Taylor series with a base point different from $0$












4












$begingroup$


What's the need for $f(x) = sum_{k=0}^{infty}frac{f^{(k)}(a)}{k!}(x-a)^{k}$ if we already have the formula at $0$?

Isn't the $(x-a)$ just making the $a$ as the new origin?

When is this formula more useful than that at $0$?






functions taylor-expansion







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    How would you find the Taylor series for $log(x)$ at $0$?
    $endgroup$
    – jjagmath
    Dec 5 '18 at 15:03
















4












$begingroup$


What's the need for $f(x) = sum_{k=0}^{infty}frac{f^{(k)}(a)}{k!}(x-a)^{k}$ if we already have the formula at $0$?

Isn't the $(x-a)$ just making the $a$ as the new origin?

When is this formula more useful than that at $0$?






functions taylor-expansion







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    How would you find the Taylor series for $log(x)$ at $0$?
    $endgroup$
    – jjagmath
    Dec 5 '18 at 15:03














4












4








4





$begingroup$


What's the need for $f(x) = sum_{k=0}^{infty}frac{f^{(k)}(a)}{k!}(x-a)^{k}$ if we already have the formula at $0$?

Isn't the $(x-a)$ just making the $a$ as the new origin?

When is this formula more useful than that at $0$?






functions taylor-expansion







share|cite|improve this question











$endgroup$




What's the need for $f(x) = sum_{k=0}^{infty}frac{f^{(k)}(a)}{k!}(x-a)^{k}$ if we already have the formula at $0$?

Isn't the $(x-a)$ just making the $a$ as the new origin?

When is this formula more useful than that at $0$?






functions taylor-expansion




functions taylor-expansion






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 5 '18 at 15:56









Andrews

3901317




3901317










asked Dec 5 '18 at 14:58







user531476















  • 4




    $begingroup$
    How would you find the Taylor series for $log(x)$ at $0$?
    $endgroup$
    – jjagmath
    Dec 5 '18 at 15:03














  • 4




    $begingroup$
    How would you find the Taylor series for $log(x)$ at $0$?
    $endgroup$
    – jjagmath
    Dec 5 '18 at 15:03








4




4




$begingroup$
How would you find the Taylor series for $log(x)$ at $0$?
$endgroup$
– jjagmath
Dec 5 '18 at 15:03




$begingroup$
How would you find the Taylor series for $log(x)$ at $0$?
$endgroup$
– jjagmath
Dec 5 '18 at 15:03










2 Answers
2






active

oldest

votes


















6












$begingroup$

Some functions, like $frac1x$, or $ln x$, do not have a Taylor series centered around $0$. Around $1$, on the other hand, we have
$$
frac1x = 1-(x-1) + (x-1)^2 - (x-1)^3 + cdots\
ln x = (x-1) - frac{(x-1)^2}{2} + frac{(x-1)^3}{3} - cdots
$$



Some times, you're interested in approximating a function value at some point $x_0$, which is far from $0$, but close to some other point $a$ where you know the coefficients. For instance, finding $sin(6.27)$ using the Taylor series of $sin$ centered at $2pi$ is going to go much faster than using the one centered at $0$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Why is it going to be faster? If you center $sin$ at $2pi$ you've shifted the whole graph such that $2pi$ is the new $0$, and thus $6.27$ is still the same distance away if that makes sens.
    $endgroup$
    – user531476
    Dec 5 '18 at 15:08










  • $begingroup$
    This is the point of my question by the way.
    $endgroup$
    – user531476
    Dec 5 '18 at 15:08






  • 2




    $begingroup$
    @archaic So your question isn't really about Taylor series at all, but rather more generally about whether to shift the coordinate system (wiht an accompanying variable substitution, but somewhat simpler expressions), or not to shift the coordinate system (with less abstraction but somewhat more complicated expressions). To that there is no answer. Do whatever you feel like, but be aware that some people feel differently than you do. Besides, if you want to shift the coordinate system, then the point of that formula is to tell you how to do it.
    $endgroup$
    – Arthur
    Dec 5 '18 at 15:12












  • $begingroup$
    Thank you for replying, but could you also tell me about how it'll be faster, perhaps an example? Thanks again!
    $endgroup$
    – user531476
    Dec 5 '18 at 15:14






  • 3




    $begingroup$
    @archaic It's faster in the sense that the partial sum of the series will converge towards the actual value much quicker i.e. if you want an approximation of the value of sin(6.27) using taylor series, choosing the series about $2pi$ will let you get your desired accuracy with calculating fewer terms than looking at the series centered at 0.
    $endgroup$
    – Shufflepants
    Dec 5 '18 at 15:32





















5












$begingroup$

If you are going to use Taylor polynomials to compute, say, an approximation of $sqrt{4+frac15}$, what you'll need is the Taylor series of $sqrt x$ centered at $4$, not centered at $0$ (which, by the way, doesn't exist, since $sqrt x$ isn't even differentiable there).






share|cite|improve this answer









$endgroup$













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    6












    $begingroup$

    Some functions, like $frac1x$, or $ln x$, do not have a Taylor series centered around $0$. Around $1$, on the other hand, we have
    $$
    frac1x = 1-(x-1) + (x-1)^2 - (x-1)^3 + cdots\
    ln x = (x-1) - frac{(x-1)^2}{2} + frac{(x-1)^3}{3} - cdots
    $$



    Some times, you're interested in approximating a function value at some point $x_0$, which is far from $0$, but close to some other point $a$ where you know the coefficients. For instance, finding $sin(6.27)$ using the Taylor series of $sin$ centered at $2pi$ is going to go much faster than using the one centered at $0$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Why is it going to be faster? If you center $sin$ at $2pi$ you've shifted the whole graph such that $2pi$ is the new $0$, and thus $6.27$ is still the same distance away if that makes sens.
      $endgroup$
      – user531476
      Dec 5 '18 at 15:08










    • $begingroup$
      This is the point of my question by the way.
      $endgroup$
      – user531476
      Dec 5 '18 at 15:08






    • 2




      $begingroup$
      @archaic So your question isn't really about Taylor series at all, but rather more generally about whether to shift the coordinate system (wiht an accompanying variable substitution, but somewhat simpler expressions), or not to shift the coordinate system (with less abstraction but somewhat more complicated expressions). To that there is no answer. Do whatever you feel like, but be aware that some people feel differently than you do. Besides, if you want to shift the coordinate system, then the point of that formula is to tell you how to do it.
      $endgroup$
      – Arthur
      Dec 5 '18 at 15:12












    • $begingroup$
      Thank you for replying, but could you also tell me about how it'll be faster, perhaps an example? Thanks again!
      $endgroup$
      – user531476
      Dec 5 '18 at 15:14






    • 3




      $begingroup$
      @archaic It's faster in the sense that the partial sum of the series will converge towards the actual value much quicker i.e. if you want an approximation of the value of sin(6.27) using taylor series, choosing the series about $2pi$ will let you get your desired accuracy with calculating fewer terms than looking at the series centered at 0.
      $endgroup$
      – Shufflepants
      Dec 5 '18 at 15:32


















    6












    $begingroup$

    Some functions, like $frac1x$, or $ln x$, do not have a Taylor series centered around $0$. Around $1$, on the other hand, we have
    $$
    frac1x = 1-(x-1) + (x-1)^2 - (x-1)^3 + cdots\
    ln x = (x-1) - frac{(x-1)^2}{2} + frac{(x-1)^3}{3} - cdots
    $$



    Some times, you're interested in approximating a function value at some point $x_0$, which is far from $0$, but close to some other point $a$ where you know the coefficients. For instance, finding $sin(6.27)$ using the Taylor series of $sin$ centered at $2pi$ is going to go much faster than using the one centered at $0$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Why is it going to be faster? If you center $sin$ at $2pi$ you've shifted the whole graph such that $2pi$ is the new $0$, and thus $6.27$ is still the same distance away if that makes sens.
      $endgroup$
      – user531476
      Dec 5 '18 at 15:08










    • $begingroup$
      This is the point of my question by the way.
      $endgroup$
      – user531476
      Dec 5 '18 at 15:08






    • 2




      $begingroup$
      @archaic So your question isn't really about Taylor series at all, but rather more generally about whether to shift the coordinate system (wiht an accompanying variable substitution, but somewhat simpler expressions), or not to shift the coordinate system (with less abstraction but somewhat more complicated expressions). To that there is no answer. Do whatever you feel like, but be aware that some people feel differently than you do. Besides, if you want to shift the coordinate system, then the point of that formula is to tell you how to do it.
      $endgroup$
      – Arthur
      Dec 5 '18 at 15:12












    • $begingroup$
      Thank you for replying, but could you also tell me about how it'll be faster, perhaps an example? Thanks again!
      $endgroup$
      – user531476
      Dec 5 '18 at 15:14






    • 3




      $begingroup$
      @archaic It's faster in the sense that the partial sum of the series will converge towards the actual value much quicker i.e. if you want an approximation of the value of sin(6.27) using taylor series, choosing the series about $2pi$ will let you get your desired accuracy with calculating fewer terms than looking at the series centered at 0.
      $endgroup$
      – Shufflepants
      Dec 5 '18 at 15:32
















    6












    6








    6





    $begingroup$

    Some functions, like $frac1x$, or $ln x$, do not have a Taylor series centered around $0$. Around $1$, on the other hand, we have
    $$
    frac1x = 1-(x-1) + (x-1)^2 - (x-1)^3 + cdots\
    ln x = (x-1) - frac{(x-1)^2}{2} + frac{(x-1)^3}{3} - cdots
    $$



    Some times, you're interested in approximating a function value at some point $x_0$, which is far from $0$, but close to some other point $a$ where you know the coefficients. For instance, finding $sin(6.27)$ using the Taylor series of $sin$ centered at $2pi$ is going to go much faster than using the one centered at $0$.






    share|cite|improve this answer









    $endgroup$



    Some functions, like $frac1x$, or $ln x$, do not have a Taylor series centered around $0$. Around $1$, on the other hand, we have
    $$
    frac1x = 1-(x-1) + (x-1)^2 - (x-1)^3 + cdots\
    ln x = (x-1) - frac{(x-1)^2}{2} + frac{(x-1)^3}{3} - cdots
    $$



    Some times, you're interested in approximating a function value at some point $x_0$, which is far from $0$, but close to some other point $a$ where you know the coefficients. For instance, finding $sin(6.27)$ using the Taylor series of $sin$ centered at $2pi$ is going to go much faster than using the one centered at $0$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 5 '18 at 15:03









    ArthurArthur

    113k7115197




    113k7115197












    • $begingroup$
      Why is it going to be faster? If you center $sin$ at $2pi$ you've shifted the whole graph such that $2pi$ is the new $0$, and thus $6.27$ is still the same distance away if that makes sens.
      $endgroup$
      – user531476
      Dec 5 '18 at 15:08










    • $begingroup$
      This is the point of my question by the way.
      $endgroup$
      – user531476
      Dec 5 '18 at 15:08






    • 2




      $begingroup$
      @archaic So your question isn't really about Taylor series at all, but rather more generally about whether to shift the coordinate system (wiht an accompanying variable substitution, but somewhat simpler expressions), or not to shift the coordinate system (with less abstraction but somewhat more complicated expressions). To that there is no answer. Do whatever you feel like, but be aware that some people feel differently than you do. Besides, if you want to shift the coordinate system, then the point of that formula is to tell you how to do it.
      $endgroup$
      – Arthur
      Dec 5 '18 at 15:12












    • $begingroup$
      Thank you for replying, but could you also tell me about how it'll be faster, perhaps an example? Thanks again!
      $endgroup$
      – user531476
      Dec 5 '18 at 15:14






    • 3




      $begingroup$
      @archaic It's faster in the sense that the partial sum of the series will converge towards the actual value much quicker i.e. if you want an approximation of the value of sin(6.27) using taylor series, choosing the series about $2pi$ will let you get your desired accuracy with calculating fewer terms than looking at the series centered at 0.
      $endgroup$
      – Shufflepants
      Dec 5 '18 at 15:32




















    • $begingroup$
      Why is it going to be faster? If you center $sin$ at $2pi$ you've shifted the whole graph such that $2pi$ is the new $0$, and thus $6.27$ is still the same distance away if that makes sens.
      $endgroup$
      – user531476
      Dec 5 '18 at 15:08










    • $begingroup$
      This is the point of my question by the way.
      $endgroup$
      – user531476
      Dec 5 '18 at 15:08






    • 2




      $begingroup$
      @archaic So your question isn't really about Taylor series at all, but rather more generally about whether to shift the coordinate system (wiht an accompanying variable substitution, but somewhat simpler expressions), or not to shift the coordinate system (with less abstraction but somewhat more complicated expressions). To that there is no answer. Do whatever you feel like, but be aware that some people feel differently than you do. Besides, if you want to shift the coordinate system, then the point of that formula is to tell you how to do it.
      $endgroup$
      – Arthur
      Dec 5 '18 at 15:12












    • $begingroup$
      Thank you for replying, but could you also tell me about how it'll be faster, perhaps an example? Thanks again!
      $endgroup$
      – user531476
      Dec 5 '18 at 15:14






    • 3




      $begingroup$
      @archaic It's faster in the sense that the partial sum of the series will converge towards the actual value much quicker i.e. if you want an approximation of the value of sin(6.27) using taylor series, choosing the series about $2pi$ will let you get your desired accuracy with calculating fewer terms than looking at the series centered at 0.
      $endgroup$
      – Shufflepants
      Dec 5 '18 at 15:32


















    $begingroup$
    Why is it going to be faster? If you center $sin$ at $2pi$ you've shifted the whole graph such that $2pi$ is the new $0$, and thus $6.27$ is still the same distance away if that makes sens.
    $endgroup$
    – user531476
    Dec 5 '18 at 15:08




    $begingroup$
    Why is it going to be faster? If you center $sin$ at $2pi$ you've shifted the whole graph such that $2pi$ is the new $0$, and thus $6.27$ is still the same distance away if that makes sens.
    $endgroup$
    – user531476
    Dec 5 '18 at 15:08












    $begingroup$
    This is the point of my question by the way.
    $endgroup$
    – user531476
    Dec 5 '18 at 15:08




    $begingroup$
    This is the point of my question by the way.
    $endgroup$
    – user531476
    Dec 5 '18 at 15:08




    2




    2




    $begingroup$
    @archaic So your question isn't really about Taylor series at all, but rather more generally about whether to shift the coordinate system (wiht an accompanying variable substitution, but somewhat simpler expressions), or not to shift the coordinate system (with less abstraction but somewhat more complicated expressions). To that there is no answer. Do whatever you feel like, but be aware that some people feel differently than you do. Besides, if you want to shift the coordinate system, then the point of that formula is to tell you how to do it.
    $endgroup$
    – Arthur
    Dec 5 '18 at 15:12






    $begingroup$
    @archaic So your question isn't really about Taylor series at all, but rather more generally about whether to shift the coordinate system (wiht an accompanying variable substitution, but somewhat simpler expressions), or not to shift the coordinate system (with less abstraction but somewhat more complicated expressions). To that there is no answer. Do whatever you feel like, but be aware that some people feel differently than you do. Besides, if you want to shift the coordinate system, then the point of that formula is to tell you how to do it.
    $endgroup$
    – Arthur
    Dec 5 '18 at 15:12














    $begingroup$
    Thank you for replying, but could you also tell me about how it'll be faster, perhaps an example? Thanks again!
    $endgroup$
    – user531476
    Dec 5 '18 at 15:14




    $begingroup$
    Thank you for replying, but could you also tell me about how it'll be faster, perhaps an example? Thanks again!
    $endgroup$
    – user531476
    Dec 5 '18 at 15:14




    3




    3




    $begingroup$
    @archaic It's faster in the sense that the partial sum of the series will converge towards the actual value much quicker i.e. if you want an approximation of the value of sin(6.27) using taylor series, choosing the series about $2pi$ will let you get your desired accuracy with calculating fewer terms than looking at the series centered at 0.
    $endgroup$
    – Shufflepants
    Dec 5 '18 at 15:32






    $begingroup$
    @archaic It's faster in the sense that the partial sum of the series will converge towards the actual value much quicker i.e. if you want an approximation of the value of sin(6.27) using taylor series, choosing the series about $2pi$ will let you get your desired accuracy with calculating fewer terms than looking at the series centered at 0.
    $endgroup$
    – Shufflepants
    Dec 5 '18 at 15:32













    5












    $begingroup$

    If you are going to use Taylor polynomials to compute, say, an approximation of $sqrt{4+frac15}$, what you'll need is the Taylor series of $sqrt x$ centered at $4$, not centered at $0$ (which, by the way, doesn't exist, since $sqrt x$ isn't even differentiable there).






    share|cite|improve this answer









    $endgroup$


















      5












      $begingroup$

      If you are going to use Taylor polynomials to compute, say, an approximation of $sqrt{4+frac15}$, what you'll need is the Taylor series of $sqrt x$ centered at $4$, not centered at $0$ (which, by the way, doesn't exist, since $sqrt x$ isn't even differentiable there).






      share|cite|improve this answer









      $endgroup$
















        5












        5








        5





        $begingroup$

        If you are going to use Taylor polynomials to compute, say, an approximation of $sqrt{4+frac15}$, what you'll need is the Taylor series of $sqrt x$ centered at $4$, not centered at $0$ (which, by the way, doesn't exist, since $sqrt x$ isn't even differentiable there).






        share|cite|improve this answer









        $endgroup$



        If you are going to use Taylor polynomials to compute, say, an approximation of $sqrt{4+frac15}$, what you'll need is the Taylor series of $sqrt x$ centered at $4$, not centered at $0$ (which, by the way, doesn't exist, since $sqrt x$ isn't even differentiable there).







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 5 '18 at 15:02









        José Carlos SantosJosé Carlos Santos

        157k22126227




        157k22126227






























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            Residenzschloss Arolsen

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            Residenzschloss Arolsen Residenzschloss Arolsen Entstehungszeit: 1710–1728 Erhaltungszustand: Vollständig Ständische Stellung: Grafen Ort: Bad Arolsen Geographische Lage 51° 22′ 51″  N , 9° 1′ 19″  O 51.380833333333 9.0219444444444 293 Koordinaten: 51° 22′ 51″  N , 9° 1′ 19″  O Höhe: 293  m ü. NHN Das Residenzschloss Arolsen ist ein barockes Schloss in Bad Arolsen im Kreis Waldeck-Frankenberg in Nordhessen (Deutschland). Das Schloss wurde als dreiflügelige Anlage gebaut, an die sich ein englischer Garten anschließt. Zentrales landschaftsarchitektonisches Gestaltungselement ist ein ausgedehntes Rondell. Inhaltsverzeichnis 1 Geschichte 1.1 Vorgängerbau 1.2 Bau und Geschichte des neuen Schlosses 1.3 Bibliothek 1.4 Finanzielle und staatsrechtliche Konsequenzen 1.5 Heutige Nutzung 2 Innenausstattung 3 Umgebung 3.1 Park 3.2 Wirtschaftsgebäude 4 Literatur 5 Weblinks 6...
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