Taylor Polynomal and Conic Section












1












$begingroup$


A real function of two variable is given by:
$f(x,y) =exp(x+y) cdot cos(x-y)$.



The approximating polynomal of $boldsymbol{2}$nd degree for $boldsymbol{f(x,y)}$ with converging point $boldsymbol{(x_0,y_0) =(0,0)}$ called $boldsymbol{P_2(x,y)}$.



a) Find $P_2(x,y)$

b) Equation $P_2(x,y)=0$ describes a conic section in the $(x,y)$ plane. Give a characteristic of the conic section.



Any help would be great on how to approach this question.






taylor-expansion conic-sections







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    A real function of two variable is given by:
    $f(x,y) =exp(x+y) cdot cos(x-y)$.



    The approximating polynomal of $boldsymbol{2}$nd degree for $boldsymbol{f(x,y)}$ with converging point $boldsymbol{(x_0,y_0) =(0,0)}$ called $boldsymbol{P_2(x,y)}$.



    a) Find $P_2(x,y)$

    b) Equation $P_2(x,y)=0$ describes a conic section in the $(x,y)$ plane. Give a characteristic of the conic section.



    Any help would be great on how to approach this question.






    taylor-expansion conic-sections







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      A real function of two variable is given by:
      $f(x,y) =exp(x+y) cdot cos(x-y)$.



      The approximating polynomal of $boldsymbol{2}$nd degree for $boldsymbol{f(x,y)}$ with converging point $boldsymbol{(x_0,y_0) =(0,0)}$ called $boldsymbol{P_2(x,y)}$.



      a) Find $P_2(x,y)$

      b) Equation $P_2(x,y)=0$ describes a conic section in the $(x,y)$ plane. Give a characteristic of the conic section.



      Any help would be great on how to approach this question.






      taylor-expansion conic-sections







      share|cite|improve this question











      $endgroup$




      A real function of two variable is given by:
      $f(x,y) =exp(x+y) cdot cos(x-y)$.



      The approximating polynomal of $boldsymbol{2}$nd degree for $boldsymbol{f(x,y)}$ with converging point $boldsymbol{(x_0,y_0) =(0,0)}$ called $boldsymbol{P_2(x,y)}$.



      a) Find $P_2(x,y)$

      b) Equation $P_2(x,y)=0$ describes a conic section in the $(x,y)$ plane. Give a characteristic of the conic section.



      Any help would be great on how to approach this question.






      taylor-expansion conic-sections




      taylor-expansion conic-sections






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 18 '18 at 10:02









      Rócherz

      2,9762821




      2,9762821










      asked Mar 3 '16 at 14:37









      MathCurious314MathCurious314

      155110




      155110






















          1 Answer
          1






          active

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          2












          $begingroup$

          Well, you know that a Taylor series about $(0,0)$ takes the form



          $$f(x,y) = sum_{m=0}^{infty} sum_{n=0}^{infty} a_{mn} x^m y^n $$



          $$a_{mn} = frac1{m! n!}left [frac{partial^{m+n} f}{partial x^m partial y^n} right ]_{x=0,y=0} $$



          You want the following $6$ terms: $a_{00}$, $a_{10}$,$a_{01}$,$a_{20}$,$a_{11}$,$a_{02}$.



          To figure out which conic section you have, look at the sign of $a_{11}^2-4 a_{20} a_{02}$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Your computations will give you $2xy+x+y+1=0 Leftrightarrow y=-dfrac{x+1}{2x+1}$ which is an hyperbola. The consequence is that you have a saddle point at the origin.
            $endgroup$
            – Jean Marie
            Mar 3 '16 at 15:58










          • $begingroup$
            @JeanMarie: Cool, but I left my answer as is so that the OP could work out the result for himself.
            $endgroup$
            – Ron Gordon
            Mar 3 '16 at 15:59










          • $begingroup$
            I understand your point. I think that giving the answer leaves some work to do... Regarding the nature of the conic section, I have found with my students, that they had not the idea to find rather easily the nature of a conic section using the "trick" of expressing one variable as a function (or as 2 functions) of the other variable. The knowledge about conic sections is now very scarce and criteria like the sign of $ a_{11}^2-4 a_{20} a_{02}$ is no longer part of the curricula, in most countries, alas...
            $endgroup$
            – Jean Marie
            Mar 3 '16 at 16:31













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          1 Answer
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          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          Well, you know that a Taylor series about $(0,0)$ takes the form



          $$f(x,y) = sum_{m=0}^{infty} sum_{n=0}^{infty} a_{mn} x^m y^n $$



          $$a_{mn} = frac1{m! n!}left [frac{partial^{m+n} f}{partial x^m partial y^n} right ]_{x=0,y=0} $$



          You want the following $6$ terms: $a_{00}$, $a_{10}$,$a_{01}$,$a_{20}$,$a_{11}$,$a_{02}$.



          To figure out which conic section you have, look at the sign of $a_{11}^2-4 a_{20} a_{02}$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Your computations will give you $2xy+x+y+1=0 Leftrightarrow y=-dfrac{x+1}{2x+1}$ which is an hyperbola. The consequence is that you have a saddle point at the origin.
            $endgroup$
            – Jean Marie
            Mar 3 '16 at 15:58










          • $begingroup$
            @JeanMarie: Cool, but I left my answer as is so that the OP could work out the result for himself.
            $endgroup$
            – Ron Gordon
            Mar 3 '16 at 15:59










          • $begingroup$
            I understand your point. I think that giving the answer leaves some work to do... Regarding the nature of the conic section, I have found with my students, that they had not the idea to find rather easily the nature of a conic section using the "trick" of expressing one variable as a function (or as 2 functions) of the other variable. The knowledge about conic sections is now very scarce and criteria like the sign of $ a_{11}^2-4 a_{20} a_{02}$ is no longer part of the curricula, in most countries, alas...
            $endgroup$
            – Jean Marie
            Mar 3 '16 at 16:31


















          2












          $begingroup$

          Well, you know that a Taylor series about $(0,0)$ takes the form



          $$f(x,y) = sum_{m=0}^{infty} sum_{n=0}^{infty} a_{mn} x^m y^n $$



          $$a_{mn} = frac1{m! n!}left [frac{partial^{m+n} f}{partial x^m partial y^n} right ]_{x=0,y=0} $$



          You want the following $6$ terms: $a_{00}$, $a_{10}$,$a_{01}$,$a_{20}$,$a_{11}$,$a_{02}$.



          To figure out which conic section you have, look at the sign of $a_{11}^2-4 a_{20} a_{02}$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Your computations will give you $2xy+x+y+1=0 Leftrightarrow y=-dfrac{x+1}{2x+1}$ which is an hyperbola. The consequence is that you have a saddle point at the origin.
            $endgroup$
            – Jean Marie
            Mar 3 '16 at 15:58










          • $begingroup$
            @JeanMarie: Cool, but I left my answer as is so that the OP could work out the result for himself.
            $endgroup$
            – Ron Gordon
            Mar 3 '16 at 15:59










          • $begingroup$
            I understand your point. I think that giving the answer leaves some work to do... Regarding the nature of the conic section, I have found with my students, that they had not the idea to find rather easily the nature of a conic section using the "trick" of expressing one variable as a function (or as 2 functions) of the other variable. The knowledge about conic sections is now very scarce and criteria like the sign of $ a_{11}^2-4 a_{20} a_{02}$ is no longer part of the curricula, in most countries, alas...
            $endgroup$
            – Jean Marie
            Mar 3 '16 at 16:31
















          2












          2








          2





          $begingroup$

          Well, you know that a Taylor series about $(0,0)$ takes the form



          $$f(x,y) = sum_{m=0}^{infty} sum_{n=0}^{infty} a_{mn} x^m y^n $$



          $$a_{mn} = frac1{m! n!}left [frac{partial^{m+n} f}{partial x^m partial y^n} right ]_{x=0,y=0} $$



          You want the following $6$ terms: $a_{00}$, $a_{10}$,$a_{01}$,$a_{20}$,$a_{11}$,$a_{02}$.



          To figure out which conic section you have, look at the sign of $a_{11}^2-4 a_{20} a_{02}$.






          share|cite|improve this answer











          $endgroup$



          Well, you know that a Taylor series about $(0,0)$ takes the form



          $$f(x,y) = sum_{m=0}^{infty} sum_{n=0}^{infty} a_{mn} x^m y^n $$



          $$a_{mn} = frac1{m! n!}left [frac{partial^{m+n} f}{partial x^m partial y^n} right ]_{x=0,y=0} $$



          You want the following $6$ terms: $a_{00}$, $a_{10}$,$a_{01}$,$a_{20}$,$a_{11}$,$a_{02}$.



          To figure out which conic section you have, look at the sign of $a_{11}^2-4 a_{20} a_{02}$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Mar 3 '16 at 15:31

























          answered Mar 3 '16 at 15:23









          Ron GordonRon Gordon

          122k14156267




          122k14156267












          • $begingroup$
            Your computations will give you $2xy+x+y+1=0 Leftrightarrow y=-dfrac{x+1}{2x+1}$ which is an hyperbola. The consequence is that you have a saddle point at the origin.
            $endgroup$
            – Jean Marie
            Mar 3 '16 at 15:58










          • $begingroup$
            @JeanMarie: Cool, but I left my answer as is so that the OP could work out the result for himself.
            $endgroup$
            – Ron Gordon
            Mar 3 '16 at 15:59










          • $begingroup$
            I understand your point. I think that giving the answer leaves some work to do... Regarding the nature of the conic section, I have found with my students, that they had not the idea to find rather easily the nature of a conic section using the "trick" of expressing one variable as a function (or as 2 functions) of the other variable. The knowledge about conic sections is now very scarce and criteria like the sign of $ a_{11}^2-4 a_{20} a_{02}$ is no longer part of the curricula, in most countries, alas...
            $endgroup$
            – Jean Marie
            Mar 3 '16 at 16:31




















          • $begingroup$
            Your computations will give you $2xy+x+y+1=0 Leftrightarrow y=-dfrac{x+1}{2x+1}$ which is an hyperbola. The consequence is that you have a saddle point at the origin.
            $endgroup$
            – Jean Marie
            Mar 3 '16 at 15:58










          • $begingroup$
            @JeanMarie: Cool, but I left my answer as is so that the OP could work out the result for himself.
            $endgroup$
            – Ron Gordon
            Mar 3 '16 at 15:59










          • $begingroup$
            I understand your point. I think that giving the answer leaves some work to do... Regarding the nature of the conic section, I have found with my students, that they had not the idea to find rather easily the nature of a conic section using the "trick" of expressing one variable as a function (or as 2 functions) of the other variable. The knowledge about conic sections is now very scarce and criteria like the sign of $ a_{11}^2-4 a_{20} a_{02}$ is no longer part of the curricula, in most countries, alas...
            $endgroup$
            – Jean Marie
            Mar 3 '16 at 16:31


















          $begingroup$
          Your computations will give you $2xy+x+y+1=0 Leftrightarrow y=-dfrac{x+1}{2x+1}$ which is an hyperbola. The consequence is that you have a saddle point at the origin.
          $endgroup$
          – Jean Marie
          Mar 3 '16 at 15:58




          $begingroup$
          Your computations will give you $2xy+x+y+1=0 Leftrightarrow y=-dfrac{x+1}{2x+1}$ which is an hyperbola. The consequence is that you have a saddle point at the origin.
          $endgroup$
          – Jean Marie
          Mar 3 '16 at 15:58












          $begingroup$
          @JeanMarie: Cool, but I left my answer as is so that the OP could work out the result for himself.
          $endgroup$
          – Ron Gordon
          Mar 3 '16 at 15:59




          $begingroup$
          @JeanMarie: Cool, but I left my answer as is so that the OP could work out the result for himself.
          $endgroup$
          – Ron Gordon
          Mar 3 '16 at 15:59












          $begingroup$
          I understand your point. I think that giving the answer leaves some work to do... Regarding the nature of the conic section, I have found with my students, that they had not the idea to find rather easily the nature of a conic section using the "trick" of expressing one variable as a function (or as 2 functions) of the other variable. The knowledge about conic sections is now very scarce and criteria like the sign of $ a_{11}^2-4 a_{20} a_{02}$ is no longer part of the curricula, in most countries, alas...
          $endgroup$
          – Jean Marie
          Mar 3 '16 at 16:31






          $begingroup$
          I understand your point. I think that giving the answer leaves some work to do... Regarding the nature of the conic section, I have found with my students, that they had not the idea to find rather easily the nature of a conic section using the "trick" of expressing one variable as a function (or as 2 functions) of the other variable. The knowledge about conic sections is now very scarce and criteria like the sign of $ a_{11}^2-4 a_{20} a_{02}$ is no longer part of the curricula, in most countries, alas...
          $endgroup$
          – Jean Marie
          Mar 3 '16 at 16:31




















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