim frac{pi}{1 - pi^2 c^2} a^{-1}$$
          while if $theta = pi/2 + a/pi$
          $$ f(a, pi/2 + a/pi) sim frac{3 pi^3}{a^3} $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            If I say that $theta=frac{pi}{2}pm epsilon$ where $epsilon$ is a very small number so $costheta$ is a small number but not exactly zero, then would $f(a)=-a~tan^2theta$ be true or not?
            $endgroup$
            – Parveen
            Dec 17 '18 at 20:35










          • $begingroup$
            If $epsilon$ is a nonzero constant, then indeed $f(a) = -frac{a}{pi} tan^2(theta) + O(a^3)$.
            $endgroup$
            – Robert Israel
            Dec 17 '18 at 20:51
















          1












          $begingroup$

          There being two variables $a$ and $theta$, you have to be careful with what $O(a)$ means, and you should probably write $f(a,theta)$ rather than just $f(a)$. It is true that $f(a,theta) sim -frac{a}{pi} tan^2(theta)$ if $theta$ is fixed and $cos(theta) ne 0$. But if $cos(theta) = 0$, $f(a,theta) = pi/a$. Various
          other possibilities exist if $cos(theta) to 0$ as $a to 0$. For example, if $theta = pi/2 + c a$ for constant $c ne 1/pi$,
          $$ f(a,pi/2 + c a) sim frac{pi}{1 - pi^2 c^2} a^{-1}$$
          while if $theta = pi/2 + a/pi$
          $$ f(a, pi/2 + a/pi) sim frac{3 pi^3}{a^3} $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            If I say that $theta=frac{pi}{2}pm epsilon$ where $epsilon$ is a very small number so $costheta$ is a small number but not exactly zero, then would $f(a)=-a~tan^2theta$ be true or not?
            $endgroup$
            – Parveen
            Dec 17 '18 at 20:35










          • $begingroup$
            If $epsilon$ is a nonzero constant, then indeed $f(a) = -frac{a}{pi} tan^2(theta) + O(a^3)$.
            $endgroup$
            – Robert Israel
            Dec 17 '18 at 20:51














          1












          1








          1





          $begingroup$

          There being two variables $a$ and $theta$, you have to be careful with what $O(a)$ means, and you should probably write $f(a,theta)$ rather than just $f(a)$. It is true that $f(a,theta) sim -frac{a}{pi} tan^2(theta)$ if $theta$ is fixed and $cos(theta) ne 0$. But if $cos(theta) = 0$, $f(a,theta) = pi/a$. Various
          other possibilities exist if $cos(theta) to 0$ as $a to 0$. For example, if $theta = pi/2 + c a$ for constant $c ne 1/pi$,
          $$ f(a,pi/2 + c a) sim frac{pi}{1 - pi^2 c^2} a^{-1}$$
          while if $theta = pi/2 + a/pi$
          $$ f(a, pi/2 + a/pi) sim frac{3 pi^3}{a^3} $$






          share|cite|improve this answer









          $endgroup$



          There being two variables $a$ and $theta$, you have to be careful with what $O(a)$ means, and you should probably write $f(a,theta)$ rather than just $f(a)$. It is true that $f(a,theta) sim -frac{a}{pi} tan^2(theta)$ if $theta$ is fixed and $cos(theta) ne 0$. But if $cos(theta) = 0$, $f(a,theta) = pi/a$. Various
          other possibilities exist if $cos(theta) to 0$ as $a to 0$. For example, if $theta = pi/2 + c a$ for constant $c ne 1/pi$,
          $$ f(a,pi/2 + c a) sim frac{pi}{1 - pi^2 c^2} a^{-1}$$
          while if $theta = pi/2 + a/pi$
          $$ f(a, pi/2 + a/pi) sim frac{3 pi^3}{a^3} $$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 17 '18 at 20:25









          Robert IsraelRobert Israel

          325k23214468




          325k23214468












          • $begingroup$
            If I say that $theta=frac{pi}{2}pm epsilon$ where $epsilon$ is a very small number so $costheta$ is a small number but not exactly zero, then would $f(a)=-a~tan^2theta$ be true or not?
            $endgroup$
            – Parveen
            Dec 17 '18 at 20:35










          • $begingroup$
            If $epsilon$ is a nonzero constant, then indeed $f(a) = -frac{a}{pi} tan^2(theta) + O(a^3)$.
            $endgroup$
            – Robert Israel
            Dec 17 '18 at 20:51


















          • $begingroup$
            If I say that $theta=frac{pi}{2}pm epsilon$ where $epsilon$ is a very small number so $costheta$ is a small number but not exactly zero, then would $f(a)=-a~tan^2theta$ be true or not?
            $endgroup$
            – Parveen
            Dec 17 '18 at 20:35










          • $begingroup$
            If $epsilon$ is a nonzero constant, then indeed $f(a) = -frac{a}{pi} tan^2(theta) + O(a^3)$.
            $endgroup$
            – Robert Israel
            Dec 17 '18 at 20:51
















          $begingroup$
          If I say that $theta=frac{pi}{2}pm epsilon$ where $epsilon$ is a very small number so $costheta$ is a small number but not exactly zero, then would $f(a)=-a~tan^2theta$ be true or not?
          $endgroup$
          – Parveen
          Dec 17 '18 at 20:35




          $begingroup$
          If I say that $theta=frac{pi}{2}pm epsilon$ where $epsilon$ is a very small number so $costheta$ is a small number but not exactly zero, then would $f(a)=-a~tan^2theta$ be true or not?
          $endgroup$
          – Parveen
          Dec 17 '18 at 20:35












          $begingroup$
          If $epsilon$ is a nonzero constant, then indeed $f(a) = -frac{a}{pi} tan^2(theta) + O(a^3)$.
          $endgroup$
          – Robert Israel
          Dec 17 '18 at 20:51




          $begingroup$
          If $epsilon$ is a nonzero constant, then indeed $f(a) = -frac{a}{pi} tan^2(theta) + O(a^3)$.
          $endgroup$
          – Robert Israel
          Dec 17 '18 at 20:51











          0












          $begingroup$

          We know that near $X=0$,



          $$frac{1}{1-X}=1+X+X^2+...$$
          $$=1+O(X)$$



          thus



          $$f(a)=-frac{api^2sin^2(theta)}{pi^2cos^2(theta)}frac{1}{1-frac{a^2}{pi^2cos^2(theta)}}$$
          $$=-atan^2(theta)Bigl(1+O(a^2)Bigr)$$
          $$=-atan^2(theta)+O(a)$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            How are treating the denominator in $1-X$ form in this case?
            $endgroup$
            – Parveen
            Dec 17 '18 at 20:12










          • $begingroup$
            @Parveen Please look again.
            $endgroup$
            – hamam_Abdallah
            Dec 17 '18 at 20:16










          • $begingroup$
            This doesnt address the case $cos(theta) to 0$.
            $endgroup$
            – Robert Israel
            Dec 17 '18 at 20:17










          • $begingroup$
            You got the same answer as I mentioned in my post. As $tan^2theta$ diverges as $theta$ goes to $pi/2$, I doubt whether this is correct thing to do!
            $endgroup$
            – Parveen
            Dec 17 '18 at 20:20










          • $begingroup$
            If $thetato pi/2$ then $f(a)=frac{pi^2}{a}$ goes to infty when $ato 0$.
            $endgroup$
            – hamam_Abdallah
            Dec 17 '18 at 20:23
















          0












          $begingroup$

          We know that near $X=0$,



          $$frac{1}{1-X}=1+X+X^2+...$$
          $$=1+O(X)$$



          thus



          $$f(a)=-frac{api^2sin^2(theta)}{pi^2cos^2(theta)}frac{1}{1-frac{a^2}{pi^2cos^2(theta)}}$$
          $$=-atan^2(theta)Bigl(1+O(a^2)Bigr)$$
          $$=-atan^2(theta)+O(a)$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            How are treating the denominator in $1-X$ form in this case?
            $endgroup$
            – Parveen
            Dec 17 '18 at 20:12










          • $begingroup$
            @Parveen Please look again.
            $endgroup$
            – hamam_Abdallah
            Dec 17 '18 at 20:16










          • $begingroup$
            This doesnt address the case $cos(theta) to 0$.
            $endgroup$
            – Robert Israel
            Dec 17 '18 at 20:17










          • $begingroup$
            You got the same answer as I mentioned in my post. As $tan^2theta$ diverges as $theta$ goes to $pi/2$, I doubt whether this is correct thing to do!
            $endgroup$
            – Parveen
            Dec 17 '18 at 20:20










          • $begingroup$
            If $thetato pi/2$ then $f(a)=frac{pi^2}{a}$ goes to infty when $ato 0$.
            $endgroup$
            – hamam_Abdallah
            Dec 17 '18 at 20:23














          0












          0








          0





          $begingroup$

          We know that near $X=0$,



          $$frac{1}{1-X}=1+X+X^2+...$$
          $$=1+O(X)$$



          thus



          $$f(a)=-frac{api^2sin^2(theta)}{pi^2cos^2(theta)}frac{1}{1-frac{a^2}{pi^2cos^2(theta)}}$$
          $$=-atan^2(theta)Bigl(1+O(a^2)Bigr)$$
          $$=-atan^2(theta)+O(a)$$






          share|cite|improve this answer











          $endgroup$



          We know that near $X=0$,



          $$frac{1}{1-X}=1+X+X^2+...$$
          $$=1+O(X)$$



          thus



          $$f(a)=-frac{api^2sin^2(theta)}{pi^2cos^2(theta)}frac{1}{1-frac{a^2}{pi^2cos^2(theta)}}$$
          $$=-atan^2(theta)Bigl(1+O(a^2)Bigr)$$
          $$=-atan^2(theta)+O(a)$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 17 '18 at 20:19

























          answered Dec 17 '18 at 20:09









          hamam_Abdallahhamam_Abdallah

          38.1k21634




          38.1k21634












          • $begingroup$
            How are treating the denominator in $1-X$ form in this case?
            $endgroup$
            – Parveen
            Dec 17 '18 at 20:12










          • $begingroup$
            @Parveen Please look again.
            $endgroup$
            – hamam_Abdallah
            Dec 17 '18 at 20:16










          • $begingroup$
            This doesnt address the case $cos(theta) to 0$.
            $endgroup$
            – Robert Israel
            Dec 17 '18 at 20:17










          • $begingroup$
            You got the same answer as I mentioned in my post. As $tan^2theta$ diverges as $theta$ goes to $pi/2$, I doubt whether this is correct thing to do!
            $endgroup$
            – Parveen
            Dec 17 '18 at 20:20










          • $begingroup$
            If $thetato pi/2$ then $f(a)=frac{pi^2}{a}$ goes to infty when $ato 0$.
            $endgroup$
            – hamam_Abdallah
            Dec 17 '18 at 20:23


















          • $begingroup$
            How are treating the denominator in $1-X$ form in this case?
            $endgroup$
            – Parveen
            Dec 17 '18 at 20:12










          • $begingroup$
            @Parveen Please look again.
            $endgroup$
            – hamam_Abdallah
            Dec 17 '18 at 20:16










          • $begingroup$
            This doesnt address the case $cos(theta) to 0$.
            $endgroup$
            – Robert Israel
            Dec 17 '18 at 20:17










          • $begingroup$
            You got the same answer as I mentioned in my post. As $tan^2theta$ diverges as $theta$ goes to $pi/2$, I doubt whether this is correct thing to do!
            $endgroup$
            – Parveen
            Dec 17 '18 at 20:20










          • $begingroup$
            If $thetato pi/2$ then $f(a)=frac{pi^2}{a}$ goes to infty when $ato 0$.
            $endgroup$
            – hamam_Abdallah
            Dec 17 '18 at 20:23
















          $begingroup$
          How are treating the denominator in $1-X$ form in this case?
          $endgroup$
          – Parveen
          Dec 17 '18 at 20:12




          $begingroup$
          How are treating the denominator in $1-X$ form in this case?
          $endgroup$
          – Parveen
          Dec 17 '18 at 20:12












          $begingroup$
          @Parveen Please look again.
          $endgroup$
          – hamam_Abdallah
          Dec 17 '18 at 20:16




          $begingroup$
          @Parveen Please look again.
          $endgroup$
          – hamam_Abdallah
          Dec 17 '18 at 20:16












          $begingroup$
          This doesnt address the case $cos(theta) to 0$.
          $endgroup$
          – Robert Israel
          Dec 17 '18 at 20:17




          $begingroup$
          This doesnt address the case $cos(theta) to 0$.
          $endgroup$
          – Robert Israel
          Dec 17 '18 at 20:17












          $begingroup$
          You got the same answer as I mentioned in my post. As $tan^2theta$ diverges as $theta$ goes to $pi/2$, I doubt whether this is correct thing to do!
          $endgroup$
          – Parveen
          Dec 17 '18 at 20:20




          $begingroup$
          You got the same answer as I mentioned in my post. As $tan^2theta$ diverges as $theta$ goes to $pi/2$, I doubt whether this is correct thing to do!
          $endgroup$
          – Parveen
          Dec 17 '18 at 20:20












          $begingroup$
          If $thetato pi/2$ then $f(a)=frac{pi^2}{a}$ goes to infty when $ato 0$.
          $endgroup$
          – hamam_Abdallah
          Dec 17 '18 at 20:23




          $begingroup$
          If $thetato pi/2$ then $f(a)=frac{pi^2}{a}$ goes to infty when $ato 0$.
          $endgroup$
          – hamam_Abdallah
          Dec 17 '18 at 20:23


















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          Are they similar matrix

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          0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
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