The error of a Taylor polynomial (relativistic kinetic energy)












1












$begingroup$


I have been trying to utilize the formula:
enter image description here



I simply cannot figure out how to determine the error when using



enter image description here



Instead of:



enter image description here



I have made a Taylor polynomial around 0 of grade 4, and I cannot find the deviation when using the Taylor polynomial instead of the original function.



The Taylor polynomial is:
enter image description here



And it was made of the Lorentz factor:



enter image description here






real-analysis taylor-expansion







share|cite|improve this question











$endgroup$












  • $begingroup$
    Did you have a look at Taylor-Lagrange formula ?
    $endgroup$
    – Damien
    Dec 16 '18 at 8:23










  • $begingroup$
    Yes I have, but I cannot figure out how to implement it. Every example of using it that I can find, uses the the natural exponential function or a trigonometric function, and their bound is easy to figure out.
    $endgroup$
    – Nikolai
    Dec 16 '18 at 10:58










  • $begingroup$
    You can get a bound on the error with Taylor-Lagrange as soon as you can calculate the Taylor series. What is blocking you ?
    $endgroup$
    – Damien
    Dec 16 '18 at 12:28










  • $begingroup$
    Well i have made the taylor series, but don't understand how i get the bound
    $endgroup$
    – Nikolai
    Dec 16 '18 at 15:58
















1












$begingroup$


I have been trying to utilize the formula:
enter image description here



I simply cannot figure out how to determine the error when using



enter image description here



Instead of:



enter image description here



I have made a Taylor polynomial around 0 of grade 4, and I cannot find the deviation when using the Taylor polynomial instead of the original function.



The Taylor polynomial is:
enter image description here



And it was made of the Lorentz factor:



enter image description here






real-analysis taylor-expansion







share|cite|improve this question











$endgroup$












  • $begingroup$
    Did you have a look at Taylor-Lagrange formula ?
    $endgroup$
    – Damien
    Dec 16 '18 at 8:23










  • $begingroup$
    Yes I have, but I cannot figure out how to implement it. Every example of using it that I can find, uses the the natural exponential function or a trigonometric function, and their bound is easy to figure out.
    $endgroup$
    – Nikolai
    Dec 16 '18 at 10:58










  • $begingroup$
    You can get a bound on the error with Taylor-Lagrange as soon as you can calculate the Taylor series. What is blocking you ?
    $endgroup$
    – Damien
    Dec 16 '18 at 12:28










  • $begingroup$
    Well i have made the taylor series, but don't understand how i get the bound
    $endgroup$
    – Nikolai
    Dec 16 '18 at 15:58














1












1








1





$begingroup$


I have been trying to utilize the formula:
enter image description here



I simply cannot figure out how to determine the error when using



enter image description here



Instead of:



enter image description here



I have made a Taylor polynomial around 0 of grade 4, and I cannot find the deviation when using the Taylor polynomial instead of the original function.



The Taylor polynomial is:
enter image description here



And it was made of the Lorentz factor:



enter image description here






real-analysis taylor-expansion







share|cite|improve this question











$endgroup$




I have been trying to utilize the formula:
enter image description here



I simply cannot figure out how to determine the error when using



enter image description here



Instead of:



enter image description here



I have made a Taylor polynomial around 0 of grade 4, and I cannot find the deviation when using the Taylor polynomial instead of the original function.



The Taylor polynomial is:
enter image description here



And it was made of the Lorentz factor:



enter image description here






real-analysis taylor-expansion




real-analysis taylor-expansion






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 16 '18 at 6:07









Moo

5,61131020




5,61131020










asked Dec 16 '18 at 1:22









Nikolai Nikolai

205




205












  • $begingroup$
    Did you have a look at Taylor-Lagrange formula ?
    $endgroup$
    – Damien
    Dec 16 '18 at 8:23










  • $begingroup$
    Yes I have, but I cannot figure out how to implement it. Every example of using it that I can find, uses the the natural exponential function or a trigonometric function, and their bound is easy to figure out.
    $endgroup$
    – Nikolai
    Dec 16 '18 at 10:58










  • $begingroup$
    You can get a bound on the error with Taylor-Lagrange as soon as you can calculate the Taylor series. What is blocking you ?
    $endgroup$
    – Damien
    Dec 16 '18 at 12:28










  • $begingroup$
    Well i have made the taylor series, but don't understand how i get the bound
    $endgroup$
    – Nikolai
    Dec 16 '18 at 15:58


















  • $begingroup$
    Did you have a look at Taylor-Lagrange formula ?
    $endgroup$
    – Damien
    Dec 16 '18 at 8:23










  • $begingroup$
    Yes I have, but I cannot figure out how to implement it. Every example of using it that I can find, uses the the natural exponential function or a trigonometric function, and their bound is easy to figure out.
    $endgroup$
    – Nikolai
    Dec 16 '18 at 10:58










  • $begingroup$
    You can get a bound on the error with Taylor-Lagrange as soon as you can calculate the Taylor series. What is blocking you ?
    $endgroup$
    – Damien
    Dec 16 '18 at 12:28










  • $begingroup$
    Well i have made the taylor series, but don't understand how i get the bound
    $endgroup$
    – Nikolai
    Dec 16 '18 at 15:58
















$begingroup$
Did you have a look at Taylor-Lagrange formula ?
$endgroup$
– Damien
Dec 16 '18 at 8:23




$begingroup$
Did you have a look at Taylor-Lagrange formula ?
$endgroup$
– Damien
Dec 16 '18 at 8:23












$begingroup$
Yes I have, but I cannot figure out how to implement it. Every example of using it that I can find, uses the the natural exponential function or a trigonometric function, and their bound is easy to figure out.
$endgroup$
– Nikolai
Dec 16 '18 at 10:58




$begingroup$
Yes I have, but I cannot figure out how to implement it. Every example of using it that I can find, uses the the natural exponential function or a trigonometric function, and their bound is easy to figure out.
$endgroup$
– Nikolai
Dec 16 '18 at 10:58












$begingroup$
You can get a bound on the error with Taylor-Lagrange as soon as you can calculate the Taylor series. What is blocking you ?
$endgroup$
– Damien
Dec 16 '18 at 12:28




$begingroup$
You can get a bound on the error with Taylor-Lagrange as soon as you can calculate the Taylor series. What is blocking you ?
$endgroup$
– Damien
Dec 16 '18 at 12:28












$begingroup$
Well i have made the taylor series, but don't understand how i get the bound
$endgroup$
– Nikolai
Dec 16 '18 at 15:58




$begingroup$
Well i have made the taylor series, but don't understand how i get the bound
$endgroup$
– Nikolai
Dec 16 '18 at 15:58










1 Answer
1






active

oldest

votes


















1












$begingroup$

Let us call $f(t)$ the function
$$f(t) = frac{1}{sqrt{1-t}} $$



The $k^{th}$ derivative function of $f()$ is equal to
$$f^{k}(t) = frac{prod_{i=1}^k{(2i-1)}}{2^k} (1-t)^{-frac{2k+1}{2}} $$
This last relation can for example be demonstrated by recurrence.



According to Taylor-Lagrange formula, there exists $xi$ between $0$ and $t$ such that



$$f(t) = 1 + frac{t}{2} + frac{3}{8}t^2 + frac{5}{16}(1-xi)^{-7/2},t^3 = 1 + frac{t}{2} + frac{3}{8}t^2 + R(t)$$



It follows



$$0 < R(t) < frac{5}{16}(1-t)^{-7/2},t^3 $$



Now, one just have to replace $t$ :
$$t = x^2 = left(frac{v}{c}right)^2$$






share|cite|improve this answer











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    1 Answer
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    1 Answer
    1






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    active

    oldest

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    1












    $begingroup$

    Let us call $f(t)$ the function
    $$f(t) = frac{1}{sqrt{1-t}} $$



    The $k^{th}$ derivative function of $f()$ is equal to
    $$f^{k}(t) = frac{prod_{i=1}^k{(2i-1)}}{2^k} (1-t)^{-frac{2k+1}{2}} $$
    This last relation can for example be demonstrated by recurrence.



    According to Taylor-Lagrange formula, there exists $xi$ between $0$ and $t$ such that



    $$f(t) = 1 + frac{t}{2} + frac{3}{8}t^2 + frac{5}{16}(1-xi)^{-7/2},t^3 = 1 + frac{t}{2} + frac{3}{8}t^2 + R(t)$$



    It follows



    $$0 < R(t) < frac{5}{16}(1-t)^{-7/2},t^3 $$



    Now, one just have to replace $t$ :
    $$t = x^2 = left(frac{v}{c}right)^2$$






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      Let us call $f(t)$ the function
      $$f(t) = frac{1}{sqrt{1-t}} $$



      The $k^{th}$ derivative function of $f()$ is equal to
      $$f^{k}(t) = frac{prod_{i=1}^k{(2i-1)}}{2^k} (1-t)^{-frac{2k+1}{2}} $$
      This last relation can for example be demonstrated by recurrence.



      According to Taylor-Lagrange formula, there exists $xi$ between $0$ and $t$ such that



      $$f(t) = 1 + frac{t}{2} + frac{3}{8}t^2 + frac{5}{16}(1-xi)^{-7/2},t^3 = 1 + frac{t}{2} + frac{3}{8}t^2 + R(t)$$



      It follows



      $$0 < R(t) < frac{5}{16}(1-t)^{-7/2},t^3 $$



      Now, one just have to replace $t$ :
      $$t = x^2 = left(frac{v}{c}right)^2$$






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        Let us call $f(t)$ the function
        $$f(t) = frac{1}{sqrt{1-t}} $$



        The $k^{th}$ derivative function of $f()$ is equal to
        $$f^{k}(t) = frac{prod_{i=1}^k{(2i-1)}}{2^k} (1-t)^{-frac{2k+1}{2}} $$
        This last relation can for example be demonstrated by recurrence.



        According to Taylor-Lagrange formula, there exists $xi$ between $0$ and $t$ such that



        $$f(t) = 1 + frac{t}{2} + frac{3}{8}t^2 + frac{5}{16}(1-xi)^{-7/2},t^3 = 1 + frac{t}{2} + frac{3}{8}t^2 + R(t)$$



        It follows



        $$0 < R(t) < frac{5}{16}(1-t)^{-7/2},t^3 $$



        Now, one just have to replace $t$ :
        $$t = x^2 = left(frac{v}{c}right)^2$$






        share|cite|improve this answer











        $endgroup$



        Let us call $f(t)$ the function
        $$f(t) = frac{1}{sqrt{1-t}} $$



        The $k^{th}$ derivative function of $f()$ is equal to
        $$f^{k}(t) = frac{prod_{i=1}^k{(2i-1)}}{2^k} (1-t)^{-frac{2k+1}{2}} $$
        This last relation can for example be demonstrated by recurrence.



        According to Taylor-Lagrange formula, there exists $xi$ between $0$ and $t$ such that



        $$f(t) = 1 + frac{t}{2} + frac{3}{8}t^2 + frac{5}{16}(1-xi)^{-7/2},t^3 = 1 + frac{t}{2} + frac{3}{8}t^2 + R(t)$$



        It follows



        $$0 < R(t) < frac{5}{16}(1-t)^{-7/2},t^3 $$



        Now, one just have to replace $t$ :
        $$t = x^2 = left(frac{v}{c}right)^2$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 16 '18 at 22:00

























        answered Dec 16 '18 at 21:52









        DamienDamien

        59714




        59714






























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